general chemistry

1

M O D U L E

1CHEMISTRY OF THE

NONMETALS

N.1 The Nonmetals

N.2 The Chemistry of Hydrogen

N.3 The Chemistry of Oxygen

Chemistry in the World Around Us: The Chemistry of the Atmosphere

N.4 The Chemistry of Sulfur

N.5 The Chemistry of Nitrogen

N.6 The Chemistry of Phosphorus

N.7 The Chemistry of the Halogens

N.8 The Chemistry of the Rare Gases

N.9 The Inorganic Chemistry of Carbon

More than 75% of the known elements have the characteristic properties of metals (seeFigure N.1). They have a metallic luster; they are malleable and ductile; and they conductheat and electricity. Eight other elements (B, Si, Ge, As, Sb, Te, Po, and At) are best de-scribed as semimetals or metalloids. They often look like metals, but they tend to be brit-tle, and they are more likely to be semiconductors than conductors of electricity.

Once the metals and semimetals are removed from the list of known elements, only 17are left to be classified as nonmetals. Six of these elements belong to the family of raregases in Group VIIIA, most of which are virtually inert to chemical reactions. Discussionsof the chemistry of the nonmetals therefore tend to focus on the following elements: H, C,N, O, F, P, S, Cl, Se, Br, I, and Xe.

One way of visualizing the difference between metals, semimetals or metalloids, andnonmetals is the plot of average valence electron energies (AVEE) in Figure N.2. Becausethe AVEE provides a measure of how tightly an atom holds onto its valence electrons, itcan be used to explore the dividing line between the metals and nonmetals. As noted inSection 3.24, elements with AVEE values below 1.06 MJ/mol are metals, whereas thosewith AVEE values above 1.26 MJ/mol are nonmetals. Elements with AVEE values in therange of 1.06–1.26 MJ/mol have properties between those of the metals and nonmetals andare therefore semimetals or metalloids.

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2 NONMETAL

N.1 THE NONMETALSThere is a clear pattern in the chemistry of the main-group metals discussed in Chapter 5:The main-group metals are oxidized in all of their chemical reactions. Aluminum, for ex-ample, is oxidized by bromine when these elements react to form aluminum bromide.

Al2Br62 Al � 3 Br2

Oxidation

Reduction

0 �3 �10

H

Li

Na

K

Rb

Cs

Fr

Be

Mg

Ca

Sr

Ba

Ra

Sc

Y

La

Ac Rf Db Sg Bh Hs Mt

Ti

Zr

Hf

V

Nb

Ta

Cr

Mo

W

Mn

Tc

Re

Fe

Ru

Os

Co

Rh

Ir

Ni

Pd

Pt

Cu

Ag

Au

Zn

Cd

Hg

B

Al

Ga

In

Tl

C

Si

Ge

Sn

Pb

N

P

As

Sb

Bi

O

S

Se

Te

Po

H

F

Cl

Br

I

At

He

Ne

Ar

Kr

Xe

Rn

Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu

Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr

Metals

Nonmetals

Semimetals

FIGURE N.1 The elements can be divided into three categories: metals, semimetals,and nonmetals.

Ne

F

O

N

C

B

Ar

Kr

Xe

CI

S

P

Si

Br

Se

As

GeAl

Ga

I

Te

Sb

Sn

In

Be

Mg

Ca

Sr

Li

Na

K

Rb FIGURE N.2 Three-dimensional plot of the average valence electronenergies (AVEE) of the main-group elements versus position in the periodic table.

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The chemistry of the nonmetals is more interesting because these elements can un-dergo both oxidation and reduction. Phosphorus, for example, is oxidized when it reactswith more electronegative elements, such as oxygen.

But it is reduced when it reacts with less electronegative elements, such as calcium.

The behavior of the nonmetals can be summarized as follows.

• Nonmetals tend to oxidize metals.

2 Mg(s) � O2(g) 88n 2 MgO(s)

• Nonmetals with relatively large electronegativities (such as oxygen and chlorine)oxidize substances with which they react.

2 H2S(g) � 3 O2(g) 88n 2 SO2(g) � 2 H2O(g)PH3(g) � 3 Cl2(g) 88n PCl3(l) � 3 HCl(g)

• Nonmetals with relatively small electronegativities (such as carbon and hydrogen)can reduce other substances.

Fe2O3(s) � 3 C(s) 88n 2 Fe(s) � 3 CO(g)CuO(s) � H2(g) 88n Cu(s) � H2O(g)

N.2 THE CHEMISTRY OF HYDROGENHydrogen is the most abundant element in the universe, accounting for 90% of the atomsand 75% of the mass of the universe. But hydrogen is much less abundant on earth. Evenwhen the enormous number of hydrogen atoms in the oceans is included, hydrogen makesup less than 1% of the mass of the planet.

The name hydrogen comes from the Greek stems hydro-, “water,” and gennan, “toform or generate.” Thus, hydrogen is literally the “water former.”

2 H2(g) � O2(g) 88n 2 H2O(g)

Although it is often stated that more compounds contain carbon than any other element,this isn’t necessarily true. Most carbon compounds also contain hydrogen, and hydrogenforms compounds with virtually all the other elements as well.

6 Ca � P4

Oxidation

Reduction

0 �2 �30

2 Ca3P2

P4O10P4 � 5 O2

Oxidation

Reduction

0 �5 �20

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Compounds of hydrogen are frequently called hydrides, even though the name hydrideliterally describes compounds that contain an H� ion. There is a regular trend in the for-mula of the hydrides across a row of the periodic table, as shown in Figure N.3. This trendis so regular that the combining power, or valence, of an element was once defined as thenumber of hydrogen atoms bound to the element in its hydride.

H2

LiH

NaH

KH

RbH

CsH

BeH2

MgH2

CaH2

SrH2

BaH2

B2H6

AlH3

GaH3

CH4

SiH4

GeH4

SnH4

PbH4

NH3

PH3

AsH3

SbH3

BiH3

H2O

H2S

H2Se

H2Te

H2Po

H2

HF

HCl

HBr

HI

HAt FIGURE N.3 Hydrogen combines with every element in the peri-odic table except those in Group VIIIA. The formulas of the hydrides of the main-group elements are shown here.

Hydrogen has three oxidation states, corresponding to the H� ion, a neutral H atom,and the H� ion.

H� � 1s0

H � 1 s1

H� � 1s2

Because hydrogen forms compounds with oxidation numbers of both �1 and �1, manyperiodic tables include the element in both Group IA (with Li, Na, K, Rb, Cs, and Fr) andGroup VIIA (with F, Cl, Br, I, and At).

There are many reasons for including hydrogen among the elements in Group IA. Itforms compounds such as HCl and HNO3 that are analogs of alkali metal compounds suchas NaCl and KNO3. Under conditions of very high pressure, hydrogen has the propertiesof a metal. (It has been argued, for example, that any hydrogen present at the center ofthe planet Jupiter is present as a metallic solid.) Finally, hydrogen combines with a hand-ful of metals, such as scandium, titanium, chromium, nickel, and palladium, to form mate-rials that behave as if they were alloys of two metals.

There are equally valid arguments for placing hydrogen in Group VIIA. It forms com-pounds such as NaH and CaH2 that are analogs of halogen compounds such as NaF andCaCl2. It also combines with other nonmetals to form covalent compounds such as H2O,CH4, and NH3, the way a nonmetal should. Finally, the element is a gas at room temper-ature and atmospheric pressure, like other nonmetals such as O2 and N2.

It is difficult to decide where hydrogen belongs in the periodic table because of thephysical properties of the element. The first ionization energy of hydrogen (1312 kJ/mol),for example, is roughly halfway between the elements with the largest (He 2372 kJ/mol)and smallest (Cs 376 kJ/mol) first ionization energies. Hydrogen also has an electro-negativity (EN � 2.30) halfway between the extremes of neon, the most electronegativeelement (EN � 4.79), and cesium, the least electronegative (EN � 0.66) element. On thebasis of electronegativity, it is tempting to classify hydrogen as a semimetal, as shown inFigure N.4.

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Hydrogen is oxidized by elements that are more electronegative to form compoundsin which it has an oxidation number of �1.

Hydrogen is reduced by elements that are less electronegative to form compounds in whichits oxidation number is �1.

At room temperature, hydrogen is a colorless, odorless gas with a density only one-fourteenth the density of air. Small quantities of H2 gas can be prepared in several ways.

• By reacting an active metal with water.

2 Na(s) � 2 H2O(l) 88n 2 Na�(aq) � 2 OH�(aq) � H2(g)

• By reacting a less active metal with a strong acid.

Zn(s) � 2 HCl(aq) 88n Zn2�(aq) � 2 Cl�(aq) � H2(g)

• By reacting an ionic metal hydride with water.

NaH(s) � H2O(l) 88n Na�(aq) � OH�(aq) � H2(g)

Reduction

Oxidation

2 NaH0 0 �1

2 Na � H2�1

Reduction

Oxidation

2 HCl0 0 �1

H2 � Cl2�1

Ne

F

O

N

C

B

Ar

Kr

Xe

CI

S

P

Si

Br

Se

As

GeAl

Ga

I

Te

Sb

Sn

In

Be

H

Mg

Ca

Sr

Li

Na

K

Rb

FIGURE N.4 Three-dimensional plot of the electronegativities of themain-group elements versus position in the periodic table.

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• By decomposing water into its elements with an electric current.

electrolysis2 H2O(l) 888888n 2 H2(g) � O2(g)

Exercise N.1

Use oxidation numbers to determine what is oxidized and what is reduced in the follow-ing reactions, which are used to prepare H2 gas.(a) Mg(s) � 2 HCl(aq) n Mg2�(aq) � 2 Cl�(aq) � H2(g)(b) Ca(s) � 2 H2O(l) n Ca2�(aq) � 2 OH�(aq) � H2(g)

Solution

(a) Magnesium metal is oxidized in this reaction and the H� ions from hydrochloric acidare reduced.

(b) Calcium metal is oxidized in this reaction and the H� ions from water are reduced.

The covalent radius of a neutral hydrogen atom is smaller than any other element. Be-cause small atoms can come very close to each other, they tend to form strong covalentbonds. H2 therefore tends to be unreactive at room temperature. In the presence of a spark,however, a fraction of the H2 molecules dissociate to form hydrogen atoms that are highlyreactive.

sparkH2(g) 888n 2 H(g) �H° � 435.30 kJ/molrxn

The heat given off when these H atoms react with O2 is enough to catalyze the dissocia-tion of additional H2 molecules. Mixtures of H2 and O2 that are infinitely stable at roomtemperature therefore explode in the presence of a spark or flame.

N.3 THE CHEMISTRY OF OXYGENOxygen is the most abundant element on this planet. The earth’s crust is 46.6% oxygen bymass, the oceans are 86% oxygen by mass, and the atmosphere is 21% oxygen by volume.The name oxygen comes from the Greek stems oxys, “acid,” and gennan, “to form orgenerate.” Thus, oxygen literally means the “acid former.” The name was introduced by

Ca2� � 2 OH� � H2

Oxidation

Reduction

0 �2 0�1

Ca � 2 H2O

Mg2� � 2 Cl� � H2

Oxidation

Reduction

0 �2 0�1

Mg � 2 HCl

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NONMETAL 7

Lavoisier, who noticed that compounds rich in oxygen, such as SO2 and P4O10, dissolve inwater to give acids.

SO2(g) � H2O(aq) 88n H2SO3(aq)P4O10(s) � 6 H2O(aq) 88n 4 H3PO4(aq)

The electron configuration of an oxygen atom—[He] 2s2 2p4—suggests that O atomscan achieve an octet of valence electrons by sharing two pairs of electrons to form an OPOdouble bond, as shown in Figure N.5.

According to the Lewis structure, all of the electrons in the O2 molecule are paired. Thecompound should therefore be diamagnetic—it should be repelled by a magnetic field. Ex-perimentally, O2 is found to be paramagnetic—it is attracted to a magnetic field. This canbe explained using molecular orbital theory, which predicts that there are two unpairedelectrons in the �* antibonding molecular orbitals of the O2 molecule.

At temperatures below �183°C, O2 condenses to form a liquid with a characteristiclight blue color that results from the absorption of light with a wavelength of 630 nm. Thisabsorption isn’t seen in the gas phase and is relatively weak even in the liquid because itrequires that three bodies—two O2 molecules and a photon—collide simultaneously, whichis a very rare phenomenon, even in the liquid phase.

The Chemistry of Ozone

The O2 molecule isn’t the only elemental form of oxygen. In the presence of lightning oranother source of a spark, O2 molecules dissociate to form oxygen atoms.

sparkO2(g) 888n 2 O(g)

The O atoms can react with O2 molecules to form ozone, O3.

O2(g) � O(g) 88n O3(g)

Ozone is a resonance hybrid of two Lewis structures each of which contains one OPOdouble bond and one OOO single bond, as shown in Figure N.6. Because the valence elec-trons on the central atom are distributed toward the corners of a triangle, the O3 moleculeis angular or bent, with a bond angle of 116.5°.

FIGURE N.5 Lewis structure of the O2 molecule.

O

O O O O

O

FIGURE N.6 Lewis structures for ozone.

O O

When an element exists in more than one form—such as oxygen (O2) and ozone (O3)—the different forms of the element are called allotropes (from a Greek word meaning “inanother manner”). Because they have different structures, allotropes have different chem-ical and physical properties, as shown in Table N.1.

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Ozone is an unstable compound with a sharp, pungent odor that slowly decomposes tooxygen.

2 O3(g) 88n 3 O2(g)

At low concentrations, ozone can be relatively pleasant. (The characteristic clean odorassociated with summer thunderstorms is due to the formation of small amounts of O3.)Exposure to O3 at higher concentrations leads to coughing, rapid beating of the heart, chestpain, and general body pain. At concentrations above 1 ppm, ozone is toxic.

The most famous characteristic of ozone is its ability to absorb high energy radiationin the ultraviolet portion of the spectrum (� � 300 nm), thereby providing a filter that pro-tects us from exposure to high energy ultraviolet radiation emitted by the sun. We can un-derstand the importance of this filter if we think about what happens when radiation fromthe sun is absorbed by our skin.

Electromagnetic radiation in the infrared, visible, and low energy portions of the ul-traviolet spectrum carries enough energy to excite an electron into a higher energy or-bital. This electron eventually falls back into the orbital from which it was excited, andenergy is given off to the surrounding tissue in the form of heat. Anyone who has suf-fered from a sunburn can appreciate the painful consequences of excessive amounts ofthis radiation.

Radiation in the high energy portion of the ultraviolet spectrum carries enough energyto ionize atoms and molecules. Because living tissue is 70–90% water by weight, the mostcommon ionization reaction involves the loss of an electron from a neutral water moleculeto form an H2O� ion.

H2O 88n H2O� � e�

The H2O� ions formed in the reaction have an odd number of electrons and are extremelyreactive. They can cause permanent damage to the cell tissue and induce processes thateventually result in skin cancer. Relatively small amounts of this radiation can thereforehave drastic effects on living tissue. Therefore, the protection by the ozone (O3) layer whichprevents high energy radiation from reaching the earth is important to the health of livingorganisms.

In 1974 Molina and Rowland pointed out that chlorofluorocarbons, such as CFCl3 andCF2Cl2, which had been used as refrigerants and as propellants in aerosol cans, were be-ginning to accumulate in the atmosphere. In the stratosphere, at altitudes of 10 to 50 kmabove the earth’s surface, chlorofluorocarbons decompose to form Cl atoms and chlorineoxides such as ClO when they absorb sunlight. Cl atoms and ClO molecules also havean unpaired electron in the valence shell of the molecule. As a result, they are unusually

TABLE N.1 Properties of Allotropes of Oxygen

Property Oxygen (O2) Ozone (O3)

Melting point �218.75°C �192.5°CBoiling point �182.96°C �110.5°CDensity (at 20°C) 1.331 g/L 1.998 g/LOOO bond order 2 1.5OOO bond length 0.1207 nm 0.1278 nm

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NONMETAL 9

reactive. In the atmosphere, they react with ozone or with the oxygen atoms that are neededto form ozone.

Cl � O3 88n ClO � O2

ClO � O 88n Cl � O2

Molina and Rowland postulated that these substances would eventually deplete the ozoneshield in the stratosphere, with dangerous implications for biological systems that wouldbe exposed to increased levels of high energy ultraviolet radiation.

Oxygen as an Oxidizing Agent

Fluorine is the only element with which oxygen reacts that is more electronegative than oxy-gen. As a result, oxygen gains electrons in virtually all of its chemical reactions. Each O2

molecule must gain four electrons to satisfy the octets of the two oxygen atoms withoutsharing electrons.

Oxygen therefore oxidizes metals to form salts in which the oxygen atoms are formallypresent as O2� ions. Rust forms, for example, when iron reacts with oxygen in the pres-ence of water to give a salt that formally contains the Fe3� and O2� ions, with an averageof three water molecules coordinated to each Fe3� ion in the solid.

H2O4 Fe(s) � 3 O2(g) 88n 2 Fe2O3�3 H2O(s)

Oxygen also oxidizes nonmetals, such as carbon, to form covalent compounds in which theoxygen has an oxidation number of �2.

C(s) � O2(g) 88n CO2(g)

Oxygen is the perfect example of an oxidizing agent because it increases the oxidation stateof almost any substance with which it reacts. In the course of its reactions, oxygen is re-duced. The substances it reacts with are therefore reducing agents.

Exercise N.2

Identify the oxidizing agents and reducing agents in the following reactions.(a) Fe2O3(s) � 3 C(s) n 2 Fe(s) � 3 CO(g)(b) CH4(g) � 2 O2(g) n CO2(g) � 2 H2O(g)

Solution

(a) In this reaction, carbon reduces Fe2O3 to iron metal, which means carbon is the re-ducing agent. Fe2O3 oxidizes carbon to CO and is therefore the oxidizing agent.

2 Fe � 3 CO

Oxidation

Reduction0 �20�3

Fe2O3 � 3 C

¼

O³¼³PO � 4 e� 2 [¼O¼]2�þ

³

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(b) Oxygen oxidizes CH4 in the reaction, so O2 is the oxidizing agent. The oxygen is re-duced by CH4, which means that CH4 is the reducing agent.

Each year between 75 and 80 quads, or quadrillion (1015) BTU (British thermal units),of energy is consumed in the United States.1 Less than 10% of this energy is provided bynuclear, solar, geothermal, or hydro power. The rest can be traced to a combustion reac-tion in which a fuel is oxidized by O2. The cars, trucks, and buses that fill our highways arepowered by gasoline engines that burn hydrocarbons such as octane, C8H18,

2 C8H18(l) � 25 O2(g) 88n 16 CO2(g) � 18 H2O(g)

or diesel engines that burn larger hydrocarbons such as cetane, C16H34.

2 C16H34(l) � 49 O2(g) 88n 32 CO2(g) � 34 H2O(g)

We heat our homes by burning the methane (CH4) in natural gas, the high molecular weighthydrocarbons in fuel oil, or the hydrocarbons in wood, or by using electricity generated ina power plant that burns either oil or coal.

The energy we use to fuel our bodies also comes from combustion reactions. Energyenters our bodies in the form of lipids, proteins, and carbohydrates. These “fuels” are con-verted into carbohydrates, such as glucose (C6H12O6), which react with oxygen to producethe energy we need to survive.

C6H12O6(aq) � 6 O2(g) 88n 6 CO2(g) � 6 H2O(l)

About 65% of the energy given off in the reaction is used to synthesize the ATP (adeno-sine triphosphate) that fuels biological processes. The remaining 35% is released as theheat that keeps our body temperatures higher than the temperature of the surroundings.

There is an ever-growing awareness that the earth contains a finite amount of fossil fuels, such as oil and coal. Nuclear, solar, and geothermal power will be increasingly im-portant sources of energy. But they won’t replace fossil fuels by themselves because theyare used to produce electrical energy, which is difficult to store. One possible solution tothis problem has been labeled the hydrogen economy.

The first step in the hydrogen economy is to use energy from nuclear, solar, or geo-thermal power to split water into its elements.

2 H2O(l) 88n 2 H2(g) � O2(g)

The oxygen is then released to the atmosphere, and the hydrogen is either burned as a fuel

2 H2(g) � O2(g) 88n 2 H2O(g) �H° � �483.64 kJ/molrxn

CO2 � 2 H2OCH4 � 2 O2

Oxidation

Reduction

0 �4�4 �2�2

1One BTU is the energy needed to raise the temperature of one pound of water by 1°F; 1 BTU � 1.055 kJ.

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NONMETAL 11

or used to reduce carbon monoxide to methanol or to gasoline, which can be stored andlater burned as a fuel.

CO(g) � 2 H2(g) 88n CH3OH(l)8 CO(g) � 17 H2(g) 88n C8H18(l) � 8 H2O(l)

Peroxides

It takes four electrons to reduce an O2 molecule to a pair of O2� ions. If the reaction stopsafter the O2 molecule has gained only two electrons, the O2

2� ion is produced.

The O22� ion has two more electrons than a neutral O2 molecule, which means that the

oxygen atoms must share only a single pair of bonding electrons to achieve an octet of va-lence electrons. The O2

2� ion is called the peroxide ion because compounds that containthe ion are unusually rich in oxygen. They are not just oxides—they are (hy-)peroxides.

The easiest way to prepare a peroxide is to react sodium or barium metal with oxygen.

2 Na(s) � O2(g) 88n Na2O2(s)Ba(s) � O2(g) 88n BaO2(s)

When the peroxides are allowed to react with a strong acid, hydrogen peroxide (H2O2) isproduced.

BaO2(s) � 2 H�(aq) 88n Ba2�(aq) � H2O2(aq)

The Lewis structure of hydrogen peroxide contains an OOO single bond.

The electron domain theory predicts that the geometry around each oxygen atom in H2O2

should be bent. But this theory can’t predict whether the four atoms should lie in the sameplane or whether the molecule should be visualized as lying in two intersecting planes. Theexperimentally determined structure of H2O2 is shown in Figure N.7.

O� 2 e� [ ]2�OOO O

H

94.8°111.5°

H

FIGURE N.7 Geometry of an H2O2 molecule.

The HOOOO bond angle in the molecule is only slightly larger than the angle between apair of adjacent 2p atomic orbitals on the oxygen atom, and the angle between the planesthat form the molecule is slightly larger than the tetrahedral angle. The geometry aroundeach oxygen atom is bent, or angular, with an OOOOH bond angle of 94.8°. In order tokeep the nonbonding electrons on the oxygen atoms as far apart as possible, the four atomsin the molecule lie in two planes that intersect at an angle of 111.5°.

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The oxidation number of the oxygen atoms in hydrogen peroxide is �1. H2O2 cantherefore act as an oxidizing agent and capture two more electrons to form a pair of hy-droxide ions, in which the oxygen has an oxidation number of �2.

H2O2 � 2 e� 88n 2 OH�

Or, it can act as a reducing agent and lose a pair of electrons to form an O2 molecule.

H2O2 88n O2 � 2 H� � 2 e�

CheckpointUse Lewis structures to explain what happens in the following reactions.

H2O2 � 2 e� 88n 2 OH�

H2O2 88n O2 � 2 H� � 2 e�

Reactions in which a compound simultaneously undergoes both oxidation and reduc-tion are called disproportionation reactions. The products of the disproportionation ofH2O2 are oxygen and water.

2 H2O2(aq) 88n O2(g) � 2 H2O(l)

Exercise N.3

Use the reactions that describe what happens when H2O2 loses a pair of electrons and whathappens when H2O2 picks up a pair of electrons to explain why the disproportionation ofhydrogen peroxide gives oxygen and water.

2 H2O2(aq) 88n O2(g) � 2 H2O(l)

Solution

Adding the half-reaction for the oxidation of H2O2 to the half-reaction for the reductionof the compound gives the following results.

H2O2 � 2 e� 88n 2 OH�

H2O2 88n O2 � 2 H� � 2 e�

2 H2O2 88n O2 � 2 H� � 2 OH�

The H� and OH� ions produced in the two halves of the reaction combine to form waterto give the following overall stoichiometry for the reaction.

2 H2O2 88n O2 � 2 H2O

The disproportionation of H2O2 is an exothermic reaction.

2 H2O2(aq) 88n O2(g) � 2 H2O(l) �H° � �189.3 kJ/molrxn

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NONMETAL 13

The reaction is relatively slow, however, in the absence of a catalyst. The principal uses ofH2O2 revolve around its oxidizing ability. It is used in dilute (3%) solutions as a disinfec-tant. In more concentrated solutions (30%), it is used as a bleaching agent for hair, fur,leather, or the wood pulp used to make paper. In very concentrated (40–70%) solutions,H2O2 has been used as rocket fuel because of the ease with which it decomposes to give O2.

Methods of Preparing O2

Small quantities of O2 gas can be prepared in a number of ways.

• By decomposing a dilute solution of hydrogen peroxide with dust or a metal sur-face as the catalyst.

2 H2O2(aq) 88n O2(g) � 2 H2O(l)

• By reacting hydrogen peroxide with a strong oxidizing agent, such as the perman-ganate ion, MnO4

�.

5 H2O2(aq) � 2 MnO4�(aq) � 6 H�(aq) 88n 2 Mn2�(aq) � 5 O2(g) � 8 H2O(l)

• By passing an electric current through water.

electrolysis2 H2O(l) 888888n 2 H2(g) � O2(g)

• By heating potassium chlorate in the presence of a catalyst until it decomposes.

MnO22 KClO3(s) 888n 2 KCl(s) � 3 O2(g)

CheckpointWhich of the following elements or compounds reacts with water to give a solutionthat could be used to produce O2?(a) Na (b) Na2O (c) Na2O2 (d) NaOH (e) NaCl

Chemistry in the World Around Us

The Chemistry of the Atmosphere

Although major changes occurred in the atmosphere during the early history of ourplanet, the chemistry of the earth’s atmosphere has been more or less constant duringthe time in which the human race evolved. This is no longer true. The amount ofmethane (CH4) in the atmosphere is increasing at a rate of more than 1% per year.The concentration of carbon dioxide has more than doubled since 1750 and seems to beincreasing at an exponential rate (see Figure N.8). In recent years, attention has beenfocused on one particular change in the atmosphere, the depletion of ozone above theAntarctic continent.

For over 25 years, a team of scientists collected data on variations in the amount ofO3 at different altitudes above the British Antarctic Survey station at Halley Bay. In1985, they reported that the O3 concentration declined after the return of solar radiation

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14 NONMETAL

each September. The effect was small when first noticed in the late 1970s, but it reachedsuch high levels by 1984 that the O3 concentration declined by 30% by the end of Octoberof that year.

The data from Halley Bay sampled only a small portion of the atmosphere over theAntarctic. Once the notion of ozone depletion was reported, however, scientists wereable to retrieve data that had been recorded by satellites over a period of years, whichconfirmed that the same effect occurred over virtually the entire Antarctic continent.Several features of the ozone depletion were particularly interesting.

• The drop in the ozone concentration occurred very rapidly, within a period of sixweeks each spring. (“Spring” occurs during September and October in the south-ern hemisphere.)

• Although the drop in O3 occurred above the Antarctic, it corresponds to a loss of3% of the total ozone concentration in the planet’s atmosphere.

• The decline in the O3 concentration became more serious each year. By 1989, theO3 concentration during the summer months dropped by 70%.

• The decline was temporary. During the winter months, the ozone level built backto normal levels.

Several possible explanations were available for the ozone holes. One of the mostpopular was the suggestion by Molina and Rowland that the ozone in the atmospherecould be destroyed by Cl and ClO radicals created when chlorofluorocarbons (CFC’s) inthe atmosphere decomposed. The question was: What evidence could be found to eithersupport or refute that hypothesis? The problem is complex, because more than 200chemical reactions have been included in the models used to explain the chemistry ofthe atmosphere. It is further complicated by the fact that ClO radicals exist in the atmo-sphere at concentrations of only about 1 part per trillion by volume (pptv).

In the 4 January 1991 issue of Science, James Anderson and co-workers reported datathat probed the link between the release of chlorofluorocarbons into the atmosphere andthe disappearance of ozone from the stratosphere above the Antarctic each spring. Thedata were obtained between 23 August and 22 September 1987, using special instrumentsmounted in high-altitude aircraft. Initial measurements after the plane took off from the tipof Chile (54° S latitude) suggested that the background concentration of ClO radicals wasat the threshold of detection: about 1 pptv. As the plane flew toward the south pole, theClO concentration increased slightly until about 65° S latitude, when it rapidly increased.

A plot of the concentration of ClO radicals versus latitude for the 16 September flightis shown in Figure N.9. Once the aircraft reached a latitude of 68° S the abundance of theradicals increased by three orders of magnitude, to a level of approximately 1200 pptv.These data by themselves are suggestive, but Figure N.9 contains more compelling

270

CO

2 c

once

ntra

tion

(pp

mv)

1700

280

290

300

310

320

330

340

360

1750 1800 1850 1900 1950 2000Year

FIGURE N.8 The CO2 concentration in the at-mosphere from the time of the industrial revo-lution to present in units of parts per million byvolume (ppmv).

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NONMETAL 15

evidence. The plot at the top of Figure N.9 shows that the concentration of O3 droppedby a factor of about 2.5 at virtually the same time that the ClO concentration increased.

Anderson and co-workers concluded as follows: “When taken independently, eachelement in the case contains a segment of the puzzle that in itself is not conclusive. Whentaken together, however, they provide convincing evidence that the dramatic reductionin O3 over the Antarctic continent would not have occurred had CFC’s not been synthe-sized and then added to the atmosphere.”2

N.4 THE CHEMISTRY OF SULFURBecause sulfur is directly below oxygen in the periodic table, these elements have similarelectron configurations.

O [He] 2s2 2p4

S [Ne] 3s2 3p4

As a result, sulfur forms many compounds that are analogs of oxygen compounds, as shownin Table N.2. The last two examples in Table N.2 show how the prefix thio- can be used todescribe compounds in which sulfur replaces an oxygen atom.

2J. G. Anderson, D. W. Toohey, and W. H. Brune, Science, 251, 45 (1991).

1200

600

062 64 66 68 70 72

0

1000

2000

3000

O3 c

once

ntra

tion

in p

pbv

ClO

• co

ncen

trat

ion

in p

ptv

Latitude (degrees south)

16 SEPT

ClOO3

FIGURE N.9 Variations in the ClO and O3 concentrations in the atmosphere during the September 16, 1987, flight toward the Antarcticcontinent.

TABLE N.2 Oxygen Compounds and Their Sulfur Analogs

Oxygen Compound Sulfur Compound

Na2O (sodium oxide) Na2S (sodium sulfide)H2O (water) H2S (hydrogen sulfide)O3 (ozone) SO2 (sulfur dioxide)CO2 (carbon dioxide) CS2 (carbon disulfide)OCN� (cyanate) SCN� (thiocyanate)OC(NH2)2 (urea) SC(NH2)2 (thiourea)

There are four principal differences between the chemistry of sulfur and oxygen.

• OPO double bonds are much stronger than SPS double bonds.• SOS single bonds are almost twice as strong as OOO single bonds.• Sulfur is much less electronegative than oxygen.• Sulfur can expand its valence shell to hold more than eight electrons; oxygen cannot.

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16 NONMETAL

These seemingly minor differences have important consequences for the chemistry of theseelements.

The Effect of Differences in the Strength of XOOX and XPPX Bonds

The radius of a sulfur atom is about 60% larger than that of an oxygen atom.

� � 1.6

As a result, it is harder for sulfur atoms to come close enough together to form doublebonds. SPS double bonds are therefore much weaker than OPO double bonds.

Double bonds between sulfur and oxygen or carbon atoms can be found in compoundssuch as SO2 and CS2. But these double bonds are much weaker than the equivalent dou-ble bonds to oxygen atoms in O3 or CO2. The CPS double bonds in CS2, for example, areabout 65% as strong as the CPO double bonds in CO2.

Elemental oxygen consists of O2 molecules in which each atom completes its octet ofvalence electrons by sharing two pairs of electrons with a single neighboring atom. Becausesulfur doesn’t form strong SPS double bonds, elemental sulfur consists of cyclic S8 mole-cules in which each atom completes its octet by forming single bonds to two different neigh-boring atoms, as shown in Figure N.10.

0.104 nm��0.066 nm

Covalent radius of sulfur��Covalent radius of oxygen

S

SS S

S

S

SS

S8 FIGURE N.10 Structure of a cyclic S8 molecule.

S8 molecules can pack to form more than one crystal. The most stable form of sulfurconsists of orthorhombic crystals of S8 molecules, which are often found near volcanos. Ifthe orthorhombic crystals are heated until they melt and the molten sulfur is then cooled,an allotrope of sulfur consisting of monoclinic crystals of S8 molecules is formed. The mon-oclinic crystals slowly transform themselves into the more stable orthorhombic structureover a period of time.

The tendency of an element to form bonds to itself is called catenation (from the Latinword catena, “chain”). Because sulfur forms unusually strong SOS single bonds, it is bet-ter at catenation than any element except carbon. As a result, the orthorhombic and mono-clinic forms of sulfur are not the only allotropes of the element. Allotropes of sulfur alsoexist that differ in the size of the molecules that form the crystal. Cyclic molecules thatcontain 6, 7, 8, 10, and 12 sulfur atoms are known.

Sulfur melts at 119.25°C to form a yellow liquid that is less viscous than water. If theliquid is heated to 159°C, it turns into a dark red liquid that can’t be poured from its con-tainer. The viscosity of the dark red liquid is 2000 times greater than that of molten sulfurbecause the cyclic S8 molecules open up and link together to form long chains of as manyas 100,000 sulfur atoms.

When sulfur reacts with an active metal, it can form the sulfide ion, S2�.

16 K(s) � S8(s) 88n 8 K2S(s)

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NONMETAL 17

This is not the only product that can be obtained, however. A variety of polysulfide ionswith a charge of �2 can be produced that differ in the number of sulfur atoms in the chain.

2 K(s) � S8(s) 88n K2S2 � [K�]2[SOS]2�

K2S3 � [K�]2[SOSOS]2�

K2S4 � [K�]2[SOSOSOS]2�

K2S5 � [K�]2[SOSOSOSOS]2�

K2S6 � [K�]2[SOSOSOSOSOS]2�

K2S8 � [K�]2[SOSOSOSOSOSOSOS]2�

Exercise N.4

Use the tendency of sulfur to form polysulfide ions to explain why iron has an oxidationnumber of �2 in iron pyrite, FeS2, one of the most abundant sulfur ores.

Solution

If the oxidation number of iron in FeS2 is �2, the sulfur must be present as the S22� ion.

FeS2 � [Fe2�][S22�]

The disulfide ion, S22�, is the sulfur analog of the peroxide ion, O2

2�, and has an analo-gous Lewis structure.

The Effect of Differences in the Electronegativities of Sulfur and Oxygen

Because sulfur is much less electronegative than oxygen, it is more likely to form com-pounds in which it has a positive oxidation number (see Table N.3).

TABLE N.3 Common Oxidation Numbers for Sulfur

Oxidation Number Examples

�2 Na2S, H2S�1 Na2S2, H2S2

0 S8

�1 S2Cl2�2 S2O3

2�

�2�12

� S4O62�

�3 S2O42�

�4 SF4, SO2, H2SO3, SO32�

�5 S2O62�

�6 SF6, SO3, H2SO4, SO42�

In theory, sulfur can react with oxygen to form either SO2 or SO3, whose Lewis struc-tures are given in Figure N.11.

SO2 is therefore a resonance hybrid of two Lewis structures analogous to the structureof ozone in Figure N.6. SO3 can be thought to result from the donation of a pair of non-bonding electrons on the sulfur atom in SO2 to an empty orbital on a neutral oxygen atomto form a covalent bond.

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18 NONMETAL

In practice, combustion of sulfur compounds gives SO2, regardless of whether sulfuror a compound of sulfur is burned.

S8(s) � 8 O2(g) 88n 8 SO2(g)CS2(l) � 3 O2(g) 88n CO2(g) � 2 SO2(g)

3 FeS2(s) � 8 O2(g) 88n Fe3O4(s) � 6 SO2(g)

Although the SO2 formed in the reactions should react with O2 to form SO3, the rate ofthat reaction is very slow. The rate of the conversion of SO2 into SO3 can be greatly in-creased by adding an appropriate catalyst.

V2O5/K2O2 SO2(g) � O2(g) 888888n 2 SO3(g)

Enormous quantities of SO2 are produced by industry each year and then convertedto SO3, which can be used to produce sulfuric acid, H2SO4. In theory, sulfuric acid can bemade by dissolving SO3 gas in water.

SO3(g) � H2O(l) 88n H2SO4(aq)

In practice, this isn’t convenient. Instead, SO3 is absorbed in 98% H2SO4, where it reactswith the water to form additional H2SO4 molecules. Water is then added, as needed, tokeep the concentration of the solution between 96% and 98% H2SO4 by weight.

Sulfuric acid is by far the most important industrial chemical. It has even been sug-gested that there is a correlation between the amount of sulfuric acid a country consumesand its standard of living.

Sulfuric acid dissociates in water to give the HSO4� ion, which is known as the hy-

drogen sulfate, or bisulfate, ion.

H2SO4(aq) 88n H�(aq) � HSO4�(aq)

Roughly 10% of the hydrogen sulfate ions dissociate further to give the SO42�, or sulfate, ion.

HSO4�(aq) 88n H�(aq) � SO4

2�(aq)

Sulfur dioxide dissolves in water to form sulfurous acid.

SO2(g) � H2O(l) 88n H2SO3(aq)

Sulfurous acid doesn’t dissociate in water to as great an extent as sulfuric acid.Sulfuric acid and sulfurous acid are examples of a class of compounds known as oxy-

acids because they are literally acids that contain oxygen. Because they are negative ions

S

O O O O

S

O O

O

SS

O O

SO2

SO3

O

S

O O

O

FIGURE N.11 Lewis structures of SO2

and SO3.

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NONMETAL 19

(or anions) that contain oxygen, the SO32� and SO4

2� ions are known as oxyanions. TheLewis structures of some of the oxides of sulfur that form oxyacids and oxyanions are givenin Figure N.12. One of the oxyanions deserves special mention. This ion, which is knownas the thiosulfate ion, is formed by the reaction between sulfur and the sulfite (SO3

2�) ion.

8 SO32�(aq) � S8(s) 88n 8 S2O3

2�(aq)

CheckpointUse the following Lewis structures to explain why the S2O3

2� ion is literally a thio-sulfate.

O

S

O

O HOH

Sulfuric acid, H2SO4

S

S

O

O HOH

Thiosulfuric acid, H2S2O3

O

S O HOH

Sulfurous acid, H2SO3

O

S

O

O OOH

O

S

O

O H

Peroxydisulfuric acid, H2S2O8

O

S

O

O OO

O

S

O

O

Peroxydisulfate, S2O82–

2–

O

S

O

S SO

O

S

O

O

Tetrathionate, S4O62–

2–

2–O

S OO

Sulfite, SO32–

S

S

O

OO

Thiosulfate, S2O32–

2–

O

S

O

OO

Sulfate, SO42–

2–

Oxyacids Oxyanions

FIGURE N.12 Oxyacids of sulfurand their oxyanions.

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20 NONMETAL

The Effect of Differences in the Abilities of Sulfur and Oxygen to Expand TheirValence Shell

Oxygen reacts with fluorine to form OF2.

O2(g) � 2 F2(g) 88n 2 OF2(g)

The reaction stops at this point because oxygen can hold only eight electrons in its valenceshell.

Sulfur, however, reacts with fluorine to form SF4 and SF6 because sulfur can expand its va-lence shell to hold 10 or even 12 electrons.

S8(s) � 16 F2(g) 88n 8 SF4(g)S8(s) � 24 F2(g) 88n 8 SF6(g)

There are 10 valence electrons on the sulfur atom in SF4, so the structure of the moleculeis based on a distorted trigonal bipyramid, as shown in Figure N.13. The 12 valence elec-trons on the central atom in SF6 are distributed toward the corners of an octahedron.

FF

F

F

F

S S

F

F

F

F F

186.9°

101.6°

SF4 SF6

N.5 THE CHEMISTRY OF NITROGENThe chemistry of nitrogen is dominated by the ease with which nitrogen atoms form dou-ble and triple bonds. A neutral nitrogen atom contains five valence electrons: 2s2 2p3. Anitrogen atom can therefore achieve an octet of valence electrons by sharing three pairsof electrons with another nitrogen atom.

Because the covalent radius of a nitrogen atom is relatively small, nitrogen atoms comeclose enough together to form very strong multiple bonds. The NqN triple bond (�H°ac ��945.41 kJ/molrxn) is almost twice as strong as the OPO double bond (�H°ac � �498.34kJ/molrxn).

The strength of the NqN triple bond makes the N2 molecule so inert that lithium isone of the few elements with which it reacts at room temperature.

6 Li(s) � N2(g) 88n 2 Li3N(s)

In spite of the fact that the N2 molecule is unreactive, compounds containing nitrogen ex-ist for virtually every element in the periodic table except those in Group VIIIA (He, Ne,

FIGURE N.13 Structures of SF4 and SF6.

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NONMETAL 21

Ar, and so on). This can be explained in two ways. First, N2 becomes significantly morereactive as the temperature increases. At high temperatures, nitrogen reacts with hydro-gen to form ammonia and with oxygen to form nitrogen oxide.

N2(g) � 3 H2(g) 88n 2 NH3(g)N2(g) � O2(g) 88n 2 NO(g)

Second, a number of catalysts found in nature can overcome the inertness of N2 at lowtemperatures.

The Synthesis of Ammonia

It is difficult to imagine a living system that doesn’t contain nitrogen, which is an essentialcomponent of the proteins, nucleic acids, vitamins, and hormones that make life possible.Animals pick up the nitrogen they need from the plants or other animals in their diet. Al-though there is an abundance of N2 in the atmosphere, plants cannot use nitrogen in itselemental form. Plants have to pick up their nitrogen from the soil or absorb it as N2 fromthe atmosphere. The concentration of nitrogen in the soil is fairly small, so the process bywhich plants reduce N2 to NH3 (nitrogen fixation) is extremely important.

Although 200 million tons of NH3 are produced by nitrogen fixation each year, plants,by themselves, cannot reduce N2 to NH3. The reaction is carried out by blue-green algaeand bacteria that are associated with certain plants. The best understood example of ni-trogen fixation involves the Rhizobium bacteria found in the root nodules of legumes suchas clover, peas, and beans. The bacteria contain a nitrogenase enzyme that is capable ofthe remarkable feat of reducing N2 from the atmosphere to NH3 at room temperature.

Ammonia is made on an industrial scale by a process first developed between 1909 and1913 by Fritz Haber. In the Haber process, a mixture of N2 and H2 gas at 200 to 300 atmand 400°C to 600°C is passed over a catalyst of finely divided iron metal.

FeN2(g) � 3 H2(g) 88n 2 NH3(g)

Almost 20 million tons of NH3 are produced in the United States each year by this process.About 80% of it, worth more than $2 billion, is used to make fertilizers for plants that can’tfix nitrogen from the atmosphere. On the basis of weight, ammonia is the second most im-portant industrial chemical in the United States. (Only sulfuric acid is produced in largerquantities.)

Two-thirds of the ammonia used for fertilizers is converted into solids such as ammo-nium nitrate, NH4NO3; ammonium phosphate, (NH4)3PO4; ammonium sulfate, (NH4)2SO4;and urea, H2NCONH2. The other third is applied directly to the soil as anhydrous (liter-ally, “without water”) ammonia. Ammonia is a gas at room temperature. It can be han-dled as a liquid when dissolved in water to form an aqueous solution. Alternatively, it canbe cooled to temperatures below �33°C, in which case the gas condenses to form the an-hydrous liquid, NH3(l).

The Synthesis of Nitric Acid

The NH3 produced by the Haber process that isn’t used as fertilizer is burned in oxygento generate nitrogen oxide.

4 NH3(g) � 5 O2(g) 88n 4 NO(g) � 6 H2O(g)

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22 NONMETAL

Nitrogen oxide—or nitric oxide, as it was once known—is a colorless gas that reacts rapidlywith oxygen to produce nitrogen dioxide, a dark brown gas.

2 NO(g) � O2(g) 88n 2 NO2(g)

Nitrogen dioxide dissolves in water to give nitric acid and NO, which can be captured andrecycled.

3 NO2(g) � H2O(l) 88n 2 HNO3(aq) � NO(g)

Thus, by a three-step process developed by Friedrich Ostwald in 1908, ammonia can beconverted into nitric acid.

4 NH3(g) � 5 O2(g) 88n 4 NO(g) � 6 H2O(g)2 NO(g) � O2(g) 88n 2 NO2(g)

3 NO2(g) � H2O(l) 88n 2 HNO3(aq) � NO(g)

The Haber process for the synthesis of ammonia combined with the Ostwald processfor the conversion of ammonia into nitric acid revolutionized the explosives industry. Ni-trates have been important explosives ever since Friar Roger Bacon mixed sulfur, saltpeter,and powdered carbon to make gunpowder in 1245.

16 KNO3(s) � S8(s) � 24 C(s) 88n 8 K2S(s) � 24 CO2(g) � 8 N2(g)�H° � �4575 kJ/molrxn

Before the Ostwald process the only source of nitrates for use in explosives was naturallyoccurring minerals such as saltpeter, which is a mixture of NaNO3 and KNO3. Once a de-pendable supply of nitric acid became available from the Ostwald process, a number of ni-trates could be made for use as explosives. Combining NH3 from the Haber process with HNO3

from the Ostwald process, for example, gives ammonium nitrate, which is both an excellentfertilizer and an inexpensive, dependable explosive commonly used in blasting powder.

2 NH4NO3(s) 88n 2 N2(g) � O2(g) � 4 H2O(g)

The destructive power of ammonium nitrate is apparent in photographs of the Alfred P.Murrah Federal Building in Oklahoma City, which was destroyed with a bomb made fromammonium nitrate on April 19, 1995.

Intermediate Oxidation Numbers

Nitric acid (HNO3) and ammonia (NH3) represent the maximum (�5) and minimum (�3)oxidation numbers for nitrogen. Nitrogen also forms compounds with every oxidation num-ber between these extremes (see Table N.4).

Negative Oxidation Numbers of Nitrogen besides �3

At about the time that Haber developed the process for making ammonia and Ostwaldworked out the process for converting ammonia into nitric acid, Raschig developed a processthat used the hypochlorite ion (OCl�) to oxidize ammonia to produce hydrazine, N2H4.

2 NH3(aq) � OCl�(aq) 88n N2H4(aq) � Cl�(aq) � H2O(l)

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NONMETAL 23

This reaction can be understood by noting that the OCl� ion is a two-electron oxidizingagent. Let’s therefore imagine a hypothetical mechanism for the reaction in which the firststep is the loss of a pair of nonbonding electrons from an ammonia molecule to form anNH3

2� ion, as shown in the first step in Figure N.14. A pair of nonbonding electrons froma second NH3 molecule could then be donated into the empty valence shell orbital on theNH3

2� ion to form an NON bond. In step 3, the product of the reaction in step 2 couldthen lose a pair of H� ions to form a hydrazine molecule.

TABLE N.4 Common Oxidation Numbers for Nitrogen


�3 NH3, NH4�, NH2

�, Mg3N2

�2 N2H4

�1 NH2OH��

13

� NaN3, HN3

�0 N2

�1 N2O�2 NO, N2O2

�3 HNO2, NO2�, N2O3, NO�

�4 NO2, N2O4

�5 HNO3, NO3�, N2O5

Hydrazine is a colorless liquid with a faint odor of ammonia that can be collected whenthe solution is heated until N2H4 distills out of the reaction flask. Many of the physicalproperties of hydrazine are similar to those of water.

H2O N2H4

Density 1.000 g/cm3 1.008 g/cm3

Melting Point 0.00°C 1.54°CBoiling Point 100°C 113.8°C

Step 1

Step 3

Step 2

H

H

H

N –2 e–H

H

H

N

2+

2+H

H

H

H

HH

N N

2+

+

+

H

H

H

N

H

H

H

N

NNH

H H

H2+H

H

H

N

H

H

H

N 2 H�

FIGURE N.14 Hydrazine is prepared by reacting NH3 with a two-electronoxidizing agent.

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24 NONMETAL

There is a significant difference between the chemical properties of the compounds, how-ever. Hydrazine burns when ignited in air to give nitrogen gas, water vapor, and largeamounts of energy.

N2H4(l) � O2(g) 88n N2(g) � 2 H2O(g) �H° � �534.3 kJ/molrxn

The principal use of hydrazine is as a rocket fuel. It is second only to liquid hydrogenin terms of the kilograms of thrust produced per kilogram of fuel burned. Hydrazine hasseveral advantages over liquid H2, however. It can be stored at room temperature, whereasliquid hydrogen must be stored at temperatures below �253°C. Hydrazine is also moredense than liquid H2 and therefore requires less storage space.

Pure hydrazine is seldom used as a rocket fuel because it freezes at the temperaturesencountered in the upper atmosphere. Hydrazine is mixed with N,N-dimethylhydrazine,(CH3)2NNH2, to form a solution that remains a liquid at low temperatures. Mixtures ofhydrazine and N,N-dimethylhydrazine were used to fuel the Titan II rockets that carriedthe Project Gemini spacecraft, and the reaction between hydrazine derivatives and N2O4

is still used to fuel the small rocket engines that enable the space shuttle to maneuverin space.

The product of the combustion of hydrazine is unusual. When carbon compounds burn,the carbon is oxidized to CO or CO2. When sulfur compounds burn, SO2 is produced.When hydrazine is burned, the product of the reaction is N2 because of the unusually strongNqN triple bond in the N2 molecule.

N2H4(l) � O2(g) 88n N2(g) � 2 H2O(g)

Positive Oxidation Numbers for Nitrogen: The Nitrogen Halides

Fluorine, oxygen, and chlorine are the only elements more electronegative than nitrogenthat form compounds with nitrogen. As a result, positive oxidation numbers of nitrogenare found in compounds that contain one or more of these elements.

In theory, N2 could react with F2 to form a compound with the formula NF3. In prac-tice, N2 is too inert to undergo the reaction at room temperature. NF3 is made by reactingammonia with F2 in the presence of a copper metal catalyst. The HF produced in the re-action combines with ammonia to form ammonium fluoride. The overall stoichiometry forthe reaction is therefore written as follows.

Cu4 NH3(g) � 3 F2(g) 88n NF3(g) � 3 NH4F(s)

The Lewis structure of NF3 is analogous to the Lewis structure of NH3, and the moleculeshave similar shapes.

Ammonia reacts with chlorine to form NCl3, which seems at first glance to be closelyrelated to NF3. But there is a significant difference between the compounds. NF3 is es-sentially inert at room temperature, whereas NCl3 is a shock-sensitive, highly explosive liq-uid that decomposes to form N2 and Cl2.

2 NCl3(l) 88n N2(g) � 3 Cl2(g)

Ammonia reacts with iodine to form a solid that is a complex between NI3 and NH3. Thismaterial is the subject of a popular, but dangerous, demonstration in which freshly pre-pared samples of NI3 in ammonia are poured onto filter paper, which is allowed to dry ona ring stand. After the ammonia evaporates, the NH3/NI3 crystals are touched with a feather

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NONMETAL 25

attached to a meter stick, resulting in detonation of the shock-sensitive solid, which de-composes to form a mixture of N2 and I2.

2 NI3(s) 88n N2(g) � 3 I2(g)

Positive Oxidation Numbers for Nitrogen: The Nitrogen Oxides

Lewis structures for seven oxides of nitrogen with oxidation numbers ranging from �1 to�5 are given in Figure N.15. These compounds all have two things in common: they con-tain NPO double bonds, and they are less stable than their elements in the gas phase.

FIGURE N.15 The oxides of nitrogen.

N N N NO O

Dinitrogen oxide, N2O(nitrous oxide)

N O

Nitrogen oxide, NO(nitric oxide)

O N

N O

Dinitrogen dioxide, N2O2

Dinitrogen trioxide, N2O3

N N

O

O

NO

O

Nitrogen dioxide, NO2

NNOO

OO

Dinitrogen tetroxide, N2O4

NN OO

O

O

O

Dinitrogen pentoxide, N2O5

O

Dinitrogen oxide, N2O, which is also known as nitrous oxide, can be prepared by care-fully decomposing ammonium nitrate.

170–200°CNH4NO3(s) 888888n N2O(g) � 2 H2O(g)

Nitrous oxide is a sweet-smelling, colorless gas best known to nonchemists as “laughinggas.” As early as 1800, Humphry Davy noted that N2O, inhaled in relatively small amounts,produced a state of apparent intoxication often accompanied by either convulsive laugh-ter or crying. When taken in larger doses, nitrous oxide provides fast and efficient relieffrom pain. N2O was therefore used as the first anesthetic. Because large doses are needed

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26 NONMETAL

to produce anesthesia, and continued exposure to the gas can be fatal, N2O is used todayonly for relatively short operations.

Nitrous oxide has several other interesting properties. First, it is highly soluble in cream;for that reason, it is used as the propellant in whipped cream dispensers. Second, althoughN2O does not burn by itself, it is better than air at supporting the combustion of other ob-jects. This can be explained by noting that N2O can decompose to form an atmospherethat is one-third O2 by volume, whereas normal air is only 21% oxygen by volume.

2 N2O(g) 88n 2 N2(g) � O2(g)

For many years, the endings -ous and -ic were used to distinguish between the lowestand highest of a pair of oxidation numbers. N2O is nitrous oxide because the oxidationnumber of the nitrogen is �1. NO is nitric oxide because the oxidation number of thenitrogen is �2.

Enormous quantities of nitrogen oxide, or nitric oxide, are generated each year by thereaction between the N2 and O2 in the atmosphere, catalyzed by a stroke of lightningpassing through the atmosphere or by the hot walls of an internal combustion engine.

N2(g) � O2(g) 88n 2 NO(g)

One of the reasons for lowering the compression ratio of automobile engines in recentyears is to decrease the temperature of the combustion reaction, thereby decreasing theamount of NO emitted into the atmosphere.

NO can be prepared in the laboratory by reacting copper metal with dilute nitric acid.

3 Cu(s) � 8 HNO3(aq) 88n 3 Cu(NO3)2(aq) � 2 NO(g) � 4 H2O(l)

The NO molecule contains an odd number of valence electrons. As a result, it is impossi-ble to write a Lewis structure for the molecule in which all of the electrons are paired (seeFigure N.15). When NO gas is cooled, pairs of NO molecules combine in a reversible re-action to form a dimer (from Greek, meaning “two parts”), with the formula N2O2, inwhich all of the valence electrons are paired, as shown in Figure N.15.

NO reacts rapidly with O2 to form nitrogen dioxide (once known as nitrogen perox-ide), which is a dark brown gas at room temperature.

2 NO(g) � O2(g) 88n 2 NO2(g)

NO2 can be prepared in the laboratory by heating certain metal nitrates until they de-compose.

2 Pb(NO3)2(s) 88n 2 PbO(s) � 4 NO2(g) � O2(g)

It can also be made by reacting copper metal with concentrated nitric acid.

Cu(s) � 4 HNO3(aq) 88n Cu(NO3)2(aq) � 2 NO2(g) � 2 H2O(l)

NO2 also has an odd number of electrons and therefore contains at least one unpaired elec-tron in its Lewis structures. NO2 dimerizes at low temperatures to form N2O4 molecules,in which all the electrons are paired, as shown in Figure N.15.

Mixtures of NO and NO2 combine when cooled to form dinitrogen trioxide, N2O3,which is a blue liquid. The formation of a blue liquid when either NO or NO2 is cooled

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NONMETAL 27

therefore implies the presence of at least a small portion of the other oxide because N2O2

and N2O4 are both colorless.By carefully removing water from concentrated nitric acid at low temperatures with a

dehydrating agent we can form dinitrogen pentoxide.

4 HNO3(aq) � P4O10(s) 88n 2 N2O5(s) � 4 HPO3(s)

N2O5 is a colorless solid that decomposes in light or on warming to room temperature. Asmight be expected, N2O5 dissolves in water to form nitric acid.

N2O5(s) � H2O(l) 88n 2 HNO3(aq)

N.6 THE CHEMISTRY OF PHOSPHORUSPhosphorus was the first element whose discovery can be traced to a single individual. In1669, while searching for a way to convert silver into gold, Hennig Brand obtained a white,waxy solid that glowed in the dark and burst spontaneously into flame when exposed toair. Brand made the substance by evaporating the water from urine and allowing the blackresidue to putrefy for several months. He then mixed the residue with sand, heated themixture in the presence of a minimum of air, and collected under water the volatile prod-ucts that distilled out of the reaction flask.

Phosphorus forms a number of compounds that are direct analogs of nitrogen-containing compounds. However, the fact that elemental nitrogen is virtually inert at roomtemperature, whereas elemental phosphorus can burst spontaneously into flame when ex-posed to air, shows that there are differences between the elements as well. Phosphorusoften forms compounds with the same oxidation numbers as the analogous nitrogen com-pounds, but with different formulas, as shown in Table N.5.

TABLE N.5 Nitrogen and Phosphorus Compounds with the Same Oxidation Numbers but Different Formulas

Nitrogen Compound Phosphorus Compound Oxidation Number

N2 P4 �0HNO2 (nitrous acid) H3PO3 (phosphorous acid) �3N2O3 P4O6 �3HNO3 (nitric acid) H3PO4 (phosphoric acid) �5NaNO3 (sodium nitrate) Na3PO4 (sodium phosphate) �5N2O5 P4O10 �5

The same factors that explain the differences between sulfur and oxygen can be usedto explain the differences between phosphorus and nitrogen.

• NqN triple bonds are much stronger than PqP triple bonds.• POP single bonds are stronger than NON single bonds.• Phosphorus is much less electronegative than nitrogen.• Phosphorus can expand its valence shell to hold more than eight electrons, but ni-

trogen cannot.

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28 NONMETAL

The Effect of Differences in the XOOX and XqqX Bond Strengths

The ratio of the radii of phosphorus and nitrogen atoms is the same as the ratio of the radiiof sulfur and oxygen atoms, within experimental error.

� � 1.6

As a result, PqP triple bonds are much weaker than NqN triple bonds, for the same rea-son that SPS double bonds are weaker than OPO double bonds, namely, phosphorusatoms are too big to come close enough together to form strong multiple bonds.

Each atom in an N2 molecule completes its octet of valence electrons by sharing threepairs of electrons with a single neighboring atom. Because phosphorus doesn’t form strongmultiple bonds with itself, elemental phosphorus consists of tetrahedral P4 molecules inwhich each atom forms single bonds with three neighboring atoms, as shown in FigureN.16.

0.110 nm��0.070 nm

Covalent radius of phosphorus��

Covalent radius of nitrogen

P4, white phosphorus

P

P

P

P

FIGURE N.16 Tetrahedral P4 molecule.

Phosphorus is a white solid with a waxy appearance, which melts at 44.1°C and boilsat 287°C. It is made by reducing calcium phosphate with carbon in the presence of silica(sand) at very high temperatures.

2 Ca3(PO4)2(s) � 6 SiO2(s) � 10 C(s) 88n 6 CaSiO3(s) � P4(s) � 10 CO(g)

White phosphorus is stored under water because the element spontaneously bursts intoflame in the presence of oxygen at temperatures only slightly above room temperature.Although phosphorus is insoluble in water, it is very soluble in carbon disulfide. Solutionsof P4 in CS2 are reasonably stable; as soon as the CS2 evaporates, however, the phospho-rus bursts into flame.

The POPOP bond angle in a tetrahedral P4 molecule is only 60°. This very small an-gle produces a considerable amount of strain in the P4 molecule, which can be relieved bybreaking one of the POP bonds. Phosphorus therefore forms other allotropes by openingup the P4 tetrahedron. When white phosphorus is heated to 300°C, one bond inside eachP4 tetrahedron is broken, and the P4 molecules link together to form a polymer (from theGreek words pol, “many,” and meros, “parts”) with the structure shown in Figure N.17.This allotrope of phosphorus is dark red, and its presence in small traces often gives whitephosphorus a light yellow color. Red phosphorus is more dense (2.16 g/cm3) than whitephosphorus (1.82 g/cm3) and is much less reactive at normal temperatures.

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NONMETAL 29

Red phosphorus

P P

P

P

P P

P

P

P P

P

P

The Effect of Differences in the Strengths of PPPX and NPPX Double Bonds

The size of a phosphorus atom also interferes with its ability to form double bonds to otherelements such as oxygen, nitrogen, and sulfur. As a result, phosphorus tends to form com-pounds that contain two POO single bonds, where nitrogen would form an NPO doublebond. Nitrogen forms the nitrate (NO3

�) ion, for example, in which it has an oxidationnumber of �5. When phosphorus forms an ion with the same oxidation number, it is thephosphate (PO4

3�) ion, as shown in Figure N.18.

O

N

O O

P

O

O

OO

– 3–

Similarly, nitrogen forms nitric acid, HNO3, which contains an NPO double bond, whereasphosphorus forms phosphoric acid, H3PO4, which contains POO single bonds, as shownin Figure N.19.

P

O

O

O HHH NO OO

O H

The Effect of Differences in the Electronegativities of Phosphorus and Nitrogen

Because phosphorus is less electronegative than nitrogen, it is more likely to exhibit pos-itive oxidation numbers. The most important oxidation numbers for phosphorus are �3,�3, and �5 (see Table N.6).

FIGURE N.17 Portion of the polymeric chain in red phosphorus.

FIGURE N.18 Lewis structures for the NO3� and PO4

3� ions.

FIGURE N.19 Lewis structures for nitric acid (HNO3) andphosphoric acid (H3PO4).

TABLE N.6 Common Oxidation Numbers of Phosphorus


�3 Ca3P2

�3 PF3, P4O6, H3PO3

�5 PF5, P4O10, H3PO4, PO43�

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30 NONMETAL

Because it is more electronegative than most metals, phosphorus reacts with metals atelevated temperatures to form phosphides, in which it has an oxidation number of �3.

6 Ca(s) � P4(s) 88n 2 Ca3P2(s)

The metal phosphides react with water to produce a poisonous, highly reactive, colorlessgas known as phosphine (PH3), which has one of the foulest odors the authors have en-countered.

Ca3P2(s) � 6 H2O(l) 88n 2 PH3(g) � 3 Ca2�(aq) � 6 OH�(aq)

Samples of PH3, the phosphorus analog of ammonia, are often contaminated by traces ofP2H4, the phosphorus analog of hydrazine. As if the toxicity and odor of PH3 were notenough, mixtures of PH3 and P2H4 burst spontaneously into flame in the presence ofoxygen.

Compounds such as Ca3P2 and PH3, in which phosphorus has a negative oxidation num-ber, are far outnumbered by compounds in which the oxidation number of phosphorus ispositive. Phosphorus burns in O2 to produce P4O10 in a reaction that gives off extraordi-nary amounts of energy in the form of heat and light.

P4(s) � 5 O2(g) 88n P4O10(s) �H° � �2984 kJ/molrxn

When phosphorus burns in the presence of a limited amount of O2, P4O6 is produced.

P4(s) � 3 O2(g) 88n P4O6(s) �H° � �1640 kJ/molrxn

P4O6 consists of a tetrahedron in which an oxygen atom has been inserted into each POPbond in the P4 molecule (see Figure N.20). P4O10 has an analogous structure, with an ad-ditional oxygen atom bound to each of the four phosphorus atoms.

O

PP

OO

O

P

O P

O

P4O6

O

OO

PP

OO

O

P

O P

O

O

OP4O10

P4O6 and P4O10 react with water to form phosphorous acid, H3PO3, and phosphoricacid, H3PO4, respectively.

P4O6(s) � 6 H2O(l) 88n 4 H3PO3(aq)P4O10(s) � 6 H2O(l) 88n 4 H3PO4(aq)

(P4O10 has such a high affinity for water that it is commonly used as a dehydrating agent.)Phosphorous acid, H3PO3, and phosphoric acid, H3PO4, are examples of a large class ofoxyacids of phosphorus. Lewis structures for some of these oxyacids and their relatedoxyanions are given in Figure N.21.

FIGURE N.20 Structures of P4O6 and P4O10.

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NONMETAL 31

Hypophosphite, H2PO2–

O

H

OH P

Phosphite, HPO32–

O

O

OH P

2–

Phosphate, PO43–

O

O

OO P

3–

–

O O

O O

OO OP PHH

O O

O O

OO OH P P

H H

O

O

O HP

HTriphosphoric acid, H5P3O10 Triphosphate, P3O10

5–

Diphosphate, P2O74–

(pyrophosphate)

5–

4–

O O

O O

OO OP P

O

O

OP

O O

O O

OO OP P

H HDiphosphoric acid, H4P2O7

(pyrophosphoric acid)

O

H

OH P H

Hypophosphorous acid, H3PO2

O

O

OH P H

HPhosphorous acid, H3PO3

O

O

OO PH H

HPhosphoric acid, H3PO4

Oxyacid Oxyanion

The Effect of Differences in the Abilities of Phosphorus and Nitrogen to ExpandTheir Valence Shell

The reaction between ammonia and fluorine stops at NF3 because nitrogen uses the 2s, 2px,2py, and 2pz orbitals to hold valence electrons. Nitrogen atoms can therefore hold a maxi-mum of eight valence electrons. Phosphorus, however, can expand the valence shell to hold10 or more electrons. Thus, phosphorus can react with fluorine to form both PF3 and PF5.Phosphorus can even form the PF6

� ion, in which there are 12 valence electrons on thecentral atom, as shown in Figure N.22.

FIGURE N.21 Oxyacids of phosphorus and their oxyanions.

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32 NONMETAL

F

FF

F

P

F

F F

F

F

P

F

F

PF6–

PF5

N.7 THE CHEMISTRY OF THE HALOGENSThere are six elements in Group VIIA, the next-to-last column of the periodic table. Asexpected, these elements have certain properties in common. They all form diatomic mol-ecules (H2, F2, Cl2, Br2, I2, and At2), for example, and they all form negatively chargedions (H�, F�, Cl�, Br�, I�, and At�).

When the chemistry of these elements is discussed, hydrogen is separated from the oth-ers and astatine is ignored because it is radioactive. (The most stable isotopes of astatinehave half-lives of less than a minute. As a result, the largest samples of astatine compoundsstudied to date have weighed less than 50 ng.) Discussions of the chemistry of the elementsin Group VIIA therefore focus on four elements: fluorine, chlorine, bromine, and iodine.These elements are called the halogens (from the Greek words hals, “salt,” and gennan,“to form or generate”) because they are literally the salt formers.

None of the halogens can be found in nature in their elemental form. They are invari-ably found as salts of the halide ions (F�, Cl�, Br�, and I�). Fluoride ions are found inminerals such as fluorite (CaF2) and cryolite (Na3AlF6). Chloride ions are found in rocksalt (NaCl), in the oceans, which are roughly 2% Cl� ion by weight, and in lakes that havea high salt content, such as the Great Salt Lake in Utah, which is 9% Cl� ion by weight.Both bromide and iodide ions are found at low concentrations in the oceans, as well as inbrine wells in Louisiana, California, and Michigan.

The Halogens in Their Elemental Form

Fluorine (F2), a highly toxic, colorless gas, is the most reactive element known—so reac-tive that asbestos, water, and silicon burst into flame in its presence. It is so reactive it evenforms compounds with Kr, Xe, and Rn, elements that were once thought to be inert. Flu-orine is such a powerful oxidizing agent that it can coax elements into unusually high ox-idation numbers, as in AgF2, PtF6, and IF7.

Fluorine is so reactive that it is difficult to find a container in which it can be stored.F2 attacks both glass and quartz, for example, and it causes most metals to burst into flame.Fluorine is handled in equipment built out of certain alloys of copper and nickel. It stillreacts with the alloys, but it forms a layer of a fluoride compound on the surface that pro-tects the metal from further reaction.

Fluorine is used in the manufacture of Teflon—or poly(tetrafluoroethylene), (C2F4)n—which is used for everything from linings for pots and pans to gaskets that are inert tochemical reactions.

Chlorine (Cl2) is a highly toxic gas with a pale yellow-green color. Chlorine is a verystrong oxidizing agent, which is used commercially as a bleaching agent and as a disinfec-tant. It is strong enough to oxidize the dyes that give wood pulp its yellow or brown color,for example, thereby bleaching out this color, and strong enough to destroy bacteria andthereby act as a germicide. Large quantities of chlorine are used each year to make solventssuch as carbon tetrachloride (CCl4), chloroform (CHCl3), dichloroethylene (C2H2Cl2), andtrichloroethylene (C2HCl3).

FIGURE N.22 Structures of PF5 and the PF6� ion.

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NONMETAL 33

Bromine (Br2) is a reddish-orange liquid with an unpleasant, choking odor. The nameof the element, in fact, comes from the Greek stem bromos, “stench.” Bromine is used toprepare flame retardants, fire-extinguishing agents, sedatives, antiknock agents for gaso-line, and insecticides.

Iodine is an intensely colored solid with an almost metallic luster. The solid is relativelyvolatile, and it sublimes when heated to form a violet-colored gas. Iodine has been used formany years as a disinfectant in “tincture of iodine.” Iodine compounds are used as catalysts,drugs, and dyes. Silver iodide (AgI) plays an important role in the photographic processand in attempts to make rain by seeding clouds. Iodide is also added to salt to protect againstgoiter, an iodine deficiency disease characterized by a swelling of the thyroid gland.

Some of the chemical and physical properties of the halogens are summarized inTable N.7.

TABLE N.7 Properties of F2, Cl2, Br2, and I2

First a

Melting Boiling Ionization Electrona Ionica

Point Point Energy Affinity Radius Density(°C) (°C) Color (kJ/mol) (kJ/mol) (nm) (g/cm3)

F2 �219.6 �188.1 Colorless 1681.0 328.0 0.136 1.513Cl2 �101 �34.0 Pale green 1251.1 348.8 0.181 1.655Br2 �7.2 59.5 Dark red-brown 1139.9 324.6 0.196 3.187I2 112.9 185.2 Dark violet, 1008.4 295.3 0.216 3.960

almost black

There is a regular increase in many of the properties of the halogens as we proceed downthe column from fluorine to iodine, including the melting point, boiling point, intensity ofthe color of the halogen, the radius of the corresponding halide ion, and the density of theelement. On the other hand, there is a regular decrease in the first ionization energy as wego down the column. As a result, there is a regular decrease in the oxidizing strength ofthe halogens from fluorine to iodine.

F2 � Cl2 � Br2 � I2oxidizing strength

This trend is mirrored by an increase in the reducing strength of the corresponding halides.

I� � Br� � Cl� � F�

reducing strength

Exercise N.5

Use the fact that Cl2 is a stronger oxidizing agent than Br2 to devise a way to prepare elemental bromine from an aqueous solution of the Br� ion.

Solution

When the chemistry of the main-group metals was introduced in Chapter 5, we found thatmetals can be prepared by reacting one of their salts with a metal that is a stronger reducingagent. Titanium metal, for example, can be prepared by reacting TiCl4 with magnesium metal.

TiCl4(l) � 2 Mg(s) 88n Ti(s) � 2 MgCl2(s)

aFor the atomic species.

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34 NONMETAL

Extending this argument to oxidizing agents suggests that we can produce Br2 by reactinga solution that contains the Br� ion with something that is an even stronger oxidizing agentthan Br2, such as Cl2 dissolved in water.

2 Br�(aq) � Cl2(aq) 88n Br2(aq) � 2 Cl�(aq)

Methods of Preparing the Halogens from Their Halides

The halogens can be made by reacting a solution of the corresponding halide ion with asubstance that is a stronger oxidizing agent than the halogen being prepared. Iodine, forexample, can be made by reacting the iodide ion with either bromine or chlorine.

2 I�(aq) � Br2(aq) 88n I2(aq) � 2 Br�(aq)

Bromine can be prepared by reacting bromide ions with a solution of Cl2 dissolved in water.

2 Br�(aq) � Cl2(aq) 88n Br2(aq) � 2 Cl�(aq)

To prepare Cl2, we need an unusually strong oxidizing agent, such as manganese dioxide(MnO2) or the permanganate ion (MnO4

�).

2 Cl�(aq) � MnO2(aq) � 4 H�(aq) 88n Cl2(aq) � Mn2�(aq) � 2 H2O(l)

The synthesis of fluorine escaped the efforts of chemists for almost 100 years. Part ofthe problem involved finding an oxidizing agent strong enough to oxidize the F� ion to F2.The task of preparing fluorine was made even more difficult by the extraordinary toxicityof both F2 and the hydrogen fluoride (HF) used to make it.

The best way of producing strong reducing agents is to pass an electric current througha salt of the metal. Sodium, for example, can be prepared by the electrolysis of moltensodium chloride.

electrolysis2 NaCl(l) 888888n 2 Na(s) � Cl2(g)

The same process can be used to generate strong oxidizing agents, such as F2.Attempts to prepare fluorine by electrolysis, however, were initially unsuccessful.

Humphry Davy, who prepared potassium, sodium, barium, strontium, calcium, and mag-nesium by electrolysis, repeatedly tried to prepare F2 by the electrolysis of fluorite (CaF2),and succeeded only in ruining his health. Joseph Louis Gay-Lussac and Louis JacquesThenard, who prepared elemental boron for the first time, also tried to prepare fluorineand suffered from very painful exposures to hydrogen fluoride. George and Thomas Knoxwere badly poisoned during their attempts to make fluorine, and both Paulin Louyet andJerome Nickles died from fluorine poisoning.

Finally, in 1886 Henri Moissan successfully isolated F2 gas from the electrolysis of amixed salt of KF and HF and noted that crystals of silicon burst into flame when mixedwith the gas. Electrolysis of KHF2 is still used to prepare fluorine today.

electrolysis2 KHF2(s) 888888n H2(g) � F2(g) � 2 KF(s)

A schematic drawing of the cell in which KHF2 is electrolyzed to produce F2 gas at the an-ode and H2 gas at the cathode is shown in Figure N.23.

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NONMETAL 35

Common Oxidation Numbers for the Halogens

Fluorine is the most electronegative element in the periodic table. As a result, it has anoxidation number of �1 in all its compounds. Because chlorine, bromine, and iodine areless electronegative, it is possible to prepare compounds in which those elements have ox-idation numbers of �1, �3, �5, and �7, as shown in Table N.8.

Anodeconnection

F2 outlet

HF inlet H2 outlet

Cellcover

Carbonanode

Coolingjacket

Steelcathode

Gas separation skirt

Molten solutionof KF dissolvedin HF

General Trends in Halogen Chemistry

There are several patterns in the chemistry of the halogens.

• The chemistry of fluorine is simplified by the fact that it is the most electronegative element in the periodic table that forms compounds and by the fact that it cannotexpand its valence shell to hold more than eight valence electrons.

• Chlorine, bromine, and iodine can expand their valence shells to hold as many as14 valence electrons.

• The chemistry of the halogens is dominated by oxidation–reduction reactions.

The Hydrogen Halides (HX)

The hydrogen halides are compounds that contain hydrogen attached to one of the halo-gens (HF, HCl, HBr, and HI). The compounds are all colorless gases that are soluble in

TABLE N.8 Common Oxidation Numbers for the Halogens


�1 CaF2, HCl, NaBr, AgI�0 F2, Cl2, Br2, I2

�1 HClO, ClF�3 HClO2, ClF3

�5 HClO3, BrF5, BrF6�, IF5

�7 HClO4, BrF6�, IF7

FIGURE N.23 Schematic drawing of the cell in which KHF2 is elec-trolyzed.

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36 NONMETAL

water. Up to 512 mL of HCl gas can dissolve in 1 mL of water at 0°C and 1 atm, for ex-ample. Each of the hydrogen halides ionizes to at least some extent when it dissolves inwater.

H2OHCl(g) 88n H�(aq) � Cl�(aq)

Exercise N.6

Explain why it is easy to believe that HCl is an ionic compound. Describe the best evi-dence that it isn’t an ionic compound.

Solution

It is easy to believe HCl is an ionic compound because it forms ions when it dissolves inwater.

H2OHCl(g) 88n H�(aq) � Cl�(aq)

At first glance, the reaction seems to be similar to the reaction that occurs when ionic com-pounds dissolve in water.

H2ONaCl(s) 88n Na�(aq) � Cl�(aq)

The best evidence that HCl isn’t an ionic compound is the fact that it is a gas at room tem-perature, and ionic compounds are invariably solids at room temperature.

Some chemists try to distinguish between the behavior of HCl and NaCl when they dis-solve in water as follows. They argue that HCl ionizes when it dissolves in water becauseions are created by the reaction. NaCl, on the other hand, dissociates in water because NaClalready consists of Na� and Cl� ions in the solid.

Several of the hydrogen halides can be prepared directly from the elements. Mixturesof H2 and Cl2, for example, react with explosive violence in the presence of light to formHCl.

H2(g) � Cl2(g) 88n 2 HCl(g)

Because chemists are usually more interested in aqueous solutions of hydrogen halidesthan they are in the pure gases, the compounds are usually synthesized in water. Aqueoussolutions of the hydrogen halides are often called mineral acids because they are literallyacids prepared from minerals. Hydrochloric acid is prepared by reacting table salt with sul-furic acid, for example, and hydrofluoric acid is prepared from fluorite and sulfuric acid.

2 NaCl(s) � H2SO4(aq) 88n 2 HCl(aq) � Na2SO4(aq)CaF2(s) � H2SO4(aq) 88n 2 HF(aq) � CaSO4(aq)

The acids are purified by taking advantage of the ease with which HF and HCl gas boilout of the solutions. The gas given off when one of the solutions is heated is collected andthen redissolved in water to give relatively pure samples of the mineral acid.

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NONMETAL 37

The Interhalogen Compounds

Interhalogen compounds are formed by reactions between different halogens. All possibleinterhalogen compounds of the type XY are known. Bromine reacts with chlorine, for ex-ample, to give BrCl, which is a gas at room temperature.

Br2(l) � Cl2(g) 88n 2 BrCl(g)

Interhalogen compounds with the general formulas XY3, XY5, and even XY7 are formedwhen pairs of halogens react. Chlorine reacts with fluorine, for example, to form chlorinetrifluoride.

Cl2(g) � 3 F2(g) 88n 2 ClF3(g)

The compounds are easiest to form when Y is fluorine. Iodine is the only halogen thatforms an XY7 interhalogen compound, and it does so only with fluorine.

ClF3 and BrF5 are extremely reactive compounds. ClF3 is so reactive that wood, as-bestos, and even water spontaneously burn in its presence. The compounds are excellentfluorinating agents, which tend to react with each other to form positive ions such as ClF2

�

and BrF4� and negative ions such as ClF4

� and BrF6�.

2 BrF5(l) 88n [BrF4�][BrF6

�](s)

Neutral Oxides of the Halogens

Under certain conditions, it is possible to isolate neutral oxides of the halogens, such asCl2O, Cl2O3, ClO2, Cl2O4, Cl2O6, and Cl2O7. Cl2O7, for example, can be obtained by de-hydrating perchloric acid, HClO4. The oxides are notoriously unstable compounds that ex-plode when subjected to either thermal or physical shock. Some are so unstable they det-onate when warmed to temperatures above �40°C.

Oxyacids of the Halogens and Their Salts

Chlorine reacts with the OH� ion to form chloride ions and hypochlorite (OCl�) ions.

Cl2(aq) � 2 OH�(aq) 88n Cl�(aq) � OCl�(aq) � H2O(l)

This is a disproportionation reaction in which one-half of the chlorine atoms are oxidizedto hypochlorite ions and the other half are reduced to chloride ions.

When the solution is hot, this reaction gives a mixture of the chloride and chlorate (ClO3�)

ions.

3 Cl2(aq) � 6 OH�(aq) 88n 5 Cl�(aq) � ClO3�(aq) � 3 H2O(l)

Under carefully controlled conditions, it is possible to convert a mixture of the chlorateand hypochlorite ions into a solution that contains the chlorite (ClO2

�) ion.

ClO3�(aq) � ClO�(aq) 88n 2 ClO2

�(aq)

Cl� � OCl� � H2O

Reduction

Oxidation

0 �1

Cl2 � 2 OH�

�1

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38 NONMETAL

The last member of this class of compounds, the perchlorate ion (ClO4�), is made by elec-

trolyzing solutions of the chlorate ion.The names of the oxyanions of the halogens use the endings -ite and -ate to indicate

low and high oxidation numbers and the prefixes hypo- and per- to indicate the very low-est and very highest oxidation numbers, as shown in Table N.9. Each of the ions can beconverted into an oxyacid, which is named by replacing the -ite ending with -ous and the-ate ending with -ic.

TABLE N.9 Oxyanions and Oxyacids of Chlorine

Oxidation StateOxyanion Name Oxyacid Name of Chlorine

ClO� Hypochlorite HClO Hypochlorous acid �1ClO2

� Chlorite HOClO Chlorous acid �3ClO3

� Chlorate HOClO2 Chloric acid �5ClO4

� Perchlorate HOClO3 Perchloric acid �7

The hypochlorite (OCl�) ion is the active ingredient in liquid bleaches, such as Clorox.Calcium salts of the ion can be found in dry bleaches, such as Clorox 2. Ca(OCl)2 is alsothe active ingredient in most commercial products used to “chlorinate” swimming pools.

N.8 THE CHEMISTRY OF THE RARE GASESIn 1892 Lord Rayleigh noted that nitrogen isolated from air was more dense than nitro-gen prepared by decomposing ammonia. William Ramsay attacked this problem by puri-fying a sample of nitrogen gas to remove any moisture, carbon dioxide, and organic cont-aminants. He then passed the purified gas over hot magnesium metal, which reacts withnitrogen to form the nitride.

3 Mg(s) � N2(s) 88n Mg3N2(s)

When he was finished, Ramsay was left with a small residue of gas that occupied roughly1/80th of the original volume. He excited the gas in an electric discharge tube and foundthat the resulting emission spectrum contained lines that differed from those of all knowngases. After repeated discussions of the results of these experiments, Rayleigh and Ramsayjointly announced the discovery of a new element, which they named argon from the Greekword meaning “lazy one” because the gas refused to react with any element or compoundthey tested.

Argon didn’t fit into any of the known families of elements in the periodic table, butits atomic weight suggested that it might belong to a new group that could be insertedbetween chlorine and potassium. Shortly after reporting the discovery of argon in 1894,Ramsay found another unreactive gas when he heated a mineral of uranium. The lines inthe spectrum of the gas also occurred in the spectrum of the sun, which led Ramsay toname the element helium (from the Greek word helios, “sun”). Experiments with liquidair led Ramsay to a third gas, which he named krypton (“the hidden one”). Experimentswith liquid argon led him to a fourth gas, neon (“the new one”), and finally a fifth gas,xenon (“the stranger”).

These elements were discovered between 1894 and 1898. Because Moissan had onlyrecently isolated fluorine for the first time and fluorine was the most active of the known

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NONMETAL 39

elements, Ramsay sent a sample of argon to Moissan to see whether it would react withfluorine. It did not. The failure of Moissan’s attempts to react argon with fluorine, coupledwith repeated failures by other chemists to get the more abundant of the gases to undergochemical reaction, eventually led to their being labeled inert gases. The development of theelectronic theory of atoms did little to dispel this notion because it was obvious that thesegases had very stable electron configurations. As a result, the elements were labeled “in-ert gases” in almost every textbook and periodic table until about 30 years ago.

In 1962 Neil Bartlett found that PtF6 was a strong enough oxidizing agent to removean electron from an O2 molecule.

PtF6(g) � O2(g) 88n [O2�][PtF6

�](s)

Bartlett realized that the first ionization energy of Xe (1170 kJ/mol) was slightly smallerthan the first ionization energy of the O2 molecule (1177 kJ/mol). He therefore predictedthat PtF6 might also react with Xe. When he ran the reaction, he isolated the first com-pound of a Group VIIIA element.

Xe(g) � PtF6(g) 88n [Xe�][PtF6�](s)

A few months later, workers at the Argonne National Laboratory near Chicago foundthat Xe reacts with F2 to form XeF4. Since that time, more than 200 compounds of Kr, Xe,and Rn have been isolated. No compounds of the more abundant elements in the group(He, Ne, and Ar) have yet been isolated. However, the fact that elements in the family canundergo chemical reactions has led to the use of the term rare gases rather than inert gasesto describe these elements.

Compounds of xenon are by far the most numerous of the rare gas compounds. Withthe exception of XePtF6, rare gas compounds have oxidation numbers of �2, �4, �6, and�8, as shown by the examples cited in Table N.10.

There is some controversy over whether the rare gases should be viewed as having theoutermost shell of electrons filled (in which case they should be labeled Group VIIIA) orempty (in which case they should be labeled Group 0). The authors believe these elementsshould be labeled Group VIIIA because they behave as if they contribute eight valenceelectrons when they form compounds.

TABLE N.10 Compounds of Xenon and Their Oxidation Numbers

Compound Oxidation Number Compound Oxidation Number

XeF� �2 XeO3 �6XeF2 �2 XeOF4 �6Xe2F3

� �2 XeO2F2 �6XeF3

� �4 XeO3F� �6XeF4 �4 XeO4 �8XeOF2 �4 XeO6

4� �8XeF5

� �6 XeO3F2 �8XeF6 �6 XeO2F4 �8Xe2F11

� �6 XeOF5� �8

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40 NONMETAL

The synthesis of most xenon compounds starts with the reaction between Xe and F2 athigh temperatures (250–400°C) to form a mixture of XeF2, XeF4, and XeF6.

3 Xe(g) � 6 F2(g) 88n XeF2(s) � XeF4(s) � XeF6(s)

The positively charged XeFn� ions are then made by reacting XeF2, XeF4, or XeF6 with

either AsF5, SbF5, or BiF5.

XeF2(s) � SbF5(l) 88n [XeF�][SbF6�](s)

2 XeF2(s) � AsF5(g) 88n [Xe2F3�][AsF6

�](s)XeF4(s) � BiF5(s) 88n [XeF3

�][BiF6�](s)

2 XeF6(s) � AsF5(g) 88n [Xe2F11�][AsF6

�](s)

Oxides of xenon, such as XeOF2, XeOF4, XeO2F2, XeO3F2, XeO2F4, XeO3, and XeO4, areprepared by reacting XeF4 or XeF6 with water. The XeO6

4� ion, for example, is producedwhen XeF6 dissolves in strong base.

2 XeF6(s) � 16 OH�(aq) 88n XeO64�(aq) � Xe(g) � O2(g) � 12 F�(aq) � 8 H2O(l)

Some xenon compounds are relatively stable. XeF2, XeF4, and XeF6, for example, arestable solids that can be purified by sublimation in a vacuum at 25°C. XeOF4 and Na4XeO6

are also reasonably stable. Others, such as XeO3, XeO4, XeOF2, XeO2F2, XeO3F2, andXeO2F4, are unstable compounds that can decompose violently.

The principal use of rare gas compounds at present is as the light-emitting componentin lasers. Mixtures of 10% Xe, 89% Ar, and 1% F2, for example, can be “pumped,” or ex-cited, with high energy electrons to form excited XeF molecules, which emit a photon witha wavelength of 354 nm.

N.9 THE INORGANIC CHEMISTRY OF CARBONFor more than 200 years, chemists have divided compounds into two categories. Those thatwere isolated from plants or animals were called organic, while those extracted from oresand minerals were inorganic. Organic chemistry is often defined as the chemistry of car-bon. But that definition would include calcium carbonate (CaCO3) and graphite, whichmore closely resemble inorganic compounds. We will therefore define organic chemistryas the study of compounds such as formic acid (HCO2H), methane (CH4), and vitamin C(C6H8O6) that contain both carbon and hydrogen. This section focuses on inorganic car-bon compounds.

Elemental Forms of Carbon: Graphite, Diamond, Coke, and Carbon Black

Carbon occurs as a variety of allotropes. There are two crystalline forms—diamond andgraphite—and a number of amorphous (noncrystalline) forms, such as charcoal, coke, andcarbon black.

References to the characteristic hardness of diamond (from the Greek adamas, “in-vincible”) date back at least 2600 years. It was not until 1797, however, that SmithsonTennant was able to show that diamonds consist solely of carbon. The properties of diamondare remarkable. It is among the least volatile substances known (MP � 3550°C, BP �4827°C), it is the hardest substance known, and it expands less on heating than any othermaterial.

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NONMETAL 41

The properties of diamond are a logical consequence of its structure. Carbon, with fourvalence electrons, forms covalent bonds to four neighboring carbon atoms arranged towardthe corners of a tetrahedron, as shown in Figure N.24. Each of the sp3-hybridized atoms isthen bound to four other carbon atoms, which form bonds to four other carbon atoms, andso on. As a result, a perfect diamond can be thought of as a single giant molecule. Thestrength of the individual COC bonds and their arrangement in space give rise to the un-usual properties of diamond.

FIGURE N.24 The simplest repeating unit in diamond.

In some ways, the properties of graphite are like those of diamond. Both compoundsboil at 4827°C, for example. But graphite is also very different from diamond. Diamond(3.514 g/cm3) is significantly more dense than graphite (2.26 g/cm3). Whereas diamond isthe hardest substance known, graphite is one of the softest. Diamond is an excellent insu-lator, with little or no tendency to carry an electric current. Graphite is such a good con-ductor of electricity that graphite electrodes are used in electrical cells.

The physical properties of graphite can be understood from the structure of the solidshown in Figure N.25. Graphite consists of extended planes of sp2-hybridized carbon atomsin which each carbon is tightly bound to three other carbon atoms. (It takes 477 kJ to breaka mole of the bonds within the planes.) The strong bonds between carbon atoms withineach plane explain the exceptionally high melting point and boiling point of graphite. Thebonds between planes of carbon atoms, however, are relatively weak. (The force of attrac-tion between planes is only 17 kJ/mol.) Because the bonds between planes are weak, it iseasy to deform the solid by allowing one plane of atoms to move relative to another. As aresult, graphite is soft enough to be used in pencils and as a lubricant in motor oil.

The characteristic properties of graphite and diamond might lead you to expect thatdiamond would be more stable than graphite. This isn’t what is observed experimentally.Graphite at 25°C and 1 atm pressure is slightly more stable than diamond. At very high

FIGURE N.25 Portion of the structure of extendedplanes of carbon atoms found in graphite.

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42 NONMETAL

temperatures and pressures, however, diamond becomes more stable than graphite. In 1955General Electric developed a process to make industrial-grade diamonds by treatinggraphite with a metal catalyst at temperatures of 2000 to 3000 K and pressures above 125,000atm. Roughly 40% of industrial-quality diamonds are now synthetic. Although gem-quality diamonds can be synthesized, the costs involved are prohibitive.

Both diamond and graphite occur as regularly packed crystals. Other forms of carbonare amorphous—they lack a regular structure. Charcoal, carbon black, and coke are allamorphous forms of carbon. Charcoal results from heating wood in the absence of oxygen.To make carbon black, natural gas or other carbon compounds are burned in a limitedamount of air to give a thick, black smoke that contains extremely small particles of car-bon, which can be collected when the gas is cooled and passed through an electrostatic pre-cipitator. Coke is a more regularly structured material, closer in structure to graphite thaneither charcoal or carbon black, which is made from coal.

Carbides: Covalent, Ionic, and Interstitial

Carbon reacts with less electronegative elements at high temperatures to form compoundsknown as carbides. When carbon reacts with an element of similar size and electronega-tivity, a covalent carbide is produced. Silicon carbide, for example, is made by treating sil-icon dioxide from quartz with an excess of carbon in an electric furnace at 2300 K.

SiO2(s) � 3 C(s) 88n SiC(s) � 2 CO(g)

Covalent carbides have properties similar to those of diamond. Both SiC and diamond areinert to chemical reactions, except at very high temperatures; both have very high melt-ing points; and both are among the hardest substances known. SiC was first synthesized byEdward Acheson in 1891. Shortly thereafter, Acheson founded the Carborundum Com-pany to market the material. Then, as now, materials in this class are most commonly used asabrasives.

Compounds that contain carbon and one of the more active metals are called ioniccarbides.

CaO(s) � 3 C(s) 88n CaC2(s) � CO(g)

It is useful to think about ionic carbides as if they contained negatively charged carbonions: [Ca2�][C2

2�] or [Al3�]4[C4�]3. This model is useful because it explains why these car-bides burst into flame when added to water. Ionic carbides such as Al4C3 that formally con-tain the C4� ion react with water to form methane, which is ignited by the heat given offin the reaction.

C4� � 4 H2O 88n CH4 � 4 OH�

The ionic carbides such as CaC2 that formally contain the C22� ion react with water to

form acetylene, which is ignited by the heat of reaction.

C22� � 2 H2O 88n C2H2 � 2 OH�

At one time, miners’ lamps were fueled by the combustion of acetylene prepared from thereaction of calcium carbide with water.

The difference between covalent carbides and ionic carbides can be understood byadding compounds such as SiC, Al4C3, and CaC2 to the bond type triangle introduced in

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NONMETAL 43

Chapter 5. When those compounds are added to Figure 5.10, for example, we find that SiCfalls well into the region expected for covalent compounds. CaC2, on the other hand, isclearly an ionic compound. Al4C3 falls on the borderline between ionic and covalent, whichis consistent with the fact that the compound is hard—as one would expect for a covalentcarbide—and yet reacts with water to form methane—as might be expected for an ioniccarbide.

Interstitial carbides, such as tungsten carbide (WC), form when carbon combineswith a metal that has an intermediate electronegativity and a relatively large atomic ra-dius. In these compounds, the carbon atoms pack in the holes (interstices) betweenplanes of metal atoms. The interstitial carbides, which include TiC, ZrC, and MoC, re-tain the properties of metals. They act as alloys, rather than as either salts or covalentcompounds.

The Oxides of Carbon

Although the different forms of carbon are essentially inert at room temperature, theycombine with oxygen at high temperatures to produce a mixture of carbon monoxide andcarbon dioxide.

2 C(s) � O2(g) 88n 2 CO(g) �H° � �221.05 kJ/molrxn

C(s) � O2(g) 88n CO2(g) �H° � �393.51 kJ/molrxn

CO can also be obtained by reacting red-hot carbon with steam.

C(s) � H2O(g) 88n CO(g) � H2(g)

Because the mixture of gases is formed by the reaction of charcoal or coke with water itis often referred to as water gas. It is also known as town gas because it was once made bytowns and cities for use as a fuel. Water gas, or town gas, was a common fuel for both homeand industrial use before natural gas became readily available. The H2 burns to form wa-ter, and the CO is oxidized to CO2. Eventually, as our supply of natural gas is depleted, itwill become economical to replace natural gas with other fuels, such as water gas, that canbe produced from our abundant supply of coal.

Both CO and CO2 are colorless gases. CO boils at �191.5°C, and CO2 sublimes (passesdirectly from the solid to the gaseous state) at �78.5°C. Although CO has no odor or taste,CO2 has a faint, pungent odor and a distinctly acidic taste. Both are dangerous substancesbut at very different levels of exposure. Air contaminated with as little as 0.002 gram ofCO per liter can be fatal because CO binds tightly to the hemoglobin that carries oxygenthrough the blood. CO2 is not lethal until the concentration in the air approaches 15%. Atthat point, it has replaced so much oxygen that a person who attempts to breathe the at-mosphere suffocates. The danger of CO2 poisoning is magnified by the fact that CO2 isroughly 1.5 times more dense than the air in our atmosphere. Thus, CO2 can accumulateat the bottom of tanks or wells.

CO2 in the Atmosphere

Carbon dioxide influences the temperature of the atmosphere by a phenomenon knownas the greenhouse effect. The glass walls and ceilings of a greenhouse absorb some of thelower energy, longer wavelength radiation from sunlight thereby inevitably raising the tem-perature inside the building. CO2 in the atmosphere does exactly the same thing, it absorbs

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44 NONMETAL

low energy, long wavelength radiation from the Sun that would otherwise be reflected backfrom the surface of the planet. Thus, CO2 in the atmosphere traps heat. Although thereare other factors at work, it is worth noting that Venus, whose atmosphere contains a greatdeal of CO2, has a surface temperature of roughly 400°C, whereas Mars, with little or noatmosphere, has a surface temperature of �50°C.

There are many sources of CO2 in the atmosphere. Over geologic time scales, the largestsource has been volcanos. Within the twentieth century, the combustion of petroleum, coal,and natural gas has made a significant contribution to atmospheric levels of CO2 (see Fig-ure N.8). Between 1958 and 1978, the average level of CO2 in the atmosphere increasedby 6%, from 315.8 to 334.6 ppm.

At one time, the amount of CO2 released to the atmosphere wasn’t a matter for con-cern because natural processes that removed CO2 from the atmosphere could compensatefor the CO2 that entered the atmosphere. The vast majority of the CO2 liberated by vol-canic action, for example, was captured by calcium oxide or magnesium oxide to form cal-cium carbonate or magnesium carbonate.

CaO(s) � CO2(g) 88n CaCO3(s)MgO(s) � CO2(g) 88n MgCO3(s)

CaCO3 is found as limestone or marble, or mixed with MgCO3 as dolomite. The amountof CO2 in deposits of carbonate minerals is at least several thousand times larger than theamount in the atmosphere.

CO2 also dissolves, to some extent, in water.

H2OCO2(g) 88n CO2(aq)

It then reacts with water to form carbonic acid, H2CO3.

CO2(aq) � H2O(l) 88n H2CO3(aq)

As a result of the reactions, the sea contains about 60 times more CO2 than the atmo-sphere.

Can the sea absorb more CO2 from the atmosphere, or is it near its level of saturation?Is the rate at which the sea absorbs CO2 greater than the rate at which we are adding it tothe atmosphere? The observed increase in the concentration of CO2 in recent years sug-gests pessimistic answers to those two questions. A gradual warming of the earth’s atmo-sphere could result from continued increases in CO2 levels, with adverse effects on the cli-mate and therefore the agriculture of at least the northern hemisphere.

The Chemistry of Carbonates: CO32� and HCO3

�

Eggshells are almost pure calcium carbonate. CaCO3 can also be found in the shells ofmany marine organisms and in both limestone and marble. The fact that none of thosesubstances dissolves in water suggests that CaCO3 is normally insoluble in water. Calciumcarbonate will dissolve in water saturated with CO2, however, because carbonated water(or carbonic acid) reacts with calcium carbonate to form calcium bicarbonate, which is sol-uble in water.

CaCO3(s) � H2CO3(aq) 88n Ca2�(aq) � 2 HCO3�(aq)

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NONMETAL 45

When water rich in carbon dioxide flows through limestone formations, part of thelimestone dissolves. If the CO2 escapes from the water, or if some of the water evapo-rates, solid CaCO3 is redeposited. When this happens as water runs across the roof of acavern, stalactites, which hang from the roof of the cave, are formed. If the water dropsbefore the carbonate reprecipitates, stalagmites, which grow from the floor of the cave,are formed.

The chemistry of carbon dioxide dissolved in water is the basis of the soft drink in-dustry. The first artificially carbonated beverages were introduced in Europe at the end ofthe nineteenth century. Carbonated soft drinks today consist of carbonated water, a sweet-ening agent (sugar, saccharin, or aspartame), an acid to impart a sour or tart taste, flavor-ing agents, coloring agents, and preservatives. As much as 3.5 liters of gaseous CO2 dis-solve in a liter of soft drink to provide the characteristic “bite” associated with carbonatedbeverages.

Carbonate chemistry plays an important role in other parts of the food industry as well.Baking soda, or bicarbonate of soda, is sodium bicarbonate, NaHCO3, a weak base, whichis used to neutralize the acidity of other ingredients in a recipe. Baking powder is a mix-ture of baking soda and a weak acid, such as tartaric acid or calcium hydrogen phosphate(CaHPO4). When mixed with water, the acid reacts with the HCO3

� ion to form CO2 gas,which causes the dough or batter to rise.

HCO3�(aq) � H�(aq) 88n H2CO3(aq) 88n H2O(l) � CO2(g)

Before commercial baking powders were available, cooks obtained the same effect by mix-ing roughly a teaspoon of baking soda with a cup of sour milk or buttermilk. The acidsthat give sour milk and buttermilk their characteristic taste also react with the bicarbon-ate ion to give CO2.

AllotropeAnhydrousCarbideDiamagneticDimerDisproportionation

reaction

Haber processHalideHalogenHydrideOstwald processOxidizing agent

OxyacidOxyanionParamagneticPeroxideRare gasesReducing agent

PROBLEMS

Metals, Nonmetals, and Semimetals

1. List the elements that are nonmetals. Describe where these elements are found in theperiodic table.

2. Explain why semimetals (such as B, Si, Ge, As, Sb, Te, Po, and At) exist and describesome of their physical properties.

3. Which member of each of the following pairs of elements is more nonmetallic?(a) As or Bi (b) As or Se (c) As or S (d) As or Ge (e) As or P

KEY TERMS

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46 NONMETAL

The Chemistry of the Nonmetals

4. Which of the following elements can exist as a triatomic molecule?(a) hydrogen (b) helium (c) sulfur (d) oxygen (e) chlorine

5. Which of the following elements should form compounds with the formulas Na2X, H2X,XO2, and XF6?(a) B (b) C (c) N (d) O (e) S

6. Which of the following elements should form compounds with the formulas XH3, XF3,and Na3XO4?(a) Al (b) Ge (c) As (d) S (e) Cl

7. Which of the following can’t be found in nature? Explain why.(a) MgCl2 (b) CaCO3 (c) F2 (d) Na3AlF6 (e) NaCl

The Role of Nonmetal Elements in Chemical Reactions

8. Explain why more electronegative elements tend to oxidize less electronegative elements.9. Which of the following ions or molecules can be oxidized?

(a) H2SO3 (b) P4 (c) Cl� (d) SiO2 (e) PO43� (f) Mg2�

10. Which of the following ions or molecules can be reduced?(a) H2O (b) H2SO3 (c) HCl (d) CO2 (e) Mg2� (f) Na

Deciding What Is Oxidized and What Is Reduced

11. For each of the following reactions, identify what is oxidized and what is reduced.(a) Fe2O3(s) � 3 CO(g) n 2 Fe(s) � 3 CO2(g)(b) H2(g) � CO2(g) n H2O(g) � CO(g)(c) CH4(g) � 2 O2(g) n CO2(g) � 2 H2O(g)(d) 2 H2S(g) � 3 O2(g) n 2 SO2(g) � 2 H2O(g)

12. For each of the following reactions, identify what is oxidized and what is reduced.(a) PH3(g) � 3 Cl2(g) n PCl3(g) � 3 HCl(g)(b) 2 NO(g) � F2(g) n 2 NOF(g)(c) 2 Na(s) � 2 NH3(l) n 2 NaNH2(s) � H2(g)(d) 3 NO2(g) � H2O(l) n 2 HNO3(aq) � NO(g)

13. Hydrazine is made by a reaction known as the Raschig process.

2 NH3(aq) � NaOCl(aq) 88n N2H4(aq) � NaCl(aq) � H2O(l)

Decide whether this is an oxidation–reduction reaction. If it is, identify the compoundoxidized and the compound reduced.

14. The thiosulfate ion, S2O32�, is prepared by boiling solutions of sulfur dissolved in

sodium sulfite.

8 SO32�(aq) � S8(s) 88n 8 S2O3

2�(aq)

Is this an oxidation–reduction reaction? If it is, identify the compound oxidized andthe compound reduced.

15. Chlorine dioxide, ClO2, is used commercially as a bleach or a disinfectant because ofits excellent oxidizing ability. ClO2 is prepared by decomposing chlorous acid as follows.

8 HOClO(aq) 88n 6 ClO2(g) � Cl2(g) � 4 H2O(l)

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NONMETAL 47

Is this an oxidation–reduction reaction? If it is, identify the oxidizing agent and the re-ducing agent.

16. Nitric acid sometimes acts as an acid (as a source of the H� ion) and sometimes as anoxidizing agent. For each of the following reactions, decide whether HNO3 acts as anacid or as an oxidizing agent.(a) Na2CO3(s) � 2 HNO3(aq) n 2 NaNO3(aq) � CO2(g) � H2O(l)(b) 3 P4(s) � 20 HNO3(aq) � 8 H2O(l) n 12 H3PO4(aq) � 20 NO(g)(c) Al2O3(s) � 6 HNO3(aq) n 2 Al(NO3)3(aq) � 3 H2O(l)(d) 3 Cu(s) � 8 HNO3(aq) n 3 Cu(NO3)2(aq) � 2 NO(g) � 4 H2O(l)

17. Which of the following would you expect to be the best oxidizing agent?(a) Na (b) H2 (c) N2 (d) P4 (e) O2

18. Which of the following would you expect to be the best reducing agent?(a) Na� (b) F� (c) Na (d) Br2 (e) Fe3�

19. For each of the following pairs of elements, determine which is the better reducing agent?(a) P4 or As (b) As or S8 (c) P4 or S8 (d) S8 or Cl2 (e) C or O2

Predicting the Products of Chemical Reactions

20. Predict the products of the following reactions.(a) Mg(s) � N2(g) n(b) Li(s) � O2(g) n(c) Br2(l) � I�(aq) n

21. Predict the products of the following reactions.(a) SO2(g) � H2O(l) n(b) Cl2(g) � OH�(aq) n(c) CO2(g) � H2O(l) n

22. Predict the products of the following reactions.(a) HCl(g) � H2O(l) n(b) P4O10(s) � H2O(l) n(c) NO2(g) � H2O(l) n

23. Predict the products of the following reactions.(a) S8(s) � O2(g) n(b) Al(s) � I2(s) n(c) P4(s) � F2(g) n

The Chemistry of Hydrogen

24. Describe three ways of preparing small quantities of H2 in the lab.25. Explain why it is not a good idea to prepare H2 by reacting sodium metal with a strong

acid.26. Give an example of a compound in which hydrogen has an oxidation number of �1;

of 0; of �1.27. Which of the following reactions produce a compound in which hydrogen has an oxi-

dation number of �1?(a) Li � H2 n(b) O2 � H2 n(c) S8 � H2 n(d) Cl2 � H2 n(e) Ca � H2 n

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48 NONMETAL

28. Use tables of first ionization energies and electronegativities to explain why it is so dif-ficult to decide whether hydrogen belongs in Group IA or Group VIIA of the peri-odic table.

29. Which of the following substances can be used as evidence for placing hydrogen inGroup IA? Which can be used as evidence for including hydrogen in Group VIIA?(a) CaH2 (b) AlH3 (c) H2S (d) H3PO4 (e) H2

30. The earth’s atmosphere once contained significant amounts of H2. Explain why onlytraces of H2 are left in the earth’s atmosphere, whereas the atmospheres of otherplanets—such as Jupiter, Saturn, and Neptune—contain large quantities of H2.

31. Use Lewis structures to explain what happens in the four reactions described in Sec-tion N.2 that can be used to prepare small quantities of H2 gas.

The Chemistry of Oxygen and Sulfur

32. Describe three ways of preparing small quantities of O2 in the lab.33. Describe the relationships among oxygen (O2), the peroxide ion (O2

2�), and the oxideion (O2�). Explain why the number of electrons shared by a pair of oxygen atomsdecreases as the oxidation number of the oxygen becomes more negative.

34. Which of the following elements or compounds could eventually produce O2 when itreacts with water?(a) Ba (b) BaO (c) BaO2 (d) Ba(OH)2 (e) BaNO3

35. Explain why the only compounds in which oxygen has a positive oxidation number arecompounds, such as OF2, that contain fluorine.

36. Explain why hydrogen peroxide can be either an oxidizing agent or a reducing agent.Describe at least one reaction in which H2O2 oxidizes another substance and one re-action in which it reduces another substance.

37. Write the Lewis structures for ozone, O3, and sulfur dioxide, SO2. Discuss the rela-tionship between the compounds.

38. Explain why elemental oxygen exists as O2 molecules, whereas elemental sulfur formsS8 molecules.

39. Explain why sulfur forms compounds such as SF4 and SF6, when oxygen can only formOF2.

40. Describe the relationship between the thiosulfate and sulfate ions and between thethiocyanate and cyanate ions. Use that relationship to predict the formula of the trithio-carbonate ion.

41. Write the Lewis structures of the following products of the reaction between sodiumand sulfur.(a) Na2S (b) Na2S2 (c) Na2S3 (d) Na2S8

42. Explain why sulfur readily forms compounds in the �2, �4, and �6 oxidation states,but only a handful of compounds exist in which oxygen is in a positive oxidation state.

43. Explain why sulfur-containing compounds such as FeS2, CS2, and H2S form SO2 in-stead of SO3 when they burn.

44. Use Lewis structures to explain why solutions of the SO32� ion react with sulfur to

form thiosulfate, S2O32�.

45. Use Lewis structures to explain why a two-electron reduction of the S2O62� ion gives

SO32�.

S2O62� � 2 e� 88n 2 SO3

2�

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NONMETAL 49

46. Use Lewis structures to explain why a two-electron oxidation of the S2O32� ion gives

the S4O62� ion.

47. Which of the following does not have a reasonable oxidation number for sulfur?(a) Na2S (b) H2S (c) SO3

2� (d) SO4 (e) SF4

48. Explain why SO2 plays an important role in the phenomenon known as acid rain.49. Explain why problems with acid rain would be much more severe if sulfur compounds

burned to form SO3 instead of SO2.

The Chemistry of Nitrogen and Phosphorus

50. Nitrogen has a reasonable oxidation number in all of the following compounds, andyet one of them is still impossible. Which one is impossible?(a) NF3 (b) NF5 (c) NO3

� (d) NO2� (e) NO

51. Earth’s atmosphere contains roughly 4 1016 tons of nitrogen, and yet the biggest prob-lem facing agriculture in the world today is a lack of “nitrogen.” Explain why.

52. Explain why elemental nitrogen is almost inert, but nitrogen compounds such asNH4NO3, NaN3, nitroglycerin, and trinitrotoluene (TNT) form some of the most dan-gerous explosives.

53. Which of the following oxides of nitrogen are paramagnetic?(a) N2O (b) NO (c) NO2 (d) N2O3 (e) N2O4 (f) N2O5

54. Use Lewis structures to explain what happens in the following reaction.

2 NO � O2 88n 2 NO2

55. Use Lewis structures to explain why NO reacts with NO2 to form N2O3 when a mix-ture of the compounds is cooled.

56. Use the fact that nitrous oxide decomposes to form nitrogen and oxygen to explainwhy a glowing splint bursts into flame when immersed in a container filled with N2O.

2 N2O(g) 88n 2 N2(g) � O2(g)

57. Describe ways of preparing small quantities of each of the following compounds in thelaboratory.(a) N2O (b) NO (c) NO2 (d) N2O4

58. Describe how to tell the difference between a flask filled with NO gas and a flask filledwith NO2.

59. Lightning catalyzes the reaction between nitrogen and oxygen in the atmosphere toform nitrogen oxide, NO.

N2(g) � O2(g) 88n 2 NO(g)

Explain how lightning acts as one source of acid rain.60. Which of the following elements or compounds is not involved at some stage in the

preparation of nitric acid?(a) O2 (b) N2 (c) NO (d) NO2 (e) H2

61. Explain why phosphorus forms both PCl3 and PCl5 but nitrogen forms only NCl3.62. Explain why nitrogen is essentially inert at room temperature, but white phosphorus

bursts spontaneously into flame when it comes into contact with air.63. Explain why red phosphorus is much less reactive than white phosphorus.

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50 NONMETAL

64. Explain why nitrogen forms extraordinarily stable N2 molecules at room temperature,but phosphorus forms P2 molecules only at very high temperatures.

65. Explain why nitric acid has the formula HNO3 and phosphoric acid has the formula H3PO4.66. Write the Lewis structures for phosphoric acid, H3PO4, and phosphorous acid, H3PO3.

Explain why phosphoric acid can lose three H� ions to form a phosphate ion, PO43�,

whereas phosphorous acid can lose only two H� ions to form the HPO32� ion.

67. Explain why only two of the four hydrogen atoms in H4P2O5 are lost when the oxy-acid forms an oxyanion.

68. Describe the role of carbon in the preparation of elemental phosphorus from calciumphosphate.

69. Predict the product of the reaction of phosphorus with excess oxygen and then predictwhat will happen when the product of the reaction is dissolved in water.

70. Explain why the most common oxidation states of antimony are �3 and �5.71. Which of the following compounds should not exist?

(a) Na3P (b) (NH4)3PO4 (c) PO2 (d) PH3 (e) POCl3

The Chemistry of the Halogens

72. Which of the halogens is the most active, or reactive? Explain why.73. Describe the difference between halogens and halides. Give examples of each.74. Fe3� ions can oxidize Br� ions to Br2, but they can’t oxidize Cl� ions to Cl2. Use this

information to determine where the Fe3� ion belongs in the following sequence of de-creasing oxidizing strength: F2 � Cl2 � Br2 � I2.

75. HBr can be prepared by reacting PBr3 with water.

PBr3(l) � 3 H2O(l) 88n 3 HBr(aq) � H3PO3(aq)

Use this information to explain what happens in the following reaction.

P4(s) � 6 Br2(s) � 12 H2O(l) 88n 12 HBr(aq) � 4 H3PO3(aq)

76. Explain why chlorine reacts with fluorine to form ClF3 but not FCl3.77. Chlorine reacts with base to form the hypochlorite ion.

Cl2(aq) � 2 OH�(aq) 88n Cl�(aq) � OCl�(aq) � H2O(l)

Use this information to explain why people who make the mistake of mixing Cloroxwith hydrochloric acid often suffer damage to their lungs from breathing chlorine gas.

The Inorganic Chemistry of Carbon

78. Use the structure of graphite to explain why the bonds between carbon atoms are sostrong that it is difficult to boil off individual carbon atoms, yet the material is so softit can be used as a lubricant.

79. Explain why silicon forms a covalent carbide but calcium forms an ionic carbide.80. Write balanced equations for the combustion of both CO and H2 that explain why a

mixture of the gases can be used as a fuel.81. Use Lewis structures to explain the following reaction.

CO2(g) � H2O(l) 88n H2CO3(aq)

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NONMETAL 51

Integrated Problems

82. A recent catalog listed the following prices: $21.40 for 450 grams of sodium, $18.00 for1 kilogram of zinc, and $52.80 for 250 grams of sodium hydride. Which reagent wouldbe the least expensive source of H2 gas?

83. A solution of hydrogen peroxide in water that is 30% hydrogen peroxide by weightsells for $15.95 per 500 g, and potassium chlorate sells for $12.75 per 500 grams. Is itless expensive to generate oxygen by decomposing H2O2 or KClO3?

84. Write a sequence of reactions for the conversion of elemental nitrogen into nitric acid.Calculate the weight of nitric acid that can be produced from a ton of nitrogen gas.

85. At 1700°C, P4 molecules decompose partially to form P2.

P4(g) 88n 2 P2(g)

If the average molecular weight of phosphorus at that temperature is 91 g/mol, whatfraction of the P4 molecules decompose?

86. Uranium reacts with fluorine to form UF6, which boils at 51°C. The relative rate ofdiffusion of 235UF6 and 238UF6 in the gas phase was used in the Manhattan project toseparate the more abundant 238U isotope from 235U. Predict which substance diffusesmore rapidly, and calculate the ratio of the rate of diffusion of the compounds.

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1

M O D U L E

2TRANSITION METAL

CHEMISTRY

TM.1 The Transition Metals

TM.2 Werner’s Theory of Coordination Complexes

TM.3 Typical Coordination Numbers

TM.4 The Electron Configurations of Transition Metal Ions

TM.5 Oxidation States of the Transition Metals

TM.6 Lewis Acid–Lewis Base Approach to Bonding in Complexes

TM.7 Typical Ligands

TM.8 Coordination Complexes in Nature

Chemistry in the World Around Us: Nitrogen Fixation

TM.9 Nomenclature of Complexes

TM.10 Isomers

TM.11 The Valence Bond Approach to Bonding in Complexes

TM.12 Crystal Field Theory

TM.13 The Spectrochemical Series

TM.14 High-Spin versus Low-Spin Octahedral Complexes

TM.15 The Colors of Transition Metal Complexes

TM.16 Ligand Field Theory

TM.1 THE TRANSITION METALSThe elements in the periodic table are often divided into the four categories shown in Fig-ure TM.1: (1) main-group elements, (2) transition metals, (3) lanthanides, and (4) actinides.The main-group elements include the active metals in the two columns on the extreme leftof the periodic table and the metals, semimetals, and nonmetals in the six columns on thefar right. The transition metals are the metallic elements that serve as a bridge, or transi-tion, between the two sides of the table. The lanthanides and the actinides at the bottomof the table are sometimes known as the inner transition metals because they have atomicnumbers that fall between the first and second elements in the last two rows of the tran-sition metals.

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2 TRANSITION METAL

There is some controversy about the classification of the elements on the boundary be-tween the main-group and transition metal elements on the right side of the table (seeFigure TM.2). The elements in question are zinc (Zn), cadmium (Cd), and mercury (Hg).The disagreement about whether those elements should be classified as main-group ele-ments or transition metals suggests that the differences between the categories aren’t clear.Transition metals resemble main-group metals in many ways: They look like metals; theyare malleable and ductile; they conduct heat and electricity; and they form positive ions.The fact that the two best conductors of electricity are a transition metal (copper) and amain-group metal (aluminum) shows the extent to which the physical properties of main-group metals and transition metals overlap.

H H He

Li Be B C N O F Ne

Na Mg Al Si P S Cl Ar

K Ca Sc Ti V Cr Mn Fe Co Ni Cu Ga Ge As Se Br KrZn

Rb Sr Y Zr Nb Mo Tc Ru Rh Pb Ag In Sn Sb Te I XeCd

Cs Ba La Hf Ta W Re Os Ir Pt Au Tl Pb Bi Po At RnHg

Fr Ra Ac Rf

Ce Pr Nd Pm Sm Eu Gd Dy Ho Er Tm Yb LuTb

Th Pa U Np Pu Am Cm Cf Es Fm Md No LrBk

Main-groupElements

TransitionMetals

Main-groupElements

Db Sg Bh Hs Mt

Lanthanides

ActinidesFIGURE TM.1 The four categories of elements in the periodic table.

H He

B C N O F Ne

Al Si P S Cl Ar

Ni Cu Ga Ge As Se Br KrZn

Pb Ag In Sn Sb Te I XeCd

Pt Au Tl Pb Bi Po At RnHg

TransitionMetals

Main-groupElements

Co

Rh

IrFIGURE TM.2 There is some debate about whether zinc, cadmium, andmercury should be classified as transition metals or main-group metals.

There are also differences between transition metals and the main-group metals. Thetransition metals are more electronegative than the main-group metals, for example, andare therefore more likely to form covalent compounds.

Another difference between the main-group metals and transition metals can be seenin the formulas of the compounds they form. The main-group metals tend to form salts(such as NaCl, Mg3N2, and CaS) in which there are just enough negative ions to balancethe charge on the positive ions. The transition metals form similar compounds [such asFeCl3, HgI2, and Cd(OH)2], but they are more likely than main-group metals to form com-plexes, such as the FeCl4

�, HgI42�, and Cd(OH)4

2� ions, that have an excess number ofnegative ions.

A third difference between main-group and transition metal ions is the ease with whichthey form stable compounds with neutral molecules, such as water and ammonia. Salts ofmain-group metal ions dissolve in water to form aqueous solutions.

H2ONaCl(s) 88n Na�(aq) � Cl�(aq)

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TRANSITION METAL 3

When we let the water evaporate, we get back the original starting material, NaCl(s). Saltsof the transition metal ions can display a very different behavior. Chromium(III) chloride,for example, is a violet compound that dissolves in liquid ammonia to form a yellow com-pound with the formula CrCl3�6 NH3 that can be isolated when the ammonia is allowed toevaporate.

CrCl3(s) � 6 NH3(l) 88n CrCl3�6 NH3(s)

The FeCl4� ion and CrCl3�6 NH3 are called coordination compounds because they

contain ions or molecules linked, or coordinated, to a transition metal. They are also knownas complex ions or coordination complexes, because they are Lewis acid–base complexes.The ions or molecules that bind to transition metal ions to form complexes are called li-gands (from Latin, meaning “to tie or bind”). The number of ligands bound to the transi-tion metal ion is called the coordination number.

Although coordination complexes are particularly important in the chemistry of thetransition metals, some main-group elements also form complexes. Aluminum, tin, andlead, for example, form complexes such as the AlF6

3�, SnCl42�, and PbI4

2� ions.

TM.2 WERNER’S THEORY OF COORDINATION COMPLEXESAlfred Werner first became interested in coordination complexes in 1892, while preparinglectures for a course on atomic theory. Six months later, he proposed a model for coordi-nation complexes that still serves as the basis for work in the field. Werner then spent theremainder of his life collecting evidence to support his theory. Before we introduce Werner’smodel of coordination complexes, let’s look at some of the experimental data available atthe turn of the twentieth century.

• Three different cobalt(III) complexes can be isolated when CoCl2 is dissolved inaqueous ammonia and then oxidized by air to the �3 oxidation state. A fourth com-plex can be made by slightly different techniques. The complexes have different col-ors and different empirical formulas.

CoCl3�6 NH3 orange-yellowCoCl3�5 NH3�H2O redCoCl3�5 NH3 purpleCoCl3�4 NH3 green

• The ammonia in the cobalt(III) complexes is much less reactive than normal. By it-self, ammonia reacts rapidly with hydrochloric acid to form ammonium chloride.

NH3(aq) � HCl(aq) 88n NH4�(aq) � Cl�(aq)

The complexes, however, don’t react with hydrochloric acid, even at 100°C.

CoCl3�6 NH3(aq) � HCl(aq) 88n

• A white precipitate of AgCl normally forms when a source of the Ag� ion is addedto a solution that contains the Cl� ion.

Ag�(aq) � Cl�(aq) 88n AgCl(s)

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4 TRANSITION METAL

When excess Ag� ion is added to solutions of the CoCl3�6 NH3 and CoCl3�5NH3�H2O complexes, 3 moles of AgCl are formed for each mole of complex in so-lution, as might be expected. However, only two of the Cl� ions in the CoCl3�5 NH3

complex and only one of the Cl� ions in CoCl3�4 NH3 can be precipitated with Ag�

ions.• The results of measurements of the conductivity of aqueous solutions of the com-

plexes suggest that the CoCl3�6 NH3 and CoCl3�5 NH3�H2O complexes dissociatein water to give a total of four ions. CoCl3�5 NH3 dissociates to give three ions, how-ever, and CoCl3�4 NH3 dissociates to give only two ions.

Werner explained the observations by differentiating between ions or molecules thatwere coordinated to the transition metal and those that were present to balance the chargeon the metal ion. He assumed that there were always six ions or molecules coordinated tothe Co3� ion. Any remaining ions were only used to balance the charge on the complexion. The formulas of the compounds described above can therefore be written as follows.

[Co(NH3)63�][Cl�]3 orange-yellow

[Co(NH3)5(H2O)3�][Cl�]3 red[Co(NH3)5Cl2�][Cl�]2 purple[Co(NH3)4Cl2

�][Cl�] green

CheckpointDetermine the number of chloride ions coordinated to the chromium atom in theCr(NH3)4(H2O)2Cl3 complex and the number of these ions that are present only tobalance the charge on the complex ion.

Exercise TM.1

Describe how Werner’s theory of coordination complexes explains both the number of ionsformed when the following compounds dissolve in water and the number of chloride ionsthat precipitate when the solutions are treated with Ag� ions.(a) [Co(NH3)6]Cl3(b) [Co(NH3)5(H2O)]Cl3(c) [Co(NH3)5Cl]Cl2(d) [Co(NH3)4Cl2]Cl

Solution

The cobalt ion is coordinated to a total of six ligands in each complex. Each complex alsohas a total of three chloride ions to balance the charge on the Co3� ion. Some of the Cl�

ions are free to dissociate when the complex dissolves in water. Others are bound to theCo3� ion and neither dissociate nor react with Ag�.(a) The three chloride ions in the complex are free to dissociate when it dissolves in wa-

ter. The complex therefore dissociates in water to give a total of four ions, and all threeCl� ions are free to react with Ag� ion.

H2O[Co(NH3)6]Cl3(s) 88n Co(NH3)63�(aq) � 3 Cl�(aq)

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TRANSITION METAL 5

(b) Once again, the three Cl� ions are free to dissociate when the complex dissolves inwater, and they precipitate when Ag� ions are added to the solution.

H2O[Co(NH3)5(H2O)]Cl3(s) 88n Co(NH3)5(H2O)3�(aq) � 3 Cl�(aq)

(c) One of the chloride ions is coordinated to the cobalt in the complex. Thus, only threeions are formed when the compound dissolves in water, and only two Cl� ions are freeto precipitate with Ag� ions.

H2O[Co(NH3)5Cl]Cl2(s) 88n Co(NH3)5Cl2�(aq) � 2 Cl�(aq)

(d) Two of the chloride ions are coordinated to the cobalt in the complex. Thus, only twoions are formed when the compound dissolves in water, and only one Cl� ion is freeto precipitate with Ag� ions.

H2O[Co(NH3)4Cl2]Cl(s) 88n Co(NH3)4Cl2�(aq) � Cl�(aq)

Werner also assumed that transition metal complexes had definite shapes. Accordingto his theory, the ligands in six-coordinate cobalt(III) complexes are oriented toward thecorners of an octahedron, as shown in Figure TM.3.

Co

3+

Co(NH3)63+

NH3 NH3

NH3

NH3

NH3

NH3

orange-yellow

Co

2+

Co(NH3)5Cl2+

NH3 NH3

NH3

Cl

Cl

Cl

NH3

NH3

purple

green

Co

3+

Co(NH3)6(H2O)3+

NH3 NH3

NH3

OH2

NH3

NH3

red

Co

+

Co(NH3)4Cl2+

NH3 NH3

NH3 NH3

FIGURE TM.3 Structures of the four cobalt com-plex ions that played a major role in the develop-ment of Werner’s theory.

TM.3 TYPICAL COORDINATION NUMBERSTransition metal complexes have been characterized with coordination numbers that rangefrom 1 to 12, but the most common coordination numbers are 2, 4, and 6. Examples ofcomplexes with these coordination numbers are given in Table TM.1.

Note that the charge on the complex is always the sum of the charges on the ions ormolecules that form the complex.

Cu2� � 4 NH388nm88 Cu(NH3)4

2�

Fe2� � 6 CN� 88nm88 Fe(CN)64�

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6 TRANSITION METAL

Also note that the coordination number of a complex often increases as the charge on themetal ion becomes larger.

Cu� � 2 NH388nm88 Cu(NH3)2

�

Cu2� � 4 NH388nm88 Cu(NH3)4

2�

Exercise TM.2

Calculate the charge on the transition metal ion in the following complexes.(a) Na2Co(SCN)4

(b) Ni(NH3)6(NO3)2

(c) K2PtCl6

Solution

(a) The complex contains the Na� and Co(SCN)42� ions. Each thiocyanate ion (SCN�)

carries a charge of �1. Since the net charge on the Co(SCN)42� ion is �2, the cobalt

ion must carry a charge of �2.(b) The complex contains the Ni(NH3)6

2� and NO3� ions. Since ammonia is a neutral mol-

ecule, the nickel must carry a charge of �2.(c) The complex contains the K� and PtCl6

2� ions. Each chloride ion carries a charge of�1. Since the net charge on the PtCl6

2� ion is �2, the platinum must carry a chargeof �4.

TM.4 THE ELECTRON CONFIGURATIONS OF TRANSITIONMETAL IONS

The electron configurations of ions formed by the main-group metals are discussed in Chap-ters 3 and 5. Aluminum, for example, loses its three valence electrons when it forms Al3�

ions.

TABLE TM.1 Examples of Common Coordination Numbers

Metal Ion Ligand Complex Coordination Number

Ag� � 2 NH388nm88 Ag(NH3)2

� 2Ag� � 2 S2O3

2� 88nm88 Ag(S2O3)23� 2

Ag� � 2 Cl� 88nm88 AgCl2� 2

Cu� � 2 NH388nm88 Cu(NH3)2

� 2Cu2� � 4 NH3

88nm88 Cu(NH3)42� 4

Zn2� � 4 CN� 88nm88 Zn(CN)42� 4

Hg2� � 4 I� 88nm88 HgI42� 4

Co2� � 4 SCN� 88nm88 Co(SCN)42� 4

Fe2� � 6 H2O 88nm88 Fe(H2O)62� 6

Fe3� � 6 H2O 88nm88 Fe(H2O)63� 6

Co3� � 6 NH388nm88 Co(NH3)6

3� 6Ni2� � 6 NH3

88nm88 Ni(NH3)62� 6

Fe2� � 6 CN� 88nm88 Fe(CN)64� 6

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TRANSITION METAL 7

Al [Ne] 3s2 3p1

Al3� [Ne]

The relationship between the electron configurations of transition metal elements and theirions is more complex. Consider the chemistry of cobalt, for example, which forms com-plexes that contain either Co2� or Co3� ions.

Before we can predict the electron configuration of the ions we have to answer thequestion: Which valence electrons are removed from a cobalt atom when the Co2� andCo3� ions are formed? The electron configuration of a neutral cobalt atom is written asfollows.

Co [Ar] 4s2 3d7

We might expect cobalt to lose electrons from only the 3d orbitals, but this is not what isobserved. The Co2� and Co3� ions have the following electron configurations.

Co2� [Ar] 3d7

Co3� [Ar] 3d6

In general, electrons are removed from the valence shell s orbitals before they are removedfrom valence d orbitals when transition metals are ionized.

This raises an interesting question: Why are electrons removed from 4s orbitals before3d orbitals if they are placed in 4s orbitals before 3d orbitals? There are several ways ofanswering this question. First, note that the difference between the energies of the 3d and4s orbitals is very small. For this reason, the electron configurations for chromium and cop-per differ slightly from what might be expected.

Cr [Ar] 4s1 3d5

Cu [Ar] 4s1 3d10

Note also that the relative energies of the 4s and 3d orbitals used when predicting electronconfigurations only hold true for neutral atoms. When transition metals form positive ions,the 3d orbitals become more stable than the 4s orbitals. As a result, electrons are removedfrom the 4s orbital before the 3d orbitals.

Exercise TM.3

Predict the electron configuration of the Fe3� ion.

Solution

We start with the configuration of a neutral iron atom.

Fe [Ar] 4s2 3d6

We then remove three electrons to form the �3 ion. Two of the electrons come from the4s orbital. The other comes from a 3d orbital.

Fe3� [Ar] 3d5

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8 TRANSITION METAL

Because the valence electrons in transition metal ions are concentrated in d orbitals,the ions are often described as having dn configurations. The Co3� and Fe2� ions, for ex-ample, are said to have a d6 configuration.

Co3� [Ar] 3d6

Fe2� [Ar] 3d6

TM.5 OXIDATION STATES OF THE TRANSITION METALSMost transition metals exhibit more than one oxidation state. Manganese, for example,forms compounds in every oxidation state from �1 to �7. Some oxidation states, however,are more common than others. The most common oxidation states of the first series oftransition metals are given in Table TM.2. Efforts to explain the apparent pattern in TableTM.2 ultimately fail for a combination of reasons. Some of the oxidation states are com-mon because they are relatively stable. Others describe compounds that aren’t necessarilystable but which react slowly. Still others are common only from a historic perspective.

TABLE TM.2 Common Oxidation States and d Subshell Electron Configurations of the FirstSeries of Transition Metals

Oxidation State Sc Ti V Cr Mn Fe Co Ni Cu Zn

�1 d10

�2 d3 d5 d6 d7 d8 d9 d10

�3 d0 d3 d5 d6

�4 d0 d3

�5 d0

�6 d0

�7 d0

One point about the oxidation states of transition metals deserves particular attention:Transition metal ions with charges larger than �3 cannot exist in aqueous solution. Considerthe following reaction, for example, in which manganese is oxidized from the �2 to the�7 oxidation state.

Mn2�(aq) � 4 H2O(l) 88n MnO4�(aq) � 8 H�(aq) � 5 e�

When the manganese atom is oxidized, it becomes more electronegative. In the �7 oxi-dation state, the atom is electronegative enough to react with water to form a covalent ox-ide, MnO4

�.It is useful to have a way to distinguish between the charge on a transition metal ion

and the oxidation state of the transition metal. By convention, symbols such as Mn2� re-fer to ions that carry a �2 charge. Symbols such as Mn(VII) are used to describe com-pounds such as KMnO4 in which manganese is in the �7 oxidation state.

Mn(VII) isn’t the only example of an oxidation state powerful enough to decomposewater. As soon as Mn2� is oxidized to Mn(IV), it reacts with water to form MnO2. A sim-ilar phenomenon can be seen in the chemistry of both vanadium and chromium. Vanadiumexists in aqueous solutions as the V2� ion. But once it is oxidized to the �4 or �5 oxida-tion state, it reacts with water to form the VO2� or VO2

� ions. The Cr3� ion can be found

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TRANSITION METAL 9

in aqueous solution. But once the ion is oxidized to Cr(VI), it reacts with water to formthe CrO4

2� and Cr2O72� ions.

TM.6 LEWIS ACID–LEWIS BASE APPROACH TO BONDING INCOMPLEXES

G. N. Lewis was the first to recognize that the reaction between a transition metal ion andone or more ligands to form a coordination complex was analogous to the reaction betweenH� and OH� ions to form water. The reaction between H� and OH� ions involves the do-nation of a pair of electrons from the OH� ion to the H� ion to form a covalent bond.

The H� ion in the reaction acts as an electron pair acceptor. The OH� ion, on the otherhand, is an electron pair donor. Lewis argued that any ion or molecule that behaves likethe H� ion should be an acid. Conversely, any ion or molecule that behaves like the OH�

ion should be a base. A Lewis acid is therefore any ion or molecule that can accept a pairof electrons. A Lewis base is an ion or molecule that can donate a pair of electrons.

When Co3� ions react with ammonia, the Co3� ion accepts pairs of nonbonding elec-trons from six NH3 ligands to form covalent cobalt–nitrogen bonds (see Figure TM.4).

OOH�

88n HOOOHH� �

Co

NH3

NH3

NH3

NH3

H3N

H3N

3+

Co3+

N

HHH

N HH

H

NH H

H

NHH

HN

HH

H

NH H

H

FIGURE TM.4 Reactionbetween Co3� ions andammonia to form theCo(NH3)6

3� complex ion.

The metal ion is therefore a Lewis acid, and the ligands coordinated to this metal ion areLewis bases.

Co3� � 6 NH388nm88 Co(NH3)6

3�

Electron pair acceptor Electron pair donor Acid–base complex(Lewis acid) (Lewis base)

The Co3� ion is an electron pair acceptor, or Lewis acid, because it has empty valence-shell orbitals that can be used to hold pairs of electrons. To emphasize the empty valenceorbitals, we can write the configuration of the Co3� ion as follows.

Co3� [Ar] 3d6 4s0 4p0

There is room in the valence shell of the ion for 12 more electrons. (Four electrons canbe added to the 3d subshell, two to the 4s orbital, and six to the 4p subshell.) The NH3

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10 TRANSITION METAL

molecule is an electron pair donor, or Lewis base, because it has a pair of nonbonding elec-trons on the nitrogen atom.

According to this model, transition metal ions form coordination complexes becausethey have empty valence-shell orbitals that can accept pairs of electrons from a Lewis base.Ligands must therefore be Lewis bases—they must contain at least one pair of nonbond-ing electrons that can be donated to a metal ion.

TM.7 TYPICAL LIGANDSAny ion or molecule with a pair of nonbonding electrons can be a ligand. Many ligandsare said to be monodentate (literally, “one-toothed”) because they “bite” the metal in onlyone place. Typical monodentate ligands are given in Table TM.3.

TABLE TM.3 Typical Monodentate Ligands

Other ligands can bind to the metal more than once. Ethylenediamine (en) is a typicalbidentate ligand.

Each end of the molecule contains a pair of nonbonding electrons that can form a cova-lent bond to a metal ion. Ethylenediamine is also an example of a chelating ligand. Theterm chelate comes from a Greek stem meaning “claw.” It is used to describe ligands thatcan grab the metal in two or more places, the way a claw would. A typical ethylenediaminecomplex is shown in Figure TM.5.

H2N NH2D

H2C CH2G Ethylenediamine (en)

Co

Cl

NN

N

Cl

NCH2

CH2

H2C

H2C

+

Co(en)2Cl2+

H

H

H H

H H

H

H

FIGURE TM.5 Structure of the Co(en)2Cl2� complex ion.

O O CO OC O

�N ][ C

O

H H

N

H HH

O H�

Cl�

Br�

I�

O2�

O

O

O

2�

SSCS N�

F�

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TRANSITION METAL 11

A number of multidentate ligands are shown in Table TM.4. Linking ethylenediaminefragments gives tridentate ligands and tetradentate ligands, such as diethylenetriamine (dien)and triethylenetetramine (trien). Adding four OCH2CO2

� groups to an ethylenediamineframework gives a hexadentate ligand known as EDTA. This ligand that can single-handedlysatisfy the coordination number of six found for so many transition metal ions, as shown inFigure TM.6.

TABLE TM.4 Typical Multidentate Ligands

Structure Name

�

O O

O

OO

O

O

OCO

C C

2�

2�

Carbonate

Oxalate (ox)

Acetylacetonate (acac)

Ethylenediamine (en)

CHCH3C CCH3

CH2CH2

H2N NH2

Bidentate Ligands

O

Diethylenetriamine (dien)

Triethylenetetramine (trien)

Nitrilotriacetate (NTA)

Ethylenediaminetetraacetate (EDTA)

CH2CH2 CH2CH2

H2N NH

O

O

OO

O

NCH2CH2N

� �

NH2

CH2CH2

CH2CO2�

CH2CO2�

CCH2 CH2C

O �CH2CO� CCH2

CH2CO2�

CH2CH2 CH2CH2

H2N

N

NH NH NH2

Tridentate Ligands

Tetradentate Ligands

Hexadentate Ligands

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12 TRANSITION METAL

TM.8 COORDINATION COMPLEXES IN NATUREComplex ions play a vital role in the chemistry of living systems, as shown by the three bi-ologically important complexes in Figure TM.7: vitamin B12, chlorophyll a, and the hemefound in the proteins hemoglobin and myoglobin that carry oxygen from the lungs to themuscles.

As early as 1926, it was known that patients who suffered from pernicious anemia be-came better when they ate liver. It took more than 20 years, however, to determine thatthe vitamin B12 in liver was one of the active factors in relieving the anemia. It took an-other 10 years to determine the structure of vitamin B12, which is a six-coordinate Co3�

complex. The Co3� ion is coordinated to four nitrogen atoms that lie in a planar moleculeknown as a corrin ring. It is also coordinated to a fifth nitrogen atom and to a sixth group,labeled R in Figure TM.7a. The R group can be a CN�, NO2

�, SO32�, or OH� ion, de-

pending on the source of the vitamin B12.Every carbon atom in our bodies can be traced back to a reaction in which plants use

the energy in sunlight to make glucose (C6H12O6) from CO2 and water.

light6 CO2(g) � 6 H2O(aq) 8888888n C6H12O6(aq) � 6 O2(g)chlorophyll

The fundamental step in the photosynthesis of glucose is the absorption of sunlight by thechlorophyll in higher plants and algae. Chlorophyll a (see Figure TM.7b) is a complex inwhich an Mg2� ion is coordinated to four nitrogen atoms in a planar chlorin ring. The chlo-rin ring in chlorophyll is similar to the corrin ring in vitamin B12, but not identical. The mi-nor differences between the rings adjust their sizes so that each ring is the right size to holdthe correct transition metal ion.

The coordination complex shown in Figure TM.7c is known as a heme, which containsan Fe2� or Fe3� ion coordinated to four nitrogen atoms in a porphyrin ring. The por-phyrin ring in a heme is slightly larger than the corrin ring in vitamin B12 and slightlysmaller than the chlorin ring in chlorophyll a, so it is just the right size to hold the Fe2�

or Fe3� ion.

O

O

N

CH2

CH2

CH2

CH2

CH2

CO

OC

O

O

C

O

O

C

CH2

Fe

N

[Fe EDTA]–

–

FIGURE TM.6 EDTA forms very strong complexes with many transi-tion metal ions.

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TRANSITION METAL 13

CH2OH

CH

CH

H

H

N

N

N

N

N

H

H

H

HH

H

H

O

H2NCOCH2

H2NCOCH2

CH2CH2CONH2

Vitamin B12

Heme

Chlorophyll a

CH2CH2CONH2

CH2CH2CO2–

Fe2+, 3+

CH3

CH

–O2CCH2CH2

CH2

CH3 CH2

CH3

CH

HC

HC

H3C

CH2CH2CONH2

CH2CONH2

H3C

HOO

O

O–O

P

CH2

CH3

CH3

CH3

CH3

CH3

C

C

R

CH3

CH3

CH3

CH3

CH2

CH2

NH

CO

CH2

CH3

CH2

CH3

CH3 CH2CH3

CH3

H2C

CH

CH

CH2

CH2

CH2

CH2

CH2

CH2

CH2

CH2

CH2

CH2

CH2

CH2

CH2

CC C

C

CH

HC

CC

N N C CH3CH3

CH

CH2

CH

CH CH2

CH

O

CH

C

C

C C

O

C C

N

HC CH

CH

Mg2+

C

NC

C C

C

CC C

C

CN

CH3

CH3

O

C

C

C

N

CH3

H3C

Co3+

N

(a)

(c)

(b)

C

N N

CC C

C CC

CC

FIGURE TM.7 Structures of the biologically important coordination complexes known as vitamin B12,chlorophyll a, and heme.

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14 TRANSITION METAL


Nitrogen Fixation

Ever since the Haber process was developed, chemists have been fascinated by the con-trast between the conditions required to reduce nitrogen to ammonia in the laboratoryand the conditions under which this reaction occurs in nature.

N2(g) � 3 H2(g) 88n 2 NH3(g)

In the laboratory, the reaction requires high pressures (200–300 atm) and high tempera-tures (400–600°C). In nature, it occurs at room temperature and atmospheric pressure.

The nitrogenase enzyme that catalyzes the reaction in nature can be separated intotwo proteins, which contain a total of 34 iron atoms and two molybdenum atoms. Oneof the proteins contains both Fe and Mo atoms, the other contains only Fe. The Feprotein contains an Fe4S4 unit that binds the ATP that provides the energy that drivesthe reaction in which nitrogen is reduced to ammonia. The MoFe protein contains fourFe4S4 subunits that can hold one of the electrons necessary to reduce N2 and a pair ofMoFe7S8 units. One end of each MoFe7S8 unit is bound to the OSH group on the sidechain of a cysteine residue in the protein; the other is bound to an analog of the citrateion known as homocitrate.

Jongsun Kim and Douglas Rees, of the California Institute of Technology, reportedX-ray crystal structures of the Fe protein from Azotobacter vinelandii and of the MoFeprotein from Clostridium pasteurianum [Science, 257, 1677 (1992)]. Their work suggeststhat the N2 molecule is bound, side-on, between a pair of iron atoms in the MoF7S8 (ho-mocitrate) complex, as shown in Figure TM.8.

- S

Cysteine

Homocitrate

Histidine

- Fe

- Mo

- C

- O

- N

These results explain why chemists have been frustrated for so many years by theirinability to understand the role of the molybdenum atoms in the MoFe protein. Con-trary to most researchers’ expectations, the N2 molecule isn’t carried by the molybdenumatoms in the protein. It apparently binds to iron atoms in both the industrial catalyst andthe nitrogenase enzyme. The primary difference between these catalysts is the ease withwhich they can pump electrons into the metal atoms that carry the N2 molecule, therebyreducing it to ammonia.

TM.9 NOMENCLATURE OF COMPLEXESThe rules for naming chemical compounds are established by nomenclature committees ofthe International Union of Pure and Applied Chemistry (IUPAC). The IUPAC nomen-clature of coordination complexes is based on the following rules.

FIGURE TM.8 Model for the binding of N2 to theMoFe protein in the nitrogenase enzyme.

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TRANSITION METAL 15

Rules for Naming Coordination Complexes

• The name of the positive ion is written before the name of the negative ion.• The names of the ligands are written before the name of the metal to which they

are coordinated.• The Greek prefixes mono-, di-, tri-, tetra-, penta-, hexa-, and so on are used to indi-

cate the number of ligands when the ligands are relatively simple. The Greek pre-fixes bis-, tris-, and tetrakis- are used with more complicated ligands.

• The names of negative ligands always end in o, as in fluoro (F�), chloro (Cl�), bromo(Br�), iodo (I�), oxo (O2�), hydroxo (OH�), and cyano (CN�).

• A handful of neutral ligands are given common names, such as aquo (H2O), am-mine (NH3), and carbonyl (CO).

• Ligands are listed in the following order: negative ions, neutral molecules, and pos-itive ions. Ligands with the same charge are listed in alphabetical order.

• The oxidation number of the metal atom is indicated by a Roman numeral in paren-theses after the name of the metal atom.

• The names of complexes with a net negative charge end in -ate. Co(SCN)42�, for

example, is the tetrathiocyanatocobaltate(II) ion. When the symbol for the metal isderived from its Latin name, -ate is added to the Latin name of the metal. Thus,negatively charged iron complexes are ferrates and negatively charged copper com-plexes are cuprates.

Exercise TM.4

Name the following coordination complexes.(a) [Cr(NH3)5(H2O)][(NO3)3](b) [Cr(NH3)4Cl2]Cl

Solution

(a) The complex contains the Cr(NH3)5(H2O)3� ion and three nitrate ions to balance thecharge on the complex ion. It is therefore known as pentaammineaquochromium(III)nitrate

(b) Two of the three chloride ions are ligands that are coordinated to the Cr3� ion, andthe other is a Cl� ion that balances the charge on the Cr(NH3)4Cl2

� ion. The com-pound is therefore known as dichlorotetraamminechromium(III) chloride

Exercise TM.5

Determine the formulas of the following compounds.(a) potassium hexacyanoferrate(II)(b) tris(ethylenediamine)chromium(III) chloride

Solution

(a) The compound contains enough K� ions to balance the charge on the negatively chargedferrate ion. Because the ferrate ion is formed from an Fe2� ion and six CN� ligands, itcarries a net charge of �4. The formula for the compound is therefore K4Fe(CN)6.

(b) The compound contains a Cr3� ion coordinated to three bidentate ethylenediamine oren ligands. It therefore must contain three Cl� ions to balance the charge on the com-plex ion. Thus, the formula for the compound is [Cr(en)3]Cl3.

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16 TRANSITION METAL

TM.10 ISOMERS

Cis/Trans Isomers

An important test of Werner’s theory of coordination complexes involved the study of co-ordination complexes that formed isomers (literally, “equal parts”). Isomers are compoundswith the same chemical formula that have different structures. There are two isomers of theCo(NH3)4Cl2

� complex ion, for example, which differ in the orientation of the two chlo-ride ions around the Co3� ion, as shown in Figure TM.9. In the trans isomer, the chloridesoccupy positions across from one another in the octahedron. In the cis isomer, they occupyadjacent positions. The difference between cis and trans isomers can be remembered bynoting that the prefix trans is used to describe things that are on opposite sides, as in trans-atlantic or transcontinental.

Co

Cl

Cl

NH3

NH3

H3N

H3N

+

trans – [Co(NH3)4Cl2]+

Co

Cl

NH3

Cl

NH3

H3N

H3N

+

cis – [Co(NH3)4Cl2]+

At the time Werner proposed his theory, only one isomer of the [Co(NH3)4Cl2]Cl com-plex was known, namely, the green complex described in Section TM.2. Werner predicted thata second isomer should exist, and his discovery in 1907 of a purple compound with the samechemical formula was a key step in convincing scientists who were still skeptical of the model.

Cis/trans isomers are also possible in four-coordinate complexes that have a square-planar geometry. Figure TM.10 shows the structures of the cis and trans isomers of dichloro-diammineplatinum(II).

G D

GD

Cl NH3

H3N

Pt

Cltrans-Pt(NH3)2Cl2

G D

GD

Cl NH3

NH3

Pt

Clcis-Pt(NH3)2Cl2

FIGURE TM.9 The trans and cis isomersof the Co(NH3)4Cl2

� complex ion.

FIGURE TM.10 The cis/trans isomers of the square-planarPt(NH3)2Cl2 complex.

The cis isomer of this compound is used as a drug to treat brain tumors, under the tradename Cisplatin. This square-planar complex inserts itself into the grooves in the doublehelix structure of the DNA in cells, which inhibits the replication of DNA. This slows downthe rate at which the tumor grows, which allows the body’s natural defense mechanisms toact on the tumor.

CheckpointWhich of the following compounds can form cis/trans isomers?(a) square-planar Rh(CO)Cl3 (c) octahedral Ni(en)2(H2O)2

2�

(b) trigonal-bipyramidal Fe(CO)4(PH3) (d) tetrahedral Ni(CO)3(PH3)

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TRANSITION METAL 17

Chiral Isomers

Another form of isomerism can best be understood by considering the difference betweengloves and mittens. One glove in each pair fits the left hand, and the other fits the righthand. Mittens usually fit equally well on either hand. To understand why, hold a glove anda mitten in front of a mirror. There is no difference between the mitten shown in FigureTM.11 and its mirror image. Each mitten is therefore said to be superimposable on its mir-ror image. The same can’t be said of the relationship between a glove and its mirror im-age. The mirror image of the glove that fits the left hand looks like the glove that fits theright hand, and vice versa.

Mirror

Left-handglove

Mitten FIGURE TM.11 Gloves are chiral; mittens are not.

For all practical purposes, the two gloves that form a pair have the same “substituents.”Each glove has four “fingers” and one “thumb.” If you clap them together, you will findeven more similarities between the gloves. The thumbs are attached at the same point, sig-nificantly below the point where the fingers start. The second “fingers” on both gloves arethe longest and the little “fingers” are the shortest.

In spite of these similarities, there is a fundamental difference between the gloves thatcan be observed by trying to place your right hand into a left-hand glove. Like your hands,each glove is the mirror image of the other, and the mirror images are not superimposable.Objects that possess a similar handedness are said to be chiral (literally, “handed”). Glovesare chiral; mittens are not. Feet and shoes are both chiral, but socks are not.

The Co(en)33� ion in Figure TM.12 is an example of a chiral molecule, which forms a

pair of isomers that are mirror images of each other. The isomers have almost identicalphysical and chemical properties. They have the same melting point, boiling point, density,and color, for example. They differ only in the way they interact with plane-polarized light.

Light consists of electric and magnetic fields that oscillate in all possible directions per-pendicular to the path of the light ray. When light is passed through a polarizer, such as aNicol prism or the lens of polarized sunglasses, the only light that emerges from the po-larizer is light whose oscillations are confined to a single plane.

In 1813 Jean Baptiste Biot noticed that the plane in which polarized light oscillates wasrotated either to the right or the left when plane-polarized light was passed through singlecrystals of quartz or aqueous solutions of tartaric acid or sugar. Because they seem to in-teract with light, compounds that can rotate plane-polarized light are said to be opticallyactive. Those that rotate the plane clockwise (to the right) are said to be dextrorotatory(from the Latin word dexter, “right”). Those that rotate the plane counterclockwise (to theleft) are called levorotatory (from the Latin word laevus, “left”).

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18 TRANSITION METAL

All compounds that are chiral are optically active; one isomer is dextrorotatory andthe other is levorotatory. Chirality is a property of a molecule that results from its struc-ture. Optical activity is a macroscopic property of a collection of the molecules that arisesfrom the way they interact with light.

CheckpointWhich of the following molecules is chiral?(a) octahedral Cr(CO)6 (c) SF4 (see Figure 4.21)(b) tetrahedral Ni(CO)4 (d) octahedral Fe(acac)3

TM.11 THE VALENCE BOND APPROACH TO BONDING INCOMPLEXES

The idea that atoms form covalent bonds by sharing pairs of electrons was first proposedby G. N. Lewis in 1902. It was not until 1927, however, that Walter Heitler and Fritz Lon-don showed how the sharing of pairs of electrons holds a covalent molecule together. TheHeitler–London model of covalent bonds was the basis of the valence bond theory. Thelast major step in the evolution of the theory was the suggestion that atomic orbitals mixto form hybrid orbitals, such as the sp, sp2, sp3, dsp3, and d2sp3 orbitals shown in FigureTM.13.

H2CH2C

Co

NH2

CH2

CH2H2N

NH2

H2N CH2

CH2

H2N

H2N

Co

NH2 CH2

NH2

H2C

H2C

NH2

NH2

H2C

H2C

CH2

3+ 3+

Mirror

NH2

NH2

FIGURE TM.12 The Co(en)33� complex ion is an example of a chiral molecule

that exists as a pair of isomers that are mirror images of each other.

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TRANSITION METAL 19

It is easy to apply the valence bond theory to some coordination complexes, such asthe Co(NH3)6

3� ion. We start with the electron configuration of the transition metal ion.

Co3� [Ar] 3d6

We then look at the valence-shell orbitals and note that the 4s and 4p orbitals are empty.

Co3� [Ar] 3d6 4s0 4p0

Experiments tell us that there are no unpaired electrons in the complex so we begin byconcentrating the 3d electrons in the dxy, dxz, and dyz orbitals. This gives the following elec-tron configuration.

The 3dx2�y2, 3dz2, 4s, 4px, 4py, and 4pz orbitals are then mixed to form a set of empty d2sp3

orbitals that point toward the corners of an octahedron as shown in TM.13. Each of theorbitals can accept a pair of nonbonding electrons from a neutral NH3 molecule to form acomplex in which the cobalt atom has a filled shell of valence electrons.

Exercise TM.6

Use valence bond theory to describe the Fe(CN)64� complex ion. This complex contains

no unpaired electrons.

Solution

We start with the electron configuration of the transition metal ion.

Fe2� [Ar] 3d6 4s0 4p0

Co(NH3)63� hg hg hg hg hg hghg hg hg

3d d2sp3

Co3� hg hg hg

3d 4s 4p

sp

sp2

sp3 sp3d2 or d2sp3

sp3d

FIGURE TM.13 The shapes of the sp, sp2, sp3, sp3d, and sp3d2 hybrid orbitals.

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20 TRANSITION METAL

By investing a little energy in the system, we can pair the six 3d electrons, thereby creat-ing empty 3dx2�y2 and 3dz2 orbitals.

The orbitals are then mixed with the empty 4s and 4p orbitals to form a set of six d2sp3

hybrid orbitals, which can accept pairs of nonbonding electrons from the CN� ligands toform an octahedral complex in which the metal atom has a filled shell of valence electrons.

At first glance, some complexes, such as the Ni(NH3)62� ion, seem hard to explain with

the valence bond theory. We start, as always, by writing the configuration of the transitionmetal ion.

Ni2� [Ar] 3d8

This configuration creates a problem because there are eight electrons in the 3d orbitals.Even if we invest the energy necessary to pair the 3d electrons, we can’t find enough empty3d orbitals to use to form a set of d2sp3 hybrids.

There is a way around this problem. The five 4d orbitals on nickel are empty, so wecan form a set of empty sp3d2 hybrid orbitals by mixing the 4dx2-y2, 4dz2, 4s, 4px, 4py, and4pz orbitals. The hybrid orbitals then accept pairs of nonbonding electrons from six am-monia molecules to form a complex ion.

The valence bond theory therefore formally distinguishes between “inner-shell” complexes,which use 3d, 4s, and 4p orbitals to form a set of d2sp3 hybrids, and “outer-shell” com-plexes, which use 4s, 4p, and 4d orbitals to form sp3d2 hybrid orbitals.

TM.12 CRYSTAL FIELD THEORYAt almost exactly the same time that chemists were developing the valence bond modelfor coordination complexes, physicists such as Hans Bethe, John Van Vleck, and LeslieOrgel were developing an alternative known as crystal field theory. Valence bond theorywas used to explain the chemical properties of coordination complexes, such as the factthat the Co3� ion forms a six-coordinate Co(NH3)6

3� complex. Crystal field theory wasdeveloped to explain some of the physical properties of the complexes, such as their colorand their behavior in a magnetic field.

hg hg hg h h

3dNi(NH3)6

2� hg hg hghg hg hg

sp3d2

Ni2� hg hg hg h h

3d 4s 4p

Fe(CN)64� hg hg hg hg hg hghg hg hg

3d d2sp3

Fe2� hg hg hg

3d 4s 4p

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TRANSITION METAL 21

Crystal-field theory can be understood by thinking about the effect of the electricalfield of neighboring ions on the energies of the valence orbitals of the transition-metal ionsin manganese(II) oxide, MnO, and copper(I) chloride, CuCl.

MnO: Octahedral Crystal Fields

Each Mn2� ion in manganese(II) oxide is surrounded by six O2� ions arranged toward thecorners of an octahedron, as shown in Figure TM.14. MnO is therefore a model for an oc-tahedral complex in which a transition metal ion is coordinated to six ligands.

O2–

O2–

O2–

O2–

O2–

O2–

Mn2+

FIGURE TM.14 The octahedral geometry of O2� ions that surround each Mn2�

ion in MnO.

Repulsion between electrons on the O2� ions and electrons in the 3d orbitals on themetal ion in MnO increases the energy of these orbitals. But this repulsion doesn’t affectthe five 3d orbitals the same way. Let’s assume that the six O2� ions that surround eachMn2� ion define an xyz coordinate system. Two of the 3d orbitals (3dx2�y2 and 3dz2) on theMn2� ion point directly toward the six O2� ions, as shown in Figure TM.15. The other threeorbitals (3dxy, 3dxz, and 3dyz) lie between the O2� ions.

x

z

3dx2 – y2

yy

x

3dz2

z

x

z

3dxy

y y

x

z

3dxz

y

x

z

3dyz

FIGURE TM.15 The five orbitals in a d subshell.

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22 TRANSITION METAL

The energy of the five 3d orbitals increases when the six O2� ions are brought close tothe Mn2� ion because of repulsion between the electrons on the O2� ions and the electronsin the d orbitals on the Mn2� ion. The energy of two of these orbitals (3dx2�y2 and 3dz2), how-ever, increases much more than the energy of the other three (3dxy, 3dxz, and 3dyz), as shownin Figure TM.16. The crystal field of the six O2� ions in MnO therefore splits the degeneracyof the five 3d orbitals. Three of the orbitals are now lower in energy than the other two.

Isolated atom or ion

dx2 – y2 dz2

dxy dxz dyz

Octahedralcrystalfield

∆o

E

eg

t2g

FIGURE TM.16 The two d orbitals that point toward the ligands in an octahedral complex are higher in en-ergy than the three d orbitals that lie between the ligands.

By convention, the dxy, dxz, and dyz orbitals in an octahedral complex are called the t2g

orbitals. The dx2�y2 and dz2 orbitals, on the other hand, are called the eg orbitals. The eas-iest way to remember this convention is to note that there are three orbitals in the t2g set.

t2g dxy, dxz, and dyz

eg dx2�y2 and dz2

The difference between the energies of the t2g and eg orbitals in an octahedral complex isrepresented by the symbol �o. The splitting of the energy of the d orbitals is not trivial; �o

for the Ti(H2O)63� ion, for example, is 242 kJ/mol. Which is roughly the same as the en-

ergy given off when one mole of water is produced by burning a mixture of H2 and O2.The magnitude of the splitting of the t2g and eg orbitals changes from one octahedral

complex to another. It depends on the identity of the metal ion, the charge on that ion,and the nature of the ligands coordinated to the metal ion.

CuCl: Tetrahedral Crystal Fields

Each Cu� ion in copper(I) chloride is surrounded by four Cl� ions arranged toward thecorners of a tetrahedron, as shown in Figure TM.17. CuCl is therefore a model for a tetra-hedral complex in which a transition metal ion is coordinated to four ligands.

Cu+

Cl–

Cl–

Cl–

Cl–FIGURE TM.17 The tetrahedral geometry of Cl� ions that surround each Cu� ionin CuCl.

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TRANSITION METAL 23

Once again, the negative ions in the crystal split the energy of the d atomic orbitals onthe transition metal ion. The tetrahedral crystal field splits these orbitals into the same t2g

and eg sets of orbitals as does the octahedral crystal field.

t2g dxy, dxz, and dyz

eg dx2�y2 and dz2

But the two orbitals in the eg set are now lower in energy than the three orbitals in the t2g

set, as shown in Figure TM.18.


dxy dxz

dx2 – y2 dz2

Tetrahedralcrystalfield

∆t4––9

= ∆o

E

dyz

eg

t2g

FIGURE TM.18 In a tetrahedral complex, the eg orbitals are lower in energy than the t2g orbitals. The differ-ence between the energies of the orbitals is smaller in tetrahedral complexes than in an equivalent octahe-dral complex.

To understand the splitting of d orbitals in a tetrahedral crystal field, imagine four li-gands lying at alternating corners of a cube to form a tetrahedral geometry, as shown inFigure TM.19. The dx2�y2 and dz2 orbitals on the metal ion at the center of the cube lie be-tween the ligands, and the dxy, dxz, and dyz orbitals point toward the ligands. As a result,the splitting observed in a tetrahedral crystal field is the opposite of the splitting in an oc-tahedral complex.

x

y

z

FIGURE TM.19 In a tetrahedral complex, dxy, dxz, and dyz orbitals point toward the ligands; the dx2�y2 and dz2 orbitals point between the ligands.

Because a tetrahedral complex has fewer ligands, the magnitude of the splitting issmaller. The difference between the energies of the t2g and eg orbitals in a tetrahedral

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24 TRANSITION METAL

complex (�t) is slightly less than half as large as the splitting in analogous octahedral com-plexes (�o).

�t � 4/9 �o

Square-Planar Complexes

The crystal field theory can be extended to square-planar complexes, such as Pt(NH3)2Cl2.The splitting of the d orbitals in these compounds is shown in Figure TM.20.


dxy

dz2

dx2 – y2

dyz dxz

Square-planarcrystalfield

∆O

2––3

∼ ∆O

E

FIGURE TM.20 The splitting of the d orbitals in a square-planar complex.

TM.13 THE SPECTROCHEMICAL SERIESThe splitting of the d orbitals in the crystal field model not only depends on the geometryof the complex, it also depends on the nature of the metal ion, the charge on that ion, andthe ligands that surround the metal. When the geometry and the ligands are held constant,the splitting decreases in the following order.

Pt4� � Ir3� � Rh3� � Co3� � Cr3� � Fe3� � Fe2� � Co2� � Ni2� � Mn2�

Strong-field ions Weak-field ions

Metal ions at one end of the continuum are called strong-field ions, because the splitting dueto the crystal field is unusually strong. Ions at the other end are known as weak-field ions.

When the geometry and the metal are held constant, the splitting of the d orbitals de-creases in the following order.

CO � CN� � NO2� � NH3 � ONCS� � H2O � OH� � F� � OSCN� � Cl� � Br�

Strong-field ligands Weak-field ligands

Ligands that give rise to large differences between the energies of the t2g and eg orbitals arecalled strong-field ligands. Those at the opposite extreme are known as weak-field ligands.

Because they result from studies of the absorption spectra of transition metal com-plexes, these generalizations are known as the spectrochemical series. The range of valuesof � for a given geometry is remarkably large. The value of �o is 100 kJ/mol in theNi(H2O)6

2� ion, for example, and 520 kJ/mol in the Rh(CN)63� ion.

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TRANSITION METAL 25

TM.14 HIGH-SPIN VERSUS LOW-SPIN OCTAHEDRALCOMPLEXES

Once we know the relative energies of the d orbitals in a transition metal complex, we haveto worry about how these orbitals are filled. Orbitals that have the same energy are calleddegenerate orbitals. Degenerate orbitals are filled according to Hund’s rules.

• One electron is added to each of the degenerate orbitals in a subshell before a sec-ond electron is added to any orbital in the subshell.

• Electrons are added to a subshell with the same value of the spin quantum numberuntil each orbital in the subshell has at least one electron.

Octahedral transition metal ions with d1, d2, and d3 configurations can therefore be de-scribed by the following diagrams.

When we try to add a fourth electron, we are faced with a problem. The electron couldbe used to pair one of the electrons in the lower energy (t2g) set of orbitals or it could beplaced in one of the higher energy (eg) orbitals. One of the configurations is called low-spin because it contains only two unpaired electrons. The other is called high-spin becauseit contains four unpaired electrons with the same spin. The same problem occurs with oc-tahedral d5, d6, and d7 complexes.

For octahedral d8, d9, and d10 complexes, there is only one way to write satisfactory con-figurations.

h hhg hg hg

hg hhg hg hg

hg hghg hg hg

d8

d9

d10

Low-spin High-spin

hg h h hh

h h

h hhg hg h h h h

h hhg hg hg hg h h

h h hhg hg hg hg hg h

d4

d5

d6

d7

h h

h h h

d2

d3

hd1

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26 TRANSITION METAL

As a result, we have to differentiate between high-spin and low-spin octahedral complexesonly when there are either four, five, six, or seven electrons in the d orbitals.

The choice between high-spin and low-spin configurations for octahedral d4, d5, d6, andd7 complexes is based on by comparing the energy it takes to pair electrons with the en-ergy it takes to excite an electron to the higher energy (eg) orbitals. If it takes less energyto pair the electrons, the complex is low-spin. If it takes less energy to excite the electron,the complex is high-spin.

The amount of energy required to pair electrons in the t2g orbitals of an octahedralcomplex is more or less constant. The amount of energy needed to excite an electron intothe higher energy (eg) orbitals, however, depends on the value of �o for the complex. Asa result, we expect to find low-spin complexes among metal ions and ligands that lie to-ward the high-field end of the spectrochemical series. High-spin complexes are expectedamong metal ions and ligands that lie toward the low-field end of the series.

Compounds in which all of the electrons are paired are diamagnetic—they are repelledby both poles of a magnet. Compounds that contain one or more unpaired electrons areparamagnetic—they are attracted to the poles of a magnet. The force of attraction betweenparamagnetic complexes and a magnetic field is proportional to the number of unpairedelectrons in the complex. Thus, we can determine whether a complex is high-spin or low-spin by measuring the strength of the interaction between the complex and a magneticfield.

Exercise TM.6

Explain why the Co(NH3)63� ion is diamagnetic, whereas the CoF6

3� ion is paramagnetic.

Solution

Both complexes contain the Co3� ion, which lies toward the strong-field end of the spec-trochemical series. NH3 also lies toward the strong-field end of the series. As a result, �o

for the Co(NH3)63� ion should be relatively large. When the splitting is large, it takes less

energy to pair the electrons in the t2g orbitals than it does to excite the electrons to the eg

orbitals. The Co(NH3)63� ion is therefore a low-spin d6 complex in which the electrons are

all paired, and the ion is diamagnetic.The F� ion lies toward the low-field end of the spectrochemical series. As a result,

�o for the CoF63� ion is much smaller. In this complex, it apparently takes less energy to

excite electrons into the eg orbitals than to pair them in the t2g orbitals. As a result, theCoF6

3� ion is a high-spin d6 complex that contains unpaired electrons and is therefore para-magnetic.

TM.15 THE COLORS OF TRANSITION METAL COMPLEXESOne of the joys of working with transition metals is the extraordinary range of colors ex-hibited by their compounds. By changing the number of ammonia, water, and chloride li-gands on a Co3� ion, for example, we can vary the color of the complex over the entire

h hhg hg hg hg h h

Low-spin d6 High-spin d6

∆ο∆ο

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TRANSITION METAL 27

visible spectrum, from red to violet, as shown in Section TM.2. We can also vary the colorof the complex by keeping the ligand constant and changing the metal ion. The charac-teristic colors of aqueous solutions of some common transition metal ions are given in TableTM.5.

TABLE TM.5 Characteristic Colors of Common Transition Metal Ion Aquo and Other Complexes

Ion Color Ion Color

Cr2� Blue Ni2� GreenCr3� Blue-violet Zn2� ColorlessCo2� Pinkish red CrO4

2� YellowCu2� Light blue Cr2O7

2� OrangeFe3� Yellow MnO4

� Deep violetMn2� Very faint pink

The color of these complexes also changes with the oxidation state of the metal atom.By reducing the VO2

� ion with zinc metal, for example, we can change the color of thesolution from yellow to green to blue and then to violet as the vanadium is reduced, oneelectron at a time, from the �5 to the �2 oxidation state.

VO2� yellow

VO2� greenV3� blueV2� violet

To explain why transition metal complexes are colored, and why the color is so sensitiveto changes in the metal, ligand, and oxidation state, we need to understand the physics ofcolor.

There are two ways of producing the sensation of color. We can add color where noneexists or subtract it from white light. The three primary additive colors are red, green, andblue. When all three are present at the same intensity, we get white light. The three pri-mary subtractive colors are cyan, magenta, and yellow. When those three colors are ab-sorbed with the same intensity, light is absorbed across the entire visible spectrum. If theobject absorbs strongly enough, or if the intensity of the light is dim enough, all of the lightcan be absorbed.

Green

MagentaRedBlue

Cyan Yellow

Yellow

BlueMagentaCyan

Green Red

Primaryadditivecolors

Primarysubtractivecolors

FIGURE TM.21 The primary additive and subtractive colors.

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28 TRANSITION METAL

The additive and subtractive colors are complementary, as shown in Figure TM.21. Ifwe subtract yellow from white light, we get a mixture of cyan and magenta, and the lightlooks blue. If we subtract blue—a mixture of cyan and magenta—from white light, the lightlooks yellow. The Cu(NH3)4

2� complex ion has a blue color because it absorbs light in theyellow portion of the spectrum. The CrO4

2� ion appears yellow because it absorbs bluelight. Table TM.6 describes what happens when light in different portions of the visiblespectrum is absorbed.

TABLE TM.6 Relationship between the Color of Transition Metal Complexesand the Wavelength of Light Absorbed

Wavelength Absorbed (nm) Color of Light Absorbed Color of Complex

410 Violet Lemon yellow430 Indigo Yellow480 Blue Orange500 Blue-green Red530 Green Purple560 Lemon yellow Violet580 Yellow Indigo610 Orange Blue680 Red Blue-green

Light is absorbed when it carries just enough energy to excite an electron from one or-bital to another. Compounds therefore absorb light when the difference between the en-ergy of an orbital that contains an electron and the energy of an empty orbital correspondsto a wavelength in the narrow band of the electromagnetic spectrum that is visible to thenaked eye.

How much energy is associated with the absorption of a typical photon, such as a pho-ton with a wavelength of 480 nanometers? Since we know the wavelength of the photon,we can calculate its frequency.

� � ��

c� ��

24.98908�

�

1100�

8

9mm

/s�� 6.25 � 1014 s�1

We can then calculate the energy of the photon from its frequency and Planck’s constant.

E � h� � (6.626 � 10�34 J�s)(6.25 � 1014 s�1) � 4.14 � 10�19 J

But this is the energy associated with only a single photon. Multiplying by Avogadro’s num-ber gives us the energy in units of kilojoules per mole.

�4.1

14

p�

ho10

to

�

n

19 J�� 249 kJ/mol

Repeating the calculation for all frequencies in the visible portion of the spectrum re-veals that the energy associated with a photon of light ranges from 160 to 300 kJ/mol. Ifwe refer back to the discussion of the spectrochemical series in Section TM.13, we find that

6.022 � 1023 photons��

1 mol

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TRANSITION METAL 29

these energies fall in the range of values of �. Anything that changes the difference be-tween the energies of the t2g and eg orbitals in a transition metal complex therefore influ-ences the color of the light absorbed by the complex. Thus, it isn’t surprising that the colorof the complex is sensitive to a variety of factors, including the geometry of the complex,the identity of the metal, the nature of the ligand, and the oxidation state of the metal.

TM.16 LIGAND FIELD THEORYThe valence bond model described in Section TM.11 and the crystal field theory describedin Section TM.12 each explain some aspects of the chemistry of the transition metals, butneither model is good at predicting all of the properties of transition metal complexes. Athird model, based on molecular orbital theory, was therefore developed that is known asligand field theory. Ligand field theory is more powerful than either the valence bond orcrystal field theories. Unfortunately, it is also more abstract.

The ligand field model for an octahedral transition metal complex such as theCo(NH3)6

3� ion assumes that the 3d, 4s, and 4p orbitals on the metal overlap with one or-bital on each of the six ligands to form a total of 15 molecular orbitals, as shown in FigureTM.22. Six of the orbitals are bonding molecular orbitals, whose energies are much lowerthan those of the original atomic orbitals. Another six are antibonding molecular orbitals,whose energies are higher than those of the original atomic orbitals. Three are best de-scribed as nonbonding molecular orbitals, because they have essentially the same energyas the 3d atomic orbitals on the metal.

FIGURE TM.22 The molecular orbitalsin this ligand field diagram forCo(NH3)6

3� were generated by allow-ing the valence-shell 3d, 4s, and 4p or-bitals on the transition metal to overlapwith an orbital on each of the six lig-ands that contribute pairs of nonbond-ing electrons to form an octahedralcomplex.

3d

∆O

E

Metalorbitals

4s

4p

Ligandorbitals

Ligand field theory enables the 3d, 4s, and 4p orbitals on the metal to overlap with or-bitals on the ligand to form the octahedral covalent bond skeleton that holds the complextogether. At the same time, the model generates a set of five orbitals in the center of the

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30 TRANSITION METAL

diagram that are split into t2g and eg subshells, as predicted by the crystal field theory. Asa result, we don’t have to worry about “inner-shell” versus “outer-shell” metal complexes.In effect, we can use the 3d orbitals in two different ways. We can use them to form thecovalent bond skeleton and then use them again to form the orbitals that hold the elec-trons that were originally in the 3d orbitals of the transition metal.

KEY TERMS

ActinideBidentate ligandChelating ligandChiralCisComplex ionCoordination compoundCoordination numberCrystal field theoryDextrorotatory

DiamagneticElectron pair

acceptor/donorHigh-spinIsomersLanthanideLevorotatoryLewis acid/baseLigandLigand field theory

Low-spinMain-group elementsMonodentateOptically activeParamagneticSpectrochemical seriesTransTransition metalsValence bond theory

PROBLEMS

Transition Metals and Coordination Complexes

1. Identify the subshell of atomic orbitals filled among the transition metals in the fifthrow of the periodic table.

2. Describe some of the ways in which transition metals differ from main-group metalssuch as aluminum, tin, and lead. Describe ways in which they are similar.

3. Use the electron configurations of Zn2�, Cd2�, and Hg2� to explain why those ions of-ten behave as if they were main-group metals.

4. Use the electronegativities of cobalt and nitrogen to predict whether the CoON bondin the Co(NH3)6

3� ion is best described as ionic, polar covalent, or covalent.5. Define the terms coordination number and ligand.

Werner’s Model of Coordination Complexes

6. Werner wrote the formula of one of his coordination complexes as CoCl3�6 NH3. To-day, we write that compound as [Co(NH3)6]Cl3 to indicate the presence of Co(NH3)6

3�

and Cl� ions. Write modern formulas for the compounds Werner described as CoCl3�5NH3, CoCl3�4 NH3, and CoCl3�5 NH3�H2O.

7. What is the charge on the cobalt ion in the following complex [Co(NH3)6]Cl3?8. Explain why aqueous solutions of CoCl3�6 NH3 conduct electricity better than aque-

ous solutions of CoCl3�4 NH3.9. Explain why Ag� ions precipitate three chloride ions from an aqueous solution of

CoCl3�6 NH3 but only one chloride ion from an aqueous solution of CoCl3�4 NH3.10. Predict the number of Cl� ions that could precipitate from an aqueous solution of the

Co(en)2Cl2 complex.

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TRANSITION METAL 31

11. Predict the number of ions formed when the NiSO4�4 NH3�2 H2O complex dissociatesin water.

Typical Coordination Numbers

12. Use the examples in Table TM.1 to identify at least one factor that influences the co-ordination number of a transition metal ion.

13. Determine both the coordination number and the charge on the transition metal ionin each of the following complexes.(a) CuF4

2� (b) Cr(CO)6 (c) Fe(CN)64� (d) Pt(NH3)2Cl2

14. Determine both the coordination number and the charge on the transition metal ionin each of the following complexes.(a) Co(SCN)4

2� (b) Fe(acac)3 (c) Ni(en)2(H2O)22� (d) Co(NH3)5(H2O)3�

The Electron Configuration of Transition Metal Ions

15. Write the electron configuration of the following transition metal ions.(a) V2� (b) Cr2� (c) Mn2� (d) Fe2� (e) Ni2�

16. Explain why the Co2� ion can be described as a d7 ion.17. Which of the following ions can be described as d5?

(a) Cr2� (b) Mn2� (c) Fe3� (d) Co3� (e) Cu�

18. Explain the difference between the symbols Cr3� and Cr(VI).19. Which of the following is not an example of a d0 transition metal complex?

(a) TiO2 (b) VO2� (c) Cr2O72� (d) MnO4

�

20. Explain why manganese becomes more electronegative when it is oxidized from Mn2�

to Mn(VII).21. Use electronegativities to predict whether the MnOO bond in MnO4

� is best describedas covalent or ionic.

Lewis Acid–Lewis Base Approach to Bonding in Complexes

22. Describe what to look for when deciding whether an ion or molecule is a Lewis acid.23. Explain how Lewis acids, such as the Co3� ion, pick up Lewis bases, or ligands, to form

coordination complexes.24. Which of the following are Lewis acids?

(a) Fe3� (b) BF3 (c) H2 (d) Ag� (e) Cu2�

25. Which of the following are Lewis bases, and therefore potential ligands?(a) CO (b) O2 (c) Cl� (d) N2 (e) NH3

26. Which of the following are Lewis bases, and therefore potential ligands?(a) CN� (b) SCN� (c) CO3

2� (d) NO� (e) S2O32�

Typical Ligands

27. Define the terms monodentate, bidentate, tridentate, and tetradentate. Give an exampleof each category of ligands.

28. Use Lewis structures to explain why carbon monoxide could act as a bridge betweena pair of transition metals, but it can’t be a chelating ligand that coordinates to thesame metal twice.

29. Draw the structures of the following coordination complexes.(a) Fe(acac)3 (b) Co(en)3

3� (c) Fe(EDTA)� (d) Fe(CN)64�

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32 TRANSITION METAL

Nomenclature of Complexes

30. Name the following complexes.(a) Cu(NH3)4

2� (b) Mn(H2O)62� (c) Fe(CN)6

4� (d) Ni(en)32� (e) Cr(acac)3

31. Name the following complexes.(a) Pt(NH3)2Cl2 (b) Ni(CO)4 (c) Co(en)3

3�

32. Name the following complexes.(a) Na3[Co(NO2)6] (b) Na2[Zn(CN)4] (c) [Co(NH3)4Cl2]Cl (d) [Ag(NH3)2]Cl

33. Write the formulas for the following compounds.(a) hexamminechromium(III) chloride(b) chloropentamminechromium(III) chloride(c) trisethylenediamminecobalt(III) chloride(d) potassium tetranitritodiamminecobaltate(III)

Isomers

34. Which of the following octahedral complexes can form cis/trans isomers?(a) Co(NH3)6

3� (b) Co(NH3)5Cl2� (c) Co(NH3)5(H2O)3�

(d) Co(NH3)4Cl2� (e) Co(NH3)4(H2O)2

3�

35. Predict the structures of the cis/trans isomers of Ni(en)2(H2O)22�.

36. The octahedral Mo(PH3)3(CO)3 complex can exist as a pair of isomers. Predict thestructures of the compounds.

37. Which of the following square-planar complexes can form cis/trans isomers?(a) Cu(NH3)4

2� (b) Pt(NH3)2Cl2 (c) RhCl3(CO) (d) IrCl(CO)(PH3)2

38. Explain why square-planar complexes with the generic formula MX2Y2 can formcis/trans isomers, but tetrahedral complexes with the same generic formula cannot.

39. Explain why square-planar complexes with the generic formula MX3Y can’t formcis/trans isomers.

40. Use the fact that Rh(CO)(H)(PH3)2 forms cis/trans isomers to predict whether thegeometry around the transition metal is square planar or tetrahedral.

41. Compounds are optically active when the mirror image of the compound cannot be su-perimposed on itself. Draw the mirror images of the following complex ions and de-termine which of the ions are chiral.(a) Cu(NH3)4

2� (a square-planar complex) (b) Co(NH3)62� (an octahedral complex)

(c) Ag(NH3)2� (a linear complex) (d) Cr(en)3

3� (an octahedral complex)42. Explain why neither of the cis/trans isomers of Pt(NH3)2Cl2 is chiral.43. Explain why the cis isomer of Ni(en)2(H2O)2

2� is chiral, but the trans isomer is not.44. Explain why the cis isomer of Ni(en)2(H2O)2

2� is chiral, but the cis isomer ofNi(NH3)4(H2O)2

2� is not.45. Which of the following octahedral complexes are chiral?

(a) Cr(acac)3 (b) Cr(C2O4)33� (c) Cr(CN)6

3� (d) Cr(CO)4(NH3)2

The Valence Bond Approach to Bonding in Complexes

46. Describe the difference between the valence bond model for the Co(NH3)63� complex

ion and the valence bond model for the Ni(NH3)62� complex ion.

47. Apply the valence bond model of bonding in transition metal complexes to Ni(CO)4,Fe(CO)5, and Cr(CO)6. (Hint: Assume that the valence electrons on transition metalsin complexes in which the metal is in the zero oxidation state are concentrated in thed orbitals.)

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TRANSITION METAL 33

48. Use the results of the previous problem to explain why the transition metal is sp3 hy-bridized in Ni(CO)4, dsp3 hybridized in Fe(CO)5, and d2sp3 hybridized in Cr(CO)6.

49. Use valence bond theory to explain the �2 charge on the Fe(CO)42� ion. Predict the

charge on the equivalent Co(CO)4x� ion. Use acid–base chemistry to predict the charge

on the HFe(CO)4x� ion.

50. Use Lewis structures to explain what happens in the following reaction.

2 CrO42�(aq) � 2 H�(aq) 88nm88 Cr2O7

2�(aq) � H2O(l)

Predict the charge on the product of the following reaction. Explain why this transition-metal compound is a gas at room temperature.

MnO42�(aq) � 2 H�(aq) 88nm88 Mn2O7

x(l) � H2O(l)

51. Apply the valence bond model of bonding in transition metal complexes to theZn(NH3)4

2� and Fe(H2O)63� complex ions.

52. Use the assumption that transition metals often pick up enough ligands to fill their va-lence shell to predict the charge on the Mn(CO)5

x� ion.53. Use the assumption that transition metals often pick up enough ligands to fill their va-

lence shell to predict the charge on the HgI4x� ion.

54. Use the assumption that transition metals often pick up enough ligands to fill their va-lence shell to predict the coordination number of the Cd2� ion in the Cd(OH)x

2� ion.55. Explain why Zn2� ions form both Zn(CN)4

2� and Zn(NH3)42� ions.

Crystal Field Theory

56. Describe what happens to the energies of the 3d atomic orbitals in an octahedral crys-tal field.

57. Describe what happens to the energies of the 3d atomic orbitals in a tetrahedral crys-tal field.

58. Which of the 3d atomic orbitals in an octahedral crystal field belong to the t2g set oforbitals? Which belong to the eg set?

59. In an octahedral field what do the orbitals in a t2g set have in common? What do theorbitals in the eg set have in common?

60. The 3d orbitals are split into t2g and eg sets in both octahedral and tetrahedral crystalfields. Is there any difference between the orbitals that go into the t2g set in octahedraland in tetrahedral crystal fields?

61. Explain why �t for a tetrahedral complex is much smaller than �o for the analogousoctahedral complex.

62. Use the splitting of the 3d atomic orbitals in an octahedral crystal field to explain thestability of the oxidation states corresponding to d3 and d6 electron configurations inthe Cr(NH3)6

3� and Fe(CN)64� complex ions.

63. The difference between the energies of the t2g and eg sets of atomic orbitals in an octa-hedral or tetrahedral crystal field depends on both the metal atom and the ligands thatform the complex. Which of the following metal ions would give the largest difference?(a) Rh3� (b) Cr3� (c) Fe3� (d) Co2� (e) Mn2�

64. Which of the following ligands would give the largest value of �o?(a) CN� (b) NH3 (c) H2O (d) OH� (e) F�

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34 TRANSITION METAL

High-Spin versus Low-Spin Complexes

65. Describe the difference between a high-spin and a low-spin d6 complex.66. What factors determine whether a complex is high-spin or low-spin?67. Explain why the Mn(H2O)6

2� ion is a high-spin complex.68. One of the Fe(H2O)6

2� and Fe(CN)64� complex ions is high-spin and the other is low-

spin. Which is which?69. Use the relative magnitudes of �o and �t to explain why there are no low-spin tetra-

hedral complexes.70. Compare the positions of the Co2� with Co3� and Fe2� with Fe3� ions in the spec-

trochemical series. Explain why the value of � generally increases with the charge onthe transition metal ion.

71. Compare the positions of the Co3�, Rh3�, and Ir3� ions in the spectrochemical series.What happens to the value of � as we go down a column among the transition metals?

The Colors of Transition Metal Complexes

72. Describe the characteristic colors of aqueous solutions of the following transition metalions.(a) Cu2� (b) Fe3� (c) Ni2� (d) CrO4

2� (e) MnO4�

73. Explain why so many of the pigments used in oil paints, such as vermilion (HgS), cad-mium red (CdS), cobalt yellow [K3Co(NO2)6], chrome yellow (PbCrO4), Prussian blue(Fe4[Fe(CN)6]3), and cobalt blue (CoO�Al2O3), contain transition metal ions.

74. Explain why Cu(NH3)42� complexes have a deep-blue color if they don’t absorb blue

light. What light do they absorb?75. CrO4

2� ions are bright yellow. In what portion of the visible spectrum do the ions ab-sorb light?

76. When CrO42� reacts with acid to form Cr2O7

2� ions, the color shifts from bright yel-low to orange. Does this mean that the light absorbed shifts toward a higher or a lowerfrequency?

77. Ni2� forms a complex with the dimethylglyoxime (DMG) ligand that absorbs light inthe blue-green portion of the spectrum. What is the color of the Ni(DMG)2 complex?

78. Explain why a white piece of paper looks as if it has a faint pink color to a person whohas been working for several hours at a computer terminal that has a green screen.

Ligand Field Theory

79. Describe how ligand field theory eliminates the difference between inner-shell com-plexes, such as the Co(NH3)6

3� ion, and outer-shell complexes, such as the Ni(NH3)62�

ion.80. Explain how ligand field theory allows the valence-shell d orbitals on the transition

metal to be used simultaneously to form the skeleton structure of the complex and tohold the electrons that were originally in the d orbitals on the transition metal.

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1

M O D U L E

3COMPLEX ION EQUILIBRIA

COM.1 Complex Ions

COM.2 The Stepwise Formation of Complex Ions

COM.3 Complex Dissociation Equilibrium Constants

COM.4 Approximate Complex Ion Calculations

COM.5 Using Complex Ion Equilibria to Dissolve an Insoluble Salt

COM.6 A Qualitative View of Combined Equilibria

COM.7 A Quantitative View of Combined Equilibria

COM.1 COMPLEX IONSThe basic assumption behind the discussion of solubility equilibria is the idea that salts dis-sociate into their ions when they dissolve in water. Copper sulfate, for example, dissociatesinto the Cu2� and SO4

2� ions in water.

H2OCuSO4(s) 88nm88 Cu2�(aq) � SO4

2�(aq)

If we add 2 M NH3 to the solution, the first thing we notice is the formation of a lightblue, almost bluish-white, precipitate. This can be explained by combining what we knowabout acid–base and solubility equilibria. Ammonia acts as a base toward water to form amixture of the ammonium and hydroxide ions.

NH3(aq) � H2O(l) 88nm88 NH4�(aq) � OH�(aq) Kb � 1.8 � 10�5

The OH� ions formed in the reaction combine with Cu2� ions in the solution to form aninsoluble Cu(OH)2 precipitate.

Cu2�(aq) � 2 OH�(aq) 88nm88 Cu(OH)2(s) Ksp � 2.2 � 10�20

In theory, the OH� ion concentration should increase when more ammonia is added to thesolution. As a result, more Cu(OH)2 should precipitate from the solution. At first, this isexactly what happens. In the presence of excess ammonia, however, the Cu(OH)2 precip-itate dissolves, and the solution turns deep blue. This raises an important question: “Whydoes the Cu(OH)2 precipitate that forms in dilute ammonia dissolve in excess ammonia?”

The first step toward answering the question involves writing the electron configura-tion of copper metal and its Cu2� ion.

1012T_mod03_1-21 1/22/05 7:38 Page 1 EQA

2 COMPLEX ION

Cu [Ar] 4s1 3d10

Cu2� [Ar] 3d9

It is sometimes useful to think about the electron configuration of the Cu2� ion in termsof the entire set of valence-shell orbitals. In addition to the nine electrons in the 3d sub-shell, the Cu2� ion has an empty 4s orbital and a set of three empty 4p orbitals.

Cu2� [Ar] 4s0 3d9 4p0

The empty 4s and 4p orbitals on the Cu2� ion are used to pick up pairs of nonbondingelectrons from four NH3 molecules to form a Cu(NH3)4

2� ion, as shown in Figure COM.1.

H

HH

H

N

N

H

H

H

HH

H

N

H

H

NH3 Cu NH3

2�

NH3

NH3

NCu2+

FIGURE COM.1 The Cu2� ion has fourempty valence-shell orbitals that canaccept a pair of nonbonding electronsfrom an NH3 molecule to form a covalentCuON bond.

G. N. Lewis was the first to recognize the similarity between this reaction andthe acid–base reaction in which an H� ion combines with an OH� ion to form a watermolecule.

Both reactions involve the transfer of a pair of nonbonding electrons from one atom to anempty orbital on another atom to form a covalent bond. Both reactions can therefore beinterpreted in terms of an electron pair acceptor combining with an electron pair donor.

Lewis suggested that we could expand our definition of acids by assuming that an acidis any substance that acts like the H� ion to accept a pair of nonbonding electrons. A Lewisacid is therefore an electron pair acceptor. A Lewis base, on the other hand, is any sub-stance that acts like the OH� ion to donate a pair of nonbonding electrons. A Lewis baseis therefore an electron pair donor.

CheckpointUse Lewis structures to predict which of the following can act as an electron pairdonor, or Lewis base.(a) H2O (b) CO (c) O2 (d) H� (e) BH3

The product of the reaction of a Lewis acid with a Lewis base is an acid–base complex.When the Cu2� ion reacts with four NH3 molecules, the product of the reaction is calleda complex ion.

Cu2�(aq) � 4 NH3(aq) 88nm88 Cu(NH3)42�(aq)

Electron pair acceptor Electron pair donor Acid–base complex(Lewis acid) (Lewis base) or complex ion

O HSOQ OOQO O O�

H HH� �

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COMPLEX ION 3

Any atom or molecule with at least one empty valence-shell orbital can be a Lewis acid.Any atom or molecule that contains at least one pair of nonbonding electrons is a Lewisbase. All of the substances whose Lewis structures are shown in Figure COM.2, for ex-ample, can act as Lewis bases to form complex ions.

H H H H H

O NO O O O C OC

S C N S C N S C N

O

O

S S O–

2–

C N–

F – Cl – Br – I –O H–

FIGURE COM.2 Lewis structures of some potential Lewis bases.

COM.2 THE STEPWISE FORMATION OF COMPLEX IONSWhen a transition metal ion binds one or more Lewis bases to form an acid–base complex,it picks up the Lewis bases one at a time. The Ag� ion, for example, combines with NH3

in a two-step reaction. It first picks up one NH3 molecule to form a one-coordinatecomplex.

Ag�(aq) � NH3(aq) 88nm88 Ag(NH3)�(aq)

This intermediate then picks up a second NH3 molecule in a separate step.

Ag(NH3)�(aq) � NH3(aq) 88nm88 Ag(NH3)2�(aq)

It is possible to write equilibrium constant expressions for each step in these complexion formation reactions. The complex formation equilibrium constant (Kf) expressions forthe two steps in the formation of the Ag(NH3)2

� complex ion are written as follows.

Kf1 � �[[AA

gg(�

N][HN

3

H)�

3]]

�

Kf2 �

The difference between Kf1 and Kf2 for the complexes between Ag� and ammonia is onlya factor of 4.

Ag�(aq) � NH3(aq) 88nm88 Ag(NH3)�(aq) Kf1 � 1.7 � 103

Ag(NH3)�(aq) � NH3(aq) 88nm88 Ag(NH3)2�(aq) Kf2 � 6.5 � 103

This means that most of the Ag� ions that pick up an NH3 molecule to form the one-coordinate Ag(NH3)� complex ion are likely to pick up a second NH3 molecule to form

[Ag(NH3)2�]

��[Ag(NH3)�][NH3]

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4 COMPLEX ION

the two-coordinate Ag(NH3)2� complex ion. Table COM.1 summarizes the concentrations

of the Ag�, Ag(NH3)�, and Ag(NH3)2� ions over a range of NH3 concentrations.

TABLE COM.1 Effect of Changes in the NH3 Concentration on the Fraction of Silver Present as the Ag�, Ag(NH3)�, and Ag(NH3)2

� Ions

[NH3] (M) Ag� (%) Ag(NH3)� (%) Ag(NH3)2� (%)

10�6 99.8 0.2 0.00110�5 98.2 1.7 0.110�4 78.1 13.3 8.610�3 7.3 12.4 80.410�2 0.09 1.5 98.410�1 0.0009 0.2 99.8

Data from Table COM.1 are plotted in Figure COM.3. Essentially all of the silver ispresent as the Ag� ion at very low concentrations of NH3. As the NH3 concentration in-creases, the dominant species soon becomes the two-coordinate Ag(NH3)2

� ion. Even atNH3 concentrations as small as 0.0010 M, most of the silver is present as the Ag(NH3)2

� ion.

100%

80%

Pro

port

ion

of t

otal

silv

er io

n co

ncen

trat

ion

60%

40%

20%

010–6 10–5 10–4 10–3 10–2 10–1

[Ag(NH3)+]

[Ag(NH3)2+][Ag+]

[NH3] (M)

FIGURE COM.3 The effect of changes in the NH3 concentration on the proportion of the total silver ion concentration present as Ag�, Ag(NH3)�, and Ag(NH3)2

� ions.

The concentration of the one-coordinate Ag(NH3)� intermediate is never very large.Either the NH3 concentration is so small that most of the silver is present as the Ag� ion,or it is large enough that essentially all of the silver is present as the two-coordinateAg(NH3)2

� complex ion.If the only important components of the equilibrium are the free Ag� ion (at low NH3

concentrations) and the two-coordinate Ag(NH3)2� complex ion (at moderate to high NH3

concentrations), we can collapse the individual steps in the reaction into the following over-all equation.

Ag�(aq) � 2 NH3(aq) 88nm88 Ag(NH3)2�(aq)

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COMPLEX ION 5

The overall complex formation equilibrium constant expression for the reaction is writtenas follows.

Kf � �[[AA

gg(�

N][HN

3

H)2

3

�

]2

]�

This expression is equal to the product of the equilibrium constant expressions for the in-dividual steps in the reaction.

�[[AA

gg(�

N][HN

3

H)2

3

�

]2

]� � �

[[AA

gg(�

N][HN

3

H)�

3]]

� �

The overall complex formation equilibrium constant is therefore equal to the product ofthe Kf values for the individual steps.

Kf � Kf1 � Kf2 � 1.1 � 107

Overall complex formation equilibrium constants for common complex ions can be foundin Appendix B.1.

Exercise COM.1

Calculate the complex formation equilibrium constant for the two-coordinate Fe(SCN)2�

complex ion from the following data.

Fe3�(aq) � SCN�(aq) 88nm88 Fe(SCN)2�(aq) Kf1 � 890Fe(SCN)2�(aq) � SCN�(aq) 88nm88 Fe(SCN)2

�(aq) Kf2 � 2.6

Solution

The difference between the stepwise formation equilibrium constants for the complex isrelatively small.

Kf1 ��[[FFee3

(�

S]C[S

NC)N

2�

�

]]

�� 890

Kf2 � � 2.6

Solutions of these complex ions are therefore best described in terms of an overall com-plex formation equilibrium when there is an excess of the ligand.

Fe3�(aq) � 2 SCN�(aq) 88nm88 Fe(SCN)2�(aq)

The equilibrium constant expression for the overall reaction is equal to the product of theexpressions for the individual steps in the reaction.

�[F[Fee3�

(S][CSNC

)N2�

�

]]2��

[[FFee3

(�

S]C[S

NC)N

2�

�

]]

��[Fe(SCN)2

�]��[Fe(SCN)2�][SCN�]

[Fe(SCN)2�]

��[Fe(SCN)2�][SCN�]

[Ag(NH3)2�]

��[Ag(NH3)�][NH3]

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6 COMPLEX ION

The overall equilibrium constant is therefore the product of Kf1 and Kf2.

Kf � Kf1 � Kf2 � 2.3 � 103

COM.3 COMPLEX DISSOCIATION EQUILIBRIUM CONSTANTSComplex ions can also be described in terms of complex dissociation equilibria. We canstart by assuming, for example, that most of the silver ions in an aqueous solution of am-monia are present as the two-coordinate Ag(NH3)2

� complex ion. We then assume thatsome of the ions dissociate to form Ag(NH3)� complex ions and then eventually Ag�

ions.

Ag(NH3)2�(aq) 88nm88 Ag(NH3)�(aq) � NH3(aq)

Ag(NH3)�(aq) 88nm88 Ag�(aq) � NH3(aq)

A complex dissociation equilibrium constant (Kd) expression can be written for each ofthe reactions.

Kd1 �

Kd2 � �[[AA

gg(

�

N][HN

3

H)�

3]]

�

Alternatively, the individual steps in the reaction can be collapsed into an overall equa-tion, which can be described by an overall equilibrium constant expression.

Ag(NH3)2�(aq) 88nm88 Ag�(aq) � 2 NH3(aq)

Kd � �[[AA

gg(

�

N][HN

3

H)2

3�

]2

]�

Exercise COM.2

Calculate the complex dissociation equilibrium constant for the Cu(NH3)42� ion from the

value of Kf for the complex. [For Cu(NH3)42�, Kf � 2.1 � 1013.]

Solution

The equilibrium expression for the complex formation reaction is written as follows.

Kf ��[[CC

uu(2

N�]

H[N

3)H42

3

�

]4

]�

The equilibrium constant expression for the complex dissociation reaction is nothing moreor less than the inverse of the complex formation equilibrium expression.

Kd � �[[CC

uu(

2

N

�]H[N

3)H

423�

]4

]�

[Ag(NH3)�][NH3]��

[Ag(NH3)2�]

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COMPLEX ION 7

The value of Kd is therefore equal to the inverse of Kf.

Kd � �K1

f� � 4.8 � 10�14

COM.4 APPROXIMATE COMPLEX ION CALCULATIONSComplex formation equilibria provide another example of the general rule that it is useful tobegin equilibrium calculations by comparing the reaction quotient for the initial conditionswith the equilibrium constant for the reaction. Consider, forexample, the following question:

What fraction of the total iron(III) concentration is pres-ent as the Fe3� ion in a solution that was initially 0.10 MFe3� and 1.0 M SCN�? [For Fe(SCN)2

�, Kf � 2.3 � 103.]

The initial conditions for the reaction can be summarized asfollows.

Initial Fe3�(aq) � 2 SCN�(aq) 88nm88 Fe(SCN)2�(aq) Kf � 2.3 � 103

Initial 0.10 M 1.0 M 0

The initial value of the reaction quotient is therefore equal to zero.

Qf ��(F(Fee3�

(S)(CSNC

)N2�

�

))2��

(0.10()0()1.0)2� � 0 << 2.3 � 103

Because the reaction quotient is very much smaller than the equilibrium constant for thereaction, it would be absurd to assume that the reaction is close to equilibrium. We there-fore define a set of intermediate conditions in which we drive the reaction as far as possi-ble toward the right.

Fe3�(aq) � 2 SCN�(aq) 88nm88 Fe(SCN)2�(aq) Kf � 2.3 � 103

Initial 0.10 M 1.0 M 0Change �0.10 M �2(0.10) M �0.10 MIntermediate 0 0.8 M 0.10 M

We then assume that the reaction comes to equilibrium from the intermediate conditions.


Intermediate 0 0.8 M 0.10 MEquilibrium �C 0.8 � 2 �C 0.10 � �C

Substituting the information into the equilibrium expression for the reaction gives thefollowing equation.

� 2.3 � 103[0.10 � �C]

��[�C][0.8 � 2 �C]2

Initial:[Fe3+] = 0.10 M[SCN–] = 1.0 M

∆(SCN–) = 2∆(Fe3+)as we form the Fe(SCN)2

+ complex

Fe3+ + SCN–

Fe(SCN)2+

Fe(SCN)2+ + SCN–

Fe(SCN)2+

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8 COMPLEX ION

Because we defined the intermediate condition so that it was reasonably close to equilib-rium, we are now ready to assume that �C is relatively small.

�[�

[C0].1[00.]8]2� � 2.3 � 103

Solving the approximate equation gives the following result.

�C � 6.8 � 10�5

The assumption that �C is relatively small is valid because the problem was defined so thatit would be valid. We can now use the value of �C to answer the original question.

[Fe3�] � 6.8 � 10�5 M

Even though the complex formation equilibrium constant is not very large, essentially allof the iron in the solution is present as the Fe(SCN)2

� complex ion. Only a negligible frac-tion is present as the Fe3� ion.

�6.8

0�

.1100�5

� � 100 � 0.068%

Exercise COM.3

Calculate the concentration of the Cu2� ion in a solution that is initially 0.10 M Cu2� and1.0 M NH3. [For Cu(NH3)4

2�, Kf � 2.1 � 1013.]

Solution

We can set up the calculation as follows.

Cu2�(aq) � 4 NH3(aq) 88nm88 Cu(NH3)42�(aq) Kf � 2.1 � 1013

Initial 0.10 M 1.0 M 0

Because there is no Cu(NH3)42� ion in the solution when we start, the initial value of Qf

for the reaction is zero and Qf is very much smaller than Kf for the reaction. Because Kf

for the complex is very large, essentially all of the Cu2� ions should form Cu(NH3)42� ions

at equilibrium. We therefore define a set of intermediate conditions in which we shift thereaction to the right until all of the Cu2� ion is converted into Cu(NH3)4

2� complex ions.

Initial:0.10 M Cu2+

1.0 M NH3

∆(NH3) = 4∆(Cu2+)as we form the Cu(NH3)4

2+ complex

Cu2+ + 4NH3Cu(NH3)4

2+


Initial 0.10 M 1.0 M 0Change �0.10 M �4(0.10) M �0.10 MIntermediate 0 0.6 M 0.10 M

We then let the reaction come to equilibrium from the intermediate conditions.

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COMPLEX ION 9

The next step in solving the problem involves writing the equilibrium constant ex-pression for the reaction.

Kf �

We then substitute what we know about the equilibrium concentrations of the three com-ponents of the reaction into the equation.

� 2.1 � 1013

The value of Kf is so large that very little Cu(NH3)42� complex ion dissociates as the re-

action comes to equilibrium. It is therefore reasonable to assume that �C is relatively small.

�[�

[C0].1[00.]6]4� � 2.1 � 1013

Solving the approximate equation gives the following result.

�C � 3.7 � 10�14

The assumption that �C is small is valid. We can therefore use the results of the calcula-tion to determine the concentration of the free (uncomplexed) Cu2� ion in the solution.

[Cu2�] � 3.7 � 10�14 M

The results of the preceding exercise can be used to explain why Cu(OH)2 dissolves inexcess ammonia. Before we can do this, however, we need to understand why Cu(OH)2

precipitates in the first place. Ammonia acts as a base toward water.

Kb � � 1.8 � 10�5

Even fairly dilute solutions of the OH� ion have more thanenough OH� ion to precipitate Cu(OH)2 from a 0.10 M Cu2�

ion solution. As the amount of NH3 added to the solutionincreases, the concentration of the OH� ion increases. Butit doesn’t become very much larger. The OH� ion concen-tration in a 0.001 M NH3 solution is 1.3 � 10�4 M. By thetime the NH3 concentration reaches 0.10 M, the OH� ionconcentration has increased by only a factor of 10, to 1.3 �10�3 M.

As the amount of NH3 added to the solution increases,the concentration of Cu2� ions rapidly decreases because the

[NH4�][OH�]

��[NH3]

[0.10 � �C]��[�C][0.6 � 4 �C]4

[Cu(NH3)42�]

��[Cu2�][NH3]4

Cu(OH)2(s)

Drops of 2 M NH3

Cu2+ + 2 OH–

Cu(NH3)42+

NH3


Intermediate 0 0.6 M 0.10 MEquilibrium �C 0.6 � 4 �C 0.10 � �C

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10 COMPLEX ION

ions are tied up as Cu(NH3)42� complex ions. According to the preceding exercise, the Cu2�

ion concentration in 1.0 M NH3 is only 3.7 � 10�14 M. Thus, the ion product for Cu(OH)2

under these conditions is about the same size as the solubility product for the compound.

Qsp � (Cu2�)(OH�)2 � (3.7 � 10�14)(1.3 � 10�3)2 � 6.3 � 10�20 � Ksp

As soon as the NH3 concentration exceeds 1 M, the Cu2� ion concentration becomes sosmall that the ion product for Cu(OH)2 is smaller than Ksp, and the Cu(OH)2 precipitatedissolves.

COM.5 USING COMPLEX ION EQUILIBRIA TO DISSOLVE ANINSOLUBLE SALT

The key to using complex ion equilibria to dissolve an insoluble salt is simple: Choose acomplex for which Kf is large enough that the concentration of the uncomplexed metal ionis too small for the ion product to exceed the solubility product. To show how this is done,let’s examine the step in the processing of photographic film in which the AgBr crystalsthat don’t capture light are washed from the film. Because this step permanently fixes theimage onto the film, the reagent used to achieve it is called a “fixer.”

We can decide whether a particular complexing agent is strong enough to be used asa fixer by calculating the solubility of AgBr in an aqueous solution of the reagent. Con-sider, for example, the following question.

Use the following equilibria to decide whether the thiosulfate ion (S2O32�) is a reasonable

choice for the complexing agent for washing unreacted AgBr from photographic film.

AgBr(s) 88nm88 Ag�(aq) � Br�(aq) Ksp � 5.0 � 10�13

Ag�(aq) � 2 S2O32�(aq) 88nm88 Ag(S2O3)2

3�(aq) Kf � 2.9 � 1013

AgBr will dissolve if the Ag(S2O3)23� complex is strong

enough to reduce the Ag� ion concentration to the point atwhich the product of the Ag� and Br� ion concentrationsat equilibrium is smaller than the Ksp for AgBr. One way ofdeciding whether Kf for the complex is large enough to over-come the solubility product equilibrium involves combiningthe equilibria in the solution to give the following overallequation.

AgBr(s) � 2 S2O32�(aq) 88nm88 Ag(S2O3)2

3�(aq) � Br�(aq)

K �[Ag(S2O3)2

3�][Br�]��

[S2O32�]2

The equilibrium constant for the reaction is equal to the product of Kf times Ksp.

K � Kf � Ksp � (2.9 � 1013)(5.0 � 10�13) � 15

Because the equilibrium constant is larger than 1, a significant amount of AgBr should dis-solve in the solution. Let’s see how much AgBr can dissolve in a liter of a 1.0 M S2O3

2�

solution.

AgBr(s)

Drops of S2O32–

Ag+ + Br–S 2

O 32–

Ag(S2O3)23–

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COMPLEX ION 11

When AgBr dissolves in pure water, the concentration of the Ag� and Br� ions mustbe the same.

In pure water [Ag�] � [Br�] � �C

In 1.0 M S2O32� essentially all of the silver ion will be present as the two-coordinate

Ag(S2O3)23� ion.

In 1.0 M S2O32� [Ag(S2O3)2

3�] � [Br�] � �C

The concentration of the S2O32� ion at equilibrium will be equal to the initial concentra-

tion of the ion minus the amount consumed when the Ag(S2O3)23� complex is formed.

[S2O32�] � 1.0 � 2 �C

Substituting the information into the equilibrium constant expression for the overallreaction gives the following equation.

�[1.

[0�

�

C][2�

�

CC]]2�� 15

Solving the equation with the quadratic formula gives the following result.

�C � 0.44 M

The Ag(S2O3)23� complex ion is strong enough to dissolve up to 0.44 moles of AgBr per

liter of solution. Thus, it isn’t surprising that the thiosulfate ion is used as the fixer in theprocessing of virtually all commercial photographic films.

COM.6 A QUALITATIVE VIEW OF COMBINED EQUILIBRIAMost of the discussion so far has focused on individual equilibria. This section will exam-ine how Le Châtelier’s principle can be applied to systems in which many chemical equi-libria exist simultaneously.

Consider what happens when solid CuSO4 dissolves in an aqueous NH3 solution. If westart with enough copper sulfate to form a saturated solution, the following solubility prod-uct equilibrium will exist in the solution.

CuSO4(s) 88nm88 Cu2�(aq) � SO42�(aq)

The SO42� ion formed when CuSO4 dissolves in water is a weak Brønsted base that can

pick up an H� ion to form the hydrogen sulfate and hydroxide ions.

SO42�(aq) � H2O(l) 88nm88 HSO4

�(aq) � OH�(aq)

There are other sources of the OH� ion in the solution. Water, of course, dissociates tosome extent to form the OH� ion.

2 H2O(l) 88nm88 H3O�(aq) � OH�(aq)

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12 COMPLEX ION

Ammonia also reacts with water, to some extent, to form the NH4� and OH� ions.

NH3(aq) � H2O(l) 88nm88 NH4�(aq) � OH�(aq)

The Cu2� ion released into solution when CuSO4 dissolves reacts with ammonia to forma series of complex ions.

Cu2�(aq) � NH3(aq) 88nm88 Cu(NH3)2�(aq)Cu(NH3)2�(aq) � NH3(aq) 88nm88 Cu(NH3)2

2�(aq)Cu(NH3)2

2�(aq) � NH3(aq) 88nm88 Cu(NH3)32�(aq)

Cu(NH3)32�(aq) � NH3(aq) 88nm88 Cu(NH3)4

2�(aq)

If the concentration of the OH� ion is large enough, a second solubility product equilib-rium will exist in the solution.

Cu(OH)2(s) 88nm88 Cu2�(aq) � 2 OH�(aq)

The simple process of dissolving copper(II) sulfate in aqueous ammonia therefore caninvolve nine simultaneous equilibria. In theory, we must consider each of the reactions ifwe want to predict what will happen under a particular set of experimental conditions. Inpractice, we can make at least one simplifying assumption. We can assume that the com-plex ion equilibria in the system can be represented by a single equation in which the Cu2�

ion combines with four NH3 molecules to form the four-coordinate Cu(NH3)42� ion. We

can therefore construct a fairly complete model of what happens in the solution if we takeinto account the equilibria summarized in Figure COM.4.

H2O

H2O

Cu2+ +aq SO42– aq

aq aq aq aq aq+

H3O+ + OH–

CuSO4(s)

H+OH–

NH3

NH4+ Cu(NH3)4

2+ HSO4–

H2O

OH–OH–+

Cu(OH2)(s)

FIGURE COM.4 Relationships amongthe equilibria that exist in a saturatedsolution of CuSO4 in ammonia.

Let’s now apply Le Châtelier’s principle to the model in Figure COM.4 to predict theeffect of adding various substances to a saturated solution of CuSO4 in aqueous NH3.

• Solid CuSO4: Adding excess CuSO4 to a saturated solution has no effect on any ofthe equilibria shown in Figure COM.4 because it has no effect on the concentrationof either the Cu2� or SO4

2� ions.• Nitric acid: Nitric acid is a strong acid that should convert most of the NH3 into

NH4� ions. Anything that removes NH3 from the solution tends to destroy the

Cu(NH3)42� complex ion. Adding nitric acid therefore tends to increase the Cu2�

ion concentration, which should cause CuSO4 to precipitate from solution. We don’texpect Cu(OH)2 to precipitate, however, because the concentration of the OH� ionin a strong acid solution is much too small.

• Sodium hydroxide: Sodium hydroxide is a strong base. In theory, it should reactwith the NH4

� ion in the solution to form more NH3. In practice, there isn’t verymuch NH4

� ion in the solution to begin with, so adding NaOH has little effect onmost of the equilibria in the solution. The presence of excess OH� ion, however, islikely to precipitate some of the Cu2� ion in solution as Cu(OH)2.

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COMPLEX ION 13

• Aqueous ammonia: Any increase in the NH3 concentration should increase theamount of the Cu2� ion tied up as Cu(NH3)4

2� complex ions. This reduces the con-centration of the free Cu2� ion, which makes the ion product for CuSO4 smallerthan the solubility product. As a result, adding more NH3 can cause more CuSO4

to dissolve. It is also likely to dissolve any residual Cu(OH)2 that might be present.

Another example of combined equilibria revolves around the aqueous chemistry of theFe3� ion. Dilute solutions of the ion prepared by dissolving an iron(III) salt in perchloricacid are essentially colorless. In the presence of the thiocyanate ion, however, a blood-redsolution is formed. The solution contains a pair of complex ions.

Fe3�(aq) � SCN�(aq) 88nm88 Fe(SCN)2�(aq) Kf � 890Fe(SCN)2�(aq) � SCN�(aq) 88nm88 Fe(SCN)2

�(aq) Kf � 2.6

The Fe3� ion also forms a complex with the citrate ion (Cit3�).

Fe3�(aq) � Cit3�(aq) 88nm88 Fe(Cit)(aq) Kf � 6.3 � 1011

Different colors are observed for the following aqueous solutions shown in Figure COM.5.(a) A dilute aqueous solution of the Fe3� ion is colorless.(b) A solution formed by mixing the Fe3� and SCN� ions is blood-red.(c) A solution formed by adding a strong acid to a mixture of the Fe3� and SCN� ions is

blood-red.(d) A solution formed by mixing the Fe3� and Cit3� ions is yellow.(e) A solution formed by adding a strong acid to a mixture of the Fe3� and Cit3� ions is

colorless.(f) A solution formed by mixing the Fe3�, SCN�, and Cit3� ions is yellow.(g) A solution formed by adding a strong acid to a mixture of the Fe3�, SCN�, and Cit3�

ions is blood-red.

FIGURE COM.5 (Photo by Andy Washnik.)

The color (or lack thereof) tells us something about each of the solutions.

• For solution (a), the color of dilute solutions of the Fe3� ion is so weak they are es-sentially colorless.

• For (b), when the SCN� ion is added to an aqueous solution of the Fe3� ion, theFe(SCN)2� and Fe(SCN)2

� complex ions are formed, and the solution turns a blood-red color.

• For (c), nothing happens when a strong acid is added to a mixture of the Fe3� andSCN� ions. This tells us something about the strength of the conjugate acid of theSCN� ion—thiocyanic acid, HSCN. If HSCN were a weak acid, adding a strong acidto the solution would tie up the SCN� ion as HSCN. If that happened, the SCN�

1012T_mod03_1-21 1/22/05 7:38 Page 13 EQA

14 COMPLEX ION

ion would no longer be free to form a complex with the Fe3� ion. This would de-crease the amount of the Fe(SCN)2� and Fe(SCN)2

� complex ions in solution,thereby decreasing the intensity of the color of the solution. Since this is not ob-served, HSCN must be a relatively strong acid. This conclusion is consistent withthe Ka for thiocyanic acid found in Appendix B.

• For (d), mixing the Fe3�(aq) and Cit3�(aq) ions forms the Fe(Cit) complex, whichhas a pale yellow color.

• For (e), because citric acid is a relatively weak acid, the Cit3� ion is a reasonablygood base. Adding a strong acid therefore destroys the Fe(Cit) complex by con-verting the citrate ion into its conjugate acid–citric acid (H3Cit). When this happens,the pale yellow color disappears.

• For (f), the Kf for the Fe(Cit) complex is much larger than Kf for the Fe(SCN)2�

complex ion. Given a choice between SCN� and Cit3� ions, the Fe3� ions form com-plexes with Cit3� ions. A solution containing a mixture of the three ions thereforehas the characteristic color of the Fe(Cit) complex.

• For (g), when a strong acid is added to a mixture of the Fe3�, Cit3�, and SCN�

ions, the Cit3� ions are converted to citric acid (H3Cit). When the Cit3� ions areremoved from solution, the only complexing agent left in the solution is the SCN�

ion. The solution therefore turns the color of the Fe(SCN)2� and Fe(SCN)2� com-

plex ions.

COM.7 A QUANTITATIVE VIEW OF COMBINED EQUILIBRIAWe can perform quantitative calculations for solutions of combined equilibria if we keepin mind how these equilibria are coupled. Consider, for example, the following question.

What is the solubility of zinc sulfide in a 0.10 M H2S solution buffered at pH 9.00?

We might start by building a representation of the equilib-ria that must be considered in order to do the calculation,such as the one shown in the diagram.

We then calculate the H3O� ion concentration in thebuffer solution.

ZnS(s)

pH = 10 buffer

Zn2+ + S2– + 2 H3O+

H2S + 2 H2O

0.10 M H2S

[H3O�] � 10�pH � 10�10.00 � 1.0 � 10�10 M

We then ask, What effect does this pH have on the dissociation of H2S?

H2S(aq) � 2 H2O(l) 88nm88 2 H3O�(aq) � S2�(aq) Kc � 1.3 � 10�20

Because it is a weak acid, we can assume that the concentration of H2S at equilibrium ismore or less equal to its initial concentration.

[H2S] � 0.10 M

We now know the H2S and H3O� concentrations at equilibrium, which means that we cancalculate the concentration of the S2� ion at equilibrium before any ZnS dissolves.

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COMPLEX ION 15

�[H3O

[H

�

2

]S

2[]S2�]

� � � 1.3 � 10�20

[S2�] � 1.3 � 10�3 M

When ZnS dissolves, we get more S2� ion. The amount of ZnS that dissolves can becalculated from the following equation, where Cs is the solubility of ZnS.

[Zn2�][S2�] � Ksp

[�C][�C � 1.3 � 10�3] � 1.6 � 10�24

If we have to, we can always solve the equation with the quadratic formula. Before wedo this, however, we might look for an assumption that makes the problem easier to solve.In this case, it seems reasonable to assume that the solubility of ZnS is so small that mostof the S2� ion at a pH of 9.00 comes from the dissociation of H2S.

�C << 1.3 � 10�3 M

When that assumption is made, we get the following approximate equation.

[�C][1.3 � 10�3] � 1.6 � 10�24

When we solve the equation, we get a solubility of ZnS that is so small that we have toconclude that the assumption used to generate the equation is valid.

�C � 1.2 � 10�21

There is no magic formula that can help us divide problems such as this into steps, ordecide the order in which steps should be handled. The following general rules, however,might guide us through these problems.

• Identify the equilibria that must be included in the model of the solution.• Draw a figure that shows how the equilibria are coupled.• Find the simplest equilibrium—the one for which all of the necessary data are avail-

able—and solve that part of the problem.• Ask, “Where did this get me?” “What can I do with the information?”• Never lose track of what was accomplished in a previous step.• Use the results of one step to solve another until all of the equilibria have been solved.

Exercise COM.4

Predict whether AgOH will precipitate from a solution buffered at pH 9.00 that is initi-ally 0.010 M AgNO3 and 1.00 M NH4NO3. (For AgOH, Ksp � 2.0 � 10�8; for NH3, Kb �1.8 � 10�5; for Ag(NH3)2

�, Kf � 1.1 � 107; for H2O, Kw � 1.0 � 10�14.)

Solution

We might start by noting that silver nitrate and ammonium nitrate are both soluble salts.

H2OAgNO3(s) 88n Ag�(aq) � NO3�(aq)

H2ONH4NO3(s) 88n NH4�(aq) � NO3

�(aq)

[1.0 � 10�9]2[S2�]��

[0.10]

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16 COMPLEX ION

We then build a representation of the equilibria present inthe solution such as the one in the accompanying drawing.

We can solve some problems by looking at the initialconditions and working toward the final answer. Others,such as this, are so complex it is useful to look at the goaland then work backwards. The goal in this problem is todecide whether AgOH precipitates from the solution. Inorder to make that decision, we need two pieces of infor-mation: the Ag� and OH� ion concentrations at equilib-rium. We can therefore divide the problem into two partswhose goals consist of determining the values of the [Ag�] and [OH�] terms when all other reactions in the system are at equilibrium.

It seems reasonable to start with the goal that is easiest to achieve: finding the OH�

ion concentration. Because we know the pH of the buffered solution, we can calculate theH3O� ion concentration at equilibrium.

[H3O�] � 10�pH � 10�9.00 � 1.0 � 10�9 M

We can then use the H3O� ion concentration to calculate the OH� ion concentration inthe buffer solution.

[OH�] � �[H

K

3Ow

�]� � �

11..00

�

�

1100

�

�

1

9

4

� � 1.0 � 10�5 M

We now turn to the second part of the problem, finding the concentration of the Ag�

ion at equilibrium. We know that the Ag� ion is in equilibrium with the Ag(NH3)2� com-

plex ion.

Kf � �[[AA

gg(�

N][HN

3

H)2

3

�

]2

]�

If we knew the concentrations of NH3 and the Ag(NH3)2� complex ion, we could calcu-

late the Ag� ion concentration at equilibrium. This gives us two new subgoals: determin-ing the Ag(NH3)2

� and NH3 concentrations.Determining the Ag(NH3)2

� ion concentration is relatively easy. We know that the ini-tial concentration of the Ag� ion in this solution is 0.010 M. We also know that the ionforms a strong complex with NH3. We can therefore assume that essentially all of the sil-ver ions in the solution will be present as Ag(NH3)2

� complex ions.

[Ag(NH3)2�] � 0.010 M

Now all we need is the NH3 concentration. We know that ammonia is in equilibrium withthe NH4

� ion.

Kb ��[NH

[4

N

�

H][O

3]H�]

�� 1.8 � 10�5

Substituting the known OH� ion concentration gives the following equation.

Kb � � 1.8 � 10�5[NH4

�][1.0 � 10�5]��

[NH3]

pH = 9.00 buffer

Ag+ + 2 NH3

NH4+

1.00 M NH4NO3

Ag(NH3)2+

AgOH———?

0.010 M AgNO3

OH–

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COMPLEX ION 17

Rearranging the equation gives the following result.

�[[NNHH

4

3

�

]]

� � 1.8

The concentration of the NH4� ion at equilibrium is therefore 1.8 times the concentration

of NH3.[NH4

�] � 1.8[NH3]

We also know that the solution was initially 1.00 M in NH4� ions. Those ions are now pres-

ent as NH3 molecules, as NH4� ions, or as part of the Ag(NH3)2

� complex ions.

[NH3] � [NH4�] � 2[Ag(NH3)2

�] � 1.00 M

We can ignore the last term in the equation because the concentration of the Ag(NH3)2�

ion can’t be larger than the initial concentration of the Ag� ion: 0.010 M. We can there-fore assume that only a negligibly small fraction of the NH4

� that was added to the solu-tion is present in the Ag(NH3)2

� complex ion.

[NH3] � [NH4�] � 1.00 M

We now have two equations in two unknowns.

[NH4�] � 1.8[NH3]

[NH3] � [NH4�] � 1.00 M

Substituting the first of the equations into the second allows us to solve for the NH3 andNH4

� concentrations at equilibrium.

[NH3] � 0.36 M[NH4

�] � 0.64 M

We can now return to the complex formation equilibrium.

Kf � �[[AA

gg(�

N][HN

3

H)2

3

�

]2

]�

and substitute into the equation the approximate values for the concentrations of NH3 andthe Ag(NH3)2

� complex ion at equilibrium.

�[Ag

[0�

.]0[100.3]6]2� � 1.1 � 107

We can then solve the equation for the Ag� ion concentration at equilibrium.

[Ag�] � 7.0 � 10�9 M

We can now look at the product of the Ag� and OH� ion concentrations when the sys-tem is at equilibrium.

[Ag�][OH�] � [7.0 � 10�9][1.0 � 10�5] � 7.0 � 10�14

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18 COMPLEX ION

The ion product for the solution (7.0 � 10�14) is very much smaller than the solubility prod-uct for AgOH (2.0 � 10�8), which means that AgOH won’t precipitate from solution.

PROBLEMS

Complex Ion Equilibria

1. Define the following terms: Lewis acid, Lewis base, complex ion, and coordinationnumber.

2. Use examples to explain the difference between Lewis acids and Brønsted acids, andbetween Lewis bases and Brønsted bases.

3. Which of the following is a Lewis acid?(a) CO (b) C2H2 (c) BeF2 (d) CH4 (e) NF3

4. Which of the following is a Lewis acid?(a) CH3

� (b) CH4 (c) NH3 (d) BF4� (e) O2�

5. Which of the following is not a Lewis acid?(a) H� (b) BF3 (c) CO (d) Cu2� (e) Fe3�

6. Which of the following is not a Lewis base?(a) NH4

� (b) OH� (c) Cl� (d) O2 (e) SCN�

7. Which of the following Lewis acids is not a Brønsted acid?(a) HF (b) HOAc (c) H3PO4 (d) NH3 (e) BF3

8. Explain why the charges on the Fe(SCN)2� and Fe(SCN)2� ions are �2 and �1, re-

spectively.9. Calculate the charge on the transition metal ions in the Zn(NH3)4

2�, Fe(SCN)2�,

Sn(OH)3�, Co(SCN)4

2�, and Ag(S2O3)23� complex ions.

The Stepwise Formation of Complex Ions

10. Explain the difference between polyprotic acids and complex ions that allows us to as-sume that polyprotic acids dissociate in steps, whereas the dissociation of complex ionscan be collapsed into a single overall reaction.

11. Cu2� forms a four-coordinate complex with ammonia.


What is the relationship between the overall complex formation equilibrium constantfor the reaction, Kf, and the stepwise formation constants, Kf1, Kf2, Kf3, and Kf4?

12. Calculate the complex formation equilibrium constant, Kf, for the following overallreaction

Ag�(aq) � 2 S2O32�(aq) 88nm88 Ag(S2O3)2

3�(aq)

Acid–base complexComplex ion

Complex dissociationequilibrium constant,Kd

Complex formation equilibrium constant,Kf

KEY TERMS

1012T_mod03_1-21 1/22/05 7:38 Page 18 EQA

COMPLEX ION 19

from the values of the stepwise formation constants.

Ag�(aq) � S2O32�(aq) 88nm88 Ag(S2O3)�(aq) Kf1 � 6.6 � 108

Ag(S2O3)�(aq) � S2O32�(aq) 88nm88 Ag(S2O3)2

3�(aq) Kf2 � 4.4 � 104

13. Calculate the complex formation equilibrium constant, Kf, for the following overallreaction

Cd2�(aq) � 4 CN�(aq) 88nm88 Cd(CN)42�(aq)

from the values of the stepwise formation constants.

Cd2�(aq) � CN�(aq) 88nm88 Cd(CN)�(aq) Kf1 � 3.0 � 105

Cd(CN)�(aq) � CN�(aq) 88nm88 Cd(CN)2(aq) Kf2 � 1.3 � 105

Cd(CN)2(aq) � CN�(aq) 88nm88 Cd(CN)3�(aq) Kf3 � 4.3 � 104

Cd(CN)3�(aq) � CN�(aq) 88nm88 Cd(CN)4

2�(aq) Kf4 � 3.5 � 103

14. Which of the following solutions has the smallest concentration of the Ag� ion?(a) 0.10 M Ag� and 1.0 M Cl� (for AgCl2

�, Kf � 1.1 � 105)(b) 0.10 M Ag� and 1.0 M NH3 (for Ag(NH3)2

�, Kf � 1.1 � 107)(c) 0.10 M Ag� and 1.0 M S2O3

2� (for Ag(S2O3)23�, Kf � 2.9 � 1013)

(d) 0.10 M Ag� and 1.0 M CN� (for Ag(CN)2�, Kf � 1.3 � 1021)

15. Which of the following solutions has the largest concentration of the Hg2� ion?(a) 0.10 M Hg2� and 1.0 M Cl� (for HgCl4

2�, Kf � 1.2 � 1015)(b) 0.10 M Hg2� and 1.0 M Br� (for HgBr4

2�, Kf � 1 � 1021)(c) 0.10 M Hg2� and 1.0 M I� (for HgI4

2�, Kf � 6.8 � 1029)(d) 0.10 M Hg2� and 1.0 M CN� (for Hg(CN)4

2�, Kf � 3 � 1041)

Complex Dissociation Equilibrium Constants

16. Which of the following equations describes the relationship between the complex for-mation equilibrium constant, Kf, and the complex dissociation equilibrium constant,Kd, for the Fe(CN)6

3� complex ion?(a) Kf � Kd (b) Kf � Kw/Kd (c) Kf � Kd/Kw (d) Kf � Kw � Kd (e) Kf � 1/Kd

17. Derive the relationships among Kf1, Kf2, Kd1, and Kd2 for the Fe(SCN)2� andFe(SCN)2

� complex ions.

Approximate Complex Ion Calculations

18. Calculate the Fe3� ion concentration at equilibrium in a solution prepared by adding0.100 mole of SCN� to 250 mL of 0.0010 M Fe(NO3)3.


19. Calculate the Cu2� ion concentration at equilibrium in a solution that is initially 0.10 Min Cu2� and 4.0 M in NH3. [For Cu(NH3)4

2�, Kf � 2.1 � 1013.]20. Calculate the Zn2� ion concentration at equilibrium in a solution prepared by dissolv-

ing 0.220 mole of ZnCl2 in 500 mL of 2.0 M ammonia. [For Zn(NH3)42�, Kf � 2.9 � 109.]

21. Calculate the Sb3� ion concentration at equilibrium in a 0.10 M Sb3� ion solution thathas been buffered at pH 8.00. [For Sb(OH)4

�, Kf � 2 � 1038.]

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20 COMPLEX ION

22. Calculate the Sb3� ion concentration at equilibrium in a solution that is initially 0.10 Min Sb3� and 6.0 M in HCl. (For SbCl4

�, Kf � 5.2 � 104.)23. Calculate the Co3� ion concentration at equilibrium in a solution that is initially 0.10 M

in Co3� and 1.0 M in SCN�. [For Co(SCN)4�, Kf � 1 � 103.]

24. Calculate the Cd2� ion concentration at equilibrium in a solution prepared by adding10.0 mL of 15 M aqueous ammonia to 100 mL of a solution of 7.00 � 10�3 grams ofCdCl2 in water. [For Cd(NH3)4

2�, Kf � 1.3 � 107.]

Using Complex Ion Equilibria to Dissolve an Insoluble Salt

25. AgCl is virtually insoluble in water, but it is reasonably soluble in 4.0 M NH3. Howmany moles of AgCl will dissolve in 1.00 liter of 4.00 M NH3? [For AgCl, Ksp �1.8 � 10�10; for Ag(NH3)2

�, Kf � 1.1 � 107.]26. AgBr is less soluble than AgCl, so we need to raise the concentration of ammonia in

the solution in order to get a reasonable amount to dissolve. How many grams of AgBrwill dissolve in 250 mL of 6.0 M NH3? [For Ag(NH3)2

�, Kf � 1.1 � 107; for AgBr,Ksp � 5.0 � 10�13.]

27. Kf for the Ag(NH3)2� complex ion is not large enough to allow silver bromide to dis-

solve in 4.0 M NH3. How many moles of AgBr will dissolve in 1.0 liter of 15 M NH3?[For AgBr, Ksp � 5.0 � 10�13; for Ag(NH3)2

�, Kf � 1.1 � 107.]28. Kf for the thiosulfate complex is not large enough to allow silver iodide to dissolve in

S2O32� solutions, but the value of Kf for the cyanide complex ion, Ag(CN)2

�, is muchlarger. How many moles of AgI will dissolve in 1.0 liter of 4.0 M CN�? [For AgI,Ksp � 8.3 � 10�17; for Ag(CN)2

�, Kf� 1.3 � 1021.]

A Qualitative View of Combined Equilibria

29. Fe3� forms blood-red complexes with the thiocyanate ion, SCN�. What is the best wayto increase the concentration of these complex ions in a solution that contains the fol-lowing equilibria?

2 H2O(l) 88nm88 H3O�(aq) � OH�(aq) Kw � 1.0 � 10�14

Fe(OH)3(s) 88nm88 Fe3�(aq) � 3 OH�(aq) Ksp � 4 � 10�38

Fe3� � SCN�(aq) 88nm88 Fe(SCN)2�(aq) Kf1 � 890Fe(SCN)2�(aq) � SCN�(aq) 88nm88 Fe(SCN)2

�(aq) Kf2 � 2.6SCN�(aq) � H2O(l) 88nm88 HSCN(aq) � OH�(aq) Ka � 71

(a) Add HNO3. (b) Add NaOH. (c) Add NaSCN. (d) Add Fe(OH)3.30. What is the effect of adding a strong acid to a solution that contains the Zn(CN)4

2�

complex ion if HCN is a weak acid (Ka � 6 � 10�10)?(a) The Zn2� ion concentration increases.(b) The Zn2� ion concentration decreases.(c) The Zn2� ion concentration remains the same.(d) The Zn2� and CN� ion concentrations both increase.(e) There is no way of predicting what will happen to the Zn2� ion concentration.

31. What is one way to increase the concentration of the Cu2� ion in a saturated solutionof CuSO4 in ammonia in which the following equilibria occur?

1012T_mod03_1-21 1/22/05 7:38 Page 20 EQA

COMPLEX ION 21

CuSO4(s) 88nm88 Cu2�(aq) � SO42�(aq)


Cu2�(aq) � 2 OH�(aq) 88nm88 Cu(OH)2(s)NH3(aq) � H2O(l) 88nm88 NH4

�(aq) � OH�(aq)2 H2O(l) 88nm88 H3O�(aq) � OH�(aq)

SO42�(aq) � H2O 88nm88 HSO4

�(aq) � OH�(aq)

(a) Add an acid, such as HNO3.(b) Add a base, such as NaOH.(c) Increase the ammonia concentration.(d) Add more CuSO4.(e) None of these increases the Cu2� ion concentration in the solution.

Integrated Problems

32. What concentration of NH3 must be present in a 0.10 M AgNO3 solution to preventAgCl from precipitating when 4.0 grams of sodium chloride is added to 250 mL of thesolution? [For Ag(NH3)2

�, Kf � 1.1 � 107; for AgCl, Ksp � 1.8 � 10�10.]33. Will Co(OH)3 precipitate from a solution that is initially 0.10 M in Co3� and 1.0 M in

SCN� if the solution is buffered at pH 7.00? [For Co(OH)3, Ksp � 1.6 � 10�44; forCo(SCN)4

�, Kf � 1 � 103.]34. It is possible to keep Co(OH)3 from precipitating from a 0.010 M CoCl3 solution by

buffering the solution at pH 9.10 with a buffer that contains NH3 and the NH4� ion.

How much 6.0 M NH3 and 6.0 M HCl must be added per liter of the solution to preventCo(OH)3 from precipitating? [For NH3, Kb � 1.8 � 10�5; for Co(OH)3, Ksp � 1.6 �10�44; for Co(NH3)6

3�, Kf � 2 � 1035.]35. Calculate the CO3

2� ion concentration in a 0.10 M HCO3� solution buffered with equal

numbers of moles of NH3 and NH4�. Is this CO3

2� concentration large enough to pre-cipitate BaCO3 when the solution is mixed with an equal volume of a 0.10 M Ba2� ionsolution? (For BaCO3, Ksp � 5.1 � 10�9; for H2CO3, Ka1 � 4.5 � 10�7, Ka2 � 4.7 �10�11; for NH3, Kb � 1.8 � 10�5.)

36. Research has shown that enough CO32� ion can be leached out of clay to buffer ground-

water at a pH of about 8. Assume that the total concentration of the HCO3� and CO3

2�

ions in the solution is 0.10 M. Calculate the maximum concentration of 60Co that couldleach into the groundwater if clay were used as a barrier to store the radioactive isotope.[For CoCO3, Ksp � 1.4 � 10�13; for Co(OH)3, Ksp � 1.6 � 10�44; for H2CO3, Ka1 �4.5 � 10�7, Ka2 � 4.7 � 10�11.]

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1

M O D U L E

4ORGANIC CHEMISTRY:

STRUCTURE ANDNOMENCLATURE OF

HYDROCARBONS

O1.1 What Is an Organic Compound?

O1.2 The Saturated Hydrocarbons, or Alkanes

O1.3 The Cycloalkanes

O1.4 Rotation around C—C Bonds

O1.5 The Nomenclature of Alkanes

O1.6 The Unsaturated Hydrocarbons: Alkenes and Alkynes

O1.7 Alkene Stereoisomers

O1.8 The Reactions of Alkanes, Alkenes, and Alkynes

O1.9 Naturally Occurring Hydrocarbons and Their Derivatives

O1.10 Aromatic Hydrocarbons and Their Derivatives

O1.11 The Chemistry of Petroleum Products

O1.12 The Chemistry of Coal

O1.13 Chiral Stereoisomers

O1.14 Optical Activity

O1.1 WHAT IS AN ORGANIC COMPOUND?When you drive up to the pump at some gas stations you are faced with a variety of choices.As you filled the tank, you might wonder, “What is ‘unleaded’ gas, and why would anyonewant to add ‘lead’ to gas?” Or, “What would I get for my money if I bought premium gas,with a higher octane number?” Or, “What would happen if I filled the tank at the pumpmarked ‘diesel’?”

You then stop to buy drugs for a sore back that has been bothering you since you helpeda friend move into a new apartment. Once again, you are faced with choices (see Fig-ure O1.1). You could buy aspirin, which has been used for more than a hundred years. OrTylenol, which contains acetaminophen. Or a more modern pain-killer, such as ibuprofen.While you are deciding which drug to buy, you might wonder, “What is the differencebetween these drugs?,” and even, “How do they work?”

1012T_mod04_1-42 1/20/05 16:13 Page 1 EQA

2 THE STRUCTURE OF HYDROCARBONS

You then drive to campus, where you sit in a “plastic” chair to eat a sandwich that hasbeen wrapped in “plastic,” without worrying about why one of the plastics is flexibile whilethe other is rigid. While you’re eating, a friend stops by and starts to tease you about theeffect of your diet on the level of cholesterol in your blood, which brings up the questions,“What is cholesterol?” and “Why do so many people worry about it?”

Answers to each of these questions fall within the realm of a field known as organicchemistry. For more than 200 years, chemists have divided materials into two categories.Those isolated from plants and animals were classified as organic, while those that traceback to minerals were inorganic. At one time, chemists believed that organic compoundswere fundamentally different from those that were inorganic because organic compoundscontained a vital force that was only found in living systems.

The first step in the decline of the vital force theory occurred in 1828, when FriederichWöhler synthesized urea from inorganic starting materials. Wöhler was trying to make am-monium cyanate (NH4OCN) from silver cyanate (AgOCN) and ammonium chloride(NH4Cl). What he expected is described by the following equation.

AgOCN(aq) � NH4Cl(aq) 88n AgCl(s) � NH4OCN(aq)

The product he isolated from the reaction, however, had none of the properties of cyanatecompounds. It was a white, crystalline material that was identical to urea, H2NCONH2,which could be isolated from urine.

C

C

C

CH3

CH3

O OH

OH

NH

O

O

O

H

E

E

B

B

B

Aspirin(acetylsalicylic acid)

Tylenol(acetaminophen)

CH

C

O

CH3 CH2

CH3

CH OH

H

CH3H H

E

E

Advil(ibuprofen)

P

O O

FIGURE O1.1 The structures of the active ingredients in three common painkillers.

HONAH

AH

OH OOCqN�

Ammonium cyanate

Expected Product Observed Product�

N

H

H

H

H

OCBO

ON

Urea

G

D G

D

Neither Wöhler nor his contemporaries claimed that his results disproved the vital forcetheory. But his results set in motion a series of experiments that led to the synthesis of avariety of organic compounds from inorganic starting materials. This inevitably led to thedisappearance of “vital force” from the list of theories that had any relevance to chemistry,although it did not lead to the death of the theory, which still had proponents more than90 years later.

1012T_mod04_1-42 1/20/05 19:24 Page 2 EQA

THE STRUCTURE OF HYDROCARBONS 3

If the difference between organic and inorganic compounds isn’t the presence of somemysterious vital force required for their synthesis, what is the basis for distinguishing be-tween these classes of compounds? Most compounds extracted from living organisms con-tain carbon. It is therefore tempting to identify organic chemistry as the chemistry of car-bon. But that definition would include compounds such as calcium carbonate (CaCO3) aswell as the elemental forms of carbon—diamond and graphite—that are clearly inorganic.We will therefore define organic chemistry as the chemistry of compounds that contain bothcarbon and hydrogen.

Even though organic chemistry focuses on compounds that contain carbon and hydro-gen, more than 95% of the compounds that have been isolated from natural sources orsynthesized in the laboratory are organic. The special role of carbon in the chemistry ofthe elements is the result of a combination of factors, including the number of valence elec-trons on a neutral carbon atom, the electronegativity of carbon, and the atomic radius ofcarbon atoms (see Table O1.1).

TABLE O1.1 Physical Properties of Carbon

Electronic configuration 1s2 2s2 2p2

Electronegativity 2.54Covalent radius 0.077 nm

Carbon has four valence electrons (2s2 2p2), and it must either gain four electrons orlose four electrons to reach a rare gas configuration. The electronegativity of carbon is toosmall for carbon to gain electrons from most elements to form C4� ions, and too large forcarbon to lose electrons to form C4� ions. Carbon therefore forms covalent bonds with alarge number of other elements, including the hydrogen, nitrogen, oxygen, phosphorus,and sulfur found in living systems.

Because they are relatively small, carbon atoms can come close enough together toform strong CPC double bonds or even CqC triple bonds. Carbon also forms strong dou-ble and triple bonds to nitrogen and oxygen. It can even form double bonds to elementssuch as phosphorus and sulfur that do not form double bonds to themselves.

When the unmanned Viking spacecraft carried out experiments designed to search forevidence of life on Mars, the experiments were based on the assumption that living systemscontain carbon, and the absence of any evidence for carbon-based life on that planet waspresumed to mean that no life existed. Several factors make carbon essential to life.

• Carbon atoms form strong bonds to other carbon atoms.• Carbon forms strong bonds to other nonmetals, such as N, O, P, and S.• Carbon forms multiple bonds to other nonmetals, including C, N, O, P, and S atoms.

These factors provide an almost infinite variety of potential structures for organic com-pounds, such as vitamin C shown in Figure O1.2. No other element can provide the vari-ety of combinations and permutations necessary for life to exist.

O1.2 THE SATURATED HYDROCARBONS, OR ALKANESCompounds that contain only carbon and hydrogen are known as hydrocarbons. Thosethat contain as many hydrogen atoms as possible are said to be saturated. The saturatedhydrocarbons are also known as alkanes.

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The simplest alkane is methane: CH4. The Lewis structure of methane can be gener-ated by combining the four electrons in the valence shell of a neutral carbon atom withfour hydrogen atoms to form a compound in which the carbon atom shares a total of eightvalence electrons with the four hydrogen atoms.

CH2OH

C

C

C

CH

CH

O

O

OH

HO

HO

O

FIGURE O1.2 Structure of vitamin C.

HHC HC

H

H

TRTHT P

HP

HT

T

Methane is an example of a general rule that carbon is tetravalent; it forms a total of fourbonds in almost all of its compounds. To minimize the repulsion between pairs of electronsin the four COH bonds, the geometry around the carbon atom is tetrahedral, as shown inFigure O1.3.

FIGURE O1.3 Ball-and-stick model of methane.

Exercise O1.1

Use the fact that carbon is usually tetravalent to predict the formula of ethane, the alkanethat contains two carbon atoms.

Solution

As a rule, compounds that contain more than one carbon atom are held together by COCbonds. If we assume that carbon is tetravalent, the formula of the compound must be C2H6.

HOCAH

AH

OCAH

AH

OH Ethane

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The alkane that contains three carbon atoms is known as propane, which has the for-mula C3H8 and the following skeleton structure.

HOCAH

AH

OCAH

AH

OCAH

AH

OH Propane

HOCAH

AH

OCAH

AH

OCAH

AH

OCAH

AH

OH Butane

The four-carbon alkane is butane, with the formula C4H10.

The names, formulas, and physical properties for a variety of alkanes with the generic for-mula CnH2n�2 are given in Table O1.2. The boiling points of the alkanes gradually increasewith the molecular weight of the compounds. At room temperature, the lighter alkanes aregases; the midweight alkanes are liquids; and the heavier alkanes are solids, or tars.

TABLE O1.2 The Saturated Hydrocarbons, or Alkanes

Molecular Melting Boiling StateName Formula Point (oC) Point (oC) at 25oC

Methane CH4 �182.5 �164 GasEthane C2H6 �183.3 �88.6 GasPropane C3H8 �189.7 �42.1 GasButane C4H10 �138.4 �0.5 GasPentane C5H12 �129.7 36.1 LiquidHexane C6H14 �95 68.9 LiquidHeptane C7H16 �90.6 98.4 LiquidOctane C8H18 �56.8 124.7 LiquidNonane C9H20 �51 150.8 LiquidDecane C10H22 �29.7 174.1 LiquidUndecane C11H24 �24.6 195.9 LiquidDodecane C12H26 �9.6 216.3 LiquidEicosane C20H42 36.8 343 SolidTriacontane C30H62 65.8 449.7 Solid

The alkanes in Table O1.2 are all examples of straight-chain hydrocarbons, in whichthe carbon atoms form a chain that runs from one end of the molecule to the other. Thegeneric formula for the compounds can be understood by assuming that they contain chainsof CH2 groups with an additional hydrogen atom capping either end of the chain. Thus,for every n carbon atoms there must be 2n � 2 hydrogen atoms: CnH2n�2.

Because two points define a line, the carbon skeleton of the ethane molecule is linear,as shown in Figure O1.4.

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Because the bond angle in a tetrahedron is 109.5°, alkane molecules that contain three orcarbon atoms can no longer be thought of as “linear,” as shown in Figure O1.5.

FIGURE O1.4 A molecule of ethane.

FIGURE O1.5 Ball-and-stick models of propane and butane.

In addition to the straight-chain examples considered so far, alkanes also form branchedstructures. The smallest hydrocarbon in which a branch can occur has four carbon atoms.The compound has the same formula as butane (C4H10), but a different structure. Com-pounds with the same formula and different structures are known as isomers (from theGreek words isos, “equal,” and meros, “parts”). When it was first discovered, the branchedisomer with the formula C4H10 was therefore given the name isobutane.

CH3OCACH3

HOCH3 Isobutane

The best way to understand the difference between the structures of butane and isobutaneis to compare the ball-and-stick models of the compounds shown in Figure O1.6.

FIGURE O1.6 The two isomers with the formula C4H10.

Butane and isobutane are called constitutional isomers because they literally differ in theirconstitution. One contains two CH3 groups and two CH2 groups; the other contains threeCH3 groups and one CH group.

There are three constitutional isomers of pentane, C5H12. The first is “normal” pen-tane, or n-pentane.

Propane Butane

ButaneIsobutane

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A branched isomer is also possible, which was originally named isopentane. When a morehighly branched isomer was discovered, it was named neopentane (the new isomer of pen-tane).

CH3OCH2OCH2OCH2OCH3 n-Pentane

CH3OCACH3

HOCH2OCH3 Isopentane

CH3OCACH3

ACH3

OCH3 Neopentane

Ball-and-stick models of the three isomers with the formula C5H12 are shown in Fig-ure O1.7.

FIGURE O1.7 The three isomers with the formula C5H12.

Exercise O1.2

The following structures all have the same molecular formula: C6H14. Which of the struc-tures represent the same molecule?

Neopentane

n-Pentane

Isopentane

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Solution

There is no difference between compounds A and B; they both contain a five-carbon chainwith a branch on the second carbon. Compound C, on the other hand, contains a four-carbon chain with two branches on the second carbon atom.

Exercise O1.3

Determine the number of constitutional isomers of hexane, C6H14.

Solution

There are five constitutional isomers of hexane. There is a straight-chain, or normal, isomer.

There are two isomers with a single carbon branch.

And there are two isomers with two branches.

CheckpointUse a set of molecular models to confirm that the following compounds are isomers,

whereas the following are different ways of writing the structure of the same compound.

CH3OCACH3

HOCACH3

HOCH3 CH3OCHACH3

AOCHOCH3

CH3

CH3OCACH3

HOCACH3

HOCH3 CH3OCACH3

ACH3

OCH2OCH3

CH3OCACH3

HOCACH3

HOCH3 CH3OCACH3

ACH3

OCH2OCH3

CH3OCACH3

HOCH2OCH2OCH3 CH3OCH2OCACH3

HOCH2OCH3

CH3OCH2OCH2OCH2OCH2OCH3

CHO

CH3

CH2 OCOCH3

CH3

CH3

CH2A B COCH3CH3 CH2 CH2

CH2 CH3

CH3CH3

CH

CH3

G

D G

A

G

A A

A

G D

G

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As we have seen, there are two constitutional isomers with the formula C4H10, threeisomers of C5H12, and five isomers of C6H14. The number of isomers of a compound in-creases rapidly with additional carbon atoms. There are over 4 billion isomers for C30H62,for example.

O1.3 THE CYCLOALKANESIf the carbon chain that forms the backbone of a straight-chain hydrocarbon is long enough,we can envision the two ends coming together to form a cycloalkane. One hydrogen atomhas to be removed from each end of the hydrocarbon chain to form the COC bond thatcloses the ring. Cycloalkanes therefore have two less hydrogen atoms than the parent alkaneand a generic formula of CnH2n.

The smallest alkane that can form a ring is cyclopropane, C3H6, in which the three car-bon atoms lie in the same plane. The angle between adjacent COC bonds is only 60°, whichis very much smaller than the 109.5° angle in a tetrahedron, as shown in Figure O1.8. Cy-clopropane is therefore susceptible to chemical reactions that can open up the three-membered ring.

FIGURE O1.8 Ball-and-stick model of cyclopropane.

Any attempt to force the four carbons that form a cyclobutane ring into a plane ofatoms would produce the structure shown in Figure O1.9, in which the angle between ad-jacent COC bonds would be 90°. One of the four carbon atoms in the cyclobutane ring istherefore displaced from the plane of the other three to form a “puckered” structure thatis vaguely reminiscent of the wings of a butterfly.

FIGURE O1.9 Cyclobutane.

The angle between adjacent COC bonds in a planar cyclopentane molecule would be108°, which is close to the ideal angle around a tetrahedral carbon atom. Cyclopentane isnot a planar molecule, as shown in Figure O1.10, because displacing two of the carbonatoms from the plane of the other three produces a puckered structure that relieves someof the repulsion between the hydrogen atoms on adjacent carbon atoms in the ring.

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By the time we get to the six-membered ring in cyclohexane, a puckered structure canbe formed by displacing a pair of carbon atoms at either end of the ring from the plane ofthe other four members of the ring. One of the carbon atoms is tilted up, out of the ring,whereas the other is tilted down to form the “chair” structure shown in Figure O1.11.

FIGURE O1.10 Structure of cyclopentane.

O1.4 ROTATION AROUND COC BONDSIt is easy to fall into the trap of thinking about the ethane molecule as if it were static.Nothing could be further from the truth. At room temperature, the average velocity of anethane molecule is about 500 m/s—more than twice the speed of a Boeing 747. While itmoves through space, the molecule is tumbling around its center of gravity like an airplaneout of control. At the same time, the COH and COC bonds are vibrating like springs atrates as fast as 9 � 1013 s�1.

There is another way in which the ethane molecule can move. The CH3 groups at ei-ther end of the molecule can rotate with respect to each other around the COC bond.When this happens, the molecule passes through an infinite number of conformations thathave slightly different energies. The highest energy conformation corresponds to a struc-ture in which the hydrogen atoms are “eclipsed.” If we view the molecule along the COCbond, the hydrogen atoms on one CH3 group would obscure those on the other, as shownin Figure O1.12.

FIGURE O1.11 Chair form of cyclohexane.

FIGURE O1.12 The eclipsed conformation of ethane.

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The lowest energy conformation is a structure in which the hydrogen atoms are “stag-gered,” as shown in Figure O1.13.

The difference between the eclipsed and staggered conformations of ethane are best illus-trated by viewing these molecules along the COC bond, as shown in Figure O1.14.

FIGURE O1.13 The staggered conformation of ethane.

The difference between the energies of these conformations is relatively small, only about12 kJ/mol. But it is large enough that rotation around the COC bond isn’t smooth. Al-though the frequency of the rotation is on the order of 1010 revolutions per second, theethane molecule spends a slightly larger percentage of the time in the staggered confor-mation.

The different conformations of a molecule are often described in terms of Newmanprojections. These line drawings show the six substituents on the COC bond as if the struc-ture of the molecule were projected onto a piece of paper by shining a bright light alongthe COC bond in a ball-and-stick model of the molecule. Newman projections for the dif-ferent staggered conformations of butane are shown in Figure O1.15.

FIGURE O1.14 The eclipsed and staggered confor-mations of ethane viewed along the COC bond.

CH3

CH3H

H HH H

CH3

CH3

HH

H H

CH3

HH3C

H H

Because of the ease of rotation around COC bonds, there are several conformationsof some of the cycloalkanes described in the previous section. Cyclohexane, for example,forms both the “chair” and “boat” conformations shown in Figure O1.16. The differencebetween the energies of the chair conformation, in which the hydrogen atoms are stag-gered, and the boat conformation, in which they are eclipsed, is about 30 kJ/mol. As a re-sult, even though the rate at which the two conformations interchange is about 1 � 105 s�1,we can assume that most cyclohexane molecules at any moment in time are in the chairconformation.

FIGURE O1.15 Newman projec-tions for the staggered conforma-tions of butane.

Eclipsed Staggered

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O1.5 THE NOMENCLATURE OF ALKANESCommon names such as pentane, isopentane, and neopentane are sufficient to differenti-ate between the three isomers with the formula C5H12. They become less useful, however,as the size of the hydrocarbon chain increases.

The International Union of Pure and Applied Chemistry (IUPAC) has developed asystematic approach to naming alkanes and cycloalkanes based on the following steps.

• Find the longest continuous chain of carbon atoms in the skeleton structure. Namethe compound as a derivative of the alkane with that number of carbon atoms. Thefollowing compound, for example, is a derivative of pentane because the longestchain contains five carbon atoms.

• Name the substituents on the chain. Substituents derived from alkanes are namedby replacing the -ane ending with -yl. This compound contains a methyl (OCH3)substituent.

• Number the chain starting at the end nearest the first substituent and specify thecarbon atoms on which the substituents are located. Use the lowest possible num-bers. This compound, for example, is 2-methylpentane, not 4-methylpentane.

OH

H

H

COCOCOCOCOHA

AH

HA

AH

HA

AHA

AAA

H

HA

HA

A

OH COH

OH

H

H

COCOCOCOCOHA

AH

HA

AH

HA

AHA

AAA

H

HA

HA

A

OH COH

FIGURE O1.16 The chair and boat conformations of cyclohexane.

Chair Boat

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• Use the prefixes di-, tri-, and tetra- to describe substituents that are found two, three,or four times on the same chain of carbon atoms.

• Arrange the names of the substituents in alphabetical order.

Exercise O1.4

Use the IUPAC system to name the following compound.

Solution

The compound is a derivative of pentane because the longest chain contains five carbonatoms. There are three identical CH3 substituents on the backbone. Two of the methylgroups are on the second carbon, and one is on the fourth. The compound is therefore2,2,4-trimethylpentane. Since it contains a total of eight carbon atoms, it is also known bythe common name isooctane. This compound is the “octane” used as a standard againstwhich to measure octane numbers.

Exercise O1.5


Solution

The longest continuous chain in the skeleton structure of the compound contains sevencarbon atoms. It is therefore named as a derivative of heptane.

OOA

CH3

1CH3

5CH2

6CH2

7CH3

2 3

4CH2 OCH2OCH3OCHCH

A

A

A

OOA

CH3

CH3

CH2 OCH2OCH3OCHCHACH2ACH2ACH3

OOA

A

ACH3

CH3

CH3

CH3

CH2 OCH3OCHC

OH

H

C1 C2 C3 C4 C5

H

O O O O OHA

AH

HA

AH

HA

AHA

AAA

H

HA

HA

A

OH COH

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The heptane chain contains two substituents: a methyl group on the second carbon atomand an ethyl group on the fourth carbon atom.

Because the substituents are listed in alphabetical order, the systematic name for the com-pound is 4-ethyl-2-methylheptane.

O1.6 THE UNSATURATED HYDROCARBONS: ALKENES ANDALKYNES

Carbon not only forms the strong COC single bonds found in alkanes, it also forms strongCPC double bonds. Compounds that contain CPC double bonds were once known asolefins (literally, “to make an oil”) because they were hard to crystallize. (They tend to re-main oily liquids when cooled.) These compounds are now called alkenes. The simplestalkene has the formula C2H4 and the following Lewis structure.

The relationship between alkanes and alkenes can be understood by thinking aboutthe following hypothetical reaction. We start by breaking the bond in an H2 molecule sothat one of the electrons ends up on each of hydrogen atoms. We do the same thing to oneof the bonds between the carbon atoms in an alkene. We then allow the unpaired electronon each hydrogen atom to interact with the unpaired electron on a carbon atom to forma new COH bond.

Thus, in theory, we can transform an alkene into the parent alkane by adding an H2 mol-ecule across a CPC double bond. In practice, the reaction occurs only at high pressuresin the presence of a suitable catalyst, such as piece of nickel metal.

HOCAH

AH

OHCAH

AH

OH

H

H

HCPC H2

Ni�

C H

H H

CH

CP HE

H H

HHE O OC HO

H HHO

A A

T HT

P P H

H H

H H

CO OC HOA A

A A

CH

CP HE

H H

HHE

OOA

CH3

CH3

CH2 OCH2OCH3OCHCH

MethylACH2ACH2ACH3

Ethyl

1012T_mod04_1-42 1/20/05 16:13 Page 14 EQA


Because an alkene can be thought of as a derivative of an alkane from which an H2 mol-ecule has been removed, the generic formula for an alkene with one CPC double bond isCnH2n.

Alkenes are examples of unsaturated hydrocarbons because they have fewer hydrogenatoms than the corresponding alkanes. They were once named by adding the suffix -ene tothe name of the substituent that carried the same number of carbon atoms.

The IUPAC nomenclature for alkenes names the compounds as derivatives of the parentalkanes. The presence of the CPC double bond is indicated by changing the -ane endingon the name of the parent alkane to -ene.

The location of the CPC double bond in the skeleton structure of the compound is indi-cated by specifying the number of the carbon atom at which the CPC bond starts.

The names of substituents are then added as prefixes to the name of the alkene.

Exercise O1.6


Solution

The compound is a derivative of hexane because the longest carbon chain contains six car-bon atoms. It contains a CPC double bond, which means it is a hexene. Because the dou-ble bond links the second and third carbon atoms, it is a 2-hexene. Because the CH3 sub-stituents are on the third and fifth carbon atoms, the compound is 3,5-dimethyl-2-hexene.

Compounds that contain CqC triple bonds are called alkynes. These compounds havefour less hydrogen atoms than the parent alkanes, so the generic formula for an alkynewith a single CqC triple bond is CnH2n�2. The simplest alkyne has the formula C2H2 andis known by the common name acetylene.

HO CC OH Acetyleneq

CH3O OCH2

CH3

CH3

OCH3OCHCH CPA

A

1-Butene

CH2 OCH2OCH3CHP2-Butene

CH3O OCH3CH CHP

CH3Ethane

Propane

CH3

CH2 CH3CH3 OO

O CH2Ethene

Propene

CH2

CH3O CH2CHP

P

CH2 EthyleneCH2PCH2 CH3 PropyleneCH2P O

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The IUPAC nomenclature for alkynes names the compounds as derivatives of the parentalkane, with the ending -yne replacing -ane.

In addition to compounds that contain one double bond (alkenes) or one triple bond(alkynes), we can also envision compounds with two double bonds (dienes), three doublebonds (trienes), or a combination of double and triple bonds.

O1.7 ALKENE STEREOISOMERSThe geometry around the CPC double bond in an alkene plays an important role in thechemistry of the compounds. To understand why, let’s return to the hypothetical interme-diate from the previous section in which we have a C2H4 molecule with an unpaired elec-tron on each of the carbon atoms.

The sigma (�) bond skeleton in the molecule is formed by the overlap of sp2 hybridizedorbitals on each carbon atom with either a 1s orbital on a hydrogen atom or the sp2 hy-bridized orbital on the other carbon atom. This leaves one unpaired electron in an empty2p orbital on each carbon atom. The orbitals that hold these electrons interact to form api (�) bond.1

The geometry around a CPC double bond is therefore different from the geometry arounda COC single bond. Because of the double bond, the six atoms in a C2H4 molecule mustall lie in the same plane, as shown in Figure O1.17. The presence of the � bond restrictsrotation around a CPC double bond. There is no way to rotate one end of the bond rel-ative to the other without breaking the � bond. Because the � bond is relatively strong(~270 kJ/mol), rotation around the CPC double bond cannot occur at room temperature.

Alkenes therefore form stereoisomers that differ in the way substituents are arrangedaround the CPC double bond. The isomer with similar substituents on the same side of

C C

pi bond

CH3CH CHCH2C 4-Hexen-1-yne

1,3-Butadiene

P CHCH2 CH2CHCHP P

CH3C CCH2CH3

CH3A

CH3C CCH2CHCH2CH3

HC CCH2CH3

5-Methyl-2-heptyne

1-Butyne

2-Pentyne

1Students sometimes ask, “Why are there three lines between the carbon atoms in this drawing if there areonly two bonds?” The answer is: Both of the curved lines are needed to represent a single � bond.

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the double bond is called the cis isomer, from a Latin stem meaning “on this side.” The iso-mer in which similar substituents are across from each other, is called the trans isomer, froma Latin stem meaning “across.” Consider, for example, the terms “transcontinental,” or “trans-Atlantic.” The cis isomer of 2-butene, for example, has both CH3 groups on the same side ofthe double bond. In the trans isomer the CH3 groups are on opposite sides of the double bond.

Cis/trans isomers have similar chemical properties but different physical properties. cis-2-Butene, for example, freezes at �138.9°C, whereas trans-2-butene freezes at �105.6°C.

Exercise O1.7

Name the straight-chain constitutional and stereoisomers of pentene (C5H10).

Solution

There are two constitutional isomers of pentene in which all the carbon atoms lie in thesame chain.

There are no cis/trans isomers of 1-pentene because there is only one way of arranging thesubstituents around the double bond.

Cis and trans isomers are possible, however, for 2-pentene.

PG

G

D

DC

H Hcis-2-Pentene trans-2-Pentene

CH3

C PG

G

D

DC

H

HCH3

C

CH2CH3

CH2CH3

PG

G

D

DC

H

H

H

C

CH2CH2CH3

1-PenteneCH2 CHCH2CH2CH3P

2-PenteneCH3CH CHCH2CH3P

PG

G

D

DC

H H

cis-2-Butene trans-2-Butene

CH3

C

CH3

PG

G

D

DC

H

HCH3

C

CH3

H

H

H

H

C

C

FIGURE O1.17 The planar C2H4 molecule.

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O1.8 THE REACTIONS OF ALKANES, ALKENES, AND ALKYNES

Alkanes

In the absence of a spark or a high-intensity light source, alkanes are generally inert tochemical reactions. However, anyone who has used a match to light a gas burner, or droppeda match onto charcoal coated with lighter fluid, should recognize that alkanes burst intoflame in the presence of a spark. It doesn’t matter whether the starting material is themethane found in natural gas,

CH4(g) � 2 O2(g) 88n CO2(g) � 2 H2O(g)

the mixture of butane and isobutane used in disposable cigarette lighters,

2 C4H10(g) � 13 O2(g) 88n 8 CO2(g) � 10 H2O(g)

the mixture of C5 to C6 hydrocarbons in charcoal lighter fluid,

C5H12(g) � 8 O2(g) 88n 5 CO2(g) � 6 H2O(g)

or the complex mixture of C6 to C8 hydrocarbons in gasoline.

2 C8H18(l) � 25 O2(g) 88n 16 CO2(g) � 18 H2O(g)

Once the reaction is ignited by a spark, the hydrocarbons burn to form CO2 and H2O andgive off between 45 and 50 kJ of energy per gram of fuel consumed.

In the presence of light, or at high temperatures, alkanes react with halogens to formalkyl halides. Reaction with chlorine gives an alkyl chloride.

lightCH4(g) � Cl2(g) 888n CH3Cl(g) � HCl(g)

Reaction with bromine gives an alkyl bromide.

lightCH4(g) � Br2(l) 888n CH3Br(g) � HBr(g)

Alkenes and Alkynes

Unsaturated hydrocarbons such as alkenes and alkynes are much more reactive than theparent alkanes. They react rapidly with bromine, for example, to add a Br2 molecule acrossthe CPC double bond.

This reaction provides a way to test for alkenes or alkynes. Solutions of bromine in CCl4have an intense red-orange color. When Br2 in CCl4 is mixed with a sample of an alkane,

CH3CH CH3CHCHCH3CHCH3 Br2

Br

Br

PA

A�

2,3-Dibromobutane

1012T_mod04_1-42 1/20/05 16:13 Page 18 EQA


no change is initially observed. When it is mixed with an alkene or alkyne, the color of Br2

rapidly disappears.The reaction between 2-butene and bromine to form 2,3-dibromobutane is just one ex-

ample of the addition reactions of alkenes and alkynes. Hydrogen bromide (HBr) addsacross a CPC double bond to form the corresponding alkyl bromide, in which the hydro-gen ends up on the carbon atom that had more hydrogen atoms to begin with. Additionof HBr to 2-butene, for example, gives 2-bromobutane.

As noted in Section O1.6, H2 adds across double (or triple) bonds in the presence of a suit-able catalyst to convert an alkene (or alkyne) to the corresponding alkane.

In the presence of an acid catalyst, it is even possible to add a molecule of water across aCPC double bond.

Addition reactions provide a way to add new substituents to a hydrocarbon chain andthereby produce new derivatives of the parent alkanes.

In theory, two products can form when an unsymmetric reagent such as HBr is addedto an unsymmetric CPC double bond. In practice, only one product is obtained. WhenHBr is added to 2-methylpropene, for example, the product is 2-bromo-2-methylpropane,not 1-bromo-2-methylpropane.

In 1870, after careful study of many examples of addition reactions, the Russian chemistVladimir Markovnikov formulated a rule for predicting the product of the reactions.Markovnikov’s rule states that the hydrogen atom adds to the carbon atom that already hasthe larger number of hydrogen atoms when HX adds to an alkene. Thus, water (H—OH)adds to propene to form the product in which the OH group is on the middle carbonatom.

CH3CH CH3CHCH2CH2 HOH

OH

H

P �

Carbon in the double bond withthe most H atoms H�

CH3 CH3CHBrA

A�P

GD

CCH3

CH3

Br

CH2

CH3

O O (not CH3 CH2Br)CHACH3

O O

CH3CH CH3CHCH2CH3CHCH3 H2O

OH

PA

�H2SO4

CH3CH CH3CH2CH2CH3CHCH3 H2P �Pt

CH3CH CH3CHCH2CH3CHCH3 HBr

Br

PA

�2-Bromobutane

1012T_mod04_1-42 1/20/05 17:14 Page 19 EQA


O1.9 NATURALLY OCCURRING HYDROCARBONS AND THEIRDERIVATIVES

Complex hydrocarbons and their derivatives are found throughout nature. Natural rubber,for example, is a hydrocarbon that contains long chains of alternating CPC double bondsand COC single bonds.

Writing the structure of complex hydrocarbons can be simplified by using a line notationin which a carbon atom is assumed to be present wherever a pair of lines intersect andenough hydrogen atoms are present to satisfy the tendency of carbon to form a total offour bonds.

There are a variety of techniques for isolating both pleasant- and foul-smelling com-pounds known as essential oils from natural sources, particularly from plants. These com-pounds are not “essential,” in the sense of being vital to life. They were given that namebecause they give off a distinct “essence,” or smell.

The essential oils are used in perfumes and medicines. Some of the compounds can beisolated by gently heating, or steam distilling, the crushed flowers of plants. Others can beextracted into nonpolar solvents or absorbed onto grease-coated cloths in which the plantsare wrapped. Many of the essential oils belong to classes of compounds known as terpenesand terpenoids. The terpenes are hydrocarbons that usually contain one or more CPCdouble bonds. The terpenoids are oxygen-containing analogs of the terpenes.

Examples of terpenes include �-pinene and �-pinene, the primary components of tur-pentine that give rise to its characteristic odor.

Camphor and menthol are examples of terpenoids.

MentholCamphor

OH

OCH3

CH3CH3

CH3

CH3-Pinene�

CH3

CH2

CH3� -Pinene

CH3

B

......

ENCC

CH2 EC ECH2

ECH2E

CH3

...CH2

HA

NC

HA

ECH2...NC

HA

ACH3A

CH3ACH

CH2H

1012T_mod04_1-42 1/20/05 17:14 Page 20 EQA


Both of the compounds have a fragrant, penetrating odor and taste cool. Camphor isused as a moth repellent. Menthol is a mild anesthetic that is added to some brands ofcigarettes.

Although the terpenes and terpenoids discussed so far have very different structures,they have one important property in common: They all contain 10 carbon atoms, neithermore nor less. Each of the compounds can be traced back to a reaction in which a pair offive-carbon molecules are fused. Thus, it isn’t surprising that we can also find sesquiter-penes (15 carbon atoms), diterpenes (20 carbons), triterpenes (30 carbons), and so on. Im-portant examples of these compounds include vitamin A and the �-carotene that gives car-rots their characteristic color.

Steroids aren’t terpenes or terpenoids in the literal sense because they don’t containthe characteristic number of carbon atoms. Consider cholesterol, for example, which is oneof the most important steroids.

Analysis of the structure suggests the formula C27H46O, which doesn’t fit the pattern ex-pected of a terpenoid. The biosynthetic precursor of the molecule, however, is a 30-carbontriterpene that is converted into cholesterol by a series of enzyme-catalyzed reactions.

By definition, the steroids are compounds that have the basic structure formed by fus-ing three six-membered rings and a five-membered ring. The most important property ofthe molecule is the fact that, with the exception of the OOH group on the lower left-handcorner of the molecule, there is nothing about the structure of the compound that wouldmake it soluble in water.

In this day of cholesterol-free products, it is useful to question the label on certain prod-ucts such as peanut butter that are advertised as cholesterol-free. That is like saying thatthe Sahara desert is rain-free. Peanut butter is made from peanuts and cholesterol isn’t acharacteristic ingredient in plants; it is synthesized by animals, particularly mammals. It is

HO

CholesterolCH3

CH3

CH3 CH3 CH3

CH3 CH3 CH3

CH3

CH3

CH3

CH3� -Carotene

CH3 CH3 CH3

OH

CH3

CH3Vitamin A

1012T_mod04_1-42 1/20/05 16:13 Page 21 EQA


also useful to note that placing someone on a cholesterol-free diet won’t reduce their cho-lesterol level to zero. Even on a low-cholesterol diet, the individual will synthesize about 0.80gram of cholesterol per day. The key question is, Is there an excess of cholesterol in thebloodstream? If there is, a diet that reduces the intake of cholesterol might be important.

O1.10 AROMATIC HYDROCARBONS AND THEIR DERIVATIVESAt the turn of the nineteenth century, one of the signs of living the good life was havinggas lines connected to your house, so that you could use gas lanterns to light the house af-ter dark. The gas burned in the lanterns was called coal gas because it was produced byheating coal in the absence of air. The principal component of coal gas was methane, CH4.

In 1825, Michael Faraday was asked to analyze an oily liquid with a distinct odor thatcollected in tanks used to store coal gas at high pressures. Faraday found that the com-pound had the empirical formula CH. Ten years later, Eilhardt Mitscherlich produced thesame material by heating benzoic acid with lime. Mitscherlich named the substance ben-zin, which became benzene when translated into English. He also determined that the mol-ecular formula of the compound was C6H6.

Benzene must be an unsaturated hydrocarbon because it has far less hydrogen than theequivalent saturated hydrocarbon: C6H14. But benzene is too stable to be an alkeneor alkyne. Alkenes and alkynes rapidly add Br2 to their CPC and CqC bonds, whereasbenzene reacts with bromine only in the presence of a catalyst, FeBr3. Furthermore, whenbenzene reacts with Br2 in the presence of FeBr3, the product of the reaction is a com-pound in which a bromine atom has been substituted for a hydrogen atom, not added tothe compound the way an alkene adds bromine.

FeBr3C6H6 � Br2 8888n C6H5Br � HBr

Other compounds were eventually isolated from coal that had similar properties. Their for-mulas suggested the presence of multiple CPC bonds, but the compounds were not reac-tive enough to be alkenes. Because they often had a distinct odor, or aroma, they becameknown as aromatic compounds.

The structure of benzene was a recurring problem throughout most of the nineteenthcentury. The first step toward solving the problem was taken by Friedrich August Kekuléin 1865. (Kekulé’s interest in the structure of organic compounds may have resulted fromthe fact that he first enrolled at the University of Giessen as a student of architecture.)One day, while dozing before a fire, Kekulé dreamed of long rows of atoms twisting in asnakelike motion until one of the snakes seized hold of its own tail. This dream led Kekuléto propose that benzene consists of a ring of six carbon atoms with alternating COC sin-gle bonds and CPC double bonds. Because there are two ways in which the bonds can al-ternate, Kekulé proposed that benzene was a mixture of two compounds in equilibrium.

HH

H H

HH

H

HHH

H

H

Kekulé’s explanation ofthe structure of benzene

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Kekulé’s structure explained the formula of benzene, but it did not explain why ben-zene failed to behave like an alkene. The unusual stability of benzene wasn’t understooduntil the development of the theory of resonance. This theory states that molecules forwhich two or more satisfactory Lewis structures can be drawn are an average, or hybrid,of the structures. Benzene, for example, is a resonance hybrid of the two Kekulé structures.

The difference between the equilibrium and resonance descriptions of benzene is subtle,but important. In the equilibrium approach, a pair of arrows is used to describe a reversiblereaction, in which the molecule on the left is converted into the one on the right, and viceversa. In the resonance approach, a double-headed arrow is used to suggest that a benzenemolecule does not shift back and forth between two different structures; it is a hybrid mix-ture of the structures.

One way to probe the difference between Kekulé’s idea of an equilibrium between twostructures and the resonance theory in which benzene is a hybrid mixture of the structureswould be to study the lengths of the carbon–carbon bonds in benzene. If Kekulé’s ideawas correct, we would expect to find a molecule in which the bonds alternate betweenrelatively long COC single bonds (0.154 nm) and significantly shorter CPC doublebonds (0.133 nm). When benzene is cooled until it crystallizes, and the structure of themolecule is studied by X-ray diffraction, we find that the six carbon–carbon bonds in the molecule are the same length (0.1395 nm). The crystal structure of benzene is there-fore more consistent with the resonance model of bonding in benzene than the originalKekulé structures.

The resonance theory does more than explain the structure of benzene—it also ex-plains why benzene is less reactive than an alkene. The resonance theory assumes that mol-ecules that are hybrids of two or more Lewis structures are more stable than those thataren’t. It is the extra stability that makes benzene and other aromatic derivatives less re-active than normal alkenes. To emphasize the difference between benzene and a simple alkene,many chemists replace the Kekulé structures for benzene and its derivatives with an aro-matic ring in which the circle in the center of the ring indicates that the electrons in thering are delocalized; they are free to move around the ring.

It is the delocalization of electrons around the aromatic ring that makes benzene less re-active than a simple alkene.

Aromatic compounds were being extracted from coal tar as early as the 1830s. As aresult, many of the compounds were given common names that are still in use today. A

H

HH

H

HH

HH

H H

HH

H

HHH

H

H

The resonance structuresfor benzene

1012T_mod04_1-42 1/20/05 16:59 Page 23 EQA


few of these compounds are shown below.

There are three ways in which a pair of substituents can be placed on an aromatic ring.In the ortho (o) isomer, the substituents are in adjacent positions on the ring. In the meta(m) isomer, they are separated by one carbon atom. In the para (p) isomer, they are onopposite ends of the ring. The three isomers of dimethylbenzene, or xylene, are shown below.

Exercise O1.8

Predict the structure of para-dichlorobenzene, one of the active ingredients in mothballs.

Solution

Para isomers of benzene contain two substituents at opposite positions in the six-membered ring. para-Dichlorobenzene therefore has the following structure.

Aromatic compounds can contain more than one six-membered ring. Naphthalene, an-thracene, and phenanthrene are examples of aromatic compounds that contain two or morefused benzene rings.

A ball-and-stick model of anthracene is shown in Figure O1.18.

H

H

H

H H

HH

H H

HH

H H H

HH

H H

HH

HH

HH

H

HHH

Naphthalene (C10H8) Anthracene (C14H10)

Phenanthrene (C14H10)

Cl Cl

p-Dichlorobenzene

CH3

CH3

CH3

CH3

CH3

CH3Para

MetaOrtho

CH3 OH OCH3 NH2 CO2H

Toluene Phenol Anisole Aniline Benzoic acid

1012T_mod04_1-42 1/20/05 16:13 Page 24 EQA


O1.11 THE CHEMISTRY OF PETROLEUM PRODUCTSThe term petroleum comes from the Latin stems petra, “rock,” and oleum, “oil.” It is usedto describe a broad range of hydrocarbons that are found as gases, liquids, or solids be-neath the surface of the earth. The two most common forms are natural gas and crude oil.

Natural gas is a mixture of lightweight alkanes. A typical sample of natural gas whenit is collected at its source contains 80% methane (CH4), 7% ethane (C2H6), 6% propane(C3H8), 4% butane and isobutane (C4H10), and 3% pentanes (C5H12). The C3, C4, and C5

hydrocarbons are removed before the gas is sold. The commercial natural gas delivered tothe customer is therefore primarily a mixture of methane and ethane. The propane andbutanes removed from natural gas are usually liquefied under pressure and sold as lique-fied petroleum gases (LPG).

Natural gas was known in England as early as 1659. But it didn’t replace coal gas as animportant source of energy in the United States until after World War II, when a networkof gas pipelines was constructed. By 1980, annual consumption of natural gas had grownto more than 55,000 billion cubic feet, which represented almost 30% of total U.S. energyconsumption.

The first oil well was drilled by Edwin Drake in 1859, in Titusville, PA. It producedup to 800 gallons per day, which far exceeded the demand for the material. By 1980, con-sumption of oil had reached 2.5 billion gallons per day. About 225 billion barrels of oilwere produced by the petroleum industry between 1859 and 1970. Another 200 billion bar-rels were produced between 1970 and 1980. The total proven world reserves of crude oilin 1970 were estimated at 546 billion barrels, with perhaps another 800 to 900 billion bar-rels of oil that remained to be found. It took 500 million years for the petroleum beneaththe earth’s crust to accumulate. At the present rate of consumption, we might exhaust theworld’s supply of petroleum by the 200th anniversary of the first oil well.

Crude oil is a complex mixture that is between 50 and 95% hydrocarbon by weight.The first step in refining crude oil involves separating the oil into different hydrocarbonfractions by distillation. A typical set of petroleum fractions is given in Table O1.3. Sincethere are a number of factors that influence the boiling point of a hydrocarbon, thesepetroleum fractions are complex mixtures. More than 500 different hydrocarbons have beenidentified in the gasoline fraction, for example.

About 10% of the product of the distillation of crude oil is a fraction known as straight-run gasoline, which served as a satisfactory fuel during the early days of the internal com-bustion engine. As the automobile engine developed, it was made more powerful by in-creasing the compression ratio.2 Straight-run gasoline burns unevenly in high-compressionengines, producing a shock wave that causes the engine to “knock,” or “ping.” As the pe-troleum industry matured, it faced two problems: increasing the yield of gasoline from eachbarrel of crude oil and decreasing the tendency of gasoline to knock when it burned.

FIGURE O1.18 Anthracene.

2Modern cars run at compression ratios of about 9�1, which means the gasoline–air mixture in the cylinder iscompressed by a factor of nine before it is ignited.

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The relationship between knocking and the structure of the hydrocarbons in gasolineis summarized in the following general rules.

• Branched alkanes and cycloalkanes burn more evenly than straight-chain alkanes.• Short alkanes (C4H10) burn more evenly than long alkanes (C7H16).• Alkenes burn more evenly than alkanes.• Aromatic hydrocarbons burn more evenly than cycloalkanes.

The most commonly used measure of a gasoline’s ability to burn without knocking is itsoctane number. Octane numbers compare a gasoline’s tendency to knock against the ten-dency of a blend of two hydrocarbons—heptane and 2,2,4-trimethylpentane, or isooctane—to knock. Heptane (C7H16) is a long, straight-chain alkane, which burns unevenly and pro-duces a great deal of knocking. Highly branched alkanes such as 2,2,4-trimethylpentaneare more resistant to knocking. Gasolines that match a blend of 87% isooctane and 13%heptane are given an octane number of 87.

There are three ways of reporting octane numbers. Measurements made at high speedand high temperature are reported as motor octane numbers. Measurements taken underrelatively mild engine conditions are known as research octane numbers. The road indexoctane numbers reported on gasoline pumps are an average of the two. Road index octanenumbers for a few pure hydrocarbons are given in Table O1.4.

TABLE O1.3 Petroleum Fractions

Fraction Boiling Range ( oC) Number of Carbon Atoms

Natural gas �20 C1 to C4

Petroleum ether 20–60 C5 to C6

Gasoline 40–200 C5 to C12, but mostly C6 to C8

Kerosene 150–260 Mostly C12 to C13

Fuel oils �260 C14 and higherLubricants �400 C20 and aboveAsphalt or coke Residue Polycyclic

TABLE O1.4 Hydrocarbon Octane Numbers

Hydrocarbon Road Index Octane Number

Heptane 02-Methylheptane 23Hexane 252-Methylhexane 441-Heptene 60Pentane 621-Pentene 84Butane 91Cyclohexane 972,2,4-Trimethylpentane (isooctane) 100Benzene 101Toluene 112

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By 1922 a number of compounds had been discovered that could increase the octanenumber of gasoline. Adding as little as 6 mL of tetraethyllead (shown in Figure O1.19) toa gallon of gasoline, for example, can increase the octane number by 15 to 20 units. Thisdiscovery gave rise to the first “ethyl” gasoline, and it enabled the petroleum industry toproduce aviation gasolines with octane numbers greater than 100.

FIGURE O1.19 Structure of the tetraethyllead used to make “leaded” gasoline.

Another way to increase the octane number is thermal reforming. At high tempera-tures (500–600°C) and high pressures (25–50 atm), straight-chain alkanes isomerize to formbranched alkanes and cycloalkanes, thereby increasing the octane number of the gasoline.Running the reaction in the presence of hydrogen and a catalyst such as a mixture of sil-ica (SiO2) and alumina (Al2O3) results in catalytic reforming, which can produce a gaso-line with even higher octane numbers. Thermal or catalytic reforming and gasoline addi-tives such as tetraethyllead increase the octane number of the straight-run gasoline obtainedfrom the distillation of crude oil, but neither process increases the yield of gasoline froma barrel of oil.

The data in Table O1.3 suggest that we could increase the yield of gasoline by “crack-ing” the hydrocarbons that end up in the kerosene or fuel oil fractions into smaller pieces.Thermal cracking was discovered as early as the 1860s. At high temperatures (500°C) andhigh pressures (25 atm), long-chain hydrocarbons break into smaller pieces. A saturatedC12 hydrocarbon in kerosene, for example, might break into two C6 fragments. Becausethe total number of carbon and hydrogen atoms remains constant, one of the products ofthe reaction must contain a CPC double bond.

CH3(CH2)10CH3 888n CH3(CH2)4CH3 � CH2PCH(CH2)3CH3

The presence of alkenes in thermally cracked gasolines increases the octane number (70)relative to that of straight-run gasoline (60), but it also makes thermally cracked gasolineless stable for long-term storage. Thermal cracking has therefore been replaced by catalyticcracking, which uses catalysts instead of high temperatures and pressures to crack long-chain hydrocarbons into smaller fragments for use in gasoline.

About 87% of the crude oil refined in 1980 went into the production of fuels such asgasoline, kerosene, and fuel oil. The remainder went for nonfuel uses, such as petroleumsolvents, industrial greases and waxes, or starting materials for the synthesis of petro-chemicals. Petroleum products are used to produce synthetic fibers such as nylon, Orlon,and Dacron, and other polymers such as polystyrene, polyethylene, and synthetic rubber.They also serve as raw materials in the production of refrigerants, aerosols, antifreeze, de-tergents, dyes, adhesives, alcohols, explosives, weed killers, insecticides, and insect repel-lents. The H2 given off when alkanes are converted to alkenes or when cycloalkanes areconverted to aromatic hydrocarbons can be used to produce a number of inorganic petro-chemicals, such as ammonia, ammonium nitrate, and nitric acid. As a result, most fertiliz-ers as well as other agricultural chemicals are also petrochemicals.

CH3

CH3

CH3

CH3

CH2

CH2

CH2CH2 Pb

G

G

DD

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O1.12 THE CHEMISTRY OF COALCoal can be defined as a sedimentary rock that burns. It was formed by the decomposi-tion of plant matter, and it is a complex substance that can be found in many forms. Coalis divided into four classes: anthracite, bituminous, subbituminous, and lignite. Elementalanalysis gives empirical formulas such as C137H97O9NS for bituminous coal and C240H90O4NSfor high-grade anthracite. A model for one portion of the extended structure for coal isshown in Figure O1.20.

FIGURE O1.20 Model for a portion of the extended structure of coal.

Anthracite coal is a dense, hard rock with a jet-black color and a metallic luster. It con-tains between 86% and 98% carbon by weight, and it burns slowly, with a pale blue flameand very little smoke. Bituminous coal, or soft coal, contains between 69% and 86% car-bon by weight and is the most abundant form of coal. Subbituminous coal contains lesscarbon and more water, and it is therefore a less efficient source of heat. Lignite coal, orbrown coal, is a very soft coal that contains up to 70% water by weight.

The total energy consumption in the United States for 1990 was 86 � 1015 kJ. Of thattotal, 41% came from oil, 24% from natural gas, and 23% from coal. Coal is unique as asource of energy in the United States, however, because none of the 2118 billion poundsused in 1990 was imported. Furthermore, the proven reserves are so large we can continueusing coal at that level of consumption for at least 2000 years.

H

H

H

H

H

H

H

H

OH

H

H

H

H O

H

O

O

O

O

O

C

O

HH H

C

S

MNH2

H2 H2

H2

H2

H2

H2

P

ifHO H

H2

H2

H2

H2

HHH

OH

H

S

N

H

HH

H

HH

Hif

C

Ci

i

N

H

H

H

H

OH

H

OH

H

HCf

f

f

f

H

H H H

H H H

H

OH

H

H

S

CH2CH3

H

CC CO

SOO

O

i

iH

H

H H

H

S

S

H

C CO

H H

H H

C HOO

H

HH

C HOOH

C HOOH

C HOOH

C HOOH

C HOOH

H

N O

H

H

H

C

H

HH

H H

NO

H

H

H

H S

O

S

H

H

H H

O

H

H

HH

O

C HOOH

S

H

H HHH H

H

CH3

1012T_mod04_1-42 1/20/05 16:13 Page 28 EQA


At the time this text was written, coal was the most cost-efficient fuel for heating. Thecost of coal delivered to the Purdue University physical plant was $1.41 per million kilo-joules of heating energy. The equivalent cost for natural gas would have been $5.22, and#2 fuel oil would have cost $7.34. Although coal is less expensive than natural gas and oil,it is more difficult to handle. As a result, there has been a long history of efforts to turncoal into either a gaseous or a liquid fuel.

Coal Gasification

As early as 1800, coal gas was made by heating coal in the absence of air. Coal gas is richin CH4 and gives off up to 20.5 kJ per liter of gas burned. Coal gas—or town gas, as it wasalso known—became so popular that most major cities and many small towns had a localgas house in which it was generated, and gas burners were adjusted to burn a fuel that pro-duced 20.5 kJ/L. Gas lanterns, of course, were eventually replaced by electric lights. Butcoal gas was still used for cooking and heating until the more efficient natural gas (38.3kJ/L) became readily available.

A slightly less efficient fuel known as water gas can be made by reacting the carbon incoal with steam.

C(s) � H2O(g) 88n CO(g) � H2(g) �Ho 131.3 kJ/molrxn

Water gas burns to give CO2 and H2O, releasing roughly 11.2 kJ per liter of gas consumed.Note that the enthalpy of reaction for the preparation of water gas is positive, which meansthat the reaction is endothermic. As a result, the preparation of water gas typically involvesalternating blasts of steam and either air or oxygen through a bed of white-hot coal. Theexothermic reactions between coal and oxygen to produce CO and CO2 provide enoughenergy to drive the reaction between steam and coal.

Water gas formed by the reaction of coal with oxygen and steam is a mixture of CO,CO2, and H2. The ratio of H2 to CO can be increased by adding water to the mixture, totake advantage of a reaction known as the water-gas shift reaction.

CO(g) � H2O(g) 88n CO2(g) � H2(g) �Ho �41.2 kJ/molrxn

The concentration of CO2 can be decreased by reacting the CO2 with coal at high tem-peratures to form CO.

C(s) � CO2(g) 88n 2 CO(g) �Ho 172.5 kJ/molrxn

Water gas from which the CO2 has been removed is called synthesis gas because it canbe used as a starting material for a variety of organic and inorganic compounds. It can beused as the source of H2 for the synthesis of ammonia, for example.

N2(g) � 3 H2(g) 88n 2 NH3(g)

It can also be used to make methyl alcohol, or methanol.

CO(g) � 2 H2(g) 88n CH3OH(l)

Methanol can then be used as a starting material for the synthesis of alkenes, aromaticcompounds, acetic acid, formaldehyde, and ethyl alcohol (ethanol). Synthesis gas can alsobe used to produce methane, or synthetic natural gas (SNG).

1012T_mod04_1-42 1/20/05 16:13 Page 29 EQA


CO(g) � 3 H2(g) 88n CH4(g) � H2O(g)2 CO(g) � 2 H2(g) 88n CH4(g) � CO2(g)

Coal Liquefaction

The first step toward making liquid fuels from coal involves the manufacture of synthesisgas (CO and H2) from coal. In 1925, Franz Fischer and Hans Tropsch developed a cata-lyst that converted CO and H2 at 1 atm and 250–300°C into liquid hydrocarbons. By 1941,Fischer–Tropsch plants produced 740,000 tons of petroleum products per year in Germany.

Fischer–Tropsch technology is based on a complex series of reactions that use H2 toreduce CO to CH2 groups linked to form long-chain hydrocarbons.

CO(g) � 2 H2(g) 88n (CH2)n(l) � H2O(g) �Ho �165 kJ/molrxn

The water produced in the reaction combines with CO in the water-gas shift reaction toform H2 and CO2.

CO(g) � H2O(g) 88n CO2(g) � H2(g) �Ho �41.2 kJ/molrxn

The overall Fischer–Tropsch reaction is therefore described by the following equation.

2 CO(g) � H2(g) 88n (CH2)n(l) � CO2(g) �Ho �206 kJ/molrxn

At the end of World War II, Fischer–Tropsch technology was under study in most indus-trial nations. The low cost and high availability of crude oil, however, led to a decline ininterest in liquid fuels made from coal. The only commercial plants using this technologytoday are in the Sasol complex in South Africa, which uses 30.3 million tons of coal peryear.

Another approach to liquid fuels is based on the reaction between CO and H2 to formmethanol, CH3OH.

CO(g) � 2 H2(g) 88n CH3OH(l)

Methanol can be used directly as a fuel, or it can be converted into gasoline with catalystssuch as the ZSM-5 zeolite catalyst developed by Mobil Oil Company.

As the supply of petroleum becomes smaller and its cost continues to rise, a gradualshift may be observed toward liquid fuels made from coal. Whether this takes the form ofa return to a modified Fischer–Tropsch technology, the conversion of methanol to gaso-line, or other alternatives, only time will tell.

O1.13 CHIRAL STEREOISOMERSThe cis/trans isomers formed by alkenes aren’t the only example of stereoisomers. To un-derstand the second example of stereoisomers, it might be useful to start by considering apair of hands. For all practical purposes, they contain the same “substituents”—four fin-gers and one thumb on each hand. If you clap them together, you will find even more sim-ilarities between the two hands. The thumbs are attached at about the same point on thehand, significantly below the point where the fingers start. The second fingers on bothhands are usually the longest, then the third fingers, then the first fingers, and finally the“little” fingers.

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In spite of the many similarities, there is a fundamental difference between a pair ofhands that can be observed by trying to place your right hand into a left-hand glove. Yourhands have two important properties: (1) each hand is the mirror image of the other, and(2) the mirror images are not superimposable. The mirror image of the left hand looks likethe right hand, and vice versa, as shown in Figure O1.21.

FIGURE O1.21 Each hand is the mirror image of the other, butneither hand is superimposable on the other.

Objects that possess a similar handedness are said to be chiral (literally, “handed”). Thosethat do not are said to be achiral. Gloves are chiral. (It is difficult, if not impossible, toplace a right-hand glove on your left hand or a left-hand glove on your right hand.) Mit-tens, however, are often achiral. (Either mitten can fit on either hand.) Feet and shoes areboth chiral, but socks are not.

In 1874 Jacobus van’t Hoff and Joseph Le Bel recognized that a compound that con-tains a single tetrahedral carbon atom with four different substituents could exist in twoforms that were mirror images of each other. Consider the CHFClBr molecule, for exam-ple, which contains four different substituents on a tetrahedral carbon atom. Figure O1.22shows one possible arrangement of the substituents and the mirror image of the structure.By convention, solid lines are used to represent bonds that lie in the plane of the paper.Wedges are used for bonds that come out of the plane of the paper toward the viewer;dashed lines describe bonds that go behind the paper.

FIGURE O1.22 One form of the CHFClBr molecule and its mirrorimage.

If we rotate the molecule on the right by 180° around the COH bond we get the struc-tures shown in Figure O1.23. Chiral molecules are said to be optically active because theyare capable of rotating a plane of polarized light.

FIGURE O1.23 To predict whether a molecule is optically active, we haveto decide whether the molecule and its mirror image can be superim-posed. We can start by rotating the mirror image by 180° around the C—H bond.

C C≥

(

A

E ≥(

A

E

H

F FCl ClBr Br

H

C ≥

(

A

E

H

F ClBr

C ≥

(

A

E

H

F BrCl

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These structures are different because they cannot be superimposed on one another, asshown in Figure O1.24.

FIGURE O1.24 The compound CHFClBr is chiral because the structures are not superim-posable.

CHFClBr is therefore a chiral molecule that exists in the form of a pair of stereoisomersthat are mirror images of each other. As a rule, any tetrahedral atom that carries four dif-ferent substituents is a stereocenter, or a stereogenic atom. Compounds that contain a sin-gle stereocenter are always chiral. Some compounds that contain two or more stereocen-ters are achiral because of the symmetry of the relationship between the stereocenters.

The prefix “en-” often means “to make, or cause to be,” as in “endanger.” It is alsoused to strengthen a term, to make it even more forceful, as in “enliven.” Thus, it isn’tsurprising that a pair of stereoisomers that are mirror images of each other are called enan-tiomers. They are literally compounds that contain parts that are forced to be across fromeach other. Stereoisomers that aren’t mirror images of each other are called diastereomers.3

The cis/trans isomers of 2-butene, for example, are stereoisomers, but they are not mirrorimages of each other. As a result, they are diastereomers.

Exercise O1.9

Which of the following compounds would form enantiomers because the molecule is chiral?

Solution

The second carbon atom in 2-bromo-2-methylbutane contains two identical CH3 sub-stituents. As a result, the compound is achiral and does not form enantiomers.

CH2CH3

CH3CH3

A

(¨ E

BrC 2-Bromo-2-methylbutane

A

AO O

CH3

CCH3 CH2CH3

Br

A

AO O

CH3

CBrCH2 CH2CH3

H

2-Bromo-2-methylbutane

1-Bromo-2-methylbutane

3The prefix “dia-” is often used to indicate “opposite directions,” or “across,” as in “diagonal.”

C ≥

A

E

H

FCl

Br

C ≥A

E

H

FBr

Cl

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The second carbon atom in 1-bromo-2-methylbutane carries four different substituents: H,BrCH2, CH3, and CH2CH3. As a result, the molecule is chiral and it forms enantiomers.

Note: Every object—with the possible exception of the vampires found on late-night TV—has a mirror image. The relevant question is whether the mirror image can be superim-posed on the source of the image. If it can, the object is not chiral (achiral). If it cannot,the object is chiral and it can exist as a pair of stereoisomers.

O1.14 OPTICAL ACTIVITY

The Difference between Enantiomers on the Macroscopic Scale

If you could analyze the light that travels toward you from a lamp, you would find the elec-tric and magnetic components of the radiation oscillating in all of the planes parallel to thepath of the light. However, if you analyzed light that has passed through a polarizer, suchas a Nicol prism or the lens of polarized sunglasses, you would find that the oscillationswere now confined to a single plane.

In 1813 Jean Baptiste Biot noticed that plane-polarized light was rotated either to theright or the left when it passed through single crystals of quartz or aqueous solutions oftartaric acid or sugar. Because they interact with light, substances that can rotate plane-polarized light are said to be optically active. Those that rotate the plane clockwise (to theright) are said to be dextrorotatory (from the Latin word dexter, “right”).4 Those that ro-tate the plane counterclockwise (to the left) are called levorotatory (from the Latin wordlaevus, “left”).

In 1848 Louis Pasteur noted that sodium ammonium tartrate forms two different kindsof crystals that are mirror images of each other, much as the right hand is a mirror imageof the left hand. By separating one type of crystal from the other with a pair of tweezershe was able to prepare two samples of the compound. One was dextrorotatory when dis-solved in aqueous solution; the other was levorotatory. Since the optical activity remainedafter the compound had been dissolved in water, it could not be the result of macroscopicproperties of the crystals. Pasteur therefore concluded that there must be some asymme-try in the structure of the compound that allowed it to exist in two forms.

Once techniques were developed to determine the three-dimensional structure of a mol-ecule, the source of the optical activity of a substance was recognized: Compounds that areoptically active contain molecules that are chiral. Chirality is a property of a molecule thatresults from its structure. Optical activity is a macroscopic property of a collection of themolecules that arises from the way they interact with light. Compounds, such as CHFClBr,that contain a single stereocenter are the simplest to understand. One enantiomer of the chi-ral compound is dextrorotatory; the other is levorotatory. To decide whether a compoundshould be optically active, we look for evidence that the molecules are chiral.

CH2CH3

CH3H

A

(¨ E

BrCH2

C

CH2CH3

HCH3

A

(¨ E

BrCH2

C

Enantiomers of 1-bromo-2-methylbutane

and

4You might remember that “dextro” means right by noting that the predominantly right-handed world inwhich we live uses words such as dextrous to mean unusually skilled at the use of one’s hands—in particular,the “right” hand.

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The instrument with which optically active compounds are studied is a polarimeter,shown in Figure O1.25. Imagine a horizontal line that passes through the zero of a coor-dinate system. By convention, negative numbers are placed on the left and positive num-bers on the right of zero. Thus levorotatory compounds are indicated with a negative sign(�) and dextrorotatory compounds with a positive sign (�).

FIGURE O1.25 A polarimeter for studying the rotation of plane-polarized light.

The magnitude of the angle through which an enantiomer rotates plane-polarized lightdepends on four quantities: (1) the wavelength of the light, (2) the length of the cell throughwhich the light passes, (3) the concentration of the optically active compound in the solu-tion through which the light passes, and (4) the specific rotation of the compound, whichreflects the relative ability of the compound to rotate plane-polarized light. The specificrotation of the dextrorotatory isomer of glucose is written as follows:

[�] 20D �3.12

When the spectrum of sunlight was first analyzed by Joseph von Fraunhofer in 1814, heobserved a limited number of dark bands in the spectrum, which he labeled A–H. We nowknow that the D band in the spectrum is the result of the absorption by sodium atoms oflight that has a wavelength of 589.6 nm. The “D” in the symbol for specific rotation indi-cates that it is light of this wavelength that was studied. The “20” indicates that the ex-periment was done at 20°C. The “�” sign indicates that the compound is dextrorotatory;it rotates light clockwise. Finally, the magnitude of the measurement indicates that whena solution of the compound with a concentration of 1.00 g/mL was studied in a 10-cm cell,it rotated the light by 3.12°.

The magnitude of the rotations observed for a pair of enantiomers is always the same.The only difference between the compounds is the direction in which they rotate plane-polarized light. The specific rotation of the levorotatory isomer of the compound wouldtherefore be �3.12°.

The Difference between Enantiomers on the Molecular Scale

An unambiguous system for differentiating between enantiomers is based on the Latinterms for left (sinister) and right (rectus).

• Arrange the four substituents in order of decreasing atomic number of the atomsattached to the stereocenter. (The substituent with the highest atomic number gets

Unpolarizedlight

Nicolprismpolarizer

Incidentplane-polarizedlight

Polarimeter tube containingsolution of an optical isomer

Emergent lightwith rotatedplane ofpolarization

Nicolprismanalyzer

α

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the highest priority.) The substituents in 2-bromobutane, for example, would belisted in the order: Br � CH3 � CH2CH3 � H.

• When two or more substituents have the same priority—such as the CH3 andCH2CH3 groups in 2-bromobutane—continue working down the substituent chainuntil you find a difference. In 2-bromobutane, we would give the CH2CH3 group ahigher priority than the CH3 group because the next point down the chain is a CH3

group in the CH2CH3 substituent and an H atom in the CH3 group. Thus, the foursubstituents on 2-bromobutane would be listed in the order: Br � CH2CH3 �CH3 � H.

• View the enantiomer from the direction that places the substituent with the lowestpriority as far from the eye as possible. In the following example, this involves ro-tating the molecule counterclockwise around the C—CH2CH3 bond and tilting itslightly around an axis that lies in the plane of the paper. When this is done, thesubstituent that has the lowest priority is hidden from the eye.

• Trace a path that links the substituents in decreasing order of priority. If the pathcurves to the right—clockwise—the molecule is the rectus or R enantiomer. If itcurves to the left—counterclockwise—it is the sinister or S enantiomer.

';

%

C

CH3

CH2CH3

Br

2nd priority

3rd priority 1st priority

(S)-2-Bromobutane

';

%CH2CH3 CH2CH3

HCH3 CH3

A

(¨ EC C

Br Br

CH2CH3

Br

CH3

H

A

AOOC

2nd priority

1st priority

3rd priority

4th priority

CH2CH3

Br

CH3

H

A

AOOC

2nd priority

1st priority

2nd priority

3rd priority

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In this example, the path curves to the left, so the enantiomer is the (S)-2-bromobutanestereoisomer.

It is important to recognize that the (R)/(S) system is based on the structure of an indi-vidual molecule and the (�)/(�) system is based on the macroscopic behavior of a largecollection of molecules. The most complete description of an enantiomer combines aspectsof both systems. The enantiomer analyzed in this section is best described as (S)-(�)-2-bromobutane. It is the (S) enantiomer because of its structure and the (�) enantiomer be-cause samples of the enantiomer with this structure are levorotatory; they rotate plane-polarized light clockwise.

KEY TERMSAchiralAddition reactionAlkaneAlkeneAlkyl bromideAlkyl chlorideAlkyl halideAlkyneBranched hydrocarbonCatalytic crackingCatalytic reformingChiralCis/trans isomerCoalCoal gasConformationConstitutional isomerCrude oil

CycloalkaneDextrorotatoryDiastereomerEnantiomerEssential oilHydrocarbonIsomerLevorotatoryMarkovnikov’s ruleNatural gasNewman projectionOctane numberOptical activityOrganic chemistrySpecific rotationStereocenterStereogenic atomSteroid

StereoisomerStraight-chainhydrocarbonStraight-run gasolineSynthesis gasTerpeneTerpenoidTetravalentThermal crackingThermal reformingTown gasUnsaturatedhydrocarbonVital forceWater gasWater-gas shift reaction

PROBLEMS

The Special Role of Carbon

1. Explain why carbon forms covalent bonds, not ionic bonds, with so many other ele-ments.

2. Carbon forms relatively strong double bonds, not only with itself, but with other non-metals such as nitrogen, oxygen, phosphorus, and sulfur. Why is this important in termsof the great diversity of carbon compounds?

3. In The Life Puzzle, Graham Cairns-Smith proposed three rules for speculating on theorigin of life: (1) take a good look at atoms and see what they can do, (2) take a goodlook at organisms and see how they work, and then (3) put an organism together inthe easiest way you can think of. He reports two hypotheses to explain the strikingfundamental biochemical similarity between life in all its forms. That “the first organ-isms on Earth had a similar composition to modern forms.” Or that “modern biochemical

1012T_mod04_1-42 1/20/05 16:13 Page 36 EQA


uniformity is a product of evolution—a tribute to the effectiveness of naturalselection—rather than an indication that life depends uniquely on amino acids, purinesand so on.” Speculate on a few of the relative merits of these hypotheses, or proposean alternative hypothesis.

The Saturated Hydrocarbons: Alkanes and Cycloalkanes

4. Use examples to explain the difference between saturated and unsaturated hydrocar-bons and between straight-chain and branched hydrocarbons.

5. Explain why it is better to describe butane as a “straight-chain hydrocarbon” than itis to describe it as a “linear hydrocarbon.”

6. Use the fact that straight-chain alkanes have a CH3 group at either end and a chain ofCH2 groups down the middle to explain why alkanes have the generic formula CnH2n�2.Write the generic formulas for cycloalkanes, alkenes, and alkynes.

7. Describe the difference between n-pentane, isopentane, and neopentane. Classify thecompounds as either stereoisomers or constitutional isomers.

8. Predict the number of constitutional isomers of heptane, C7H16.9. Write the molecular formula for the saturated hydrocarbon that has the following car-

bon skeleton and name the compound.

10. Write the molecular formula for the saturated hydrocarbon that has the following car-bon skeleton and name the compound.

11. Explain why it is possible to isolate different constitutional isomers of butane, but notdifferent conformations of butane.

12. Two of the conformations of butane can be described by the following diagrams, knownas Newman projections, which look down the C2—C3 bond in butane. Which of theseconformations is more favorable?

13. Provide the systematic (IUPAC approved) name for the following compound.

CH3CH2CHCHCH2CHCH3

A

A A

CH3

CH3 CH2CH3

CH3H

H HH

CH3CH3

H

H HCH3

H

A B

C C

C

O

CO CO CO CO

CO CO

CO

ACA

CA

A A

A

O O O OC C

C C

C

C C C

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14. Write the formula for the compound known as 1-methyl-1-chlorocyclopentane.15. Draw the structure of 2,3,4-trimethyl-4-ethyloctane.16. Provide the systematic for the compound in the following Newman projection.

17. One way to decide whether a pair of structures represent different compounds is to as-sign a systematic name to each structure. Use this approach to decide whether the fol-lowing are isomers or different descriptions of the same compound.

The Unsaturated Hydrocarbons: Alkenes and Alkynes

18. Draw the structures of all the alkenes that have the formula C6H12 and name thecompounds.

19. Draw the structures of all the alkynes that have the formula C5H8 and name thecompounds.

20. Explain why it is a mistake to name a compound 3-pentene. What would be the cor-rect name of the compound?

21. Use the VSED theory to predict the shape of the tetrafluoroethylene (C2F4) moleculethat is the starting material used to make Teflon.

22. Explain why alkenes can form both constitutional isomers and stereoisomers. Whatcharacteristic feature of a pair of alkenes can be used to decide whether they are con-stitutional isomers or stereoisomers?

23. Explain why alkynes can have constitutional isomers but not stereoisomers.24. Describe the hybridization of each of the carbon atoms in the following compound.

CH3CHPCHCH2CqCH

25. Which of the following compounds does not have the same molecular formula as theothers?(a) cyclopentane (b) methylcyclobutane (c) 1-pentene (d) pentane(e) 1,1-dimethylcyclopropane

26. Which of the following compounds have cis/trans isomers?(a) CHCl3 (b) F2CPCF2 (c) Cl2CPCHCH3 (d) FClCPCFCl (e) H2CPCHF

27. Which of the following compounds have cis/trans isomers?(a) 1-pentene (b) 2-pentene (c) 2-methyl-2-butene(d) 1-chloro-2-butene (e) 2-pentyne

28. Which of the following pairs of compounds would be constitutional isomers?

CH3

CH3

A AO

OG

DCH

CH

CH2

CH3 CH3CHCH2CHCH3

CH3

i

f

CH3 CH3

CH2CH3

H HCH3

BrBr

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29. Determine whether the following compounds are isomers. Explain why or why not.

30. Draw the structure for cis-2,3-dichloro-2-hexene.31. Draw the structure for 2,6-dimethyl-3-heptyne.32. Draw the structure for trans-6-methyl-2-heptene.33. Draw the structure for cyclohexene.34. Draw the structure for 2,3-dimethyl-2-pentene. Explain why the compound can’t exist

as a pair of cis/trans isomers.

The Reactions of Alkanes, Alkenes, and Alkynes

35. Write a balanced equation for the reaction in which isooctane, C8H18, burns in oxygento form CO2 and H2O vapor.

36. Define the term addition reaction and give an example of an addition reaction of analkene.

37. Predict the product of the addition reaction between Br2 and 2-pentene.38. Predict the product of the addition reaction between water and 1-pentene in the pres-

ence of sulfuric acid.39. Predict the product of the addition reaction between excess H2 and 2-pentyne in the

presence of a platinum metal catalyst.40. Which of the following is a product of the reaction between chlorine and ethylene?

(a) CH3CHCl2 (b) CH3CH2Cl (c) ClCH2CH2Cl (d) Cl2CHCHCl241. Which of the following does not rapidly react with Br2 dissolved in CCl4?

(a) pentane (b) 1-pentene (c) 2-pentene (d) 1-pentyne (e) 2-pentyne42. Assume that HI adds to a CPC double bond according to Markovnikov’s rule. Draw

the structure and name the product of the reaction between hydrogen iodide and 2-methyl-2-pentene.

43. Initial attempts to confirm Markovnikov’s rule for the addition of HBr to an alkenewere complicated by the fact that the reactions did not always give the same product.We now know that HBr adds to a CPC double according to Markovnikov’s rule unless the solvent is contaminated with a source of free radicals. Draw the structuresand name the products of both the Markovnikov addition of HBr to 1-pentene andthe anti-Markovnikov addition of HBr that occurs in the presence of a free radicalcontaminant.

(a) CH3CH2OCH2CH3 and CH3CH2CH2CH2OH(b) (CH3)2CHCH3 and CH3CH2CH2CH3

(c) CH3CH2CH2CH3 and CH2CH2

(d) andH H

Cl Cl

CH3

CH3

A

A

H Cl

Cl H

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The Aromatic Hydrocarbons and Their Derivatives

44. Draw the structures of the following aromatic compounds: aniline, anisole, benzene,and benzoic acid.

45. Draw the structures of ortho-, meta-, and para-bromotoluene.46. TNT is an abbreviation for 2,4,6-trinitrotoluene. Toluene is a derivative of benzene in

which a methyl (CH3) group is substituted for one of the hydrogen atoms. Trinitro-toluene is a derivative of toluene in which NO2 groups have replaced three more hy-drogen atoms on the benzene ring. Draw the structures of all the possible isomers oftrinitrotoluene. Label the isomer that is 2,4,6-trinitrotoluene.

47. On an exam, a student described benzene as a mixture of two structures that are rapidlybeing converted from one to the other. What is wrong with that answer?

48. Explain why the 12 atoms in a benzene molecule all lie in the same plane.

The Chemistry of Petroleum Products and Coal

49. Explain why a mixture of CO and H2 can be used as a fuel. What are the products ofthe combustion of the mixture, which was once known as “water gas.”

50. Natural gas, petroleum ether, gasoline, kerosene, and asphalt are all different forms ofhydrocarbons that give off energy when burned. Describe how the substances differ.What happens to the boiling points of the mixtures as the average length of the hy-drocarbon chain increases?

51. Which of the following compounds has the largest octane number?(a) n-butane (b) n-pentane (c) n-hexane (d) n-octane

52. Which of the following won’t increase the octane number of gasoline?(a) increasing the concentration of branched-chain alkanes (b) increasing the con-centration of cycloalkanes (c) increasing the concentration of aromatic hydrocar-bons (d) increasing the average length of the hydrocarbon chains.

53. Describe how thermal cracking and catalytic cracking increase the amount of high-octane gasoline that can be obtained from a barrel of oil. Explain why neither thermalreforming nor catalytic reforming can achieve this.

54. Coal gas can be obtained when coal is heated in the absence of air. Water gas can beobtained when coal reacts with steam. Describe the difference between the gases andexplain why much more water gas can be extracted from a ton of coal.

Optical Activity

55. Describe the difference between constitutional isomers and stereoisomers. What char-acteristic feature of a molecule could be used to distinguish between the two forms ofisomers?

56. Objects that cannot be superimposed on their mirror images are said to be chiral, andchiral molecules are optically active. Which of the following molecules are opticallyactive?(a) CH4 (b) CH3Cl (c) CHCl3 (d) CHFCl2 (e) CHFClBr

57. Which of the following compounds are optically active?CH3A

(a) C2H4 (b) C6H6 (c) C6H4Cl2 (d) CH3CH2CHCH3

58. Coniine is the active ingredient in the poison known as hemlock that was given to

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Socrates in 399 B.C. Use the following diagram of the structure of coniine to predictwhether the compound is optically active.

59. Use the structure of caffeine shown below to predict whether the compound is opti-cally active.

60. Determine the number of centers of chirality in the structure of vitamin C shown inFigure O1.2.

61. Predict whether the following compound is chiral.

62. Which of the following compounds, which play an important role in the chemistry ofbiological systems, is chiral?

63. Which (if any) of the carbon atoms in 2-bromo-3-methylbutane are stereocenters?64. Determine whether the following isomers are enantiomers or diastereomers. What

characteristic feature of the molecule can be used to make the decision?

A

D CH3

CH2CH3

C

BrH(

ACH2CH3

C

H CH3

D Br(

HOOCACH2

ACO2H

ACH2ACO2H

OCO2H HOCACH2

ACO2H

AHOOCH

ACO2H

OCO2H HOOCACH2

ACH2

ACO2H

ACH2ACO2H

OCO2H

Citric acid Isocitric acid Homocitric acid

CH3ONACH2CH2CH2CH3

CH3ACH(CH3)2

OCH2

�

O

N

N N

NO

CH3

CH3CH3

N

HA

1012T_mod04_1-42 1/20/05 16:13 Page 41 EQA


65. Determine whether the following isomers are enantiomers or diastereomers. Whatcharacteristic feature of the molecule can be used to make the decision?

66. Determine whether the following compound is the R or S enantiomer of 2-bromo-butane.

A

DCH3

CH2CH3

C

HBr(

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1

M O D U L E

5ORGANIC CHEMISTRY:FUNCTIONAL GROUPS

O2.1 Functional Groups

O2.2 Oxidation–Reduction Reactions

O2.3 Alkyl Halides

O2.4 Alcohols and Ethers

O2.5 Aldehydes and Ketones

O2.6 Reactions at the Carbonyl Group

O2.7 Carboxylic Acids and Carboxylate Ions

O2.8 Esters

O2.9 Amines, Alkaloids, and Amides

O2.10 Grignard Reagents

Chemistry in the World Around Us: The Chemistry of Garlic

O2.1 FUNCTIONAL GROUPSBromine reacts with 2-butene to form 2,3-dibromobutane.

It also reacts with 3-methyl-2-pentene to form 2,3-dibromopentane.

Instead of trying to memorize both equations, we can build a general rule that bromine re-acts with compounds that contain a CPC double bond to give the product expected fromaddition across the double bond. This approach to understanding the chemistry of organiccompounds presumes that certain atoms or groups of atoms known as functional groupsgive these compounds their characteristic properties.

CH3CHPC CH3CH2 � Br2 CH3CA ABr

HCABr

CH3

CH3ACH3

CH2

CH3CHPCHCH3 � Br2 CH3CABr

HCABr

HCH3

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2 FUNCTIONAL GROUP CHEMISTRY

Functional groups focus attention on the important aspects of the structure of a mol-ecule. We don’t have to worry about the differences between the structures of 1-buteneand 2-methyl-2-hexene, for example, when the compounds react with hydrogen bromide.We can focus on the fact that both compounds are alkenes that add HBr across the CPCdouble bond in the direction predicted by Markovnikov’s rule, introduced in Section O1.8.

Some common functional groups are given in Table O2.1.

CH3CACH3

ACH3

PCHCH2CH2CH HBr�3 CH3CABr

CH2CH2CH2CH3

CH2 CHP CH2CH3 � HBr CH3CABr

H CH3CH2

Functional Group Name Example

AOCO Alkane CH3CH2CH3 (propane)

ACPC Alkene CH3CHPCH2 (propene)

CH3C CH (propyne)CqC Alkyne q

F, Cl, Br, or I Alkyl halide CH3Br (methyl bromide)OOH Alcohol CH3CH2OH (ethanol)OOO Ether CH3OCH3 (dimethyl ether)ONH2 Amine CH3NH2 (methylamine)

TABLE 02.1 Common Functional Groups

The CPO group plays a particularly important role in organic chemistry. This groupis called a carbonyl, and some of the functional groups based on a carbonyl are shown inTable O2.2.

Functional Group Name Example

OBO

COH Aldehyde CH3CHO (acetaldehyde)

OBO

CO Ketone CH3COCH3 (acetone)

OBO

COCl

Carboxylic acid

CH3COCl

(acetic acid)OBO

COOH

Acyl chloride

CH3CO H2

(acetyl chloride)

OBO

COOO Ester CH3CO2CH3 (methyl acetate)

OBO

CONH2 Amide CH CO3 NH2 (acetamide)

TABLE 02.2 Functional Groups That Contain a Carbonyl

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FUNCTIONAL GROUP CHEMISTRY 3

Exercise O2.1

Root beer hasn’t tasted the same since the use of sassafras oil as a food additive was out-lawed because sassafras oil is 80% safrole, which has been shown to cause cancer in ratsand mice. Identify the functional groups in the structure of safrole.

Solution

Safrole is an aromatic compound because it contains a benzene ring. It is also an alkenebecause it contains a CPC double bond. The most difficult functional group to recognizein the molecule might be the two ether linkages (OOO).

Exercise O2.2

The following compounds are the active ingredients in over-the-counter drugs used as anal-gesics (to relieve pain without decreasing sensibility or consciousness), antipyretics (to re-duce the body temperature when it is elevated), and/or anti-inflammatory agents (to coun-teract swelling or inflammation of the joints, skin, and eyes). Identify the functional groupsin each molecule.

Solution

All three compounds are aromatic. Aspirin is also a carboxylic acid (OCO2H) and an es-ter (OCO2CH3). Tylenol is also an alcohol (OOH) and an amide (OCONHO). Ibupro-fen contains alkane substituents and a carboxylic acid functional group.

Exercise O2.3

The discovery of penicillin in 1928 marked the beginning of what has been called the“golden age of chemotherapy,” in which previously life-threatening bacterial infections

C

O

O

C CH3

O OHM D

O OB

Aspirin(acetylsalicylic acid)

OH

NHC

O

CH3G DB

Tylenol(acetaminophen)

CH OH

CHCH3 CH2

CH3

G D

CH3 CG GD

AAdvil

(ibuprofen)

BO

O

OOCH2

CH2

CH2

i

CHG

P

Safrole

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were transformed into little more than a source of discomfort. For those who are allergicto penicillin, a variety of antibiotics, including tetracycline, are available. Identify the nu-merous functional groups in the tetracycline molecule.

Solution

The compound contains an aromatic ring fused to three six-membered rings. It is also analcohol (with five OOH groups), a ketone (with CPO groups at the bottom of the secondand fourth rings), an amine [the ON(CH3)2 substituent at the top of the fourth ring], andan amide (the OCONH2 group at the bottom right-hand corner of the fourth ring.)

O2.2 OXIDATION–REDUCTION REACTIONSFocusing on the functional groups in a molecule allows us to recognize patterns in the be-havior of related compounds. Consider what we know about the reaction between sodiummetal and water, for example.

We can divide the reaction into two half-reactions. One involves the oxidation of sodiummetal to form sodium ions.

The other involves the reduction of an H� ion in water to form a neutral hydrogen atomthat combines with another hydrogen atom to form an H2 molecule.

Once we recognize that water contains an OOH functional group, we can predict whatmight happen when sodium metal reacts with an alcohol that contains the same functionalgroup. Sodium metal should react with methanol (CH3OH), for example, to give H2 gasand a solution of the Na� and CH3O� ions dissolved in the alcohol.

2Na(s) � � �2 CH3OH( 2 CH3O�(l) H2(g) 2 Na�(alc) alc)

H2 HReduction �

�

2 H� H�2 OSOQ

OGDOS

2 e�2 H� 2 H H2T

O

NaOxidation �Na� e�

2Na(s) � � �2 H2O(l) H2(g) 2 Na�(aq) 2 OH�(aq)

OH OH

HO

OH

OH

OC OP

HN(CH3)2CH3

fNH2

/∑

Tetracycline

BOB

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Because they involve the transfer of electrons, the reactions between sodium metal andeither water or an alcohol are examples of oxidation–reduction reactions. But what aboutthe following reaction, in which hydrogen gas reacts with an alkene in the presence of atransition metal catalyst to form an alkane?

There is no change in the number of valence electrons on any of the atoms in the reaction.Both before and after the reaction, each carbon atom shares a total of eight valence elec-trons and each hydrogen atom shares two electrons. Instead of electrons, the reaction involves the transfer of atoms—in this case, hydrogen atoms. There are so many atom-transfer reactions that chemists developed the concept of oxidation number (see Chapter 5)to extend the idea of oxidation and reduction to reactions in which electrons aren’t neces-sarily gained or lost.

Oxidation involves an increase in the oxidation number of an atom. Reduction oc-curs when the oxidation number of an atom decreases.

During the transformation of ethene into ethane, there is a decrease in the oxidation num-ber of the carbon atom. The reaction therefore involves the reduction of ethene to ethane.

Reactions in which none of the atoms undergo a change in oxidation number are calledmetathesis reactions. Consider the reaction between a carboxylic acid and an amine, forexample,

CH3CO2H � CH3NH2 88n CH3CO2� � CH3NH3

�

or the reaction between an alcohol and hydrogen bromide.

CH3CH2OH � HBr 88n CH3CH2Br � H2O

These are metathesis reactions because there is no change in the oxidation number of anyatom in either reaction.

The oxidation numbers of the carbon atoms in a variety of compounds are given inTable O2.3. The oxidation numbers can be used to classify organic reactions as either oxidation–reduction reactions or metathesis reactions.

HH2 C HC

H

HH

H

H

H H

H

CCG

G

D

DP �

Ni

�2

�3

HH2 C HC

H

HH

H

H

H H

H

CCG

G

D

DP �

Ni

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Exercise O2.4

Classify the following as either oxidation–reduction or metathesis reactions.H�

(a) 2 CH3OH 8n CH3OCH3 � H2OO

H�

OB B

(b) HCOH � CH3OH 8n HCOCH3 � H2O(c) CO � 2 H2 n CH3OH(d) CH3Br � 2 Li n CH3Li � LiBr

Solution(a) This is a metathesis reaction because there is no change in the oxidation number of

the carbon atoms when an alcohol is converted to an ether.

2 CH3OH 88n CH3OCH3 � H2O�2 �2

(b) This is a metathesis reaction because there is no change in the oxidation number of anyof the carbon atoms when a carboxylic acid reacts with an alcohol to form an ester.

HCO2H � CH3OH 88n HCO2CH3 � H2O�2 �2 �2 �2

(c) This is an oxidation–reduction reaction because the carbon atom is reduced from the�2 to the �2 oxidation state when CO combines with H2 to form methanol.

CO � 2 H2 88n CH3OH�2 �2

(d) This is an oxidation–reduction reaction because the carbon atom is reduced from the�2 to the �4 oxidation state when CH3Br reacts with lithium metal to form CH3Li.

CH3Br � 2 Li 88n CH3Li � LiBr�2 �4

TABLE O2.3 Typical Oxidation Numbers of Carbon

Oxidation Number of theClasses of Compounds Example Carbon

Alkane CH4 �4Alkyllithium CH3Li �4Alkene H2CPCH2 �2Alcohol CH3OH �2Ether CH3OCH3 �2Alkyl halide CH3Cl �2Amine CH3NH2 �2Alkyne HCqCH �1Aldehyde H2CO 0Carboxylic acid HCO2H 2

— CO2 4

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Because electrons are neither created nor destroyed, oxidation can’t occur in the absenceof reduction, and vice versa. It is often useful, however, to focus attention on one compo-nent of the reaction and ask, is that substance oxidized or reduced?

Exercise O2.5

Determine whether the following transformations involve the oxidation or the reductionof the carbon atom.

Solution

Each of the transformations involves the oxidation of the carbon atom. The first reactioninvolves oxidation of the carbon atom from the �2 to the 0 oxidation state. In the secondreaction the carbon atom is oxidized to the �2 state, and the third reaction involves oxi-dation of the carbon atom to the �4 oxidation state.

Assigning oxidation numbers to the individual carbon atoms in a complex moleculecan be difficult. Fortunately, there is another way to recognize oxidation–reduction reac-tions in organic chemistry:

Oxidation occurs when hydrogen atoms are removed from a carbon atom or whenan oxygen atom is added to a carbon atom.

Reduction occurs when hydrogen atoms are added to a carbon atom or when anoxygen atom is removed from a carbon atom.

H OHC

H

H

H HC

OB

H OHC

O

H HC

OB B

H OHC

O

O OCB

P P

�2

0

0

2�

2� 4�

H OHC

H

H

H HC

OB

H OHC

O

H HC

OB B

H OHC

O

O OCB

P P

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The first reaction in Exercise O2.5 involves oxidation of the carbon atom becausea hydrogen atom is removed from that atom when the alcohol is oxidized to analdehyde.

The second reaction in the exercise is an example of oxidation because an oxygen atom isadded to the carbon atom when an aldehyde is oxidized to a carboxylic acid.

Reduction, on the other hand, occurs when hydrogen atoms are added to a carbon atomor when an oxygen atom is removed from a carbon atom. An alkene is reduced, for ex-ample, when it reacts with H2 to form the corresponding alkane.

NiCH2PCHCH3 � H2 88n CH3CH2CH3

Figure O2.1 provides a useful guide to the oxidation–reduction reactions of organiccompounds. Each of the arrows in this figure involves a two-electron oxidation of a car-bon atom along the path toward carbon dioxide. A line is drawn through the first arrowbecause it is impossible to achieve this transformation in a single step.

HCBO

BO

H HCOH

CH3OH 88n HCBO

H

FIGURE O2.1 The stepwise oxidationof carbon.

O2.3 ALKYL HALIDESImagine that a pair of crystallizing dishes are placed on an overhead projector as shown inFigure O2.2. An alkene is added to the dish in the upper left corner of the projector, andan alkane is added to the dish in the upper right corner. A few drops of bromine dissolvedin chloroform (CHCl3) are then added to each of the crystallizing dishes.

FIGURE O2.2 Demonstrating the rela-tive rate of bromination reactions.

CH4 CH3OH HCH HCOH CO2+4+2–2–4 0

B BO O

The characteristic red-orange color of bromine disappears the instant this reagent isadded to the alkene in the upper left corner as the Br2 molecules add across the CPC dou-ble bond in the alkene.

CH3(CH2)4CH3CH2=CH(CH2)3CH3

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BrA

CH2PCH(CH2)3CH3 � Br2 88n BrCH2CH(CH2)3CH3

The other crystallizing dish picks up the characteristic color of a dilute solution of brominebecause the reagent doesn’t react with alkanes under normal conditions.

If the crystallizing dish in the upper right corner is moved into the center of the pro-jector, however, the color of the bromine slowly disappears. This can be explained by not-ing that alkanes react with halogens at high temperatures or in the presence of light toform alkyl halides, as noted in Section O1.8.

BrLight

ACH3(CH2)4CH3 � Br2 88n CH3CH(CH2)3CH3 � HBr

The light source in an overhead projector is intense enough to initiate the reaction, al-though the reaction is still much slower than the addition of Br2 to an alkene.

The reaction between an alkane and one of the halogens (F2, Cl2, Br2, or I2) can beunderstood by turning to a simpler example.

CH4(g) � Cl2(g) 88n CH3Cl(g) � HCl(g)

This reaction has the following characteristic properties.

• It doesn’t take place in the dark or at low temperatures.• It occurs in the presence of ultraviolet light or at temperatures above 250ºC.• Once the reaction gets started, it continues after the light is turned off.• The products of the reaction include CH2Cl2 (dichloromethane), CHCl3 (chloro-

form), and CCl4 (carbon tetrachloride), as well as CH3Cl (chloromethane).• The reaction also produces some C2H6.

These facts are consistent with a chain reaction mechanism that involves three processes:chain initiation, chain propagation, and chain termination.

Chain Initiation

A Cl2 molecule can dissociate into a pair of chlorine atoms by absorbing energy in the formof either ultraviolet light or heat.

Cl2 88n 2 Cl� �H° � 243.4 kJ/molrxn

The chlorine atom produced in the reaction is an example of a free radical—an atom ormolecule that contains one or more unpaired electrons.

Chain Propagation

Free radicals, such as the Cl� atom, are extremely reactive. When a chlorine atom collideswith a methane molecule, it can remove a hydrogen atom to form HCl and a CH3�radical.

CH4 � Cl� 88n CH3� � HCl �H° � �16 kJ/molrxn

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If the CH3� radical then collides with a Cl2 molecule, it can remove a chlorine atom to formCH3Cl and a new Cl� radical.

CH3� � Cl2 88n CH3Cl � Cl� �H° � �87 kJ/molrxn

Because a Cl� atom is generated in the second reaction for every Cl� atom consumed inthe first, the reaction continues in a chainlike fashion until the radicals involved in the chainpropagation steps are destroyed.

Chain Termination

If a pair of the radicals that keep the chain reaction going collide, they combine in a chainterminating step. Chain termination can occur in three ways.

2 Cl� 88n Cl2 �H° � �243.4 kJ/molrxn

CH3� � Cl� 88n CH3Cl �H° � �330 kJ/molrxn

2 CH3� 88n CH3CH3 �H° � �350 kJ/molrxn

Because the concentration of the radicals is relatively small, the chain termination reac-tions are relatively infrequent.

The chain reaction mechanism for free radical reactions explains the observations listedfor the reaction between CH4 and Cl2.

• The reaction doesn’t occur in the dark or at low temperatures because energy mustbe absorbed to generate the free radicals that carry the reaction.

Cl2 88n 2 Cl� �H° � 243.4 kJ/molrxn

• The reaction occurs in the presence of ultraviolet light because a UV photon hasenough energy to dissociate a Cl2 molecule to a pair of Cl� atoms. The reaction oc-curs at high temperatures because Cl2 molecules can dissociate to form Cl� atomsby absorbing thermal energy.

• The reaction continues after the light has been turned off because light is neededonly to generate the Cl� atoms that start the reaction. The chain reaction then con-verts CH4 into CH3Cl without consuming the Cl� atoms.

CH4 � Cl� 88n CH3� � HClCH3� � Cl2 88n CH3Cl � Cl�

• The reaction doesn’t stop at CH3Cl because the Cl� atoms can abstract additionalhydrogen atoms to form CH2Cl2, CHCl3, and eventually CCl4.

CH3Cl � Cl� 88n CH2Cl� � HClCH2Cl� � Cl2 88n CH2Cl2 � Cl�, and so on

• The formation of C2H6 is a clear indication that the reaction proceeds through afree radical mechanism. When two CH3� radicals collide, they combine to form anethane molecule.

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2 CH3� 88n CH3CH3

Free radical halogenation of alkanes provides us with another example of the role ofatom transfer reactions in organic chemistry. The net effect of the reaction is to oxidize acarbon atom by removing a hydrogen from the atom.

The reaction, however, doesn’t involve the gain or loss of electrons. It occurs by the trans-fer of a hydrogen atom in one direction and a chlorine atom in the other.

O2.4 ALCOHOLS AND ETHERSAlcohols contain an OOH group attached to a saturated carbon. The common names foralcohols are based on the name of the alkyl group.

The systematic nomenclature for alcohols adds the ending -ol to the name of the parentalkane and uses a number to identify the carbon that carries the OOH group. The sys-tematic name for isopropyl alcohol, for example, is 2-propanol.

Exercise O2.6

More than 50 organic compounds have been isolated from the oil that gives rise to thecharacteristic odor of a rose. One of the most abundant of the compounds is known by thecommon name citronellol. Use the systematic nomenclature to name this alcohol, whichhas the following structure.

Solution

The longest chain of carbon atoms in the compound contains eight atoms.

CH3

CH3CH3

CH2

CH2

H2C

H2C

CH

CHOH

CB

G

G

D

GD

A

A AO

CH3OHCH3CH2OHCH3CHOHCH3

MethanolEthanol2-Propanol

CH3OHCH3CH2OHCH3CHOHCH3

Methyl alcoholEthyl alcoholIsopropyl alcohol

CH4 � �Cl2 CH3Cl HCl�2�4

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The longest chain contains the OOH group, which means the compound is named as a de-rivative of octane. Because it is an alcohol, it would be tempting to name it as an octanol.But it contains a CPC double bond, which means it must be an octenol. We now have toindicate that the OOH group is on one end of the chain and the CPC double bond oc-curs between the sixth and seventh carbon atoms of the chain, which can be done by con-sidering the compound to be a derivative of 6-octen-1-ol. Finally, we have to indicate thepresence of a pair of CH3 groups on the third and seventh carbon atoms. The compoundis therefore given the systematic name 3,7-dimethyl-6-octen-1-ol.

Methanol, or methyl alcohol, is also known as wood alcohol because it was originallymade by heating wood until a liquid distilled. Methanol is highly toxic, and many peoplehave become blind or have died from drinking it. Ethanol, or ethyl alcohol, is the alcoholassociated with “alcoholic” beverages. It has been made for at least 6000 years by addingyeast to solutions that are rich in either sugars or starches. The yeast cells obtain energyfrom enzyme-catalyzed reactions that convert sugar or starch to ethanol and CO2.

C6H12O6(aq) 88n 2 CH3CH2OH(aq) � 2 CO2(g)

When the alcohol reaches a concentration of 10% to 12% by volume, the yeast cells die.Brandy, rum, gin, and the various whiskeys that have a higher concentration of alcohol areprepared by distilling the alcohol produced by the fermentation reaction. Ethanol isn’t astoxic as methanol, but it is still dangerous. Most people are intoxicated at blood-alcohollevels of about 0.1 gram per 100 mL. An increase in the level of alcohol in the blood tobetween 0.4 and 0.6 g/100 mL can lead to coma or death.

The method of choice for determining whether an individual is DUI—driving underthe influence—or DWI—driving while intoxicated—is the Breathalyzer, for which a patentwas issued to R. F. Borkenstein in 1958. The chemistry behind the Breathalyzer is basedon the reaction between alcohol in the breath and the chromate or dichromate ion.

3 CH3CH2OH(g) � 2 Cr2O72�(aq) � 16 H�(aq) 88n

3 CH3CO2H(aq) � 4 Cr3�(aq) � 11 H2O(l)

The instrument contains two ampules that hold small samples of potassium dichromate dis-solved in sulfuric acid. One of the ampules is used as a reference. The other is opened andthe breath sample to be analyzed is added. If alcohol is present in the breath, it reducesthe yellow-orange Cr2O7

2� ion to the green Cr3� ion. The extent to which the color bal-ance between the two ampules is disturbed is a direct measure of the amount of alcoholin the breath sample. Measurements of the alcohol on the breath are then converted into

CH3

CH3CH3

CH2

CH2

H2C

H2C

CH

CHOH

CB

G

G

D

GD

A

A AO5

6

7

8

1

23

4

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estimates of the concentration of the alcohol in the blood by assuming that 2100 mL of airexhaled from the lungs contains the same amount of alcohol as 1 mL of blood.

Measurements taken with the Breathalyzer are reported in units of percent blood-alcohol concentration (BAC). In most states, a BAC of 0.10% is sufficient for a DUI orDWI conviction. (This corresponds to a blood-alcohol concentration of 0.10 gram of alco-hol per 100 mL of blood.)

Ethanol is oxidized to CO2 and H2O by the alcohol dehydrogenase enzymes in thebody. This reaction gives off 30 kilojoules per gram, which makes ethanol a better sourceof energy than carbohydrates (17 kJ/g), and almost as good a source of energy as fat(38 kJ/g). An ounce of 80-proof liquor can provide as much as 3% of the average dailycaloric intake, and drinking alcohol can contribute to obesity. Many alcoholics are mal-nourished, however, because of the absence of vitamins in the calories they obtain fromalcoholic beverages.

As a general rule, polar or ionic substances dissolve in polar solvents; nonpolar sub-stances dissolve in nonpolar solvents. As a result, hydrocarbons don’t dissolve in water.They are often said to be immiscible (literally, “not mixable”) in water. Alcohols, as mightbe expected, have properties between the extremes of hydrocarbons and water. When thehydrocarbon chain is short, the alcohol is soluble in water. There is no limit on the amountof methanol (CH3OH) and ethanol (CH3CH2OH), for example, that can dissolve in a givenquantity of water. As the hydrocarbon chain becomes longer, the alcohol becomes less sol-uble in water, as shown in Table O2.4. One end of the alcohol molecules has so much non-polar character it is said to be hydrophobic (literally, “water-hating”). The other end con-tains an OOH group that can form hydrogen bonds to neighboring water molecules andis therefore said to be hydrophilic (literally, “water-loving”). As the hydrocarbon chain be-comes longer, the hydrophobic character of the molecule increases, and the solubility ofthe alcohol in water gradually decreases until it becomes essentially insoluble in water.

TABLE O2.4 Solubilities of Alcohols in Water

Formula Name Solubility in Water (g/100 g)

CH3OH Methanol Infinitely solubleCH3CH2OH Ethanol Infinitely solubleCH3(CH2)2OH Propanol Infinitely solubleCH3(CH2)3OH Butanol 9CH3(CH2)4OH Pentanol 2.7CH3(CH2)5OH Hexanol 0.6CH3(CH2)6OH Heptanol 0.18CH3(CH2)7OH Octanol 0.054CH3(CH2)9OH Decanol Insoluble in water

Alcohols are classified as either primary (1°), secondary (2°), or tertiary (3°) on thebasis of their structures.

CH3CH2OH CH3CHCH3 CH3CCH3

OH

CH3

AOHA

A

3�2�1�

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Ethanol is a primary alcohol because there is only one alkyl group attached to the carbonthat carries the OOH substituent. The structure of a primary alcohol can be abbreviatedas RCH2OH, where R stands for an alkyl group. The isopropyl alcohol found in rubbingalcohol is a secondary alcohol, which has two alkyl groups on the carbon atom with theOOH substituent (R2CHOH). An example of a tertiary alcohol (R3COH) is tert-butyl (ort-butyl) alcohol or 2-methyl-2-propanol.

Another class of alcohols are the phenols, in which an OOH group is attached to anaromatic ring, as shown in Figure O2.3. Phenols are potent disinfectants. When antiseptictechniques were first introduced in the 1860s by Joseph Lister, it was phenol (or carbolicacid, as it was then known) that was used. Phenol derivatives, such as o-phenylphenol, arestill used in commercial disinfectants such as Lysol.

FIGURE O2.3 The structures of phenol and o-phenylphenol.

OH

Phenol o-Phenylphenol

OH

Water has an unusually high boiling point because of the hydrogen bonds between theH2O molecules. Alcohols can form similar hydrogen bonds, as shown in Figure O2.4.

FIGURE O2.4 A hydrogen bond between a pair of methanol molecules.

H

H

CH3

CH3

,, O

O

As a result, alcohols have boiling points that are much higher than alkanes with similarmolecular weights. The boiling point of ethanol, for example, is 78.5°C, whereas propane,with about the same molecular weight, boils at �42.1°C.

Alcohols are Brønsted acids in aqueous solution.

CH3CH2OH(aq) � H2O(l) 88nm88 H3O�(aq) � CH3CH2O�(aq)

Alcohols therefore react with sodium metal to produce sodium salts of the correspondingconjugate base, as noted in Section O2.2.

2 Na(s) � 2 CH3OH(l) 88n 2 Na�(alc) � 2 CH3O�(alc) � H2(g)

The conjugate base of an alcohol is known as an alkoxide.

[Na�][CH3O�] [Na�][CH3CH2O�]Sodium methoxide Sodium ethoxide

As we saw in Section O1.8, alcohols can be prepared by adding water to an alkene inthe presence of a strong acid such as concentrated sulfuric acid. The reaction involvesadding an H2O molecule across a CPC double bond. Because the reactions followMarkovnikov’s rule, the product of the reaction is often a highly substituted 2° or 3°alcohol.

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Less substituted 1° alcohols can be prepared by substitution reactions that occur when aprimary alkyl halide is allowed to react with the OH� ion.

CH3CH2CH2Br � OH� 88n CH3CH2CH2OH � Br�

Alcohols (ROH) can be thought of as derivatives of water in which one of the hydro-gen atoms has been replaced by an alkyl group. If both of the hydrogen atoms are replacedby alkyl groups, we get an ether (ROR). These compounds are named by adding the wordether to the names of the alkyl groups.

CH3CH2OCH2CH3 diethyl ether

Diethyl ether, often known by the generic name “ether,” was once used extensively as ananesthetic. Because mixtures of diethyl ether and air explode in the presence of a spark,ether has been replaced by safer anesthetics.

There are important differences between both the physical and chemical properties ofalcohols and ethers. Consider diethyl ether and 1-butanol, for example, which are consti-tutional isomers with the formula C4H10O.

CH3CH2OCH2CH3 CH3CH2CH2CH2OHBP � 34.5°C BP � 117.2°C

density � 0.7138 g/mL density � 0.8098 g/mLInsoluble in water Soluble in water

The shapes of the molecules are remarkably similar, as shown in Figure O2.5.

CH3 CHCHP CH3 � H2O CH3CAOH

H CH3CH2H2SO4

FIGURE O2.5 The structures of diethyl ether and 1-butanol.OHO

The fundamental difference between the compounds is the presence of OOH groups inthe alcohol that are missing in the ether. Because hydrogen bonds can’t form between themolecules in the ether, the boiling point of the compound is more than 80°C lower thanthat of the corresponding alcohol. Because there are no hydrogen bonds to organize thestructure of the liquid, the ether is significantly less dense than the corresponding alcohol.

Ethers can act as a hydrogen bond acceptor, as shown in Figure O2.6, but they can’tact as hydrogen bond donors. As a result, ethers are less soluble in water than alcoholswith the same molecular weight.

FIGURE O2.6 Water acting as a hydrogen bond donor toward an ether.CH3

CH3

H

O

O

H

The absence of an OOH group in an ether also has important consequences for itschemical properties. Unlike alcohols, ethers are essentially inert to chemical reactions. They

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don’t react with most oxidizing or reducing agents, and they are stable to most acids andbases, except at high temperatures. They are therefore frequently used as solvents for chem-ical reactions.

Compounds that are potential sources of an H� ion, or proton, are often described asbeing protic. Ethanol, for example, is a protic solvent.

Substances that can’t act as a source of a proton are said to be aprotic. Because they don’tcontain an OOH group, ethers are aprotic solvents.

Ethers can be synthesized by splitting out a molecule of water between two alcoholsin the presence of heat and concentrated sulfuric acid.

They can also be formed by reacting a primary alkyl halide with an alkoxide ion.

O2.5 ALDEHYDES AND KETONESThe connection between the structures of alkenes and alkanes was established in SectionO1.6, which noted that we can transform an alkene into an alkane by adding an H2 mole-cule across the CPC double bond.

The driving force behind the reaction is the difference between the strengths of the bondsthat must be broken and the bonds that form in the reaction. In the course of the hydro-genation reaction, a relatively strong HOH bond (435 kJ/mol) and a moderately strong carbon–carbon � bond (~270 kJ/mol) are broken, but two strong COH bonds (439 kJ/mol)are formed. The reduction of an alkene to an alkane is therefore an exothermic reaction.

What about the addition of an H2 molecule across a CPO double bond?

Once again, a significant amount of energy has to be invested in the reaction to break theHOH bond (435 kJ/mol) and the carbon–oxygen � bond (~375 kJ/mol). The overall re-action is still exothermic, however, because of the strength of the COH bond (439 kJ/mol)and the OOH bond (498 kJ/mol) that are formed.

The addition of hydrogen across a CPO double bond raises several important points.First, and perhaps foremost, it shows the connection between the chemistry of primary al-cohols and aldehydes. But it also helps us understand the origin of the term aldehyde. If a

HH2 C HC

H

H

OH

H

H C HC

H

H

O

�Pt

B

HH2 C HC

H

HH

H

H

H H

H

CCG

G

D

DP �

Ni

CH3CH2CH2 3Br � �CH O 88n CH3CH2CH2 3OCH Br��

2 CH3CH2OH CH3CH2OCH2CH3H�

CH3CH2OH(aq) CH3CH2O�(aq)� �H2O(l) H3O�(aq)

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reduction reaction in which H2 is added across a double bond is an example of a hydro-genation reaction, then an oxidation reaction in which an H2 molecule is removed to forma double bond might be called dehydrogenation. Thus, using the symbol [O] to representan oxidizing agent, we see that the product of the oxidation of a primary alcohol is liter-ally an “al-dehyd” or aldehyde. It is an alcohol that has been dehydrogenated.

The reaction also illustrates the importance of differentiating between primary, sec-ondary, and tertiary alcohols. Consider the oxidation of isopropyl alcohol, or 2-propanol,for example.

The product of this particular reaction was originally called aketone, although the namewas eventually softened to azetone and finally acetone. Thus, it is not surprising that anysubstance that exhibited chemistry that resembled “aketone” became known as a ketone.

Aldehydes can be formed by oxidizing a primary alcohol; oxidation of a secondary al-cohol gives a ketone. What happens when we try to oxidize a tertiary alcohol? The an-swer is simple: Nothing happens.

There aren’t any hydrogen atoms that can be removed from the carbon atom carrying theOOH group in a 3° alcohol. And any oxidizing agent strong enough to insert an oxygenatom into a COC bond would oxidize the alcohol all the way to CO2 and H2O.

A variety of oxidizing agents can be used to transform a secondary alcohol to a ketone.A common reagent for the reaction is some form of chromium(VI)—chromium in the �6oxidation state—in acidic solution. The reagent can be prepared by adding a salt of thechromate (CrO4

2�) or dichromate (Cr2O72�) ion to sulfuric acid. Or it can be made by

adding chromium trioxide (CrO3) to sulfuric acid. Regardless of how it is prepared, the ox-idizing agent in the reactions is chromic acid, H2CrO4.

The choice of oxidizing agents to convert a primary alcohol to an aldehyde is muchmore limited. Most reagents that can oxidize the alcohol to an aldehyde carry the reactionone step further—they oxidize the aldehyde to the corresponding carboxylic acid.

CH3 OHCH2 CH3CH

OBH2CrO4 CH3COH

OB

CHCH3 CH3CH2

OHA

CCH3 CH3CH2

OBH2CrO4

CCH3 CH3

CH3

OHA

A

[O]

CHCH3 CH3 CH3 CH3C

OB

OHA [O]

CH2CH3 OH CH3C

O

HB[O]

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A weaker oxidizing agent, which is just strong enough to prepare the aldehyde from theprimary alcohol, can be obtained by dissolving the complex that forms between CrO3 andpyridine, C6H5N, in a solvent such as dichloromethane that doesn’t contain any water.

The Nomenclature of Aldehydes and Ketones

The common names of aldehydes are derived from the names of the corresponding car-boxylic acids.

The systematic names for aldehydes are obtained by adding -al to the name of the parentalkane.

The presence of substituents is indicated by numbering the parent alkane chain from theend of the molecule that carries the OCHO functional group. For example,

The common name for a ketone is constructed by adding ketone to the names of thetwo alkyl groups on the carbon of the CPO double bond, listed in alphabetical order.

The systematic name is obtained by adding -one to the name of the parent alkane and us-ing numbers to indicate the location of the CPO group.

Common Aldehydes and Ketones

Formaldehyde has a sharp, somewhat unpleasant odor. The aromatic aldehydes in FigureO2.7, on the other hand, have a very pleasant “flavor.” Benzaldehyde has the characteristic

CH3CCH2CH3

BO

2-Butanone

CH3CCH2CH3

BO

Ethyl methyl ketone

BrCH2CH2CHBO

3-Bromopropanal

HCH CH3CHBO

BO

EthanalMethanal

HCOHBO

HCHBO

CH3COH CH3CHBO

BO

Formic acid Formaldehyde

AcetaldehydeAcetic acid

CH3 OHCH2CrO3/pyridine

CH3CH

OB

CH2Cl2

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odor of almonds, vanillin is responsible for the flavor of vanilla, and cinnamaldehyde makesan important contribution to the flavor of cinnamon.

FIGURE O2.7 Aromatic alde-hydes with characteristic odors.

CH

O

Benzaldehyde Vanillin CinnamaldehydeCH3O

HOB

CH

OB

CH

OB

O2.6 REACTIONS AT THE CARBONYL GROUPIt is somewhat misleading to write the carbonyl group as a covalent CPO double bond.The difference between the electronegativities of carbon and oxygen is large enough tomake the CPO bond moderately polar. As a result, the carbonyl group is best describedas a hybrid of the following resonance structures.

We can represent the polar nature of the hybrid by indicating the presence of a slight neg-ative charge on the oxygen (��) and a slight positive charge (��) on the carbon of theCPO double bond.

Reagents that attack the electron-rich �� end of the CPO bond are called electrophiles(literally, “lovers of electrons”). Electrophiles include ions (such as H� and Fe3�) and neu-tral molecules (such as AlCl3 and BF3) that are Lewis acids, or electron pair acceptors.Reagents that attack the electron-poor �� end of the bond are nucleophiles (literally,“lovers of nuclei”). Nucleophiles are Lewis bases (such as NH3 or the OH�ion).

The polarity of the CPO double bond can be used to explain the reactions of carbonylcompounds. Aldehydes and ketones react with a source of the hydride ion (H�) becausethe H� ion is a Lewis base, or nucleophile, that attacks the �� end of the CPO bond.When this happens, the two valence electrons on the H� ion form a covalent bond to thecarbon atom. Since carbon is tetravalent, one pair of electrons in the CPO bond is dis-placed onto the oxygen to form an intermediate with a negative charge on the oxygen atom.

C

OB

GD

OS Electrophiles

(H�, Fe3�, BF3, etc.)

Nucleophiles(OH�, NH3, etc.)

C

OOOB

GD

��

��

C

O OSOSOS

BCA

GD GD�

�

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The alkoxide ion can then remove an H� ion from water to form an alcohol.

Common sources of the H� ion include lithium aluminum hydride (LiAlH4) and sodiumborohydride (NaBH4). Both compounds are ionic.

LiAlH4: [Li�][AlH4�]

NaBH4: [Na�][BH4�]

The aluminum hydride (AlH4�) and borohydride (BH4

�) ions act as if they were com-plexes between an H� ion, acting as a Lewis base, and neutral AlH3 or BH3 molecules,acting as a Lewis acid.

LiAlH4 is such a good source of the H� ion that it reacts with the H� ions in water orother protic solvents to form H2 gas. The first step in the reduction of a carbonyl withLiAlH4 is therefore carried out using an ether as the solvent. The product of the hydridereduction reaction is then allowed to react with water in a second step to form the corre-sponding alcohol.

NaBH4 is less reactive toward protic solvents, which means that borohydride reductionsare usually done in a single step, using an alcohol as the solvent.

O2.7 CARBOXYLIC ACIDS AND CARBOXYLATE IONSWhen one of the substituents on a carbonyl group is an OH group, the compound is a car-boxylic acid with the generic formula RCO2H. These compounds are acids, as the namesuggests, which form carboxylate ions (RCO2

�) by the loss of an H� ion.

CH2 CH2CHCH3 CH3C CH3 CH3

OB

OHANaBH4

CH3CH2OH

CH2 CH2CH2CH3 CH OHCH3

OB 1. LiAlH4

in ether

2. H2O

H

H

H

S�

�

Al HOA

A

H

H

AlH HOOA

A

H

O

CCH3 CH3

O

O OA

ASO

H

O�

CCH3 CH3 H2OOOA

ASOS

� �

H

OH�

C

OB

GD

OS

HH

O

S�

�

CH3 CH3

CCH3 CH3OOA

ASOS

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The carboxylate ion formed in the reaction is a hybrid of two resonance structures.

Resonance delocalizes the negative charge in the carboxylate ion, which makes the ionmore stable than the alkoxide ion formed when an alcohol loses an H� ion. By increasingthe stability of the conjugate base, resonance increases the acidity of the acid that formsthe base. Carboxylic acids are therefore much stronger acids than the analogous alcohols.The Ka value for a typical carboxylic acid is about 10�5, whereas alcohols have Ka valuesonly 10�16.

Carboxylic acids were among the first organic compounds to be discovered. As a re-sult, they have well-established common names that are often derived from the Latin stemsof their sources in nature. Formic acid (Latin formica, “an ant”) and acetic acid (Latin ace-tum, “vinegar”) were first obtained by distilling ants and vinegar, respectively. Butyric acid(Latin butyrum, “butter”) is found in rancid butter, and caproic, caprylic, and capric acids(Latin caper, “goat”) are all obtained from goat fat. A list of common carboxylic acids isgiven in Table O2.5.

The systematic nomenclature of carboxylic acids is easy to understand. The ending -oic acid is added to the name of the parent alkane to indicate the presence of the OCO2Hfunctional group.

HCO2H Methanoic acidCH3CO2H Ethanoic acidCH3CH2CO2H Propanoic acid

Unfortunately, because of the long history of carboxylic acids and their importance in bi-ology and biochemistry, you are more likely to encounter these compounds by their com-mon names.

Formic acid and acetic acid have sharp, pungent odors. As the length of the alkyl chainincreases, the odor of carboxylic acids becomes more unpleasant. Butyric acid, for exam-ple, is found in sweat, and the odor of rancid meat is due to carboxylic acids released asthe meat spoils.

The solubility data in Table O2.5 show that carboxylic acids also become less solublein water as the length of the alkyl chain increases. The OCO2H end of the molecule is po-lar and therefore soluble in water. As the alkyl chain gets longer, the molecule becomesmore nonpolar and less soluble in water.

CH3C CH3CO

OHG

JH��

O

O�G

J

CH3OH CH3O� H��

Ka � 1.8 10�5

Ka � 8 10�16

CH3 CO

OG

JCH3 C

O

O�

O�

OM

D

CH3 CO

OHO

GCH3 H��C

O

O�

OG

J J

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Compounds that contain two OCO2H functional groups are known as dicarboxylicacids. A number of dicarboxylic acids (see Table O2.6) can be isolated from natural sources.Tartaric acid, for example, is a by-product of the fermentation of wine, and succinic, fu-maric, malic, and oxaloacetic acids are intermediates in the metabolic pathway used to ox-idize sugars to CO2 and H2O.

TABLE O2.5 Common Carboxylic Acids

Solubility in H2OCommon Name Formula (g/100 mL)

Saturated carboxylic acids and fatty acids

Formic acid HCO2H �

Acetic acid CH3CO2H �

Propionic acid CH3CH2CO2H �

Butyric acid CH3(CH2)2CO2H �

Caproic acid CH3(CH2)4CO2H 0.968Caprylic acid CH3(CH2)6CO2H 0.068Capric acid CH3(CH2)8CO2H 0.015Lauric acid CH3(CH2)10CO2H 0.0055Myristic acid CH3(CH2)12CO2H 0.0020Palmitic acid CH3(CH2)14CO2H 0.00072Stearic acid CH3(CH2)16CO2H 0.00029

Unsaturated fatty acids

Palmitoleic acid CH3(CH2)5CHPCH(CH2)7CO2H —Oleic acid CH3(CH2)7CHPCH(CH2)7CO2H —Linoleic acid CH3(CH2)4CHPCHCH2CHPCH(CH2)7CO2H —Linolenic acid CH3CH2CHPCHCH2CHPCHCH2CHPCH(CH2)7CO2H —

HO2CCO2H Oxalic acidHO2CCH2CO2H Malonic acid

Maleic acid

Fumaric acid

HO2CCH2CH2CO2H Succinic acidHO2C CO2H

C CH H

OHA

HO2CCH2CHCO2H Malic acidOHOHA A

HO2CCHCHCO2H Tartaric acidOB

HO2CCH2CCO2H Oxaloacetic acid

Structure Common Name Structure Common Name

TABLE 02.6 Common Dicarboxylic Acids

GGDD

P

HO2C

CO2HC C

H

HG

GDD

P

Several tricarboxylic acids also play an important role in the metabolism of sugar. Themost important example of this class of compounds is the citric acid that gives so manyfruit juices their characteristic acidity.

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O2.8 ESTERSCarboxylic acids (OCO2H) can react with alcohols (ROH) in the presence of either acidor base to form esters (OCO2R). Acetic acid, for example, reacts with ethanol to formethyl acetate.

This isn’t an efficient way of preparing an ester, however, because the equilibrium constantfor the reaction is relatively small (Kc � 3). Chemists therefore tend to synthesize estersfrom carboxylic acids in a two-step process. They start by reacting the acid with a chlori-nating agent such as thionyl chloride (SOCl2) to form the corresponding acyl chloride.

They then react the acyl chloride with an alcohol in the presence of base to form the ester.

The base absorbs the HCl given off in the reaction, thereby driving it to completion.As might be expected, esters are named as if they were derivatives of a carboxylic acid

and an alcohol. The ending -ate or -oate is added to the name of the parent carboxylic acid,and the alcohol is identified using the “alkyl alcohol” convention. The following ester, forexample, can be named as a derivative of acetic acid (CH3CO2H) and ethyl alcohol(CH3CH2OH).

Or it can be named as a derivative of ethanoic acid (CH3CO2H) and ethyl alcohol(CH3CH2OH).

The term ester is commonly used to describe the product of the reaction of any strongacid with an alcohol. Sulfuric acid, for example, reacts with methanol to form a diesterknown as dimethyl sulfate.

CH3 CH3CH2CO

OB

Ethyl ethanoate

CH3 CH3CH2CO

OB

Ethyl acetate

CH3CCl � �CH3CH2OHB

CH3CBO

BO

OCH2CH BH�Cl�3

CH3CBO

OH � � �SOCl2 CH3CBO

Cl SO2 HCl

CH3CBO

OH � �CH3CH2OH CH3CBO

OCH2CH3H �

H2O

HOCACH2CO2H

ACH2CO2H

OCO2H Citric acid

1012T_mod05_1-36 1/25/05 22:24 Page 23 EQA


Phosphoric acid reacts with alcohols to form triesters such as trimethyl phosphate.

Compounds that contain the OCO2R functional group might therefore best be called car-boxylic acid esters, to indicate the acid from which they are formed.

Carboxylic acid esters with low molecular weights are colorless, volatile liquids that of-ten have a pleasant odor. They are important components of both natural and syntheticflavors (see Figure O2.8).

3 CH3OHHO P OH

OH

O

�A

A A3 H2OCH3O P OCH3

OCH3

O

�A

2 CH3OHHO S OH

O

O

� 2 H2OCH3O S OCH3

O

O

�A

A A

A

FIGURE O2.8 Typical carboxylic acid esters with pleasant odors or flavors.

Methyl salicylate(oil of wintergreen)

COB

B

CH3O

CH3 CH3

CH3

CH2

CH2 CH2

CH2

C CH

O

OIsoamyl butyrate

(chocolate)

BCH3

CH3

CH3 CH2

CH2

C CH

O

OIsoamyl acetate

(apple)

BCH3 CH2

CH2 CH3

CH2

C

O

OEthyl butyrate

(pineapple)

EOH

O2.9 AMINES, ALKALOIDS, AND AMIDESAmines are derivatives of ammonia in which one or more hydrogen atoms are replaced byalkyl groups. We indicate the degree of substitution by labeling the amine as either pri-mary (RNH2), secondary (R2NH), or tertiary (R3N). The common names of the compoundsare derived from the names of the alkyl groups.

The systematic names of primary amines are derived from the name of the parent alkaneby adding the prefix amino- and a number specifying the carbon that carries the ONH2

group.

CH

CH3 CH2

NH2

CH3

CH

CHP

OG

G

D D5-Amino-2-hexene

(CH3)2CHNH2 CH3CH2NH CH3 3NCHIsopropylamine Ethylmethylamine Trimethylamine

CH3A

CH3A

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The chemistry of amines mirrors the chemistry of ammonia. Amines are weak basesthat pick up a proton to form ammonium salts. Trimethylamine, for example, reacts withacid to form the trimethylammonium ion.

The salts are more soluble in water than the corresponding amines, and the reaction canbe used to dissolve otherwise insoluble amines in aqueous solution.

The difference between amines and their ammonium salts plays an important role inboth over-the-counter and illicit drugs. Cocaine, for example, is commonly sold as the hy-drochloride salt, which is a white, crystalline solid. By extracting the solid into ether, it ispossible to obtain the “free base.” A glance at the side panel of almost any over-the-counter medicine will provide examples of ammonium salts of amines that are used to en-sure that the drug dissolves in water (see Figure O2.9).

CH3 H��

CH3

CH3

NOA

AS CH3 H

CH3

CH3

NO OA

A

�

FIGURE O2.9 Over-the-counter drugs sold as amine hydrochloride or hydrobromide salts.

CH3

CH3

CH3

NH2�Cl�CH3O

HBr�N

Dextromethorphanhydrobromide

(cough suppressant)Pseudoephedrine

hydrochloride(decongestant)

CH

CH

HO�

Amines that are isolated from natural sources, especially from plants, are known as al-kaloids. They include poisons such as nicotine, coniine, and strychnine (shown in FigureO2.10). Nicotine has a pleasant, invigorating effect when taken in minuscule quantities butis extremely toxic in larger amounts. Coniine is the active ingredient in hemlock, a poisonthat has been used since the time of Socrates. Strychnine is another toxic alkaloid that hasbeen a popular poison in murder mysteries.

FIGURE O2.10 The alkaloids are complex amines that canbe isolated from plants.

CH3

Nicotine

Strychnine

Coniine

N

N

N

H

O

O

N

N

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FIGURE O2.11 The structures of some alkaloids used as drugs.

Morphine

Cocaine

Quinine

Lysergic aciddiethylamide

(LSD)

CH3

N

NH

N

O

HO

HO

O

CH3CO

CH3CO

CO2CH3

C

O

Heroin

CH3N

O

O

OC

P

O

O

O

O

N

HOCH

CH CH2

CH3O

N

O

N

NCH3CH2

CH3CH2

CH3

HB

B

E

E E

O

HCH3

OH

H

H

A

E

P

HE

O

A

B

The alkaloids also include a number of drugs, such as morphine, quinine, and cocaine.Morphine is obtained from poppies; quinine can be found in the bark of the chinchonatree; and cocaine is isolated from coca leaves. This family of compounds also includes syn-thetic analogs of naturally occurring alkaloids, such as heroin and LSD (see Figure O2.11).

To illustrate the importance of minor changes in the structure of a molecule, threeamines with similar structures are shown in Figure O2.12. One of the compounds is caf-feine, which is the pleasantly addictive substance that makes a cup of coffee an importantpart of the day for so many people. Another is theobromine, which is the pleasantly ad-dictive substance that draws so many people to chocolate. The third compound is theo-phylline, which is a prescription drug used as a bronchodilator for people with asthma.

FIGURE O2.12 The caffeine in coffee, the theobromine in chocolate, and the active in-gredient in a common bronchodilator are central nervous system stimulants with verysimilar structures.

Caffeine

CH3

CH3

CH3O

O N N

NN

CH3

Theobromine Theophylline

H

CH3

O

O N N

NN

H

CH3

CH3

O

O N N

NNH f

A A A

fH H i

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It is tempting to assume that carboxylic acids will react with amines to form the classof compounds known as amides. In practice, when aqueous solutions of carboxylic acidsand amines are mixed all we get is an acid–base reaction.

The best way to prepare an amide is to react the appropriate acyl chloride with an amine.

Excess amine is used to drive the reaction to completion by absorbing the HCl given offin the reaction.

O2.10 GRIGNARD REAGENTSSo far, we have built a small repertoire of reactions that can be used to convert one func-tional group to another. We have briefly discussed converting alkenes to alkanes; alkanesto alkyl halides; alkyl halides to alcohols; alcohols to ethers, aldehydes, and ketones; andaldehydes to carboxylic acids. We have also shown how carboxylic acids can be convertedto esters and amides. We have yet to encounter a reaction, however, that addresses a basicquestion: How do we make COC bonds? One answer resulted from the work that FrancoisAuguste Victor Grignard started as part of his Ph.D. research at the turn of the twentiethcentury.

Grignard noted that alkyl halides react with magnesium metal in diethyl ether (Et2O)to form compounds that contain a metal–carbon bond. Methyl bromide, for example, formsmethylmagnesium bromide.

Because carbon is considerably more electronegative than magnesium, the metal–carbonbond in the product has a significant amount of ionic character. Grignard reagents such asCH3MgBr are best thought of as hybrids of ionic and covalent Lewis structures.

Grignard reagents are therefore a potential source of carbanions (literally, “anions ofcarbon”). The Lewis structure of the CH3

� ion suggests that carbanions can be Lewis bases,or electron pair donors.

H

H

H

COA

AS

�

CH3OMgOBr [CH3 ][Mg2�][Br�]�S

CH3Br � Mg 88nEt2O CH3MgBr

CH3CBO

Cl � �2 CH3NH2 CH3CBO

NHCH3 CH3NH3�Cl�

CH3CBO

OH � CH3NH2 CH3CBO

O�CH3NH3�

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Grignard reagents such as methylmagnesium bromide are therefore sources of a nucle-ophile that can attack the �� end of the CPO double bond in aldehydes and ketones.

If we treat the product of the reaction with water, we get a tertiary alcohol.

If we wanted to make a secondary alcohol, we could add the Grignard reagent to an alde-hyde, instead of a ketone.

By reacting a Grignard reagent with formaldehyde we can add a single carbon atom toform a primary alcohol.

The alcohol can then be oxidized to the corresponding aldehyde.

The Grignard reagent therefore provides us with a way of performing the following over-all transformation.

A single carbon atom can also be added if the Grignard reagent is allowed to reactwith CO2 to form a carboxylic acid.

Perhaps the most important aspect of the chemistry of Grignard reagents is the easewith which the reaction allows us to couple alkyl chains. Isopropylmagnesium bromide, for

CH3 CH2CACH3

ACH3

MgBr � �CO2 CH3CACH O3

ACH3

CH2CO� �H2OBCH3C

ACH O3

ACH3

CH2COHB

OH

CH3CACH3

ACH3

CH2MgBr3. CrO in CH2Cl2

3 /pyridineCH3 CH2C

ACH3

ACH3

CH

OB1. H2CO

2. H2O

CH3CACH3

ACH3

CH2CH2OHCrO3 /pyridine

CH3 CH2CACH3

ACH3

CH

OB

CH2Cl2

CH3 CH2CACH3

ACH3

MgBr � H2CPO CH3CACH3

ACH3

CH2CH2O� H2OCH3C

ACH3

ACH3

CH2CH2OH

CH3CBO

H1. CH3MgBr

2. H2OCH3C

AOH

HCH3

CH3OCA

ACH3

CH3

OO� � �H2O CH3OCACH3

ACH3

OOH OH�

AO

G

DOCCH3

�

CH3 AOCH3

CH3

�OPC

CH3CH3

S

1012T_mod05_1-36 1/22/05 11:32 Page 28 EQA


example, can be used to graft an isopropyl group onto the hydrocarbon chain of an ap-propriate ketone, as shown in Figure O2.13.

FIGURE O2.13 Grignard reagents provide a powerful way to introduce alkyl groups onto a hydrocarbon chain.

CHMgBr CH3CH2CCH3

CH3

CH3

�

CH3CH2CCH3

CH3 CH3

�

CH

O

CH3CH2CCH3

CH3 CH3

H

CH

H2O

EG AB

A

A

A

GG DD


The Chemistry of Garlic

The volatile materials that can be distilled from plants were named essential oils byParacelsus in the sixteenth century because they were thought to be the quintessence(literally, the “fifth essence,” or vital principle) responsible for the odor and flavor of theplants from which they were isolated.

The Egyptians extracted essential oils from fragrant herbs more than 5000 years agoby pressing the herb or by extracting the fragrant material with olive or palm oil. Someessential oils are still obtained by pressing. Others are extracted into a nonpolar solvent.The most common method for isolating essential oils, however, is steam distillation. Theground botanical is immersed in water which is heated to boiling, or boiling water is al-lowed to pass through a sample of the ground botanical. The oil and water vapor passinto a condenser, where the oil separates from the water vapor.

The function of the essential oils isn’t fully understood. Some act as attractants forthe insects involved in pollination. Most are either bacteriostats (which stop the growthof bacteria) or bactericides (which kill bacteria). In some cases, they can be a source ofmetabolic energy. In other cases, they appear to be waste products of plant metabolism.

The essential oils are mixtures of up to 200 organic compounds, many of which areeither terpenes (with 10 carbon atoms) or sesquiterpenes (with 15 carbon atoms). Al-though the three components shown in Figure O2.14 represent almost 60% of the massof a sample of rose oil, 50 other components of that essential oil have been identified.

FIGURE O2.14 The three most abun-dant components of rose oil.

CH2OH

CH2OH

Citronellol(3-7-dimethyl-6-octen-1-ol)

Geraniol(E-3,7-dimethyl-

2,6-octadien-1-ol)

Neral(Z-3,7-dimethyl-2,6-octadienal)

C O

HAP

Garlic, onions, and mustard seed differ from most other sources of essential oils.In each case, the fragrance producing part of the plant must be crushed before thevolatile components are released. For more than 100 years, chemists have knownthat the principal component of the oil that distills from garlic is diallyl disulfide [F. W. Semmler, Archiv der Pharmazie, 230, 434–448 (1892)].

CH2PCHCH2SSCH2CHPCH2 Diallyl disulfide

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Only recently, however, have chemists explained how the compound is produced when aclove of garlic is crushed [E. Block, Angewandte Chemie, International Edition in Eng-lish, 31, 1101–1264 (1992)].

Before garlic is crushed, the intact cell contains S–2-propenyl-L-cysteine S-oxide—oralliin—which can be found in the cell cytoplasm.

Within the cell there are vacuoles that contain an enzyme known as alliinase. When thecell is crushed, the enzyme is released. The enzyme transforms the natural product alliininto an intermediate that reacts with itself to form a compound known as allicin.

Allicin has been described as an odoriferous, unstable antibacterial substance that poly-merizes easily and must be stored at low temperatures. When heated, it breaks down togive a variety of compounds, including the diallyl disulfide obtained when oil of garlic isdistilled from the raw material.

An alliinase enzyme can also be found in onions, where it converts an isomer of al-liin known as S-(E)-1-propenyl-L-cysteine S-oxide into propanethial S-oxide.

The product of the reaction is known as the lacrimator factor of onion because it is thesubstance primarily responsible for the tears generated when onions are cut.

A great deal of progress has been made in recent years in identifying the variousorganosulfur compounds formed when garlic and onion are cut and in understanding theprocess by which the compounds are formed. The structures of some of the principalorganosulfur compounds associated with garlic are shown in Figure O2.15. In spite ofthe progress made so far, much still has to be learned about the compounds that can beisolated from extracts of the genus Allium, which includes both garlic and onion.

CH3CH2CHPS�OO�Propanethial S-oxide

CH2PCHCH2SBO

SCH2CHPCH2 Allicin

CH2PCHCH2SBO

CH2CANH3

�

HCO2�

Alliin

SOH

O

SS

S

S O

S

O

S

SS

SS

S

SS

S

S

S

SS

SS

SS

S

SS

S

S

S

S

SS

HOS

FIGURE O2.15 Some of the organosulfur compounds associated with garlic.

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KEY TERMS

Acyl chlorideAlcoholAldehydeAlkaloidAlkoxideAmideAmineAproticCarbanionCarbonyl

Carboxylate ionCarboxylic acidCarboxylic acid esterChain initiationChain propagationChain reactionChain terminationDicarboxylic acidElectrophileEster

EtherFree radicalFunctional groupGrignard reagentHydrophilicHydrophobicKetoneNucleophileOxidation numberProtic

PROBLEMSFunctional Groups

1. Give examples of compounds that contain each of the following functional groups.(a) an alcohol (b) an aldehyde (c) an amine (d) an amide (e) an alkyl halide(f) an alkene (g) an alkyne

2. Describe the difference between the members of each of the following pairs of func-tional groups.(a) an alcohol and an alkoxide ion(b) an alcohol and an aldehyde(c) an amine and an amide(d) an ether and an ester(e) an aldehyde and a ketone

3. Classify the following compounds as an alcohol, an aldehyde, an ether, or a ketone.O OB B

(a) CH3CH2CH (b) CH3CH2OH (c) CH3CH2OCH2CH3 (d) CH3CH2CCH3

4. Classify the following compounds as primary, secondary, or tertiary alcohols.CH3 CH3 OHA A A

(a) CH3CH2OH (b) CH3CH2CHOH (c) CH3CH2COH (d) CH3CH2CHCH2CH3ACH3

5. Cortisone is an adrenocortical steroid that is used as an anti-inflammatory agent.Use the fact that carbon is tetravalent to determine the molecular formula of the com-pound from the following stick structure. Identify the functional groups present in themolecule.

J

OP

A

A

A

AM

C OOHO

O

CH2OH

CH3

CH3

1012T_mod05_1-36 1/22/05 11:32 Page 31 EQA


6. Piperine {(E,E)-1-[5-(1,3-benzodioxol-5-yl)-1-oxo-2,4-pentadienyl]piperidine} can beextracted from black pepper. Identify the functional groups in the structure of thecompound.

7. PGE2 is a member of the family of compounds known as prostaglandins, which havevery significant physiological effects in even small quantities. They can affect bloodpressure, heart rate, body temperature, blood clotting, and conception. Some induceinflammation; others relieve it. Aspirin, which is both an anti-inflammatory and an an-tipyretic (fever-reducing) drug, acts by blocking the synthesis of prostaglandins. Iden-tify the functional groups in the structure of the PGE2 molecule and calculate its mol-ecular formula.

8. Pseudoephedrine hydrochloride is the active ingredient in a variety of decongestants,including Sudafed. Use the structure of the compound shown in Figure O2.9 to iden-tify the functional groups in the molecule.

9. Cocaine was once used in Coca Cola. Quinine is still added to tonic water. Use thestructures of the alkaloids shown in Figure O2.11 to identify the functional groups inthe compounds.

OXIDATION–REDUCTION REACTIONS

10. Arrange the following substances in order of increasing oxidation number of the car-bon atom.(a) C (b) HCHO (c) HCO2H (d) CO (e) CO2 (f) CH4 (g) CH3OH

11. Classify the following reactions as examples of either metathesis or oxidation–reductionreactions.(a) CH4 � Cl2 n CH3Cl � HCl(b) CH3OH n HCHO(c) HCHO n HCO2H(d) CH3OH � HBr n CH3Br � H2O(e) (CH3)2CO n (CH3)2CHOH

12. Classify the following transformations as examples of either oxidation or reduction.(a) CH3CH2OH n CH3CHO(b) CH3CHO n CH3CO2H(c) (CH3)2CPO n (CH3)2CHOH(d) CH3CHPCHCH3 n CH3CH2CH2CH3

(e) (CH3)2CHCqCH n (CH3)2CHCH2CH3

O

OH

H

OH

CO2

N

O

O

O

B

1012T_mod05_1-36 1/22/05 11:32 Page 32 EQA


13. Which of the following compounds can be oxidized to form an aldehyde?(a) CH3CH2OH (b) CH3CHOHCH3 (c) CH3OCH3 (d) (CH3)2CPO

14. Which of the following compounds should be the most difficult to oxidize?(a) CH3CH2OH (b) (CH3)2CHOH (c) (CH3)3COH (d) CH3CHO

15. Which of the following compounds can be prepared by reducing CH3CHO?(a) CH3CH3 (b) CH3CH2OH (c) CH3CO2H (d) CH3CH2CO2H

16. Which of the following compounds can be prepared by oxidizing CH3CHO?(a) CH3CH3 (b) CH3CH2OH (c) CH3CO2H (d) CH3CH2CO2H

17. Predict the product of the oxidation of 2-methyl-3-pentanol.18. Predict the product of the reduction of 2-methyl-2-pentene with hydrogen gas over a

nickel metal catalyst.

Alkyl Halides

19. Use Lewis structures to describe the free radical chain reaction mechanism involvedin the bromination of methane to form methyl bromide. Clearly label the chain initi-ation, chain propagation, and chain termination steps.

20. Use Lewis structures to describe the difference between a CH3� ion, a CH3� radical,

and a CH3� ion. Which of these Lewis structures describes a carbanion? Which describes

a carbocation?21. How many different products could be formed by the free radical chlorination of pen-

tane? If attack at the different hydrogen atoms in the compound were more or lessrandom, what would be the relative abundance of the different isomers of chloropen-tane formed in the reaction?

22. Consider the reaction between a Cl� atom and a pentane molecule. Classify the dif-ferent intermediates that could be produced in the reaction as either primary, secondary,or tertiary free radicals.

Alcohols and Ethers

23. Describe the differences between the structures of water, methyl alcohol, and dimethylether.

24. Draw the structures of the seven constitutional isomers that have the formula C4H10O.Classify the isomers as either alcohols or ethers.

25. Ethyl alcohol and dimethyl ether have the same chemical formula, C2H6O. Explainwhy one of the compounds reacts rapidly with sodium metal but the other does not.

26. Predict the product of the dehydration of ethyl alcohol with sulfuric acid at 130°C.27. Use examples to explain why it is possible to oxidize either a primary or secondary al-

cohol, but not a tertiary alcohol.28. Draw the structure of o-phenylphenol, the active ingredient in Lysol.29. Explain why alcohols are Brønsted acids in water.30. Write the structures of the conjugate bases formed when the following alcohols act as

a Brønsted acid.(a) CH3CH2OH (ethyl alcohol) (b) C6H5OH (phenol)(c) (CH3)2CHOH (isopropyl alcohol)

31. Use the fact that there are no hydrogen bonds between ether molecules to explain whydiethyl ether (C4H10O) boils at 34.5°C, whereas its constitutional isomer—1-butanol (C4H10O)—boils at 118°C.

1012T_mod05_1-36 1/22/05 11:32 Page 33 EQA


32. Assign a systematic name to the following alcohols.

33. Assign a systematic name to the compound known by the common name menthol.

Aldehydes and Ketones

34. At which end of a carbonyl group will each of the following substances react?(a) H� (b) H� (c) OH� (d) NH3 (e) BF3

35. Explain why mild oxidation of a primary alcohol gives an aldehyde, whereas oxidationof a secondary alcohol gives a ketone.

36. Explain why strong oxidizing agents can’t be used to convert a primary alcohol to analdehyde.

37. Assign systematic names to the following compounds.

38. Assign a systematic name to the compound known by the common name geranial.

Carboxylic Acids, Carboxylate Ions, and Carboxylic Acid Esters

39. Explain the difference between a carboxylic acid, a carboxylate ion, and a carboxylicacid ester.

40. What major differences between carboxylic acids (such as butyric acid,CH3CH2CH2CO2H) and esters (such as ethyl butyrate, CH3CH2CH2CO2CH2CH3)help explain why butyric acid gives rise to the odor of rotten meat but ethyl butyrategives rise to the pleasant odor of pineapple?

41. Explain why fats, which are esters of long-chain carboxylic acids, are insoluble in wa-ter. Explain what happens during saponification (see Chapter 8) that makes the re-sulting compounds marginally soluble in water.

O

A

O

P

B

O

H

P

OH

OH

OH

BA

1012T_mod05_1-36 1/22/05 11:32 Page 34 EQA


42. Explain the difference between CH3CH2OH and CH3CO2H that makes one of theCOOH bonds in the compounds more than 10 orders of magnitude more acidic thanthe other when values of Ka for the compounds are compared.

43. Use the structures shown in Figure O2.11 to propose a series of reactions that couldbe used to convert morphine into heroin.

Amines, Alkaloids, and Amides

44. Classify each of the compounds in Figures O2.10 and O2.11 as either a primary amine,a secondary amine, a tertiary amine, and/or an amide.

45. Explain why reacting a complex amine, such as pseudoephedrine, with an acid makesthe compounds more soluble in water.

46. Draw the structure of caffeine and label each of the nitrogen atoms in the compoundas either an amine or an amide.

Grignard Reagents

47. Use Lewis structures to decide whether the organic component of a Grignard reagentis a nucleophile (Lewis base) or an electrophile (Lewis acid). Predict which end of aCPO bond this portion of the Grignard reagent will attack.

48. If CH3CH2MgBr can be thought of as containing the CH3CH2�, Mg2�, and Br� ions,

what would be the product of the reaction between the Grignard reagent and dilutehydrochloric acid?

49. Explain why Grignard reagents are prepared in an aprotic solvent, such as an ether,rather than a protic solvent such as an alcohol.

50. Write a step-by-step mechanism for the reaction between CH3MgBr and H2CPO toform CH3CH2OH.

51. Predict the product of the reaction between (CH3)2CHMgBr and 2-pentanone.O

52. If CH3CH2CB

CH3 is used as the starting material, show how a tertiary alcohol can besynthesized by a Grignard reaction.

53. Select from the reagents given below to design a sequence of reactions that would leadto the following compound.

(a) (CH3)2CHMgBr (b) CH3CHO (c) CH3MgBr (d) CH3COCH2CH3

(e) CH3COCH3 (f) CH3CH2MgBr (g) CH3COCH3

Qualitative Organic Analysis

54. Describe a way of determining whether a compound is an alkane or an alkene.55. Describe a way of determining whether a compound is a carboxylic acid or an ester.56. Describe a way of determining whether a compound is an alcohol or an ether.57. Describe a way of determining whether a compound is a primary or a tertiary alcohol.

CH3CAOH

HCACH3

HCH3

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58. Describe a way of determining whether a compound is an alcohol or an alkyl halide.59. Describe a way of determining whether a compound is a carboxylic acid or an amide.60. Describe a way of determining whether a compound is an alcohol or an amine.61. Describe a way of determining whether a compound is an aldehyde or ketone.

Integrated Problems

62. Identify the Brønsted acids and the Brønsted bases in the following reaction.

CH3CqCH � NH2� 88n CH3CqC� � NH3

63. Succinic acid plays an important role in the Krebs cycle, malic acid (apple acid) is foundin apples, and tartaric acid ( fruit acid) is found in many fruits. Which of these dicar-boxylic acids is chiral? See Table O2.6.

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1

M O D U L E

6ORGANIC CHEMISTRY:

REACTION MECHANISMS

O3.1 Acids and Bases

O3.2 Carbanion Attack at a Carbonyl Group

O3.3 The Mechanism of Reduction Reactions

O3.4 Nucleophilic Attack by Water

O3.5 Nucleophilic Attack by an Alcohol

O3.6 Addition/Elimination Reactions of Carboxylic Acid Derivatives

O3.7 Free Radical Reactions

O3.8 Bimolecular Nucleophilic Substitution or SN2 Reactions

O3.9 Unimolecular Nucleophilic Substitution or SN1 Reactions

O3.10 Elimination Reactions

O3.11 Substitution versus Elimination Reactions

O3.1 ACIDS AND BASESAs noted in Section 11.1, for more than 300 years, chemists have classified substances thatbehave like vinegar as acids, while those that have properties like the ash from a wood firehave been called alkalies or bases. Today, when chemists use the words “acid” or “base”they refer to a model developed independently by Brønsted, Lowry, and Bjerrum. Sincethe most explicit statement of this theory was contained in the writings of Brønsted, it ismost commonly known as the “Brønsted acid–base” theory.

Brønsted Acid–Base Theory

Brønsted argued that all acid–base reactions involve the transfer of an H� ion, or proton.Water reacts with itself, for example, by transferring an H� ion from one molecule to an-other to form an H3O� ion and an OH� ion.

H2O H2O OH�� H3O�

�H�

�H�

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2 ORGANIC REACTION MECHANISMS

According to the theory, a Brønsted acid is a proton donor and a Brønsted base is a pro-ton acceptor.

Acids are often divided into categories such as “strong” and “weak.” One measure ofthe strength of an acid is the acid dissociation equilibrium constant, Ka, for that acid.

Ka �

When Ka is relatively large, we have a strong acid.

HCl Ka � 1 � 106

When it is small, we have a weak acid.

CH3CO2H Ka � 1.8 � 10�5

When it is very small, we have a very weak acid.

H2O Ka � 1.8 � 10�16

In 1909, S. P. L. Sørenson suggested that the enormous range of concentrations of theH3O� and OH� ions in aqueous solutions could be compressed into a more manageableset of data by taking advantage of logarithmic mathematics and calculating the pH or pOHof the solution.

pH � �log[H3O�]pOH � �log[OH�]

The “p” in pH and pOH is an operator that indicates that the negative of the logarithmshould be calculated for any quantity to which it is attached. Thus, pKa is the negative ofthe logarithm of the acid dissociation equilibrium constant.

pKa � �log Ka

The only disadvantage of using pKa as a measure of the relative strengths of acids is thefact that large numbers now describe weak acids, and small (negative) numbers describestrong acids.

HCl pKa � �6CH3CO2H pKa � 4.7

H2O pKa � 15.7

An important feature of the Brønsted theory is the relationship it creates between acidsand bases. Every Brønsted acid has a conjugate base, and vice versa.

HCl H2O Cl�� H3O�

BaseAcid BaseAcid

[H3O�][A�]��

[HA]

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ORGANIC REACTION MECHANISMS 3

Just as the magnitude of Ka is a measure of the strength of an acid, the value of Kb reflectsthe strength of its conjugate base. Consider what happens when we multiply the Ka ex-pression for a generic acid (HA) by the Kb expression for its conjugate base (A�).

� � [H3O�][OH�]

If we now replace each term in the equation by the appropriate equilibrium constant, weget the following result.

KaKb � Kw � 1 � 10�14

Because the product of Ka times Kb is a relatively small number, either the acid or its con-jugate base can be “strong.” But if one is strong, the other must be weak. Thus, a strongacid must have a weak conjugate base.

HCl � H2O 88n H3O+ � Cl�

Strong Weakacid base

A strong base, on the other hand, must have a weak conjugate acid.

NH4� � OH� 88n NH3 � H2O

Strong Weakbase acid

Brønsted Acids and Bases in Nonaqueous Solutions

Water has a limiting effect on the strength of acids and bases. All strong acids behave thesame in water—1 M solutions of the strong acids all behave as 1 M solutions of the H3O�

ion—and very weak acids may not act as acids in water. Acid–base reactions don’t have tooccur in water, however. When other solvents are used, the full range of acid–base strengthshown in Table O3.1 can be observed.

The strongest acids are in the upper left corner of Table O3.1; the strongest basesare in the bottom right corner. Each base is strong enough to deprotonate the acid inany line above it. The hydride ion (H�), for example, can convert an alcohol into itsconjugate base,

CH3OH � H� 88nm88 CH3O� � H2

and the amide ion (NH2�) can deprotonate an alkyne.

CH3CqCH � NH2� 88nm88 CH3CqC� � NH3

[HA][OH�]��

[A�][H3O�][A�]��

[HA]

NH3 H2O OH�� NH4�

AcidBase BaseAcid

1012T_mod06_1-30 1/24/05 12:46 Page 3 EQA


Lewis Acid–Base Theory

In 1923, G. N. Lewis introduced a theory of acids and bases that is even more power-ful than the Brønsted theory. As a result, it is important to differentiate between theterms “acid” and “base” as they have been used so far and the terms “Lewis acid” and“Lewis base.”

Lewis noted that the Brønsted theory was limited because it focused exclusively on thetransfer of a proton (H�). He noted that a more general definition of acid–base reactionscould be obtained by looking at what happens when an H� ion combines with an OH� ionto form water.

Lewis pointed out that the H� ion picks up (or accepts) a pair of electrons from the OH�

ion to form a new covalent bond. As a result, any substance that can act as an electronpair acceptor is a Lewis acid.

A Lewis acid is an electron pair acceptor, such as the H� ion.

The pair of electrons that went into the new covalent bond were donated by the OH� ion.Lewis therefore argued that any substance that can act as an electron pair donor is a Lewisbase.

A Lewis base is an electron pair donor, such as the OH� ion.

The Lewis acid–base theory doesn’t affect the category of compounds we have called “bases”because any Brønsted base must have a pair of nonbonding electrons in order to accept a

H� O H� H HO

TABLE O3.1 Typical Brønsted Acids and Their Conjugate Bases

ConjugateCompound Ka pKa Base Kb pKb

HI 3 � 109 �9.5 I� 3 � 10�24 23.5HCl 1 � 106 �6 Cl� 1 � 10�20 20H2SO4 1 � 103 �3 HSO4

� 1 � 10�17 17H3O� 55 �1.7 H2O 1.8 � 10�16 15.7HNO3 28 �1.4 NO3

� 3.6 � 10�16 15.4H3PO4 7.1 � 10�3 2.1 H2PO4

� 1.4 � 10�12 11.9CH3CO2H 1.8 � 10�5 4.7 CH3CO2

� 5.6 � 10�10 9.3H2S 1.0 � 10�7 7.0 HS� 1 � 10�7 7H2O 1.8 � 10�16 15.7 OH� 55 �1.7CH3OH 1 � 10�18 18 CH3O� 1 � 104 �4HCqCH 1 � 10�25 25 HCqC� 1 � 1011 �11NH3 1 � 10�33 33 NH2

� 1 � 1019 �19H2 1 � 10�35 35 H� 1 � 1021 �21CH2PCH2 1 � 10�44 44 CH2PCH� 1 � 1030 �30CH4 1 � 10�49 49 CH3

� 1 � 1035 �35

1012T_mod06_1-30 1/24/05 19:57 Page 4 EQA


FIGURE O3.1 The reaction between BF3 acting as a Lewis acid andNH3 acting as a Lewis base to form an acid–base complex.


In this case, a pair of nonbonding electrons from each of the four NH3 molecules is do-nated into an empty orbital on the Cu2� ion to form a covalent CuON bond.

Curved Arrow Symbolism

The flow of electrons from a Lewis base to a Lewis acid is often indicated with a curvedarrow. The arrow starts on a pair of nonbonding electrons on the Lewis base and pointstoward the Lewis acid with which it reacts. Because adding a pair of electrons to one pointon a molecule often displaces electrons in the molecule, combinations of curved arrows areoften used to describe even simple chemical reactions. Consider the following example, inwhich a pair of electrons on an NH2

� ion are donated to the H� ion formed when the elec-trons in one of the COH bonds in acetylene are given to the carbon atom instead of be-ing shared by the C and H atoms in the bond.

O3.2 CARBANION ATTACK AT A CARBONYL GROUPThe discussion of acids and bases in the previous section helps us understand the chem-istry of the Grignard reagents introduced in Section O2.10. Grignard reagents are madeby reacting an alkyl bromide with magnesium metal in diethyl ether.

C CH H H� CN �H

H�HN

H

HC

2�

NH3

N

HHH

NH3

NH3

NH3

CuCu2�NH

H

H

NH

H

HN

HHH

proton. The Lewis theory, however, vastly expanded the family of compounds that can becalled “acids.” Anything that has one or more empty valence-shell orbitals can act as an acid.

The theory explains why BF3 reacts instantaneously with NH3. The nonbonding elec-trons on the nitrogen in ammonia are donated into an empty orbital on the boron to forma new covalent bond, as shown in Figure O3.1. It also explains why Cu2� ions pick up ammonia to form the four-coordinate Cu(NH3)4

2� ion.

Q A

A

B OOO

O O

FF

F

BOO FO

FF

N

H HHA

O ON

H HH

1012T_mod06_1-30 1/24/05 13:22 Page 5 EQA


An analogous reagent, known as an alkyllithium, can be prepared by reacting the alkylbromide with lithium metal in diethyl ether.

In the course of the reactions the carbon atom is reduced from the �2 to the �4oxidation state. Whereas the starting material contains a carbon atom with a partialpositive charge, the carbon atom in the products of these reactions carries a partialnegative charge.

CH3Li and CH3MgBr can therefore be thought of as a source of the CH3� ion.

The CH3� ion is the conjugate base of methane, which is the weakest Brønsted acid in

Table O3.1.

The CH3� ion is therefore the strongest Brønsted base in the table.

Exercise O3.1

A graduate student once tried to run the following reaction to prepare a Grignard reagent.Explain what he did wrong, state why the yield of the desired product was zero, and pre-dict the product he obtained.

Solution

There is nothing wrong with the starting material (CH3CH2Br) or the product of the re-action (CH3CH2MgBr). Nor is there anything wrong with using magnesium metal to forma Grignard reagent. The only place where a mistake could have been made is in the choiceof the solvent. The solvent that was used was ethanol (CH3CH2OH), whereas the usualsolvent for a Grignard reagent is diethyl ether.

Ethanol is a protic solvent (see Section O2.4), which would react instantaneouslywith the carbanion in a Grignard reagent and thereby destroy the reagent to form ethane.

H�CH3CH2� CH3CH3�

CH3CH2Br CH3CH2MgBrCH3CH2OH

Mg

H�CH4 CH3� Ka � 1 � 10�49�

Li�CH3Li CH3� �

C Br��H

H

H

C Li��H

H

H

2 Li LiBrCH3Br CH3Li� �Et2O

Mg

Et2OCH3Br CH3MgBr

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A subtle but important point must be made before we can extend our understandingof acid–base chemistry to the reaction between a Grignard or alkyllithium reagent and acarbonyl group. The data in Table O3.1 reflect the strengths of common acids and baseswhen they act as Brønsted acids or bases. The data predict that methyllithium should re-act with acetylene to form methane and an acetylide ion, for example.

The reaction should occur because it converts the stronger of a pair of Brønsted acids andthe stronger of a pair of Brønsted bases into a weaker acid and a weaker base.

The reaction between a carbonyl and CH3Li or CH3MgBr, on the other hand, involves at-tack by a CH3

� ion acting as a Lewis base or nucleophile at the positive end of the car-bonyl group.

This raises an interesting question: Is the stronger of a pair of Brønsted bases always the strongerof a pair of Lewis bases? Unfortunately, the answer is no, it isn’t. At times, the stronger of apair of Brønsted bases is the weaker Lewis base or nucleophile. As a rule, however, strongBrønsted bases are strong nucleophiles, and weak Brønsted bases are weak nucleophiles.1

Despite the enormous utility of Grignard reagents in organic chemistry, the exact mech-anism of the reaction between the reagents and a carbonyl is not known. There is reasonto believe that two molecules of the Grignard reagent are involved in the reaction. Themagnesium atom of one molecule of the reagent acts as a Lewis acid that interacts withthe oxygen atom of the carbonyl group. The alkyl group of the other reagent molecule thenacts as a Lewis base, attacking the positive end of the carbonyl.

In essence, the reaction involves the attack by a negatively charged CH3� ion at the

positively charged end of the carbonyl group. When this happens, the pair of nonbondingelectrons on the CH3

� ion are used to form a C—C bond. This, in turn, displaces the pairof electrons in the � bond onto the other end of the carbonyl group.

O

C

Mg

CH3

Br

CH3 Mg Br

C O CH3 C

CH3

CH3

H3C

CH3

CH3

O��

HC CH CCH4� � CH�CH3�

Strongerbase

Strongeracid

Weakeracid

Weakerbase

HC CH Li� CCH3Li CH4� � � CH�

1See T. B. McMahon, T. Heinis, G. Nicol, J. K. Hovey, and P. Kebarle, J. Am. Chem. Soc., 110, 7591 (1988).

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The second molecule of the Grignard reagent, which binds at the oxygen end of the car-bonyl, isn’t consumed in the reaction. Its function is simple. When it acts as a Lewis acid,binding to the oxygen atom in the CPO double bond, it increases the polarity of the bond.By making the bond more polar, it increases the rate at which the CH3

� ion attacks thepositive end of the CPO bond.

O3.3 THE MECHANISM OF REDUCTION REACTIONSTwo fundamentally different reducing agents have been used in the previous chapters onorganic chemistry to add hydrogen across a double bond. In Section O1.6, a metal was usedto catalyze the reaction between hydrogen gas and the CPC double bond in an alkene.

A source of the hydride ion (H�), on the other hand, was used to reduce CPO doublebonds in Section O2.6.

The first step toward understanding the difference between these reactions involves notingthat the first reaction uses a nonpolar reagent to reduce a nonpolar double bond. The atomson the surface of a metal are different from those buried in the body of the solid because theycannot satisfy their tendency to form strong metal-metal bonds. Some metals can satisfy aportion of their combining power by binding hydrogen atoms and/or alkenes to the surface.

Transferring one of the hydrogen atoms on the metal surface to the alkene bound to themetal surface forms an alkyl group, which can bond to the metal until the second hydrogenatom can be added to form the alkane.

H

CHC H

CHC

H C CH

CH3CH2CH

O

CH3CH2CH2OH2. H2O

1. LiAlH4 in ether

C

H

HH

H

H

HCC C HH

H

HH2

Ni

O

C C

O

CH3CH3�

�

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Although the hydrogen atoms are transferred one at a time, the reaction is fast enoughthat both of the atoms usually end up on the same side of the CPC double bond. Thiscan’t be seen in most alkanes produced by the reaction because of the free rotation aroundC—C bonds. Reduction of a cycloalkene, however, gives a stereoselective product.

Reduction of an alkyne with hydrogen on a metal catalyst gives the correspondingalkane. By selectively “poisoning” the catalyst it is possible to reduce an alkyne to analkene. Once again, the reaction is stereoselective, adding both hydrogen atoms from thesame side of the CqC bond to form the cis-alkene.

Because it is a polar reagent, LiAlH4 won’t react with a CPC double bond. It acts asa source of the H� ion, however, which is a strong Brønsted base and a strong nucleophile.The H� ion can therefore attack the �� end of a polar CPO double bond.

The neutral AlH3 molecule formed when an AlH4� ion acts as a hydride donor is a Lewis

acid that coordinates to the negatively charged oxygen atom in the product of the reac-tion. When a protic solvent is added to the reaction in a second step, an alcohol is formed.

O3.4 NUCLEOPHILIC ATTACK BY WATERIn the 1930s and 1940s, Dashiell Hammett (1894–1961) created the genre of the “hard-boiled” detective novel with books such as The Maltese Falcon and The Thin Man. A com-mon occurrence in this literature was a character who “slipped someone a Mickey Finn”—a dose of the sedative known as chloral hydrate dissolved in a drink that contains alcohol.

C

Cl

Cl

C H Chloral hydrateCl

OH

OH

C

O

H

CH3 CH3

H

C

O AlH3

H

CH3 CH3 H2O� OH�� AlH3�

�

O

C C

O

HH

CH3 CH3CH3 CH3

�

�

CCH3 CH3

CH3CH3

HH

CC CH2

Pd on CaCO3

CH3

CH3

HH2 /Ni

CH3

H

CH3

1012T_mod06_1-30 1/24/05 12:46 Page 9 EQA


Chloral hydrate is a white solid formed by adding a molecule of water across the CPOdouble bond in the corresponding aldehyde.

The equilibrium constant for the reaction is sensitive to the substituents on the carbonylcarbon. Electron-withdrawing substituents, such as the Cl3C group in chloral, drive the re-action toward the dialcohol, or diol (K �� 1). Electron-donating substituents, such as thepair of CH3 groups in acetone, pull the equilibrium back toward the aldehyde (K �2 � 10�3).

The rate of the reaction can be studied by following the incorporation of isotopicallylabeled water. The vast majority (99.76%) of water molecules contain 16O, but some con-tain 17O (0.04%) or 18O (0.2%). When acetone is dissolved in a sample of water that hasbeen enriched in 18O, it gradually picks up the 18O isotope.

The rate of the reaction is infinitesimally slow in a neutral solution (pH � 7). But in thepresence of a trace of acid (or base), the reaction occurs very rapidly.

The role of the acid catalyst can be understood by noting that protonation of the oxygenatom increases the polarity of the carbonyl bond.

This increases the rate at which a water molecule can act as a nucleophile toward the pos-itive end of the CPO double bond.

The product of the reaction then loses an H� ion to form the diol.

C

O

H

H

OH

B

E

HE

O

�

�

CH3CH3CH3CH3Acid-catalyzed hydration, step 1

OA

SO

O

C

HO

H

H

O

OOA

A

SO

OA

S

C

OH

B

E

HE C

O H

HE

AOSOO

�

�

CH3 CH3 H218OC

OB

O O CH3 CH3C

18OB

O OCH3 CH3C

18OH

OH

O O� H2O�A

A

C

Cl

Cl

C H H2OCl

O

C

Cl

Cl

C HCl�

HO

OH

1012T_mod06_1-30 1/24/05 12:46 Page 10 EQA


The role of the base catalyst can be understood by noting that the OH� ion is a muchstronger nucleophile than water, strong enough to attack the carbonyl by itself.

The product of the reaction then picks up a proton from a water molecule to form the dioland regenerate the OH� ion.

There is a fundamental relationship between the mechanisms of the reactions at thecarbonyl group introduced so far. In each case, a nucleophile or Lewis base attacks thepositive end of the carbonyl group. And, in each case, the rate of reaction can be increasedby coordinating a Lewis acid or electrophile at the other end of the carbonyl.

There is a subtle difference between the reactions, however. Very strong nucleophiles, suchas Grignard reagents and the hydride ion, add to the carbonyl in an irreversible reaction.

Attack by a weaker nucleophile, such as water, is a reversible reaction that can occur ineither direction.

CH3CH3

OH

COOA

AOH

CH3 H2O�CH3

O

COOB

C

OB

HE CH3

CH3

CH3CH3

CH3CH3

S

O

COOA

A

SOOS

�

�S

C

Nucleophiles(CH3

�, H�, H2O, OH�, etc.)

Electrophiles(H�, Mg2�, AlH3, etc.)

OB

HE

OS

C

O H

OB

HE

O

�

�

CH3CH3CH3CH3Base-catalyzed hydration, step 1

OSO

O

C

HO

OOA

A

SO

OS

QQ

S

S

CH3 H�CH3

O

C

HO

HO

OOA

A

SO

OS

�

� Q

CH3CH3Acid-catalyzed hydration, step 2

O

C

HO

HO

OOA

A

A

SO

O

H

S

CH3CH3

O

C

HO

OOA

A

SO

O

HO

SQ

�OSOS

OH��CH3 H2OCH3Base-catalyzed hydration, step 2 C

HO

OOA

AOSQ

�

1012T_mod06_1-30 1/24/05 13:22 Page 11 EQA


O3.5 NUCLEOPHILIC ATTACK BY AN ALCOHOLWhat would happen if we dissolved an aldehyde or ketone in an alcohol, instead of wa-ter? We would get a similar reaction, but now an ROH molecule is added across the CPOdouble bond.

Once again, the reaction is relatively slow in the absence of an acid or base catalyst. If webubble HCl gas through the solution, or add a small quantity of concentrated H2SO4, weget an acid-catalyzed reaction that occurs by a mechanism analogous to that described inthe previous section.2

The product of the reaction is known as a hemiacetal (literally, “half of an acetal”). If ananhydrous acid is added to a solution of the aldehyde in a large excess of alcohol, the re-action continues to form an acetal.

Hemiacetals can be recognized by looking for a carbon atom that has both an OOHand an OOR group.

Acetals, on the other hand, contain a carbon atom that has two OOR groups.

CH3

CH3

A

ACH3CH2OCOCH3 An acetal

CH3A

CH3CH2OCHOH A hemiacetal

HCH3HCl

2 CH3OH COOA

AOCH3

OCH3

H � H2O�CH3

O

COOB

C

O H

OH

Acid-catalyzed reaction of an alcohol with a carbonyl

B

E

HE

O

�

�

CH3

CH3

CH3CH3CH3

OA CH3

ASO

O

C

HO

HO

OOA

A

SO

OS

CH3

CH3

H�CH3

O

C

HO

O

OOA

A

SO

OS

�

Q

HCH3CH3OH

O

CO

HO

OA

AOCH3

H �CH3

O

COOB

2Gaseous HCl and concentrated H2SO4 give an acidic solution without contaminating the solution with wa-ter, which could react with the carbonyl.

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Hemiacetals and acetals play an important role in the chemistry of carbohydrates. Con-sider what would happen, for example, if the OOH group on the fifth carbon atom in aglucose molecule attacked the aldehyde at other end of the molecule.

The product of the reaction is a hemiacetal that contains a six-membered ring known as apyranose. Two isomers of glucopyranose can be formed, depending on whether the OOHgroup attacks from above or below the CPO group.

An analogous intramolecular reaction can occur within a fructose molecule.

CH2OH

CH2OHA

A

CACACAC

O

HO OHO

HO OHO

HOO HO

P

CH2OHCH2OH

H

HO

HO

H

HOH

HOO

H H

HO HOH

H

H or

� -Glucopyranose-D

OH

OH

H

OH

CH2OHCH2OH

H

HO

HO

H

HOH

HO

HO

H

HO

H

OH

H

H or

-Glucopyranose-D

OHOH

HOH

CH2OH

A

A

C

CH

ACACAC

OB

HO OHO

HO OHO

HO OHO

HOO HO

1012T_mod06_1-30 1/24/05 12:46 Page 13 EQA


In this case, a hemiacetal is formed that contains a five-membered furanose ring. Onceagain, there are two isomers, depending on how the OOH group attacks the CPO group.

Sugars, such as glucose and fructose, can be linked to form complex carbohydrates byforming an acetal linkage between the OOH group on one sugar and the hemiacetal onthe other. Sucrose, or cane sugar, for example, is an acetal formed by linking �-D-glucopyranose and �-D-fructofuranose residues.

O3.6 ADDITION/ELIMINATION REACTIONS OF CARBOXYLICACID DERIVATIVES

The following reaction was used in Section O2.8 to illustrate the synthesis of an ester froma carboxylic acid.

These reactions occur very slowly in the absence of a strong acid. When gaseous HCl isbubbled through the solution, or a small quantity of concentrated H2SO4 is added, the re-actions reach equilibrium within a few hours. Once again, the acid protonates the oxygenof the CPO double bond, thereby increasing the polarity of the carbonyl group, whichmakes it more susceptible to attack by a nucleophile.

As might be expected, the first step in the reaction involves attack by a nucleophile atthe positively charged end of the CPO double bond. A pair of nonbonding electrons onthe oxygen atom of the alcohol is donated to the carbon atom of the carbonyl to form aCOO bond. As the bond forms, the electrons in the � bond of the carbonyl are displacedonto the oxygen atom. A proton is then transferred back to the solvent to give a tetrahe-dral addition intermediate.

C

OH

Nucleophilic addition

B

E

O O

O

�

�

CH3

HOCH2CH3

OH

S

SQ

OQ CO OCH3

HOCH2CH3

OHO

OHO

Q

QOQ

A

A

S

CO OCH3

OCH2CH3

OHO

OHO

Q

A

AH��

CH3COCH2CH3CH3CH2OH�CH3COH

OH� B

OB

HOCH2

CH2OHSucrose

O

HO

CH2OHHO

OH OH

O

O

OH

OHCH2

CH2OHHH HO

HOH

OHO OHCH2 CH2OH

HH HO

HOH

OH

O

� -D-Fructofuranose -Fructofuranose-D

1012T_mod06_1-30 1/24/05 12:46 Page 14 EQA


One of the OOH groups in the intermediate picks up a proton, loses a molecule of water,and then transfers a proton back to the solvent to give the ester.

The combination of addition and elimination reactions has the overall effect of substi-tuting one nucleophile for another—in this case, substituting an alcohol for water. The rateof these nucleophilic substitution reactions is determined by the ease with which the elim-ination step occurs. As a rule, the best leaving groups in nucleophilic substitution reactionsare weak bases. The most reactive of the carboxylic acid derivatives are the acyl chloridesbecause the leaving group is a chloride ion, which is a very weak base (Kb � 10�20).

Esters are less reactive because the leaving group is an alcohol, which is a slightly betterbase (Kb � 10�14).

Amides are even less reactive because the leaving group is ammonia or an amine, whichare significantly more basic (Kb � 10�5).

O3.7 FREE RADICAL REACTIONSThe starting point for reactions at a carbonyl involves attack by a nucleophile on the car-bon atom of the CPO double bond.

Or it involves the heterolytic splitting of a bond to form a nucleophile that can attack thecarbonyl group.

C

O

H

H

OH

B

E

HE

O

�

�

CH3CH3CH3CH3

OA

SO

O

C

HO

H

H

O

OOA

A

SO

OA

S

CH3CONHCH3 CH3OH� CH3NH2�

O

CH3COCH3

O

CH3COCH2CH3 CH3OH� CH3CH2OH�

O

CH3COCH3

O

CH3CCl CH3OH� Cl��

O

CH3COCH3

O

Nucleophilic elimination

C

OH

O O�

CH3

OCH2CH3

OH

S

OQ CO O OCH3

OCH2CH3

O H

OHO

Q

AHAA

A ASQ

COCH3 CO OCH3 OCH2CH3

OCH2CH3

OH�O

ASQSQ

SO

H�� H��H2O� BOBO

S

OQ

1012T_mod06_1-30 1/24/05 13:22 Page 15 EQA


In either case, the reaction is carried out by a reagent that donates a pair of electrons toa carbon atom to form a new covalent bond.

Section O2.3 introduced a reaction that occurs by a very different mechanism: free rad-ical halogenation of an alkane. The first step in these reactions is the homolytic splittingof a bond to give a pair of free radicals.

A series of reactions then occurs that involves a chain reaction. Consider the chlorinationof propane, for example. A Cl atom can attack the CH3 group at one end of the molecule.

Or it can attack the CH2 group in the center of the molecule.

The free radicals generated in the reactions then react with chlorine to form either 1-chloro-propane or 2-chloropropane and regenerate a Cl radical.

Chain propagation H

H

C SClOQ ClOQO OCOC OA

HA

C

HA

H

H H H

A

A A A

H

COHO OC ClOA

H

H H H

A

A A A�T S SClOQT

H H

H

C SClOQ ClOQO OCO OC OA

C

ClA

H

H H H

A

A A A

H

COHO OC HOA

H

H H H

A

A A A�S SClOQTP

Chain propagation

SCl

H

H

C

OQT

SClOQO OCOC HO HOA

HA

H

H H H

A

A A A

H

COHO COC HOA

H

H H H

A

A A A� P

Chain propagation SCl H

H

COQT SClOQO OCOC HO HO

AHA

H

H H H

A

A A A

H

COCOC HOA

HA

H

H H H

A

A A A�T

ClChain initiation 2SOQ Cl SOOSQ ClOQT

HACAH

HO LiO

HACAH

H Li�O S

�

CB

HE CH3

H3C

CH3

CH3

CH3CH3

COOA

A

OO S

S�

�OSOS

1012T_mod06_1-30 1/24/05 12:46 Page 16 EQA


There are six hydrogen atoms in the two CH3 groups and two hydrogens in the CH2

group in propane. If attack occurred randomly, six-eighths (or three-quarters) of the prod-uct of the reaction would be 1-chloropropane. The distribution of products of the reaction,however, suggests that 1-chloropropane is formed slightly less often than 2-chloropropane.

This can be explained by noting that the 2° radical formed by removing a hydrogen atomfrom the CH2 group in the center of the molecule is slightly more stable than the 1° radi-cal produced when a hydrogen atom is removed from one of the CH3 groups at either endof the molecule.

The difference between the radicals can be appreciated by considering the energy ittakes to break the COH bond in the following compounds.

These data suggest that it takes less energy to break a COH bond as the number of alkylgroups on the carbon atom that contains the bond increases. This can be explained by as-suming that the products of the bond-breaking reaction become more stable as the num-ber of alkyl groups increases. In other words, 3° radicals are more stable than 2° radicals,which are more stable than 1° radicals.

The activation energy for the chain propagation steps in free radical bromination re-actions is significantly larger than the activation energy for the steps during chlorination.As a result, free radical bromination reactions are more selective than chlorination reac-tions. Bromination reactions are far more likely to give the product predicted from therelative stability of the free radical intermediate. Bromination of 2-methylpropane, for ex-ample, gives almost exclusively 2-bromo-2-methylpropane, not the statistically more likely1-bromo-2-methylpropane.

CH3

CH3

CH33°

C >OA

AT CH3

CH3

H2°

C >OA

AT H

CH3

H1°

COA

AT

CH3

CH3

CH3

C HOOA

ACH3 H� �H° � �381 kJ/molrxn

CH3

CH3

COA

AT T

CH3

CH3

H

C HOOA

ACH3 H� �H° � �395 kJ/molrxn

CH3

H

COA

AT T

H

CH3

H

C HOOA

AH H� �H° � �410 kJ/molrxn

CH3

H

COA

AT T

ClCH2CH2CH31-Chloropropane

(45%)

CH3CHCH3

Cl

2-Chloropropane(55%)

Cl2� �CH3CH2CH3

A

1012T_mod06_1-30 1/24/05 12:46 Page 17 EQA


O3.8 BIMOLECULAR NUCLEOPHILIC SUBSTITUTION OR SN2REACTIONS

Most of our knowledge of the mechanisms of chemical reactions has come from the studyof the factors that influence the rate of these reactions. The type of reaction that has beenstudied more than any other involves attack by a nucleophile on a saturated carbon atom.Consider the following reaction, for example, which converts an alkyl bromide into an al-cohol.

CH3Br � OH� 88n CH3OH � Br�

In the course of the reaction, one nucleophile (the OH� ion) is substituted for another (theBr� ion). This is therefore a nucleophilic substitution reaction.

The rate of the reaction is first-order in both CH3Br and the OH� ion, and second-order overall.

Rate � k(CH3Br)(OH�)

In the 1930s, Sir Christopher Ingold proposed a mechanism for the reaction in which boththe alkyl halide and the hydroxyl ion are involved in the rate-limiting or slowest step ofthe reaction. The OH� ion attacks the “backside” of the CH3Br molecule. (It attacks thecarbon atom at a point directly opposite the Br substituent or leaving group.) When thishappens, a pair of nonbonding electrons on the OH� ion are used to form a covalent bondto the carbon atom at the same time that the carbon–bromine is broken, as shown in Fig-ure O3.2. Because the rate-limiting step in the reaction involves both the CH3Br and OH�

molecules, it is called a bimolecular nucleophilic substitution reaction, or SN2 reaction.

CH3CCH3

99%

HBrBr2� �CH3CHCH3

CH3A

CH3

Br

A

A

FIGURE O3.2 The mechanism for nucleophilic attack on CH3Br.

The most important point to remember about the mechanism of SN2 reactions is thatthey occur in a single step. The species in the middle of Figure O3.2 is known as a transi-tion state. If you envision the reaction as an endless series of snapshots that capture theinfinitesimally small changes which occur as one bond forms and the other bond breaks,the transition state is the snapshot in the series that has the highest energy—and is there-fore the least stable. The transition state has an infinitesimally short lifetime, on the orderof 10�12 second.

In the course of an SN2 reaction, the other three substituents on the carbon atom are“flipped” from one side of the atom to the other. This inevitably leads to inversion of the

AC O OO Br

HH

H H

H

H

HH

#

{O

#{

#{HO

HO�

HOC Br C � Br�

1012T_mod06_1-30 1/24/05 12:46 Page 18 EQA


configuration at a stereocenter. Consider the following reaction, for example, in which cis-1-bromo-3-methylcyclopentane is converted to trans-3-methylcyclopentanol.

Or consider the reaction in which the R isomer of 2-bromobutane is transformed into theS isomer of 2-butanol.

O3.9 UNIMOLECULAR NUCLEOPHILIC SUBSTITUTION OR SN1REACTIONS

The kinetics of nucleophilic substitution reactions have been studied in greater detail thanany other type of reaction because they don’t always proceed through the same mecha-nism. Consider the reaction between the OH� ion and t-butyl bromide, for example.

(CH3)3CBr � OH� 88n (CH3)3COH � Br�

The rate of the reaction depends only on the concentration of the alkyl bromide. (Addingmore OH� ion to the solution has no effect on the rate of reaction.)

Rate � k((CH3)3CBr)

Ingold and co-workers argued that this rate law is consistent with a mechanism in whichthe rate-limiting or slowest step involves the breaking of the carbon–bromine bond to forma pair of ions. As one might expect, the pair of electrons in the COBr bond end up on themore electronegative bromine atom.

Because the bromine atom has formally gained an electron from the carbon atom, it is nowa negatively charged Br� ion. Because the carbon atom has formally lost an electron, it isnow a “carbocation.”

The first step in the mechanism is a relatively slow reaction because the activation energyfor this step is roughly 80 kJ/molrxn. If the reaction is done in water, the next step is extremelyfast. The (CH3)3C� ion is a Lewis acid because it has an empty orbital that can be used toaccept a pair of electrons. Water, on the other hand, is a reasonably good Lewis base. ALewis acid-base reaction therefore rapidly occurs in which a pair of nonbonding electronson a water molecule are donated to the carbocation to form a covalent COO bond.

C BrCH3Rate-limiting step

CH3

CH3

C� Br�CH3

CH3

CH3

�

HOHO� Br�� Br CCCH3CH3

CH2CH3CH3CH2

HH

CH3 Br CH3

� Br�

OH�OH

1012T_mod06_1-30 1/24/05 12:46 Page 19 EQA


The product of the reaction is a stronger acid than water. As a result, it transfers a protonto water.

Because the slowest step of the reaction only involves t-butyl bromide, the overall rate ofreaction depends only on the concentration of that species. The reaction is therefore a uni-molecular nucleophilic substitution reaction, or SN1 reaction.

The central carbon atom in the t-butyl carbocation formed in the first step of the re-action is planar, as shown in Figure O3.3. This means that water can attack the carbo-cation in the second step with equal probability from either side of the carbon atom. Thishas no effect on the products of the reaction, because the starting material is not opti-cally active. But what would happen if we started with an optically active halide, such asObromobutane?

C � H2OCH3

CH3

CH3

CCH3

CH3

CH3

OH2� � H3O�OH

C� � H2OCH3

CH3

CH3

C OH2�CH3

CH3

CH3

FIGURE O3.3 The geometry of the (CH3)3C� ion.

Regardless of whether we start with the R or S isomer of 2-bromobutane, we get thesame intermediate when the COBr bond breaks.

The intermediate formed in the first step in the SN1 mechanism is therefore achiral.Mixtures of equal quantities of the �/� or R/S stereoisomers of a compound are said

to be racemic. The term traces back to the Latin word racemus, which means “a cluster ofgrapes.” Just as there is an equal probability of finding grapes on either side of the stemin a cluster of grapes, there is an equal probability of finding the R and S enantiomers ina racemic mixture. SN1 reactions are therefore said to proceed with racemization. If westart with a pure sample of (R)-2-bromobutane, for example, we expect the product of theSN1 reaction with the OH� ion to be a racemic mixture of the two enantiomers of 2-butanol.

We are now ready to address a pair of important questions. First, why does CH3Br re-act with the OH� ion by the SN2 mechanism if (CH3)3CBr does not? The SN2 mechanismrequires direct attack by the OH� ion on the carbon atom that carries the COBr bond. Itis much easier for the OH� ion to get past the small hydrogen atoms in CH3Br than it isfor the ion to get past the bulkier CH3 groups in (CH3)3CBr.

Br Br�C

CH3

CH3CH2 CH2CH3

CH3 H

CHBr C

CH3

CH2CH3

H

�

A

O#

{

CH3CH3

CH3C

�

1012T_mod06_1-30 1/24/05 13:22 Page 20 EQA


Thus, SN2 reactions at the 1° carbon atom in CH3Br are much faster than the analogousreaction at the 3° carbon atom in (CH3)3CBr.

Why, then, does (CH3)3CBr react with the OH� ion by the SN1 mechanism if CH3Brdoes not? The SN1 reaction proceeds through a carbocation intermediate, and the stabil-ity of such ions decreases in the following order.

Organic chemists explain this by noting that alkyl groups are slightly “electron releasing.”They can donate electron density to a neighboring group. This tends to delocalize the chargeover a larger volume of the molecule, which stabilizes the carbocation.

When we encountered a similar phenomenon in the chemistry of free radicals in Sec-tion O3.7 we noted that 3° radicals are roughly 30 kJ/mol more stable than 1° radicals. Inthis case, the difference is much larger. A 3° carbocation is 340 kJ/mol more stable than a1° carbocation!3 As a result, it is much easier for (CH3)3CBr to form a carbocation inter-mediate than it is for CH3Br to undergo a similar reaction.

In theory, both starting materials could undergo both reaction mechanisms. But therate of SN2 reactions for CH3Br are much faster than the corresponding SN1 reactions,whereas the rate of SN1 reactions for (CH3)3CBr are very much faster than SN2 reactions.

O3.10 ELIMINATION REACTIONSWhy do we need to worry about whether a nucleophilic substitution reaction occurs by anSN1 or SN2 mechanism? At first glance, it would appear that the same product is obtainedregardless of the mechanism of the reaction. Consider the following substitution reaction,for example.

The only apparent difference between the two mechanisms is the stereochemistry of theproduct. If the reaction proceeds through an SN2 mechanism, it gives inversion of config-uration—conversion of an R starting material into an S product, or vice versa. If the reac-tion proceeds through a carbocation intermediate via an SN1 mechanism, we get a racemicmixture.

The importance of understanding the mechanism of nucleophilic substitution reactionscan best be appreciated by studying the distribution of products of the example given above.

CH3CHCH2CH3 Br�CH3O�� CH3CHCH2CH3

BrA

OCH3A

CH3

CH3

CH3

C� >OA

ACH3

CH3

H

C� >OA

AH

CH3

H

C�OA

A> H

H

H

C�OA

A

Br HO Br Br�

HO��C

HH

C

H

HH

HHO C

H

H

H

A, ,

3J. E. Bartmess, Mass Spectrometry Review, 8, 297–343 (1989).

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When 2-bromobutane is allowed to react with the methoxide ion in methanol, less thanhalf of the starting material is converted into methyl isopropyl ether; the rest is transformedinto 2-propene.

The reaction that produces the alkene involves the loss of an HBr molecule to form a CPCdouble bond. It is therefore an example of an elimination reaction.

Starting materials that are likely to undergo a bimolecular SN2 reaction undergo elim-ination reactions by a bimolecular elimination mechanism, or E2 reaction. This is a one-step reaction in which the nucleophile attacks a COH bond on the carbon atom adjacentto the site of SN2 reaction.

Starting materials that are likely to undergo a unimolecular SN1 reaction undergo elim-ination reactions by a unimolecular elimination mechanism, or E1 reaction. As might beexpected, the rate-limiting step is the formation of the carbocation.

The solvent then acts as a base, removing an H� ion from one of the alkyl groups adja-cent to the carbocation. The electrons in the COH bond that is broken are donated to theempty orbital on the carbocation to form a double bond.

O3.11 SUBSTITUTION VERSUS ELIMINATION REACTIONSThere are three ways of pushing the reaction between an alkyl halide and a nucleophiletoward elimination instead of substitution.

• Start with a highly substituted substrate, which is more likely to undergo elimination.Only 10% of a primary alkyl bromide undergoes elimination to form an alkene, forexample, when it reacts with an alkoxide ion dissolved in alcohol. The vast majorityof the starting material goes on to form the product expected for an SN2 reaction.

CH2 CH2

CH3

CH3

CCH2

CH3

CH3OH

CH3

C� �O C�OA

HA

A

CH3

CH3

A

AS P

G

D

CH3

CH3

CH3

C� � Br�OAA

A A

CH3

CH3

CH3Rate-limiting step C BrOO

CH2 CHCH2CH3 Br��CH2CHCH2CH3

CH3O�Br

CH3OHH

APO

CH3CHCH2CH3CH3O�� CH3CHCH2CH3

BrA

OCH3A

CH2 CHCH2CH3Pca. 40% ca. 60%

1012T_mod06_1-30 1/24/05 12:46 Page 22 EQA


More than half of a secondary alkyl bromide undergoes elimination under the sameconditions, as we have already seen.

When the starting material is a tertiary alkyl halide, more than 90% of the prod-uct is formed by an E1 elimination reaction.

• Use a very strong base as the nucleophile. When we use a relatively weak base, suchas ethyl alcohol, only about 20% of t-butyl bromide undergoes elimination.

In the presence of the ethoxide ion, which is a much stronger base, the product ofthe reaction is predominantly the alkene.

• Increase the temperature at which the reaction is run. Because both E1 and E2

reactions lead to an increase in the number of particles in the system, they are as-sociated with a positive entropy term. Thus, increasing the temperature of the re-action makes the overall free energy of reaction more negative, and the reactionbecomes more favorable.

Summary of Substitution/Elimination Reactions

• Methyl halides and primary alkyl halides such as CH3CH2Br undergo nucleophilicsubstitution reactions.

CH3CH2Br CH3CH2CN� CN� Br��

CH3CH2OH

CH3CH2O�A

A

CH3

CH3

CH3 C BrOO CH2

CH3

CH3

CPG

D

~95%

CH3CH2OHA

A

CH3

CH3

CH3 C BrOOA

A

CH3

CH3

CH3 C OCH2CH3

~80%

OO CH2

~20%

�

CH3

CH3

CPG

D

CH3OH

CH3O�A

A

CH3

CH3

CH3 C BrOO CH2

CH3

CH3

CPG

D

CH3CHCH2CH3 CH3CHCH2CH3

BrA

OCH3A

� CH3O� � CH2 CHCH2CH3P

CH3CH2CH2Br CH3CH2CH2OCH3CH3OH

CH3O�

1012T_mod06_1-30 1/24/05 12:46 Page 23 EQA


• Secondary alkyl halides undergo SN2 reactions when handled gently—at low tem-peratures and with moderately strong nucleophiles.

• At high temperatures, or in the presence of a strong base, secondary alkyl halidesundergo E2 elimination reactions.

• Tertiary alkyl halides undergo a combination of SN1 and E1 reactions. If the reac-tion is kept cool, and the nucleophile is a relatively weak base, it is possible to getnucleophilic substitution. At high temperatures, or with strong bases, eliminationreactions predominate.

KEY TERMS

CH3O�A

A

CH3

CH3

CH3 C BrOO CH2

CH3

CH3

CPG

D

CH3CHCH3 CH2 CHCH3

BrA

heat

(CH3)3CO�

P

CH3CHCH3 CH3CHCH3

BrA

SHA

� HS� � Br�

AcetalAcidAcid dissociation

equilibrium constant,Ka

AlkaliBaseBimolecular nucleophilic

substitution reactionBrønsted acid

Brønsted baseDiolE1 reactionE2 reactionElimination reactionHemiacetalHeterolyticHomolyticLeaving group

Lewis acidLewis baseNucleophilic

substitution reactionSN1 reactionSN2 reactionTransition stateUnimolecular nucleophilic

substitution reaction

PROBLEMSAcids and Bases

1. Predict the products of the reaction between ethanol (CH3CH2OH) and sodium hy-dride (NaH).

2. Write the Lewis structure for the conjugate base of ethanol (CH3CH2OH).3. Which of the following is the conjugate acid of ethanol (CH3CH2OH)?

(a) CH3CH2OH (b) CH3CH2O� (c) CH3CH2OH2� (d) CH3CH2

� (e) H3O�

4. Some beginning chemistry students get confused when acetic acid is described as aweak acid because they see four hydrogen atoms in a CH3CO2H molecule. Explainwhy only one of the H atoms dissociates when acetic acid is dissolved in water.

1012T_mod06_1-30 1/24/05 12:46 Page 24 EQA


5. Which of the following are Lewis acids but not a Brønsted acid?(a) H� (b) NH4

� (c) BF3 (d) CH3CH2OH (e) Mg2 �

6. Arrange the following compounds in order of increasing basicity.(a) NH3 (b) NH2

� (c) NH4� (d) N3�

7. Identify the Brønsted acids in the following reaction.

HCqCH � CH3Li 88n HCqC� � CH4 � Li�

8. Arrange the following hydrocarbons in order of increasing acidity.(a) C2H2 (b) C2H4 (c) C2H6

9. Which of the following reagents is a strong enough base to generate the ethoxide ion(CH3CH2O�) from ethanol (CH3CH2OH)?(a) NaOH (b) NaH (c) NaNH2 (d) CH3MgBr (e) HCqCNa

10. Which of the following acid–base reactions should occur as written?(a) CH3CH2OH � NaOH n Na� � CH3CH2O� � H2O(b) CH4 � NaNH2 n CH3

� � Na� � NH3

(c) CH3Li � HCqCH n CH4 � HCqCLi(d) CH3CH2OH � HS� n CH3CH2O� � H2S

Attack at a Carbonyl

11. Use Lewis structures and the concept of oxidation–reduction reactions to explain whyit takes two moles of lithium metal to reduce a mole of methyl bromide to form methyl-lithium.

CH3Br � 2 Li 88n CH3Li � LiBr

12. Explain why ethers, but not alcohols, are used as solvents for reactions that involveGrignard or alkyllithium reagents.

13. Explain why the bonding of a Lewis acid at the oxygen atom of a carbonyl group in-creases the rate at which nucleophilic attack occurs at the carbon atom.

14. Predict the products of the following reactions.

15. Predict the products of the following reactions.

16. Design a two-step reaction sequence using either a Grignard or alkyllithium reagentthat could be used to produce the following compound.

A

A

OH

CH3

C CH2CH3O

CH3 CH3MgBrC � A BH�

OBO

Li � CH3CCH2CH3 A BH�

OB

1012T_mod06_1-30 1/24/05 12:46 Page 25 EQA


The Mechanism of Reduction Reactions

17. Both H2 on a metal catalyst and LiAlH4, can be used to reduce a carbonyl to an al-cohol. Explain why H2 on a metal catalyst can be used to reduce a CPC double bondbut not LiAlH4.

18. Explain why the yield of reactions that try to reduce a carbonyl with LiAlH4 in ethanolis effectively zero.

19. Reduction of a CPC double bond with H2 gas on a metal catalyst gives a product inwhich the two hydrogen atoms are added to the same side of the double bond. Whatdoes this tell us about the relative rates of the first and second steps in the reaction?

20. Why is it important to “poison” the metal catalyst before trying to reduce an alkyneto an alkene with H2?

21. The reagent that attacks the carbon atom is the same when either LiAlH4 or NaBH4

is used to reduce a CPO double bond. (In each case, it is the H� ion.) What does thefact that LiAlH4 is significantly more reactive than NaBH4 tell us about the relativeacidity of AlH3 and BH3 as Lewis acids?

22. Predict the product of the following reaction.

23. Predict the product of the following reaction.

Nucleophilic Attack by Water or an Alcohol

24. Explain why the rate of nucleophilic attack on a carbonyl by either water or an alco-hol is relatively slow in the absence of an acid or base catalyst.

25. Identify the starting materials that would give the following product.

26. Identify the starting materials that would give the following acetal, which has a de-lightful odor of ‘Bing’ cherries.

27. Maltose, or malt sugar, is an important component of the barley malt used to brew beer.This disaccharide is formed when the OOH group on C-4 of one �-D-glucopyranoseforms an acetal by reacting with C-1 of a second �-D-glucopyranose residue. Draw thestructure of the compound.

A

A

OCH2CH3

OCH2CH3

CH

A

A

OH

CH3

C OCH2CH3O

Pd on CaCO3

H2CH3C CCH3q

2. H3O�

1. LiAlH4 in ether

CH

CH3

CH3

CH3

C

O

O OBG

D

1012T_mod06_1-30 1/27/05 14:58 Page 26 EQA


Addition/Elimination Reactions

28. Explain why acyl chlorides are more reactive than carboxylic acid esters toward addi-tion/elimination reactions.

29. Use the fact that the best leaving groups for addition/elimination reactions are weakbases to explain why attack on a carbonyl by the H� ion or a source of the CH3

� ionare not reversible reactions.

30. Which of the following would be the best leaving group?(a) NH2

� (pKb � �19) (b) CH3O� (pKb � �4) (c) HS� (pKb � 7)(d) CH3CO2

� (pKb � 9.3)

Free Radical Reactions

31. Draw the structures of all possible products of the free radical chlorination of 2-methylbutane. Predict the relative abundance of the products if the reaction is so fastthat it is not selective.

32. Draw the structures of all possible products of the free radical bromination of methyl-cyclobutane. Predict the relative abundance of the products if the reaction is equallylikely to occur at each of the hydrogen atoms.

33. Assume that the reaction in the previous problem is slow enough to give almost ex-clusively the product that would be formed from the most stable free radical interme-diate. Predict the structure of the product.

34. Calculate the value of �H° for the reaction in which H2, F2, Cl2, Br2, and I2 dissociateto form free radicals.

Nucleophilic Substitution Reactions

35. Which of the following starting materials is most likely to undergo an SN2 reaction withthe OCH3

� ion?

36. Predict the products of the following reactions.(a) CH3CH2CH2I � OH� n (b) (CH3)2CHBr � I� n(c) CH3Br � CN� n (d) CH3CH2I � HS� n

37. The relative rates of nucleophilic substitution reactions often decrease in the orderCH3

� � OH� � Cl�. Is this consistent with what we know about the relative strengthsof the ions as Brønsted bases?

38. The relative rates of nucleophilic substitution reactions often decrease in the orderI� � Br� � Cl� � F�. Is this consistent with what we know about the relative aciditiesof HI (Ka � 3 � 109), HBr (Ka � 1 � 109), HCl (Ka � 106), and HF (Ka � 7.2 � 10�4)?

BrA

BrB

BrC

CH3

Br2

1012T_mod06_1-30 1/24/05 12:46 Page 27 EQA


Or does it suggest that there is a difference between trends in nucleophilicity and ba-sicity as we go down a column of the periodic table? (In other words, the X� ion be-comes less basic and more nucleophilic as we go down the column.)

39. Write the mechanism for the following reaction.

40. Identify a set of starting materials that would give the following compound as the prod-uct of a nucleophilic substitution reaction.

CH3CH2NH2

41. Assume that you had a choice of two reagents to react with the following starting ma-terial, CH3OH and the CH3O� ion.

Which nucleophile would be more likely to give a racemized product?

Elimination Reactions

42. Write the mechanism for the following reaction.

43. Which of the following would be the most likely to undergo an E1 elimination reac-tion with a very strong base, such as the (CH3)3CO� ion?

44. Predict the product of the following elimination reaction.

45. Elimination reactions can often give two different products, depending on which car-bon atom adjacent to the COX bond is attacked. As a rule, the dominant product of

Br� (CH3)3CO�

BrA

BrB

BrC

CH2

CH3

CH3

CPG

D� OCH3

�A

A

CH3

CH3

CH3 C BrOO

A

A

CH3

H

C BrO

CH3 OCH3

CH3

CH3

C � Br�� OCH3� O O

AA

A A

CH3

CH3

CH3 C BrOO

1012T_mod06_1-30 1/24/05 12:46 Page 28 EQA


the reactions is the most highly substituted alkene. Predict the product of the follow-ing elimination reaction.

Substitution versus Elimination Reactions

46. Assume that 2-bromopropane undergoes a combination of E2 and SN2 reactions whenit reacts with the CH3O� ion in methanol. Predict the products of the reactions.

47. Predict the most likely product or products of the following reactions.

48. Predict whether the following reactions are more likely to undergo elimination or sub-stitution. Identify the mechanism of the dominant reaction (E1 versus E2; SN1 versusSN2).

49. Explain why the following reaction will not give the indicated product. Predict the prod-uct that would form.

50. Explain why the following reaction will not give the indicated product. Predict the prod-uct that would form.

H2OBr

A

A

CH3

CH3

CH3 C BrOOA

A

CH3

CH3

CH3 C O�OOA

A AA

CH3

CH3

CH3 CH3

CH3

CH3

C OO OOO�

C Br

(a)

(b)

(c)

I�

�NH2A

AOO

Br

H2O

Br

Br

Br

(a)

(b)

(c)

CH3OH

CH3NH2

CH3O�

heatBr

CHOG

DCH

CH3

CH3

Br

CH3G

D� (CH3)3CO�

1012T_mod06_1-30 1/24/05 12:46 Page 29 EQA


Integrated Problems

51. The following Lewis structures were drawn correctly by an organic chemistry studentwho forgot to indicate whether the molecules carry a positive or negative charge. Cor-rect the work by specifying whether each molecule is negatively charged, positivelycharged, or neutral.

A

A

H

HAH

H C OO HOO OA

A

HAH

HAH

H C NO HOOA

A

H

H

H C C NOO Sq

1012T_mod06_1-30 1/24/05 12:46 Page 30 EQA

1

M O D U L E

7POLYMER CHEMISTRY

P.1 Polymers

P.2 Definition of Terms

P.3 Elastomers

P.4 Free Radical Polymerization Reactions

P.5 Ionic Polymerization Reactions

P.6 Coordination Polymerization

P.7 Addition Polymers

P.8 Condensation Polymers

P.9 Properties of Polymers

Chemistry in the World Around Us: The Search for Synthetic Fibers

P.1 POLYMERSImagine that one evening you decide to go bowling. Wearing a pair of jeans, a bowlingshirt, and a lightweight jacket, you walk into the local bowling alley. After you change intoan appropriate pair of shoes, you pick up a ball, stride to the line, smoothly release theball, watch as it rolls down the lane until it collides with a set of wooden pins, and thenuse a pencil to record your score on a sheet of paper.

The fabric of your cotton jeans, your polyester shirt, your nylon jacket, and your leathershoes have something in common with the rubber bowling ball, the polyurethane coatingon the bowling lane, the wooden pins, the mixture of graphite and clay in the “lead” pen-cil, and the sheet of paper on which you wrote your score. Each of these substances is apolymer.

In 1833, Jöns Jakob Berzelius suggested that compounds with the same molecular for-mula but different structures should be called isomers (literally, “equal parts”). He thenproposed the term polymer (literally, “many parts”) to describe compounds that had thesame empirical formula but different molecular weights. Ethylene (C2H4) and butene(C4H8) are compounds that Berzelius would classify as polymers. Each compound has thesame empirical formula (CH2) but they have different molecular weights. Acetylene (C2H2)and benzene (C6H6) are another example of compounds Berzelius would call polymers.

The term polymer eventually came to mean compounds such as cellulose and naturalrubber that have unusually large molecular weights, in the range of 10,000 to 100,000 ormore grams per mole. These molecules are so large they are often called macromolecules—literally, molecules large enough to be seen with the naked eye. A perfect diamond, for ex-ample, can be thought of as a single molecule containing an array of COC bonds arranged

1012T_mod07_1-23 1/24/05 9:11 Page 1 EQA

2 POLYMERS

toward the corners of a tetrahedron around each carbon atom in the crystal. The plasticcase that enclosed one of the radios that your grandparents listened to can be thought ofas a single molecule. So can a 14-pound bowling ball.

The first explanation of why polymer molecules are so heavy was offered by HermannStaudinger in 1920. Staudinger argued that polymers contain long chains of relatively sim-ple repeating units, or monomers. Natural rubber, for example, is a polymer that containslarge numbers of OCH2C(CH3)PCHCH2O units. The number of monomers in a poly-mer can differ from one chain to the next. Rubber, for example, is a mixture of polymerchains whose mass differs by a factor of 10 or more, but whose average molecular weightis 100,000 grams per mole.

The term rubber was first used in 1770 by Joseph Priestley to describe the gum from aSouth American tree that could be used to “rub out” pencil marks. Because natural rub-ber is tacky, strong-smelling, perishable, too soft when warm, and too hard when cold, ithad only limited uses. Nathaniel Hayward was the first to note that rubber loses some ofits sticky properties when treated with sulfur. It was Charles Goodyear, however, who ac-cidentally dropped a mixture of rubber and sulfur onto a hot stove and discovered “vul-canized” rubber, which is stable over a wide range of temperatures and is far more durablethan natural rubber.

Cellulose is another example of a polymer that contains many copies of a simple repeatingunit: C6H10O5. Wood is about 50% cellulose by weight; cotton is almost 90% cellulose. Cel-lulose is used to make paper from wood pulp and cloth from cotton. In the last hundredyears, cellulose has also served as the starting material for the synthesis of the first plastics—cellulose nitrate and celluloid—and the first synthetic fibers—Chardonnet silk, or rayon.

Each OC6H10O5O repeating unit in cellulose contains three OOH groups that can re-act with nitric acid to form nitrate esters known as cellulose nitrate. In 1869 John WesleyHyatt found that mixtures of cellulose nitrate and camphor dissolve in alcohol to producea plastic substance he named celluloid. Cellulose nitrate, or celluloid, was used as a sub-stitute for ivory in the manufacture of a variety of items ranging from billiard balls to moviefilm. Because it is extremely flammable, cellulose nitrate has been replaced by other plas-tics for almost all uses except Ping-Pong balls. No other plastic has been found that hasquite the same “bounce” as celluloid.

The cellulose from wood pulp contains too many impurities to be used to make fibers.It can be purified, however, by dissolving the polymer in a mixture of NaOH and carbondisulfide (CS2). When the viscous solution is forced through tiny holes in a nozzle into anacid bath, the cellulose fiber is regenerated. When this process was introduced in 1885, theproduct was the first synthetic fiber. It was originally called Chardonnet silk, but soon be-come known as rayon. A similar process is still used to make a thin film of regeneratedcellulose known as cellophane.

P.2 DEFINITIONS OF TERMSLinear, Branched, and Cross-linked Polymers

The term polymer is used to describe compounds with relatively large molecular weightsformed by linking together many small monomers. Polyethylene, for example, is formedby polymerizing ethylene molecules.

Polyethylene is called a linear or straight-chain polymer because it consists of a long stringof carbon–carbon bonds. Those terms are misleading, however, because the geometry

n CH2 CH2 [...(CH2CH2)n...]PEthylene Polyethylene

1012T_mod07_1-23 1/24/05 9:11 Page 2 EQA

POLYMERS 3

around each carbon atom is tetrahedral and the chain is neither linear nor straight, as shownin Figure P.1.

As the polymer chain grows, it folds back on itself in a random fashion to form structuressuch as the one shown in Figure P.2.

FIGURE P.1 Small portion of a straight-chain polyethylenemolecule.

Linear polymer

FIGURE P.2 Random structure formed by a straight-chain polymer as it folds backon itself.

Polymers with branches at irregular intervals along the polymer chain are calledbranched polymers (see Figure P.3). The branches make it difficult for the polymer mole-cules to pack in a regular array, and therefore make the polymer less crystalline. Cross-linked polymers contain branches that connect polymer chains, as shown in Figure P.4. Atfirst, adding cross-links between polymer chains makes the polymer more elastic. The vul-canization of rubber, for example, results from the introduction of short chains of sulfuratoms that link the polymer chains in natural rubber. As the number of cross-links in-creases, the polymer becomes more rigid.

Branched polymer Cross-linked polymer

FIGURE P.3 Branched polymers con-tain short side chains that extendfrom the backbone of the polymer.

FIGURE P.4 Cross-linked polymershave branches that connect chains.

The decision to classify a polymer as branched or cross-linked is based on the extentto which the side chains on the polymer backbone link adjacent polymer chains. The eas-iest way to distinguish between these categories is to study the effect of various solvents onthe polymer. Branched polymers are often soluble in one or more solvents because it ispossible to separate the polymer chains. Cross-linked polymers are insoluble in all solventsbecause the polymer chains are tied together by strong covalent bonds.

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4 POLYMERS

Linear and branched polymers form a class of materials known as thermoplastics. Thesematerials flow when heated and can be molded into a variety of shapes which they retainwhen they cool. Heavy cross-linking produces materials known as thermoset plastics. Oncethe cross-links form, the polymers take on a shape that cannot be changed without de-stroying the plastic. The polypropylene used in the plastic chairs that fill so many class-rooms is a thermoplastic; as you lean back on the chair you can feel it give. The plasticcase in which early radios were placed is an example of a thermoset plastic; it had a ten-dency to shatter rather than bend if the radio was dropped on the floor.

Exercise P.1

Polyethylene can be obtained in two different forms. High-density polyethylene (0.96 g/cm3)is a linear polymer. Low-density polyethylene (0.92 g/cm3) is a branched polymer with shortside chains on 3% of the atoms along the polymer chain. Explain how the structures ofthese polymers give rise to the difference in their densities.

Solution

Linear polymers are more regular than branched polymers. Linear polymers can thereforepack more tightly, with less wasted space. As a result, linear polymers are slightly moredense than branched polymers.

Homopolymers and Copolymers

Polyethylene is an example of a homopolymer that is formed by polymerizing a singlemonomer. Copolymers are formed by polymerizing more than one monomer. Ethylene(CH2PCH2) and propylene (CH2PCHCH3) can be copolymerized, for example, to pro-duce a polymer that has two kinds of repeating units.

Copolymers are classified on the basis of the way monomers are arranged along the poly-mer chain, as shown in Figure P.5. Random copolymers contain repeating units arranged ina purely random fashion. Regular copolymers contain a sequence of regularly alternating re-peating units. The repeating units in block copolymers occur in blocks of different lengths.Graft copolymers have a chain of one repeating unit grafted onto the backbone of another.

y CH2 CHCH3 ...(CH2CH2)x CH2CH y...Px CH2 CH2P �ACH3

FIGURE P.5 Random, regular, block, and graft copolymers.

[...OAOBOBOAOAOAOBOAOBOBO...]Random copolymer

[...OAOBOAOBOAOBOAOBOAOBO...]Regular copolymer

[...OAOAOAOAOBOBOBOBOAOAOBOBOBOBOBOBOAOAOA...]Block copolymer

...OAOAOAOAOAOAOAOAOAOAO...

BOBOBOBOBO...Graft copolymer

A

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POLYMERS 5

Tacticity

Polymers with regular substituents on the polymer chain possess a property known as tac-ticity (from the Latin word tacticus, “fit for arranging”). Tacticity results from the differ-ent ways in which the substituents can be arranged on the polymer backbone (see Fig-ure P.6). When the substituents are arranged in an irregular, random fashion, the polymeris atactic (literally, “no arrangement”). When the substituents are all on the same side ofthe chain, the polymer is isotactic (literally, “the same arrangement”). If the substituentsalternate regularly from one side of the chain to the other, the polymer is syndiotactic.

... OCH2OCHOCH2OCHOCH2OCHOCH2OCHOCH2OCHO...ACH3 CH3 CH3

CH3 CH3

A

A

A

A

... OCH2OCHOCH2OCHOCH2OCHOCH2OCHOCH2OCHO...

... OCH2OCHOCH2OCHOCH2OCHOCH2OCHOCH2OCHO...A

Atactic polypropylene

Syndiotactic polystyrene

Isotactic poly(vinyl chloride)

ClACl

ACl

ACl

ACl

FIGURE P.6 Atactic, isotactic, and syndiotactic polymers.

Exercise P.2

Atactic polypropylene is a soft, rubbery material with no commercial value. The isotacticpolymer is a rigid substance with an excellent resistance to mechanical stress. Explain thedifference between the physical properties of the two forms of polypropylene.

Solution

Isotactic polypropylene is easier to pack in a regular fashion than the atactic form of thepolymer. As a result, isotactic polypropylene is more crystalline, which makes the solidmore rigid.

Addition versus Condensation Polymers

Polyethylene, polypropylene, and poly(vinyl chloride) are addition polymers formed byadding monomers to a growing polymer chain. Addition polymers can be recognized bynoting that the repeating unit always has the same formula as the monomer from whichthe polymer is formed.

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6 POLYMERS

To condense means to make something more dense, or compact. Polymers formed whena small molecule condenses out during the polymerization reaction are therefore calledcondensation polymers. Silicone, for example, is a condensation polymer formed by poly-merizing (CH3)2Si(OH)2. Each time a monomer is added to the polymer chain, a moleculeof water is condensed out, as shown in Figure P.7. Note that the repeating unit in a con-densation polymer is inevitably smaller than the monomer from which it is made.

Poly(vinyl chloride)n CH2 CHCl ... CH2CH n...PACl

n CH2 CH2 [...(CH2CH2)n...]P Polyethylene

Polypropylenen CH2 CHCH3 ... CH2CH n...PACH3

... HOOOSiOOOH � HOOOSiOOOH � HOOOSiOOOH ... A

ACH3

CH3A

ACH3

CH3A

ACH3

CH3

...OOSiOOOSiOOOSiOO... A

ACH3

CH3A

ACH3

CH3A

ACH3

CH3

�H2O

Silicone

FIGURE P.7 Silicone is a condensation polymer produced by eliminating a molecule of water each time a bond between monomers is formed.

Exercise P.3

Classify the products of the following reactions as either addition or condensation poly-mers.(a) poly(methyl methacrylate), sold as Lucite or Plexiglas

(b) nylon 6

Solution

(a) Poly(methyl methacrylate) is an addition polymer because every atom in the monomerends up in the repeating unit of the polymer.

n H2N(CH2)5CCl ... HN(CH2)5C n... n HCl

P P

OO

�

n CH2 CCO2CH3 ... CH2C n...

CO2CH3

PA

A

CH3ACH3

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POLYMERS 7

(b) Nylon 6 is a condensation polymer because a molecule of HCl is eliminated each timea monomer is added to the chain.

P.3 ELASTOMERSElastomers are polymers that have the characteristic properties of rubber—they are bothflexible and elastic. To be elastic, a polymer must meet the following criteria.

• It must contain long, flexible molecules that are coiled in the natural state and thatcan be stretched without breaking, as shown in Figure P.8.

• It must contain a few cross-links between polymer chains so that one chain does notslip past another when the substance is stretched.

• It cannot contain too many cross-links, or else it would be too rigid to be stretched.• The force of attraction between chains must be relatively small, so that the polymer

can curl back into its coiled shape after it has been stretched.

Stretch

Relax FIGURE P.8 The effect of stretching and relaxing the cross-linked chainsin an elastomer.

We can understand these requirements by taking a closer look at the chemistry of nat-ural rubber, which is a polymer of a C5H8 hydrocarbon known as isoprene.

The double bonds in natural rubber are all in the cis form. The force of attraction betweenpolymer chains is relatively small, so the polymer can curl back into its original shape af-ter the molecules have been oriented by stretching. By adding sulfur to natural rubber itis possible to introduce a small number of cross-links between polymer chains that holdthe chains together when the polymer is stretched.

At first glance, it might seem easy to make synthetic rubber. All we have to do is finda suitable catalyst that can polymerize isoprene. The task is made more difficult by the factthat the cis isomer of isoprene rearranges into the trans isomer during polymerization, andthe trans isomer of polyisoprene, which is known as gutta percha, is not elastic. It is there-fore important to control the geometry around the CPC double bond during polymeriza-tion to make sure that as few of the bonds as possible are converted to the trans geome-try. Until recently, this wasn’t possible, and other approaches to making synthetic rubberwere necessary.

The first solution to the problem involved polymerizing 2-chloro-1,3-butadiene, or“chloroprene,” to form the first major synthetic rubber, neoprene.

n CH2 CCH CH2 ... CH2C CHCH2 n...

ClCl

P P PA A

Chloroprene Neoprene

n CH2 CCH CH2 ... CH2C CHCH2 n...

CH3CH3

P P PA A

Isoprene Natural rubber

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8 POLYMERS

This approach is still used to produce a copolymer of 75% butadiene and 25% styreneknown as styrene–butadiene rubber (SBR). Roughly 40% of the rubber used in the worldtoday is SBR; another 35% is natural rubber that has been treated with sulfur.

The effect of cross-linking on elastomers can be demonstrated with a pair of rubberballs available from Flinn Scientific. One of the balls is a polybutadiene rubber that con-tains an unusually large amount of sulfur. Because the polymer chains are extensively cross-linked, this ball dissipates very little energy in the form of heat when it bounces. It is there-fore extremely resilient when bounced on the floor.

The other ball is a styrene–butadiene copolymer with much less cross-linking. Whendropped on the floor, the ball seems to “die.” This copolymer is used in applications wherean energy-absorbing medium is desired, such as automobile tires which must absorb someof the energy associated with the bumps we encounter as we drive down the highway.

P.4 FREE RADICAL POLYMERIZATION REACTIONSIt isn’t difficult to form addition polymers from monomers containing CPC double bonds;many of these compounds polymerize spontaneously unless polymerization is actively in-hibited. One of the problems with early techniques for refining gasoline, for example, wasthe polymerization of alkene components when the gasoline was stored. Even with mod-ern gasolines, deposits of “gunk” can form when a car or motorcycle is stored for extendedperiods without draining the gas tank.

The simplest way to catalyze the polymerization reaction that leads to an addition poly-mer is to add a source of a free radical to the monomer. The term free radical is used todescribe a family of very reactive, short-lived components of a reaction that contain oneor more unpaired electrons. In the presence of a free radical, addition polymers form bya chain reaction mechanism that contains chain initiation, chain propagation, and chaintermination steps.

Chain Initiation

A source of free radicals is needed to initiate the chain reaction. The free radicals are usu-ally produced by decomposing a peroxide such as di-tert-butyl peroxide or benzoyl perox-ide, shown below. In the presence of either heat or light, the peroxides decompose to forma pair of free radicals that contain an unpaired electron.

Chain Propagation

The free radical produced in the chain initiation step adds to an alkene to form a new freeradical.

C 2

P

O

P

O

C OO O OOOO

P

O

C OO O

CH3 CH3C O

CH3

CH3

O O COOA

A

CH3

CH3

A

A2 CH3 C O

CH3

CH3

O OA

AOO

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POLYMERS 9

The product of this reaction can then add additional monomers in a chain reaction.

CH3 CH2 CH2C O �

CH3

CH3

O CH2POA

ACH3 C O

CH3

CH3

O O O CH2OA

A

CH2CH2 � CH2PCH3 C O

CH3

CH3

O O O CH2OA

ACH2CH3 C O

CH3

CH3

O O O CH2 CH2O CH2O OA

A

Chain Termination

Whenever pairs of radicals combine to form a covalent bond, the chain reactions carriedby these radicals are terminated.

(CH2CH2)n (CH2CH2)m�CH3 C O

CH3

CH3

O O OA

ACH3CO

CH3

CH3

OO OA

A

(CH2CH2)n�m CH3CH3 CO

CH3

CH3

OOO OOA

AC

CH3

CH3

A

AOO

The Formation of Branched Polymers

We might expect the product of the free radical polymerization of ethylene to be a straight-chain polymer. As the chain grows, however, it begins to fold back on itself. This allows anintramolecular reaction to occur in which the site at which polymerization occurs is trans-ferred from the end of the chain to a carbon atom along the backbone.

When this happens, branches are introduced onto the polymer chain. Free radical poly-merization of ethylene produces a polymer that contains branches on between 1% and 5%of the carbon atoms. Of these branches, 10% contain two carbon atoms, 50% contain fourcarbon atoms, and 40% are longer side chains.

P.5 IONIC POLYMERIZATION REACTIONSAddition polymers can also be made by chain reactions that proceed through intermedi-ates that carry either a negative or positive charge.

(CH3)3C O

CH2CH2

CH2

CH2

CH2 CH2

AO O

H

H

E

E (CH3)3C O

CH2CH3

CH2

CH2

CH CH2

AOO

H

H

E

E

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10 POLYMERS

Anionic Polymerization

When the chain reaction is initiated and carried by negatively charged intermediates, thereaction is known as anionic polymerization. Like free radical polymerizations, these chainreactions take place via chain initiation, chain propagation, and chain termination steps.

The reaction is initiated by a Grignard reagent or alkyllithium reagent, which can bethought of as a source of a negatively charged CH3

� or CH3CH2� ion.

The CH3� or CH3CH2

� ion from one of these metal alkyls can attack an alkene toform a carbon–carbon bond.

The product of the chain initiation reaction is a new carbanion that can attack anotheralkene in a chain propagation step.

The chain reaction is terminated when the carbanion reacts with traces of water in the sol-vent in which the reaction is run.

Cationic Polymerization

The intermediate that carries the chain reaction during polymerization can also be a pos-itive ion, or cation. In this case, the cationic polymerization reaction is initiated by addinga strong acid to an alkene to form a carbocation.

The ion produced in this reaction adds monomers to produce a growing polymer chain.

� CH2 CH

XA

XA

CH3CHCH2CH�CH3CH�

XA

XA

P

� H�CH2 CH

XA

CH3CH�

XA

P

H2O� OH��CH3CH2(CH2CH)nCH2CH

XA

XA

CH3CH2(CH2CH)nCH2CH2

XA

XA

�

C

H

C �A

H

H H

H

H

H

X

A

A AOC

HA

HAOC

HA

HAOHO O C

H

CA

H

H H

A

A AO C

H

H

A

AO C

H

X

A

AOC CP HO� C C

H

H

A

AO C

H

X

A

AO �D

GD

G

C

H

C �A

H

H H

H

H

H

X

A

A AOHO C

H

CA

H

H H

A

A AO C

H

H

A

AO C

H

X

A

AOC CP HO� �D

GD

G

C

H

C LiA

H

H H

A

A AOHO O C

H

C � Li�A

H

H H

A

A AOHO

1012T_mod07_1-23 1/24/05 9:11 Page 10 EQA

POLYMERS 11

The chain reaction is terminated when the carbonium ion reacts with water that contami-nates the solvent in which the polymerization is run.

Advantages of Free-Radical versus Ionic Polymerization

The initiation step of ionic polymerization reactions has a much smaller activation energythan the equivalent step for free radical polymerizations. As a result, ionic polymerizationreactions are relatively insensitive to temperature, and they can be run at temperatures aslow as �70°C. Ionic polymerization therefore tends to produce a more regular polymer,with less branching along the backbone, and more controlled tacticity.

Because the intermediates involved in ionic polymerization reactions can’t combinewith one another, chain termination occurs only when the growing chain reacts with im-purities or with reagents that can be specifically added to control the rate of chain growth.It is therefore easier to control the average molecular weight of the product of ionic poly-merization reactions.

Ionic polymerizations are more difficult to carry out on an industrial scale than free rad-ical polymerizations. Ionic polymerization is therefore only used for monomers that don’tpolymerize by the free radical mechanism or to prepare polymers with a regular structure.

P.6 COORDINATION POLYMERIZATIONIn 1963 Karl Ziegler and Giulio Natta received the Nobel Prize in chemistry for their dis-covery of coordination compound catalysts for addition polymerization reactions.Ziegler–Natta catalysts provide the opportunity to control both the linearity and tacticityof the polymer.

Free radical polymerization of ethylene produces a low-density, branched polymer withside chains of one to five carbon atoms on up to 3% of the atoms along the polymer chain.Ziegler–Natta catalysts produce a more linear polymer, which is more rigid, with a higherdensity and a higher tensile strength. Polypropylene produced by free radical reactions, forexample, is a soft, rubbery, atactic polymer with no commercial value. Ziegler–Natta cat-alysts provide an isotactic polypropylene, which is harder, tougher, and more crystalline.

A typical Ziegler–Natta catalyst can be produced by mixing solutions of titanium(IV)chloride (TiCl4) and triethylaluminum [Al(CH2CH3)3] dissolved in a hydrocarbon solventfrom which both oxygen and water have been rigorously excluded. The product of the re-action is an insoluble olive-colored complex in which the titanium has been reduced to theTi(III) oxidation state.

The catalyst formed in the reaction can be described as coordinately unsaturated be-cause there is an open coordination site on the titanium atom. This allows an alkene to actas a Lewis base toward the titanium atom, donating a pair of � electrons to form a tran-sition metal complex.

CH2CH3CHH

HHCB

DG

D G

TiDG

D G

H2O� H��CH3CH(CH2CH)nCH2CH�

XA

XA

XA

CH3CH(CH2CH)nCH2CHOH

XA

XA

XA

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12 POLYMERS

The alkene is then inserted into a TiOCH2CH3 bond to form a growing polymer chain anda site at which another alkene can bond.

Thus, the titanium atom provides a template on which a linear polymer with carefully con-trolled stereochemistry can grow.

P.7 ADDITION POLYMERSAddition polymers such as polyethylene, polypropylene, poly(vinyl chloride), and poly-styrene are linear or branched polymers with little or no cross-linking. As a result, they arethermoplastic materials, which flow easily when heated and can be molded into a varietyof shapes. The structures, names, and trade names of some common addition polymers aregiven in Table P.1.

Polyethylene

Low-density polyethylene (LDPE) is produced by free-radical polymerization at high tem-peratures (200°C) and high pressures (above 1000 atm). The high-density polymer (HDPE)is obtained using Ziegler–Natta catalysis at temperatures below 100°C and pressures lessthan 100 atm. More polyethylene is produced each year than any other plastic. About 7800million pounds of low-density and 4400 million pounds of high-density polyethylene weresold in 1980. Polyethylene has no taste or odor and is lightweight, nontoxic, and relativelyinexpensive. It is used as a film for packaging food, clothing, and hardware. Most com-mercial trash bags, sandwich bags, and plastic wrapping are made from polyethylene films.Polyethylene is also used for everything from seat covers to milk bottles, pails, pans, anddishes.

Polypropylene

The isotactic polypropylene from Ziegler–Natta-catalyzed polymerization is a rigid, ther-mally stable polymer with an excellent resistance to stress, cracking, and chemical reac-tion. Although it costs more per pound than polyethylene, it is much stronger. Thus, bot-tles made from polypropylene can be thinner, contain less polymer, and cost less thanconventional polyethylene products. Polypropylene’s most important impact on today’s col-lege student takes the form of the plastic stackable chairs that abound on college campuses.

Poly(tetrafluoroethylene)

Tetrafluoroethylene (CF2PCF2) is a gas that boils at �76°C and is therefore stored in cylin-ders at high pressure. In 1938 Roy Plunkett received a cylinder of tetrafluoroethylene thatdidn’t deliver as much gas as it should have. Instead of returning the cylinder, he cut itopen with a hacksaw and discovered a white, waxy powder that was the first polytetraflu-oroethylene polymer. After considerable effort, a less fortuitous route to the polymer wasdiscovered, and polytetrafluoroethylene, or Teflon, became commercially available.

CH2CH3CHH

HHCB

DG

D G

TiDG

D G

CH2CH2CH2CH3

TiDG

D G

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POLYMERS 13

Teflon is a remarkable substance. It has the best resistance to chemical attack of anypolymer, and it can be used at any temperature between �73°C and 260°C with little ef-fect on its properties. It also has a very low coefficient of friction. (In simpler terms, it hasa waxy or slippery touch.) Even materials as “sticky” as crude rubber, adhesives, breaddough, and candy won’t stick to a Teflon-coated surface. Teflon is so slippery that it haseven been sprayed on plants, so that insects that might prey on the plants fall off.

TABLE P.I

Trade Name orStructure Chemical Name Common Name

(OCH2OCH2O)n polyethylene(OCF2OCF2O)n poly(tetrafluoroethylene) Teflon(OCH2O

ACH3

CHO)n polypropylene Herculon

(OCH2OACH3

ACH3

CO)n polyisobutylene butyl rubber

(OCH2OCHO)n polystyrene

(OCH2OACN

CHO)n polyacrylonitrile Orlon

(OCH2OACl

CHO)n poly(vinyl chloride) PVC

(OCH2OACO2CH3

CHO)n poly(methyl acrylate)

(OCH2OACH3

ACO2CH3

CO)n poly(methyl methacrylate) Plexiglas, Lucite

(OCH2OAH

CPAH

COCH2O)n polybutadiene

(OCH2OACl

CPCHOCH2O)n polychloroprene neoprene

(OCH2OAH

CP

ACH3

COCH2O)

n

poly(cis-1,4-isoprene) natural rubber

(OCH2OAH

ACH3

COCH2O) poly(trans-1,4-isoprene) gutta percha

CP n

Common Addition Polymers

1012T_mod07_1-23 1/24/05 9:11 Page 13 EQA

14 POLYMERS

Poly(vinyl Chloride) and Poly(vinylidene Chloride)

Chlorine is one of the top ten industrial chemicals in the United States—more than 20 billionpounds are produced annually. About 20% of the chlorine is used to make vinyl chloride(CH2PCHCl) for the production of poly(vinyl chloride), or PVC. The chlorine sub-stituents on the polymer chain make PVC more fire-resistant than polyethylene orpolypropylene. They also increase the force of attraction between polymer chains, whichincreases the hardness of the plastic. The properties of PVC can be varied over a widerange by adding plasticizers, stabilizers, fillers, and dyes, making PVC one of the most ver-satile plastics.

A copolymer of vinyl chloride (CH2PCHCl) and vinylidene chloride (CH2PCCl2) issold under the trade name Saran. The same increase in the force of attraction betweenpolymer chains that makes PVC harder than polyethylene gives thin films of Saran a ten-dency to “cling.”

Acrylics

Acrylic acid is the common name for 2-propenoic acid: CH2PCHCO2H. Acrylic fiberssuch as Orlon are made by polymerizing a derivative of acrylic acid known as acrylonitrile.

Other acrylic polymers are formed by polymerizing an ester of 2-propenoic acid, such asmethyl acrylate.

One of the most important acrylic polymers is poly(methyl methacrylate), or PMMA, whichis sold under the trade names Lucite and Plexiglas.

PMMA is a lightweight, crystal-clear, glasslike polymer used in airplane windows, taillightlenses, and light fixtures. Because it is hard, stable to sunlight, and extremely durable,PMMA is also used to make the reflectors embedded between lanes of interstate highways.

The unusual transparency of PMMA makes the polymer ideal for hard contact lenses.Unfortunately, PMMA is impermeable to oxygen and water. Oxygen must therefore betransported to the cornea of the eye in the tears and then passed under the contact lenseach time the eye blinks. Soft plastic lenses that pass both oxygen and water are made byusing ethylene glycol dimethacrylate to cross-link poly(2-hydroxyethyl methacrylate).

Poly(2-hydroxyethyl methacrylate) ... CH2C n...ACO2CH2CH2OH

ACH3

Poly(methyl methacrylate), PMMAn CH2 CCOCH3 ... CH2C n...PA

A

CO2CH3BO

CH3

ACH3

Poly(methyl acrylate)n CH2 CHCOCH3 ... CH2CH PACO2CH3

BO

n...

Polyacrylonitrilen CH2 CHCN ... CH2CH n...PACN

1012T_mod07_1-23 1/24/05 9:11 Page 14 EQA

POLYMERS 15

An interesting polymer can be prepared by copolymerizing a mixture of acrylic acidand the sodium salt of acrylic acid. The product of the reaction has the following structure.

The difference between the Na+ ion concentration inside the polymer network and in thesolution in which the polymer is immersed generates an osmotic pressure that draws wa-ter into the polymer. The amount of liquid that can be absorbed depends on the ionicstrength of the solution—the total concentration of positive and negative ions in the solu-tion. The polymer can absorb 800 times its own weight of distilled water, but only 300 timesits weight of tap water. Because the ionic strength of urine is equivalent to a 0.1 M NaClsolution, the superabsorbent polymer, which can be found in disposable diapers, can ab-sorb up to 60 times its weight of urine.

P.8 CONDENSATION POLYMERSThe first plastic (celluloid) and the first artificial fiber (rayon) were produced from cellu-lose, as noted in Section P.1. The first truly synthetic plastic was Bakelite, developed byLeo Baekland between 1905 and 1914. The synthesis of Bakelite starts with the reactionbetween formaldehyde (H2CO) and phenol (C6H5OH) to form a mixture of ortho- andpara-substituted phenols. At temperatures above 100°C, the phenols condense to form apolymer in which the aromatic rings are bridged by either OCH2OCH2O or OCH2O link-ages. The cross-linking in the polymer is so extensive that it is a thermoset plastic. Once apiece is formed, any attempt to change the shape of the plastic is doomed to failure.

Research started by Wallace Carothers and co-workers at Du Pont in the 1920s and1930s eventually led to the discovery of the families of condensation polymers known aspolyamides and polyesters. The polyamides were obtained by reacting a diacyl chloridewith a diamine.

The polyesters were made by reacting the diacyl chloride with a dialcohol.

While studying polyesters, Julian Hill found that he could wind a small amount of thepolymer on the end of a stirring rod and draw it slowly out of solution as a silky fiber. Oneday, when Carothers wasn’t in the lab, Hill and his colleagues tried to see how long a fiber

� �n HO(CH2)xOH n ClCBO

(CH2)yCBO

BO

BO

Cl O(C... H2)xOC(CH2)yC n... 2n HClDialcohol Diacyl chloride Polyester

n H2N n ClC(CH2)yCCl(CH2)xNH2 � � ... NH(CH2)xNHC(CH2)yC n...

OB

OB

OB

OB

2n HClDiamine Diacyl chloride Polyamide

Sodium polyacrylate ... CH2CH x CH2CH y...ACO2

�Na�

ACO2H

Ethylene glycol dimethacrylateCH2

CH3

CH2CCOCH2CH2OCCP PBO

B

ACH3

A

O

1012T_mod07_1-23 1/24/05 9:11 Page 15 EQA

16 POLYMERS

they could make by stretching a sample of the polymer as they ran down the hall. Theysoon realized that this playful exercise had oriented the polymer molecules in two dimen-sions and produced a new material with superior properties. They then tried the same thingwith one of the polyamides and produced a sample of what became the first synthetic fiber:nylon.

The polymerization process can be demonstrated by carefully pouring a solution ofhexamethylenediamine in water on top of a solution of adipoyl chloride in CH2Cl2.

A thin film of polymer forms at the interface between the two phases. By grasping the filmwith a pair of tweezers, we can draw a continuous string of nylon from the solution. Theproduct of the reaction is known as Nylon 6,6 because the polymer is formed from a di-amine that has six carbon atoms and a derivative of a dicarboxylic acid that has six carbonatoms.

The effect of pulling on the polymer with the tweezers is much like that of stretchingan elastomer—the polymer molecules become oriented in two dimensions. Why don’t thepolymer molecules return to their original shape when we stop pulling? Section P.3 sug-gested that polymers are elastic when there is no strong force of attraction between thepolymer chains. Polyamides and polyesters form strong hydrogen bonds between the poly-mer chains that keep the polymer molecules oriented, as shown in Figure P.9.

� �n H2N(CH2)6NH2 n ClCBO

(CH2)4CBO

BO

BO

Cl ... NH(CH2)6NHC(CH2)4C n ... 2n HClHexamethylene Adipoyl Nylon 6,6

diamine chloride

CON

NOC

A

A

B

H

H

OY

A

B

H

OY

NOCBOY

FIGURE P.9 The hydrogen bonds that form between the polymerchains when polyamides and polyesters are stretched help to keepthe chains oriented in a two-dimensional fiber.

Exercise P.4

A synthetic fiber known as Nylon 6 has the following structure.

Explain how the polymer is made.

BO

... NH(CH2)5C n ...

1012T_mod07_1-23 1/24/05 9:11 Page 16 EQA

POLYMERS 17

Solution

The polyamide must be made from a monomer that contains both an ONH2 and a OCOClfunctional group. The polymer is therefore made by the following condensation reaction.

The first polyester fibers were produced by reacting ethylene glycol and either tereph-thalic acid or one of its esters to give poly(ethylene terephthalate). This polymer is stillused to make thin films (Mylar) and textile fibers (Dacron and Fortrel).

Phosgene (COCl2) reacts with alcohols to form esters that are analogous to thoseformed when acyl chlorides react with alcohols.

The product of the reaction is called a carbonate ester because it is the diester of carbonicacid, H2CO3. Polycarbonates are produced when one of the esters reacts with an appro-priate alcohol, as shown in Figure P.10. The polycarbonate shown in Figure P.10 is knownas Lexan. It has a very high resistance to impact and is used in safety glass, bulletproofwindows, and motorcycle helmets.

��ClCCl 2 HOR 2 HClROCORBO

BO

�

�

n HOCH2 COCH3CH3CH2OH n

n

OCBO

BO

BO

BO

... OCH2CH2OC CH3OHC n ... 2

�n H2N(CH2)5CBO

BO

Cl ... NH(CH2)5C n ... n HCl

FIGURE P.10 Polycarbonates are condensation polymers formed by reacting a diester of carbonic acid with a dialcohol.

OOCOOBO

OOC n...BO

C OH

OH

A

A

CH3

CH3

CA

A

CH3

CH3

n HOn

�

�

2n ... O

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18 POLYMERS

P.9 PROPERTIES OF POLYMERSThe following variables can be controlled when producing a polymer.

• The monomer polymerized or the monomers copolymerized.• The reagent used to initiate the polymerization reaction.• The identity and amount of the reagent used to cross-link the polymer chains.• The temperature and pressure at which the polymerization occurs.• The solvent in which the monomer is polymerized.

Trade Name orStructure Common Name

Polyamides

(ONHO(CH2)6ONHOBO

CO(CH2)4OBO

CO)n Nylon 6, 6

(ONHO(CH2)6ONHOBO

CO(CH2)3OBO

CO)n Nylon 6, 10

(ONHO(CH2)5OBO

CO)n Nylon 6

(ONH CH2 NHOBO

CO(CH2)10OBO

CO)n Qiana

Polyaramides

(ONHBO

CO)n Kevlar

Polyesters

(OOOCH2CH2OOOBO

CBO

CO)n Dacron, Mylar

(OOOCH2 CH2OOOBO

CBO

CO)n Kodel

Polycarbonates

(OOACH3

ACH3

C OOBO

CO)n Lexan

Silicones

(OOOACH3

ACH3

SiO)n silicone rubber

TABLE P.2 Common Condensation Polymers

The structures and names of some common condensation polymers are given in TableP.2.

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POLYMERS 19

• The way the polymer is collected, which can produce either a more or less randomalignment of the polymer chains or a fabric in which the chains are aligned in onedirection.

Changing one or more of these parameters can affect the linearity of the polymer, its av-erage molecular weight, the tacticity of side chains on the polymer backbone, and the den-sity of the product.

It is also possible to change the properties of a polymer by adding either stabilizers orplasticizers. Stabilizers are used to increase the ability of a plastic to resist oxidation, tomake it less sensitive to either heat or light, or to make the material flame retardant. Plas-ticizers increase the flexibility of a plastic by acting as a lubricant, decreasing the frictionbetween molecules as one polymer chain moves past another. They also increase the amountof empty space—the so-called free volume—within the polymer by opening up space be-tween the polymer chains to increase the ease with which the chain ends, the side chains,and the main chain can move.

The result of all of these manipulations can be a polymer as strong as Kevlar, which isused to make bulletproof vests, or a material as easy to rip as a piece of paper. It can beas hard as a bowling ball or as soft as a piece of tissue paper. It can be as brittle as the dis-posable polystyrene glasses used at parties or as elastic as a Styrofoam coffee cup.

The following list describes some of the important properties of a polymer.

Heat capacity/heat conductivity: The extent to which the plastic or polymer acts as aneffective insulator against the flow of heat. (The polystyrene in disposable plasticglasses isn’t a very good insulator. However, blowing air through styrene while it isbeing polymerized gives the Styrofoam used for disposable coffee cups, which is amuch better insulator.)

Thermal expansion: The extent to which the polymer expands or contracts when heatedor cooled. (Silicone is often used to seal glass windows to their frames because it hasa very low coefficient of thermal expansion.) Thermal expansion is also concernedwith the question of whether the polymer expands or contracts by the same amountin all directions. (Polymers are usually anisotropic. They contain strong covalentbonds along the polymer chain and much weaker dispersive forces between the poly-mer chains. As a result, polymers can expand by differing amounts in differentdirections.)

Crystallinity: The extent to which the polymer chains are arranged in a regular struc-ture instead of a random fashion. (Some polymers, such as Silly Putty and Play Dough,are too amorphous and lack the rigidity needed to make a useful product. Polymersthat are too crystalline often are also too brittle.)

Permeability: The tendency of a polymer to pass extraneous materials. (Polyethy-lene is used to wrap foods because it is 4000 times less permeable to oxygen thanpolystyrene.)

Elastic modulus: The force it takes to stretch the plastic in one direction.Tensile strength: The strength of the plastic (i.e., the force that must be applied in one

direction to stretch the plastic until it breaks).Resilience: The ability of the plastic to resist abrasion and wear.Refractive index: The extent to which the plastic affects light as it passes through the

polymer. (Does it pass light the way PMMA does, or does it absorb light likePVC?)

Resistance to electric current: Is the material an insulator, like most polymers, or doesit conduct an electric current? (There is a growing interest in conducting polymers,

1012T_mod07_1-23 1/24/05 9:11 Page 19 EQA

20 POLYMERS

which can be charged and discharged, and photoconducting polymers that can pickup an electric charge when exposed to light.)


The Search for Synthetic Fibers

Synthetic fibers can be traced back to 1885, when the French Count Hilaire de Chardonnetreceived a patent for a synthetic silk. The first step in making Chardonnet silk, as it wasknown, involved dissolving cellulose from wood pulp in nitric acid to form cellulosenitrate. Anyone who has watched what happens when a flame is held close to a Ping-Pong ball (see Section P.1) should understand that cellulose nitrate, by itself, is far tooflammable to be used as a fiber. The next step in the production of Chardonnet silktherefore involved decomposing cellulose nitrate back to cellulose under conditions thatgenerated a continuous fiber. This was achieved by extruding a viscous solution of cellu-lose nitrate in alcohol through small holes into water.

In 1903, a slightly different process was patented in the United Kingdom. It involveddissolving cellulose from wood pulp in a mixture of caustic soda and carbon disulfide(CS2). The resulting viscous material was then extruded through small holes into a solu-tion of sulfuric acid to form a fiber that was sold under the trade name of Rayon. By1980, the demand for cellulose-based fibers had reached the point where more than3,250,000 metric tonnes were produced each year.

More than 100 years after the development of the first synthetic fiber, some peopletake comfort from the fact that the most important fiber is still cotton, which accountsfor almost 50% of the total world production of textile fibers. Advances in syntheticfibers, however, have generated materials that not only compete with, but surpass, nat-ural fibers.

Until recently, the standard against which all fibers were compared as potential insu-lators was the soft, fluffy down obtained from ducks and geese. Down is simultaneouslylightweight and an excellent thermal insulator. Unfortunately, it also readily absorbs wa-ter, and, when wet, loses much of its ability to act as an insulator. Thus, even if the sup-ply of down were plentiful—which it is not—and even if down were inexpensive—whichit is not—there would be a potential demand for a synthetic fiber that had the insulatingproperties of down but did not “wet.”

Several years ago, a synthetic insulator known as primaloft was prepared under acontract issued by the U.S. Army Research, Development and Engineering Center atNatick [Chemical and Engineering News, Oct. 16, 1989, p. 25]. Scanning electron mi-croscopy has shown that natural down is a mixture of relatively large-diameter fibers,which make it stiff, and very thin fibers that ensure the presence of many small pocketsof air. The result is a lightweight but stiff material that is a good insulator. Primaloftachieves the same effect by combining a small number of large-diameter polyester fibers with many more small fibers. The small fibers have a diameter of 7 �m, roughlyone-fourth the diameter of a human hair.

Primaloft is just as “warm” as down and even has the same feel when enclosed in ajacket or parka. More importantly, primaloft absorbs much less water when wet, andtherefore it doesn’t lose its insulating capacity in the rain. Furthermore, it is considerablyless expensive than down.

A variety of esoteric measurements are done to compare different textile fibers.Questions that are asked include the following: Does the fabric swell when the fibersabsorb water? Does the fabric have a tendency to build up static electric charge? Is the

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POLYMERS 21

fiber even, or does the yarn vary in diameter as it is spun? Does the fiber shrink on ex-posure to water? How well does the fiber “breathe?” (Is it permeable to air?) What isthe bursting strength of the fiber? What is its tear strength? How well does it resistsnag? How well does it stand up to abrasion and other forms of wear? How well does itconduct heat? (Is it a good insulator, like natural wool?) How well does it pass mois-ture in the form of water vapor? Is it water repellent, or, at least, can it be made waterrepellent? How well does it stretch? How well does it accept various dyes? Is it color-fast, once dyed? And, of course, perhaps as important as any other property—is itflammable?

Research in the 1990s will continue the search for new fabrics, such as those recentlyformulated by the Hoechst Celanese Corporation for athletic clothes that have the re-markable ability to pass water vapor but not liquid water through the fabric. These fab-rics allow perspiration to evaporate and still provide the protection against rain or snowexpected for water-repellant fabrics.

KEY TERMS

AcrylicAddition polymerAnionic polymerizationAtacticBlock copolymerBranched polymerCationic polymerizationCondensation polymerCoordination

polymerizationCopolymerCross-linked polymerElastomer

Free radicalFree radical

polymerizationGraft copolymerHomopolymerIsomerIsotacticLinear polymerMacromoleculeMonomerNylonPolyamidePolycarbonate

PolyesterPolyethylenePolymerPolypropylenePoly(vinyl chloride),

PVCRandom copolymerRegular copolymerStraight-chain polymerSyndiotacticTacticityTeflonZiegler–Natta catalyst

PROBLEMSPolymers

1. Use Berzelius’ definition of a polymer to sort the following compounds into groups ofpolymers.(a) formaldehyde, H2CO (b) ethylene, C2H4 (c) glyceraldehyde, C3H6O3

(d) 2-butene, C4H8 (e) cyclohexane, C6H12 (f ) glucose, C6H12O6

2. Which of the following substances are polymers?(a) Teflon (b) propane (c) acetic acid (d) poly(vinyl chloride) (e) polyethylene

Definition of Terms

3. Describe the difference between linear, branched, and cross-linked polymers.

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22 POLYMERS

4. Explain the following observations.(a) Linear polymers, such as the polyethylene in garbage bags, tear when stretched.(b) Lightly cross-linked polymers, such as rubber, return to their original shape when

stretched.(c) Highly cross-linked polymers, such as Bakelite, break when “stretched.”

5. Describe the difference between a homopolymer (such as polystyrene) and a copoly-mer (such as styrene–butadiene rubber).

6. Describe the difference between random, regular, block, and graft copolymers. Givean example of the sequence of monomer units that would be found in a short lengthof each of the polymers.

7. Describe the differences in the structures of atactic, isotactic, and syndiotactic formsof poly(vinyl chloride). Which of those would you expect to be the easiest to make?Which might be the most useful?

8. Calculate the range of molecular weights of polyethylene molecules in a sample inwhich individual chains contain between 500 and 50,000 (OCH2CH2O) units.

9. Individual chains in polystyrene polymers typically weigh between 200,000 and 300,000amu. If the formula for the monomer is C6H5CHPCH2 and the polymer is an exam-ple of an addition polymer, how many monomers does the typical chain contain?

10. Calculate the average molecular weight of the polymer chains in the cellulose nitratein Ping-Pong balls if the formula for the monomer is C6H9NO7 and the average poly-mer chain contains 1500 monomers.

11. Explain why it is possible to measure the average molecular weight of linear polymerssuch as polyethylene, but not cross-linked polymers such as those found in rubber.

12. Thermoplastic polymers flow when heated and can be molded into shapes they retainon cooling. Explain how increasing the amount of cross-linking between polymer chainscan transform a thermoplastic polymer into a rigid thermoset polymer.

13. Describe the difference between addition and condensation polymers. Give an exam-ple of each.

14. Classify the following as either addition or condensation polymers.(a) polyethylene, [...(CH2CH2)n...](b) poly(vinyl chloride), [...(CH2CHCl)n...](c) Nylon 6,6, [...(CO(CH2)4CONH(CH2)6NH)n...](d) polyester, [...(OCH2CH2OCOC6H4CO)n...](e) silicone, [...(OSi(CH3)2)n...](f ) Plexiglas, [...(CH2C(CH3)(CO2CH3))n...]

Elastomers

15. What are the characteristic properties of an elastomer?16. Explain why elastomers must contain long, flexible molecules that are coiled in the

natural state. Explain why polymers have to have some cross-links to be elastomersbut can’t have too many.

17. Natural rubber becomes too soft when heated and too hard when cooled to be of muchuse. Explain what happens when natural rubber is treated with sulfur that turns thematerial into a commercially useful product.

Free Radical Polymerization Reactions

18. Describe what happens during the chain initiation, chain propagation, and chain termi-nation steps when propylene (CH3CHPCH2) is polymerized by a free radical mechanism.

1012T_mod07_1-23 1/24/05 9:11 Page 22 EQA

POLYMERS 23

19. Write the Lewis structures of the intermediates in the free radical polymerization ofvinyl chloride, CH2PCHCl.

Ionic and Coordination Polymerization Reactions

20. Describe what happens during the chain initiation, chain propagation, and chain ter-mination steps in the anionic polymerization of vinyl chloride (CH2PCHCl) whenmethyllithium (CH3Li) is used as the chain initiator. Write the Lewis structures of allintermediates.

21. Describe what happens during the chain initiation, chain propagation, and chain ter-mination steps in the cationic polymerization of vinylidene chloride (CH2PCCl2) whenhydrobromic acid (HBr) is used to initiate the reaction. Write the Lewis structures ofall intermediates.

22. A typical Ziegler–Natta catalyst consists of a mixture of TiCl3 and Al(C2H5)3. Explainhow the formation of a complex between a molecule of ethylene and the titanium atomin the catalyst can be thought of as an example of a Lewis acid-base reaction.

Addition Polymers

23. Polypropylene can be made in both atactic and isotactic forms. The atactic form is softand rubbery, with no commercial value. The isotactic form is much more crystalline—it is hard enough, for example, to be used for furniture. Explain the difference betweenthe physical properties of the two forms of the polymer.

24. Polyethylene can be made in both high-density and low-density forms. One of the poly-mers has a linear structure; the other is branched, with short side chains of up to fivecarbon atoms attached to the polymer backbone. Which structure would you expectto give the denser polymer? Which structure would give the more crystalline polymer?

25. Teflon, (CF2CF2)n, is a waxy polymer to which practically nothing sticks. It is also themost chemically inert of all polymers. Describe at least five ways in which a polymerwith these properties can be used.

26. Use the concept of van der Waals forces to explain why plastic wrap made from Saranclings to itself, whereas plastic wrap made from polyethylene does not.

Condensation Polymers

27. Predict the formula of the repeating unit in the condensation polymers formed by thefollowing reactions.

28. Explain the difference between the reactions used to prepare Nylon 6,6 and Nylon 6.

�HCl ...

�HCl ...

�(a) Dacron: HOCH2 CClClCCH2OHBO

BO

�HCl ...(d) polyaramide: H2N CClBO

�HCl ...(e) silicone rubber: (CH3)2Si(OH)2

�(b) Nylon 6,6: H2 ClC(CH2 4CClN(CH2 6NH2) )BO

BO

� ...�(c) polycarbonate: HOCH2CH2OH (CH3O)2C CH3OHOP

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1

M O D U L E

8BIOCHEMISTRY

BIO.1 Biochemistry

BIO.2 The Amino Acids

BIO.3 The Acid–Base Chemistry of Amino Acids

BIO.4 The Chemistry of Hemoglobin and Myoglobin

BIO.5 Peptides and Proteins

BIO.6 The Structure of Proteins

BIO.7 Carbohydrates: The Monosaccharides

BIO.8 Carbohydrates: The Disaccharides and Polysaccharides

BIO.9 Lipids

Chemistry in the World Around Us: The Search for New Drugs

BIO.10 Nucleic Acids

BIO.11 Protein Biosynthesis

BIO.1 BIOCHEMISTRYBiochemistry is the study of the chemistry of living organisms. Biomolecules are oftensorted into four categories: (1) peptides and proteins, (2) carbohydrates, (3) nucleic acids,and (4) lipids.

The term protein comes from the Greek word proteios, which means “of first impor-tance.” Biochemistry, for many years, was almost synonymous with the study of proteinsbecause these compounds serve the broadest array of functions of any class of biomole-cules, including the following:

Structure: The actin and myosin in muscles, the collagen in skin and bone, and the ker-atins in hair, horn, and hoof are examples of proteins whose primary function is toproduce the structure of the organism.

Catalysis: Most chemical reactions in living systems are catalyzed by enzymes, whichare proteins.

Control: Many proteins regulate or control biological activity. Insulin, for example, con-trols the rate at which sugar is taken into the cell.

Energy: Some proteins, including the casein in milk and the albumin in eggs, are usedprimarily to store food energy.

Transport: O2 is carried through the bloodstream by hemoglobin. Other proteins trans-port sugars, amino acids, and ions across cell membranes.

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2 BIOCHEMISTRY

Protection: The first line of defense against viruses and bacteria are the antibodies pro-duced by the immune system, which are based on proteins.

The term carbohydrate reflects the fact that many of the compounds in that categoryhave the empirical formula CH2O—they are literally “hydrates of carbon.” Carbohydratesare the primary source of food energy for most living systems. They include simple sugarssuch as glucose (C6H12O6) and sucrose (C12H22O11) as well as polymers of these sugarssuch as starch, glycogen, and cellulose. Carbohydrates are produced from CO2 and H2Oduring photosynthesis and are therefore the end products of the process by which plantscapture the energy in sunlight.

The name nucleic acid was originally given to a class of relatively strong acids that werefound in the nuclei of cells. As monomers, nucleic acids such as adenosine triphosphate(ATP) are involved in the process by which cells capture food energy and make it avail-able to fuel the processes that keep cells alive. As polymers, they store and process the in-formation that allows the organism to grow and eventually reproduce.

Compounds are classified as proteins, carbohydrates, and nucleic acids on the basis ofsimilarities in their structures. Lipids, on the other hand, are defined on the basis of theirphysical properties. Any molecule in a biological system that is soluble in nonpolar sol-vents is classified as a lipid (from the Greek word lipos, “fat”). The lipid known as cho-lesterol, for example, is virtually insoluble in water, but it is soluble in a variety of nonpo-lar solvents—including the nonpolar region between the inner and outer surfaces of a cellmembrane.

BIO.2 THE AMINO ACIDSProteins are formed by polymerizing monomers that are known as amino acids becausethey contain an amine (ONH2) and a carboxylic acid (OCO2H) functional group. Withonly one exception,1 the amino acids used to synthesize proteins are primary amines withthe following generic formula.

These compounds are known as �-amino acids because the ONH2 group is on the carbonatom next to the OCO2H group, the so-called � carbon atom of the carboxylic acid.

The chemistry of amino acids is complicated by the fact that the ONH2 group is a baseand the OCO2H group is an acid. In aqueous solution, an H� ion is therefore transferredfrom one end of the molecule to the other to form a zwitterion (from a German wordmeaning “mongrel ion,” or hybrid ion).

Zwitterions are simultaneously electrically charged and electrically neutral. They containpositive and negative charges, but the net charge on the molecule is zero.

H2NCAR

HCO2H 88n H3N�CAR

HCO2�

H2NCAR

HCO2HAn amino acid

1The sole exception is the amino acid proline, which is a secondary amine.

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BIOCHEMISTRY 3

More than 300 amino acids are listed in the Practical Handbook of Biochemistry andMolecular Biology,2 but only the 20 amino acids in Table BIO.1 are used to synthesize pro-teins. Most of the amino acids differ only in the nature of the R substituent. The standardamino acids are therefore classified on the basis of the R groups. Amino acids with non-polar substituents are said to be hydrophobic (“water-hating”). Amino acids with polar Rgroups that form hydrogen bonds to water are classified as hydrophilic (“water-loving”).The remaining amino acids have substituents that carry either negative or positive chargesin aqueous solution at neutral pH and are therefore strongly hydrophilic.

Exercise BIO.1

Use the structures of the following amino acids in Table BIO.1 to classify the compoundsas either nonpolar/hydrophobic, polar/hydrophilic, negatively charged/hydrophilic, or pos-itively charged/hydrophilic.(a) valine, R � OCH(CH3)2 (b) serine, R � OCH2OH(c) aspartic acid, R � OCH2CO2

� (d) lysine, R � O(CH2)4NH3�

Solution

(a) Valine is a nonpolar, hydrophobic amino acid.(b) Serine is a polar, hydrophilic amino acid.(c) Aspartic acid is a hydrophilic amino acid with a negatively charged R group.(d) Lysine is a hydrophilic amino acid with a positively charged R group.

With the exception of glycine, the common amino acids all contain at least one chiralcarbon atom. The amino acids therefore exist as pairs of stereoisomers. The structures ofthe D and L isomers of alanine are shown in Figure BIO.1. Although D amino acids can befound in nature, only the L isomers are used to form proteins. The D isomers are most of-ten found attached to the cell walls of bacteria and in antibiotics that attack bacteria. Thepresence of the D isomers protects the bacteria from enzymes the host organism uses toprotect itself from bacterial infection by hydrolyzing the proteins in the bacterial cell wall.

A few biologically important derivatives of the standard amino acids are shown in Fig-ure BIO.2. Anyone who has used an “antihistamine” to alleviate the symptoms of expo-sure to an allergen can appreciate the role that histamine—a decarboxylated derivative ofhistidine—plays in mediating the body’s response to allergic reactions. L-DOPA, which isa derivative of tyrosine, has been used to treat Parkinson’s disease. The compound receivednotoriety in the film Awakening, which documented its use as a treatment for other neu-rological disorders. Thyroxine, which is an iodinated ether of tyrosine, is a hormone thatacts on the thyroid gland to stimulate the rate of metabolism.

FIGURE BIO.1 The D and L stereoisomers of the amino acid alanine.

2CRC Press, Boca Raton, Florida, 1989.

ACH3

CO2�NH3

�HC

#

{{

D-Alanine

ACH3

CO2�

NH3�

HC

#

{{

L-Alanine

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4 BIOCHEMISTRY

TABLE BIO.1 The 20 Standard Amino Acids

Name Structure (at neutral pH) Name Structure (at neutral pH)

Nonpolar (Hydrophobic) R Groups Polar (Hydrophilic) R Groups

Glycine (Gly) Serine (Ser)

Alanine (Ala)Threonine (Thr)

Valine (Val)

Tyrosine (Tyr)

Leucine (Leu)

Cysteine (Cys)

Asparagine (Asn)

Glutamine (Gln)

Methionine (Met)

Phenylalanine (Phe)

Aspartic acid (Asp)

Tryptophan (Trp)

Glutamic acid (Glu)

(continued)

Negatively Charged R Groups

Isoleucine (Ile)

Proline (Pro)

H3N�OCACH3

HOCO2�

H3N�OCA

GCH3 CH3D

HOCO2�

CH

H3N�OCA

CG

CH3

HD

CH3

2

HOCO2�

CHA

CO2

H

H2C CH2

CH�N

H

H

H2C

H3N�OCA

CH2

2

HOCO2�

CHA

CH3ASA

H3N�OCA

2

HOCO2�

CHA

ON H

H3N�OC

ACH2OH

HOCO2�

H3N�OCA

GCH3 D

OH

HOCO2�

CH

H3N�OCAH

HOCO2�

H3N�OCA

GCH3 D

CH2CH3

HOCO2�

CH

H3N�OCA

2

HOCO2�

CH

H3N�OC

ACH2

HOCO2�

AOH

A

H3N�OC

ACH2SH

HOCO2�

H3N�OC

ONH2

ACH2

ACBO

HOCO2�

H3N�OC

ACH2

ONH2

ACH2

ACBO

HOCO2�

H3N�OC

ACH2

A

HOCO2�

CO2�

H3N�OC

ACH2

ACH2

A

HOCO2�

CO2�

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BIOCHEMISTRY 5

BIO.3 THE ACID–BASE CHEMISTRY OF AMINO ACIDSAcetic acid and ammonia often play an important role in the discussion of the chemistryof acids and bases. One of the compounds is a weak acid; the other is a weak base.

CH3CO2H � H2O 88nm88 CH3CO2� � H3O� Ka � 1.8 � 10�5

NH3 � H2O 88nm88 NH4� � OH� Kb � 1.8 � 10�5

Thus, it is not surprising that an H� ion is transferred from one end of the molecule to theother when an amino acid dissolves in water.

The zwitterion is the dominant species in aqueous solutions at physiological pH (pH � 7).

H2NCAR

HCO2H 88n H3N�CAR

HCO2�

Name Structure (at neutral pH) Name Structure (at neutral pH)

Positively Charged R Groups Positively Charged R Groups

Lysine (Lys)

Histidine (His)

FIGURE BIO.2 Three biologically important derivatives of amino acids.

Arginine (Arg) H3N�OC

ACH2

ACH2

HOCO2�

ACH2

ANHA

AC NH2

�B

NH2

H3N�OC

ACH2

ACH2

HOCO2�

ACH2

ACH2

ANH3

�

A

A

AA A

A

A

N

N

H

CH2CH2NH3� CH2CH

NH3�

CO2�

A

ACH2CH

NH3�

CO2�

Histamine L-DOPAHO

HO

A

A

A A A AHO

I

I

O Thyroxine

H3N�OC

ACH2

HOCO2�

N�H

NH

H

DH

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6 BIOCHEMISTRY

The zwitterion can undergo acid–base reactions, however, if we add either a strong acid ora strong base to the solution.

Imagine what would happen if we add a strong acid to a neutral solution of an aminoacid in water. In the presence of a strong acid, the OCO2

� end of the molecule picks upan H� ion to form a molecule with a net positive charge.

In the presence of a strong base, the ONH3� end of the molecule loses an H� ion to form

a molecule with a net negative charge.

Figure BIO.3 shows what happens to the pH of an acidic solution of glycine when theamino acid is titrated with a strong base, such as NaOH.

H3N�CAR

HCO2� 88nOH�

H2NCAR

HCO2�

H3N�CAR

HCO2� 88nH�

H3N�CAR

HCO2H

FIGURE BIO.3 Titration curve for the titration of glycinewith a strong base.

To understand the titration curve, let’s start with the equation that describes the aciddissociation equilibrium constant expression for an acid, HA.

Ka � �[H3O

[H

�

A][

]A�]

�

Let’s now rearrange the Ka expression,

[H3O�] � Ka � �[[HA

A�]

]�

0.50 1.0

12

1.5 2.

10

8

4

2

pH 6

H+ ions dissociated/molecule

pK2

pK1

pI

H3N

CH2CO

O

–H 2

NCH2COO

– + H+

H 3NCH2

COOHH 3

NCH 2CO

O–+

H+

+

+

+

1012T_mod08_1-40 1/25/05 12:45 Page 6 EQA

BIOCHEMISTRY 7

take the log to the base 10 of both sides of the equation,

log[H3O�] � log Ka � log�[[HA

A�]

]�

and then multiply both sides of the equation by �1.

�log[H3O�] � �log Ka � log�[[HA

A�]

]�

By definition, the term on the left side of the equation is the pH of the solution, andthe first term on the right side is the pKa of the acid.

pH � pKa � log�[[HA

A�]

]�

The negative sign on the right side of this equation is often viewed as “inconvenient.” Thederivation therefore continues by taking advantage of the following feature of logarithmicmathematics

�log�[[HA

A�]

]� � log�

[[HA

A

�]]

�

to give the following form of the equation.

pH � pKa � log�[[HA

A

�]]

�

This equation is known as the Henderson–Hasselbalch equation, and it can be used to cal-culate the pH of the solution at any point in the titration curve in Figure BIO.3.

The following occurs as we go from left to right across the titration curve.

• The pH initially increases as we add base to the solution because the base depro-tonates some of the positively charged H3N�CH2CO2H ions that were present inthe strongly acidic solution.

• The pH then levels off because we form a buffer solution in which we have rea-sonable concentrations of both an acid, H3N�CH2CO2H, and its conjugate base,H3N�CH2CO2

�.• When virtually all of the H3N�CH2CO2H molecules have been deprotonated, we

no longer have a buffer solution and the pH rises rapidly when more NaOH is addedto the solution.

• The pH then levels off as some of the neutral H3N�CH2CO2� molecules lose pro-

tons to form negatively charged H2NCH2CO2� ions. When the ions are formed, we

once again get a buffer solution in which the pH remains relatively constant untilessentially all of the H3N�CH2CO2

� molecules have been converted intoH2NCH2CO2

� ions.• At this point, the pH rises rapidly until it reaches the value observed for a strong

base.

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8 BIOCHEMISTRY

The pH titration curve tells us the volume of base required to titrate the positivelycharged H3N�CH2CO2H molecule to the H3N�CH2CO2

� zwitterion. If we only add halfas much base, only half of the positive ions would be titrated to zwitterions. In other words,the concentration of the H3N�CH2CO2H and H3N�CH2CO2

� ions would be the same.Or, using the symbolism in the Henderson–Hasselbalch equation:

[HA] � [A�]

Because the concentrations of the ions are the same, the logarithm of the ratio of theirconcentrations is zero.

log�[[HA

A

�]]

� � 0

Thus, at this particular point in the titration curve, the Henderson–Hasselbalch equationgives the following equality.

pH � pKa

We can therefore determine the pKa of an acid by measuring the pH of a solution in whichthe acid has been half-titrated.

Because there are two titratable groups in glycine, we get two points at which the aminoacid is half-titrated. The first occurs when half of the positive H3N�CH2CO2H ions havebeen converted to neutral H3N�CH2CO2

� molecules. The second occurs when half of theH3N�CH2CO2

� zwitterions have been converted to negatively charged H2NCH2CO2�

ions.The following results are obtained when this technique is applied to glycine.

Let’s compare these values with the pKa values of acetic acid and the ammonium ion.

The acid–base properties of the �-amino group in an amino acid are very similar to theproperties of ammonia and the ammonium ion. The �-amine, however, has a significanteffect on the acidity of the carboxylic acid. The �-amine increases the value of Ka for thecarboxylic acid by a factor of about 100.

The inductive effect of the �-amine can only be felt at the �-CO2H group. If we lookat the chemistry of glutamic acid, for example, the OCO2H group on the R substituent hasan acidity that is close to that of acetic acid.

H3N�C

pKa � 4.07

HCO2�

pKa 9.47 pKa �� 2.10

ACH2

ACH2

ACO2

�

CH3CO2H pKa � 4.74NH4

� pKa � 9.24

2.359.78

H3N�CH2CO2

�

pKa� pKa�

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BIOCHEMISTRY 9

When we titrate an amino acid from the low end of the pH scale (pH � 1) to the highend (pH � 13), we start with an ion that has a net positive charge and end up with an ionthat has a net negative charge.

Somewhere between these extremes, we have to find a situation in which the vast major-ity of the amino acids are present as the zwitterion—with no net electric charge. This pointis called the isoelectric point (pI) of the amino acid.

For simple amino acids, in which the R group doesn’t contain any titratable groups, theisoelectric point can be calculated by averaging the pKa values for the �-carboxylic acidand �-amino groups. Glycine, for example, has a pI of about 6.

At pH � 6, more than 99.98% of the glycine molecules in the solution are present as theneutral H3N�CH2CO2

� zwitterion.When calculating the pI of an amino acid that has a titratable group on the R side

chain, it is useful to start by writing the structure of the amino acid at physiological pH(pH � 7). Lysine, for example, could be represented by the following diagram.

At physiological pH, lysine has a net positive charge. Thus, we have to increase the pH ofthe solution to remove positive charge in order to reach the isoelectric point. The pI forlysine is simply the average of the pKa values of the two ONH3

� groups.

At pH � 10, all of the carboxylic acid groups are present as OCO2� ions, and the total

population of the ONH3� groups is equal to one. Thus, the net charge on the molecule at

that pH is zero.If we apply the same technique, given above, to the pKa data for glutamic acid, we get

a pI of 3.09. The three amino acids in this section therefore have very different pI values.

Glutamic acid (R � OCH2CH2CO2�) pI � 3.09

Glycine (R � OH) pI � 6.07Lysine (R � OCH2CH2CH2CH2NH3

�) pI � 9.99

Thus, it isn’t surprising that a common technique for separating amino acids (or the pro-teins they form) involves placing a mixture in the center of a gel and then applying a strongvoltage across the gel. This technique, which is known as gel electrophoresis, is based on

9.18 � 10.79pI � ≈ 9.99

2

H3N�CA

(CANH3

�pKa � 10.79

H2)4

HCO2�

pKa � 9.18 pKa � 2.16

2.35 � 9.78pI � � 6.07

2

H3N�CHRCO2H 88nOH �

H2NCHRCO2�

pH�� 7 pH 7

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10 BIOCHEMISTRY

the fact that amino acids or proteins that carry a net positive charge at the pH at whichthe separation is done will move toward the negative electrode, whereas those with a netnegative charge will move toward the positive electrode.

BIO.4 THE CHEMISTRY OF HEMOGLOBIN AND MYOGLOBINAt one time or another, everyone has experienced the momentary sensation of having tostop, to “catch one’s breath,” until enough O2 can be absorbed by the lungs and trans-ported through the bloodstream. Imagine what life would be like if we had to rely only onour lungs and the water in our blood to transport oxygen through our bodies.

O2 is only marginally soluble (�0.0001 M) in blood plasma at physiological pH. If wehad to rely on the oxygen that dissolved in blood as our source of oxygen, we would getroughly 1% of the oxygen to which we are accustomed. (Consider what life would be likeif the amount of oxygen you received was equivalent to only one breath every 5 min, in-stead of one breath every 3 s.) The evolution of forms of life even as complex as an earth-worm required the development of a mechanism to actively transport oxygen through thesystem.

Our bloodstream contains about 150 g/L of the protein known as hemoglobin (Hb),which is so effective as an oxygen carrier that the concentration of O2 in the bloodstreamreaches 0.01 M—the same concentration as in air. Once the Hb–O2 complex reaches thetissue that consumes oxygen, the O2 molecules are transferred to another protein—myo-globin (Mb)—which transports oxygen through the muscle tissue.

The site at which oxygen binds to both hemoglobin and myoglobin is the heme shownin Figure BIO.4. At the center of the heme is an Fe(II) atom. Four of the six coordinationsites around the atom are occupied by nitrogen atoms from a planar porphyrin ring. The

FIGURE BIO.4 The heme groupin hemoglobin.

N

A

A

A

A

A

A

A

A

AA

A

AA

A

AA

B

AA B

NN

HNCH2

CHN

HC

O

Fe(II)

N N

CH2

H3C

H3C

H2C CH3

CH3

A ACH2 CH2AA CH2 CH2 COO��OOC

CH

B

OO

B CH

A B

CD

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BIOCHEMISTRY 11

fifth coordination site is occupied by a nitrogen atom from a histidine side chain on one ofthe amino acids in the protein. The last coordination site is available to bind an O2 mole-cule. The heme is therefore the oxygen-carrying portion of the hemoglobin and myoglo-bin molecules. This raises the question, What is the function of the globular protein or“globin” portion of the molecules?

The structure of sperm whale myoglobin, shown in Figure BIO.5, suggests that theoxygen-carrying heme group is buried inside the protein portion of the molecule, whichkeeps pairs of heme groups from coming too close together. This is important, because theproteins need to bind O2 reversibly and the Fe(II) heme, by itself, cannot do this. Whenthere is no globin to protect the heme, it reacts with oxygen to form an oxidized Fe(III)atom instead of an Fe(II)–O2 complex.

Hemoglobin consists of four protein chains, each about the size of a myoglobin mole-cule, which fold to give a structure that looks very similar to myoglobin. Thus, hemoglo-bin has four separate heme groups that can bind a molecule of O2. Even though the dis-tance between the iron atoms of adjacent hemes in hemoglobin is very large—between 250and 370 nm—the act of binding an O2 molecule at one of the four hemes in hemoglobinleads to a significant increase in the affinity for O2 binding at the other hemes.

The cooperative interaction between different binding sites makes hemoglobin an un-usually good oxygen-transport protein because it enables the molecule to pick up as muchoxygen as possible once the partial pressure of the gas reaches a particular threshold level,and then give off as much oxygen as possible when the partial pressure of O2 drops signifi-cantly below the threshold level. The hemes are much too far apart to interact directly. How-ever, changes that occur in the structure of the globin that surrounds a heme when it picksup an O2 molecule are mechanically transmitted to the other globins in the protein. Thesechanges carry the signal that facilitates the gain or loss of an O2 molecule by the other hemes.

Drawings of the structures of proteins often convey the impression of a fixed, rigidstructure, in which the side chains of individual amino acid residues are locked into posi-tion. Nothing could be further from the truth. The changes that occur in the structure ofhemoglobin when oxygen binds to the hemes are so large that crystals of deoxygenated he-moglobin shatter when exposed to oxygen. Further evidence for the flexibility of proteinscan be obtained by noting that there is no path in the crystal structures of myoglobin andhemoglobin along which an O2 molecule can travel to reach the heme group. The fact that

FIGURE BIO.5 Schematic drawing of the structure ofsperm whale myoglobin. The heme group is in apocket formed by the protein in the top left of the diagram.

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12 BIOCHEMISTRY

these proteins reversibly bind oxygen suggests that they must undergo simple changes intheir conformation—changes that have been called breathing motions—that open up andthen close down the pathway along which an O2 molecule travels as it enters the protein.Computer simulations of the motion within proteins suggests that the interior of a proteinhas a significant “fluidity,” with groups moving within the protein by as much as 20 nm.

BIO.5 PEPTIDES AND PROTEINSMyoglobin and hemoglobin are important examples of the class of compounds known asproteins, which are linear polymers of between 40 and 10,000 (or more) amino acids. Theaverage molecular weight of an amino acid is about 110 amu. As a result, a modestly sizedprotein with only 300 amino acids has a molecular weight of 33,000 g/mol, and very largeproteins can have molecular weights as high as 1,000,000 g/mol.

Proteins are formed by joining the OCO2H end of one amino acid with the ONH2 endof another to form an amide. The OCONHO bond between amino acids is known as apeptide bond because relatively short polymers of amino acids are known as peptides.

The same OCONHO bond forms the backbone of both proteins and synthetic fibers suchas nylon. This raises an interesting question: How do we explain the enormous range ofstructures and functions of proteins when nylon has such regular properties?

Nylon has a regular structure that repeats monotonously from one end of the polymerto the other because the monomers from which it is made are symmetrical. The two endsof an amino acid, on the other hand, are different. Each monomer has both an ONH2 headand a OCO2H tail. Thus, four different dipeptides can be formed from only two aminoacids. Aspartic acid (Asp) and phenylalanine (Phe), for example, can give two symmetri-cal dipeptides—Phe-Phe and Asp-Asp—and two unsymmetrical dipeptides—Phe-Asp andAsp-Phe—as shown in Figure BIO.6. When the full range of amino acids is considered, itis possible to make 400 (202) different dipeptides, 64 million (206) different hexapeptides,and 1052 (2040) different proteins that contain only 40 amino acids.

BO

A dipeptide

�H3N�CHCO�

RBO

H3N�CHCO�

RBO

H3N�CHCNHCHCO�

RBOR

FIGURE BIO.6 The four dipeptides that can be formed by phenylalanine and aspartic acid.

A

BA

A AA ACH

CH2

A

A

ACH

CH2

CO2�H3N� C NH

O

Phe-Phe

A

BA

A AA ACH

CH2

A

A

ACH

CH2

CO2�

CO2�

H3N� C NH

O

Asp-Phe

A

BA

A AA ACH

CH2

A

A

ACH

CH2

CO2�

CO2�

H3N� C NH

O

Phe-Asp

A

BA

A AA ACH

CH2

A

A

ACH

CH2

CO2�

CO2�CO2

�

H3N� C NH

O

Asp-Asp

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BIOCHEMISTRY 13

Differences between the structures of even closely related dipeptides such as Asp-Pheand Phe-Asp give rise to significant differences in their properties. The methyl ester ofAsp-Phe, for example, has a very sweet taste and is sold as an artificial sweetener underthe name aspartame.

The ester of the dipeptide with the opposite arrangement of amino acids, Phe-Asp, doesnot taste sweet and has no commercial value. As the length of the polymer chain increasesand the number of possible combinations of R groups increases, polymer chains with analmost infinite variety of structures and properties are produced.

In recent years, a group of naturally occurring peptides that mimic painkilling drugssuch as morphine has been discovered in human brain cells. These enkephalins (from Greekand meaning “in the head”) hold the promise of a synthetic painkiller that is both safe andnonaddictive. One of the enkephalins is a pentapeptide that contains four different aminoacids—tyrosine (Tyr), glycine (Gly), phenylalanine (Phe), and methionine (Met). The firststep in describing the structure of the peptide is to list the amino acids in the order in whichthey are found on the peptide chain: Tyr-Gly-Gly-Phe-Met. We then have to identify theamino acid at the OCO2H end of the chain and the amino acid at the ONH2 end. By con-vention, proteins are listed from the N-terminal amino acid residue toward the C-terminalend. The structure of the enkephalin is shown in Figure BIO.7.

BCO2CH3

O

Aspartame

H3N�O OOOCH

CH2

CO2�

NHOCHCA

CH2A

A

FIGURE BIO.7 The structure of the naturally occurring pentapeptide Tyr-Gly-Gly-Phe-Met that binds to thesame sites in the brain as synthetic painkillers such as morphine.

The pentapeptide in Figure BIO.7 illustrates the perils that face anyone who tries tosynthesize peptides or proteins from amino acids. In order to make the enkephalin in largequantities, we would have to overcome the following problems.

• Peptide bonds are not easy to form. In theory, a carboxylic acid could react with anamine to form an amide.

88nOCBO

OOH ... ... ...... � H2NO OCBO

ONHO � H2O

A

BA

A AA ACH

CH2

CH2

A

H3N� C NH

O B

A AA CH2C NH

O B

A AC NH

O

N-terminalamino acid

C-terminalamino acidTyr-Gly-Gly-Phe-Met

OH

A

BA

AA ACH

CH2

A

A

A CH

CH2

A

CH2

CH3A

S

C NH

O

A CO2�

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14 BIOCHEMISTRY

In practice, carboxylic acids are more likely to react with amines in a simpleacid–base reaction to form a salt.

We therefore have to find a way to force the reaction to form the amide.• Forming a peptide bond is also an uphill process (�G° � �17 kJ/molrxn), so a way

must be found to drive the reaction forward.• Because the sequence of amino acids is important, they must be added to the chain

one at a time, in a carefully controlled fashion. Thus, a significant entropy factormust be overcome during the synthesis of peptides or proteins.

• The R groups of certain amino acids must be protected during polymerization sothat no reactions take place on the side chains.

In 1984 R. B. Merrifield received the Nobel Prize in chemistry for developing an au-tomated approach to the synthesis of peptides. The first step involves attaching the aminoacid that will become the C-terminal residue to an inert, insoluble polystyrene resin. Aminoacids are then incorporated, one at a time, by coupling them onto the growing peptidechain. Because the product of each step in the reaction is a solid, it can be easily collected,washed, and purified before the next step in the reaction.

The Merrifield synthesis uses a dehydrating agent known as dicyclohexylcarbodiimide(DCC) to drive the reaction that forms the peptide bond. To prevent reactions at the wrongsite, appropriate blocking groups are added to reactive sites on the side chains of the aminoacids before they are polymerized. If we use the symbol B to indicate an appropriate block-ing group, the synthesis of a dipeptide can be represented by the following equation.

The blocking group on the N-terminal end of the dipeptide is then removed, and a thirdblocked amino acid residue is added to give a tripeptide. The process of adding one aminoacid at a time is continued until the polypeptide or protein synthesis is complete. Thepolypeptide or protein chain is then removed by reacting the resin with HBr in a suitablesolvent.

When it was first introduced, the peptide synthesis process was automated on an ap-paratus that required about 4 hours to add an amino acid residue to the peptide chain.Thus, insulin could be synthesized in approximately 8 days, while ribonuclease, with 124amino acids, required more than a month. The beauty of the Merrifield synthesis is theyield of each step, which is essentially 99%. The synthesis of ribonuclease, for example,took 369 chemical reactions and 11,931 automated steps, and yet still had an overall yieldof 17%.

OC... ... ... ...BO

OOH � H2NO 88n OCBO

OOJ H3N�O

BONHOCARAB

HOCBO

OOH � H2NOCARAB

HOCBO

Oresin 8nDCC

BONHOCARAB

HOCBO

ONHOCARAB

HOCBO

Oresin

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BIOCHEMISTRY 15

BIO.6 THE STRUCTURE OF PROTEINSThe Primary Structure of Proteins

The primary structure of a protein is nothing more than the sequence of amino acids, readoff one at a time, as if printed on ticker tape. Insulin obtained from cows, for example, con-sists of two chains (A and B) with the primary structures shown in Figure BIO.8. There ismore to the structure of a protein, however, than the sequence of amino acids. The polypep-tide chain folds back on itself to form a secondary structure. Interactions between aminoacid side chains then produce a tertiary structure. For some proteins, such as hemoglobin,interactions between individual polypeptide chains give rise to a quaternary structure.

FIGURE BIO.8 Insulin is a relatively small protein that consists of two chains held together by covalentSOS bonds between side chains of cysteine residues.

The Secondary Structure of Proteins

The peptide bond is a resonance hybrid of the two Lewis structures shown below. TheLewis structure on the left implies that the geometry around the carbon atom is trigonalplanar and that the carbon atom and its three nearest neighbors lie in the same plane. TheLewis structure on the right suggests a trigonal planar geometry for the nitrogen atom aswell. Because the peptide bond is a hybrid of the resonance forms, the six atoms involvedall lie in the same plane.

Since the NOH and CPO bonds are relatively polar, hydrogen bonds form betweenadjacent peptide chains.

R

GN

DR

C

HG

CC

DPO HON

HG

DG

COH OP,

,

,

,G

G

DD

D

�

CC

N�

H

C

O

......

SOS

OA

BSO

HE EC

CN

H

C

O

......

A

A

E E

Gly-Ile-Val-Glu-Gln-Cys-Cys-Ala-Ser-Val-Cys-Ser-Leu-Tyr-Gln-Leu-Glu-Asn-Tyr-Cys-AsnA chain

Phe-Val-Asn-Gln-His-Leu-Cys-Gly-Ser-His-Leu-Val-Glu-Ala-Leu-Tyr-Leu-Val-Cys-GlyGlu

ArgGly

PheAla-Lys-Pro-Thr-Tyr-Phe

B chain

N-terminalamino acid

C-terminalamino acid

S

S

S SO

S SO

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16 BIOCHEMISTRY

The fact that the six atoms in the peptide bond must lie in the same plane limits the num-ber of ways in which a polypeptide can be arranged in space. By building models, LinusPauling and Robert Corey discovered two ways in which a polypeptide chain could maxi-mize the hydrogen bonds between peptides. In one of the structures, the chain forms the� helix shown in Figure BIO.9. The other structure is the �-pleated sheet in Figure BIO.10.

FIGURE BIO.9 The hydrogen bonds between adjacent peptide bonds allow polypeptidesto form a right-handed helix. One turn along the helix contains 3.6 amino acid residues.

FIGURE BIO.10 The polypeptide chaincan fold back on itself to form a structurethat looks something liked a pleatedsheet. Two different �-pleated sheet struc-tures are found in nature that differ inwhether the adjacent polypeptide chainsrun in the same or opposite directions.This drawing shows the antiparallel structure found in silk.

The Tertiary Structure of Proteins

Most proteins have structures that lie between the extremes of ideal � helixes and �-pleatedsheets because other factors influence the way proteins fold to form three-dimensionalstructures. Particular attention must be paid to interactions between the side chains of theamino acids that form the backbone of the protein. Figure BIO.11 shows four ways in whichthe amino acid side chains can interact to form the tertiary structure of a protein.

Disulfide (SOS) linkages: If the folding of a protein brings two cysteine residues to-gether, the two OSH side chains can be oxidized to form a covalent SOS bond. Thesedisulfide bonds cross-link the polypeptide chain.

Hydrogen bonding: In addition to the hydrogen bonds between atoms involved in pep-tide bonds that give rise to the secondary structure of the protein, hydrogen bondscan form between amino acid side chains.

Hydrogen

Oxygen

Nitrogen

Carbon

R group

Hydrogen

Oxygen

Nitrogen

Carbon

R group

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BIOCHEMISTRY 17

Ionic bonding: The structure of a protein can be stabilized by the force of attractionbetween amino acid side chains of opposite charge, such as the ONH3

� side chainof Lys and the OCO2

� side chain of Asp.Hydrophobic interactions: Proteins often fold so that the hydrophobic side chains of

the amino acids Gly, Ala, Val, Leu, Ile, Pro, Met, Phe, and Trp are buried within theprotein, where they can interact to form hydrophobic pockets. These hydrophobicinteractions stabilize the structure of the protein.

Human hair is composed primarily of proteins known as the �-keratins that are about14% cysteine. Hair curls as it grows because of the disulfide (SOS) links between cysteineresidues on adjacent protein molecules. The first step in changing the way hair curls in-volves shaping the hair to our satisfaction and then locking it into place with curlers. Thehair is then treated with a mild reducing agent that reduces the SOS bonds to pairs ofOSH groups. This relaxes the proteins in the hair, allowing them to pick up the structuredictated by the curlers. The OSH side chains on cysteine residues that are now adjacentto each other are then oxidized by the O2 in air. New SOS linkages form, locking the hairpermanently in place (at least until new hair grows).

The �-keratins are divided into two categories, “hard” and “soft,” on the basis of theamount of cysteine they contain. The �-keratins in skin are soft because they contain rel-atively small amounts of sulfur, and disulfide cross-links are uncommon. Although hair isclassified as a hard keratin, horn and hoof, which contain even more sulfur, are much lesspliable because of the extensive disulfide cross-links that form.

The Quaternary Structure of Proteins

As we have seen, hemoglobin is the protein that carries O2 through the bloodstream to themuscles. This protein consists of four polypeptide chains—two � chains that contain 141amino acids and two � chains that contain 146 amino acids. Hemoglobin is therefore anexample of a protein that has a quaternary structure. It consists of four polymer chains thatmust be assembled to form the complete protein.

The polymer chains in a quaternary protein are not linked by covalent bonds such asthe SOS bonds that hold together the polypeptide chains in insulin. The primary force ofattraction between the � and � chains in hemoglobin is the result of interactions betweenhydrophobic substituents on the polymer chains. In other quaternary proteins, hydrogenbonding or ionic interactions between amino acid side chains on the surfaces of adjacentpolymer chains also contribute to the process by which the polymer chains are held together.

FIGURE BIO.11 Four factors are responsible for the tertiary structure of proteins: (1) disulfide linkages, (2) hydrogen bonding, (3) electrostatic interactions, and (4) hydrophobic interactions.

1

23

4

OH C

CH2

CH2

CH3 CH3

CH3

CH

CH3CH2CH3

CH

H2C

H2C

(CH2)4

CO 2–

H 3N+

CH2

S

S

O

O–

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18 BIOCHEMISTRY

The Denaturation of Proteins

Proteins are fragile molecules that are remarkably sensitive to changes in structure. Thereplacement of a polar Glu residue by a nonpolar Val at the sixth position on the � chainsof hemoglobin, for example, gives rise to the disease known as sickle-cell anemia. The in-troduction of a hydrophobic Val residue at that position changes the quaternary structureof hemoglobin. The “sticky,” nonpolar side chain on a valine residue at that position causeshemoglobin molecules to cluster together in an abnormal fashion, interfering with theirfunction as oxygen-carrying proteins.

Sickle-cell anemia is the result of a change in the way the protein is assembled fromamino acids. The structure of a protein can also be changed after it has been made. Any-thing that causes a protein to leave its normal, or natural, structure is said to denature theprotein. Factors that can lead to denaturation include the following.

• Heating, which disrupts the secondary and tertiary structure of the protein. (Thechanges we observe when we fry an egg result from denaturation caused by heating.)

• Changes in pH that interfere with ionic bonding between amino acid side chains.• Detergents, which make nonpolar amino acid side chains soluble and thereby de-

stroy the hydrophobic interactions that give rise to the tertiary and quaternary struc-ture of the protein.

• Oxidizing or reducing agents that either create or destroy SOS bonds.• Reagents such as urea (H2NCONH2) that disrupt the hydrogen bonds which form

the secondary structure of the protein.

BIO.7 CARBOHYDRATES: THE MONOSACCHARIDESThe term carbohydrate was originally used to describe compounds that were literally “hy-drates of carbon” because they had the empirical formula CH2O. In recent years, carbo-hydrates have been classified on the basis of their structures, not their formulas. They arenow defined as polyhydroxy aldehydes and ketones. Among the compounds that belong tothe family are cellulose, starch, glycogen, and most sugars.

There are three classes of carbohydrates: monosaccharides, disaccharides, and poly-saccharides. The monosaccharides are white, crystalline solids that contain a single aldehydeor ketone functional group. They are subdivided into two classes—aldoses and ketoses—on the basis of whether they are aldehydes or ketones. They are also classified as a triose,tetrose, pentose, hexose, or heptose on the basis of whether they contain three, four, five,six, or seven carbon atoms.

With only one exception,3 the monosaccharides are optically active compounds. Al-though both D and L isomers are possible, most of the monosaccharides found in natureare in the D configuration. Structures for the D and L isomers of the simplest aldose, glyc-eraldehyde, are shown below.

O

AC

H

CHO

D ( HCH2OO

AC

HH

CHO

DCH2OH(

HL-GlyceraldehydeD-Glyceraldehyde

3Dihydroxyacetone is the only carbohydrate that is not optically active.

HOCH2CCH2OH

O B

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BIOCHEMISTRY 19

The structures of many monosaccharides were first determined by Emil Fischer in the1880s and 1890s and are still written according to a convention he developed. The Fischerprojection represents what the molecule would look like if its three-dimensional structurewere projected onto a piece of paper. By convention, Fischer projections are written verti-cally, with the aldehyde or ketone at the top. The OOH group on the second-to-last carbonatom is written on the right side of the skeleton structure for the D isomer and on the leftfor the L isomer. Fischer projections for the two isomers of glyceraldehyde are shown below.

The Fischer projections can be obtained from the skeleton structures shown above by imag-ing what would happen if you placed a model of each isomer on an overhead projector sothat the CHO and CH2OH groups rested on the glass and then looked at the images ofthe models that would be projected on a screen.

Fischer projections for some of the more common monosaccharides are given in Fig-ure BIO.12.

CA

HOCOOHACHO

H2OH CA

HOOCOHACHO

H2OHD-Glyceraldehyde L-Glyceraldehyde

FIGURE BIO.12 Fischer projections for some of the common monosaccharides.

Exercise BIO.2

Glucose and fructose have the same formula: C6H12O6. Glucose is the sugar with the high-est concentration in the bloodstream; fructose is found in fruit and honey. Use the Fischerprojections in Figure BIO.12 to explain the difference between the structures of the com-pounds. Predict what an enzyme would have to do to convert glucose to fructose, or viceversa.

Solution

Both compounds have the same structure for the third, fourth, fifth, and sixth carbon atoms.One compound, however, is an aldehyde (glucose), while the other is a ketone (fructose).Glucose could be converted to fructose by simultaneously reducing the first carbon atomfrom an aldehyde to a primary alcohol and oxidizing the second carbon from a secondaryalcohol to a ketone. Fructose can be converted to glucose, on the other hand, by simulta-neously oxidizing the first carbon atom and reducing the second.

If the carbon chain is long enough, the alcohol at one end of a monosaccharide can at-tack the carbonyl group at the other end to form a cyclic compound. When a six-membered

A

A

CH2OH

C

A H

AA OHH

AC AA OHH

AC AA OHH

B

C

O

D-Ribose

A

A

CH2OH

C

A H

AA OHH

AC AA HHO

AC AA OHH

B

C

O

D-Xylose

A

A

CH2OH

C

A H

A H

AC

AHO

AC AA OH

A OH

H

AH

B

C

O

D-Arabinose

Aldoses Ketoses

A

A

CH2OH

C O

AC

AC AA OH

A OH

H

AH

B

BCH2OH

D-Ribulose

A

A

A

CH2OH

C

O

A H

AC

AHO

AC AA OH

A OH

H

AH

C

CH2OH

D-Fructose

CH2OH

AC

AC AA OH

A OH

H

AH

AHO

A

A

C

A H

A HAC

A OHAH

B

C

O

D-Glucose

CH2OH

AC

AC AA OHH

AHO

AHO

A

A

C

A H

A H

A H

AC

A OHAH

B

C

O

D-Galactose

CH2OH

AC

AC AA OH

A OH

H

AH

AHO

AHO A

A

C

A H

A H

A HAC

B

C

O

D-Mannose

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20 BIOCHEMISTRY

ring is formed, the product of the reaction is called a pyranose (see Figure BIO.13). Whena five-membered ring is formed, it is called a furanose (see Figure BIO.14). There are twopossible structures for the pyranose and furanose forms of a monosaccharide, which arecalled the � and � anomers. These anomers are isomers that differ in the orientation ofthe H and OH on the first carbon.

FIGURE BIO.13 The � and � anomers of D-glucopyranose.

FIGURE BIO.14 The � and � anomers of fructofuranose.

The reactions that lead to the formation of a pyranose or a furanose are reversible.Thus, it doesn’t matter whether we start with a pure sample of �-D-glucopyranose or �-D-glucopyranose. Within minutes, the anomers are interconverted to give an equilibrium mix-ture that is 63.6% of the � anomer and 36.4% of the � anomer. The 2�1 preference for the� anomer can be understood by comparing the structures of the molecules shown in Fig-ure BIO.13. In the � anomer, all of the bulky OOH and OCH2OH substituents lie moreor less within the plane of the six-membered ring. In the � anomer, one of the OOH groupsis perpendicular to the plane of the six-membered ring, in a region where it feels strongrepulsive forces from the hydrogen atoms that lie in similar positions around the ring. Asa result, the � anomer is slightly more stable than the � anomer.

CH2OH

AC

AC AA OH

A OH

H

AH

AHO

A

A

C

A H

A HAC

A OHAH

B

C

O

H

HH

OH

OH

HO

HOOH

H

H

O

H

HH

OH

HO

HO

HOOH

H

H

O

-D-Glucopyranose�

-D-Glucopyranose�

CH2OH

AC

AC AA OH

A OH

H

AH

AHO

A B O

A HAC

A

C

-D-Fructofuranose�

-D-Fructofuranose�

OH

H

H

OH

OH

CH2OH CH2OH

OH

OH

H

H

OH

OH

CH2OH

CH2OH

CH2OH OH

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BIOCHEMISTRY 21

BIO.8 CARBOHYDRATES: THE DISACCHARIDES ANDPOLYSACCHARIDES

Disaccharides are formed by condensing a pair of monosaccharides. The structures of threeimportant disaccharides with the formula C12H22O11 are shown in Figure BIO.15. Maltose,or malt sugar, which forms when starch breaks down, is an important component of thebarley malt used to brew beer. Lactose, or milk sugar, is a disaccharide found in milk. Veryyoung children have a special enzyme known as lactase that helps digest lactose. As theygrow older, many people lose the ability to digest lactose and cannot tolerate milk or milkproducts. Because human milk has twice as much lactose as milk from cows, young chil-dren who develop lactose intolerance while they are being breast-fed are switched to cows’milk or a synthetic formula based on sucrose.

FIGURE BIO.15 The structures of three common disaccharides.

H

HH

H

HH

H

Maltose

OHO

HO

OH

OH

HH

H

OO

HO

OH

OH

CH2 OH

CH2

O

OHH

HH

HH

Lactose

OH

HO

OH

OH

HH

H

O

H

HO

OHOH

CH2

OH

CH2

OH H

H

CH2OH

CH2OH

H

Sucrose

O

H

HOHO

OH

H

H OH

CH2

O

O

H

H

OH

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22 BIOCHEMISTRY

The substance most people refer to as “sugar” is the disaccharide sucrose, which is ex-tracted from either sugar cane or beets. Sucrose is the sweetest of the disaccharides. It isroughly three times as sweet as maltose and six times as sweet as lactose. In recent years,sucrose has been replaced in many commercial products by corn syrup, which is obtainedwhen the polysaccharides in cornstarch are broken down. Corn syrup is primarily glucose,which is only about 70% as sweet as sucrose. Fructose, however, is about two and one-halftimes as sweet as glucose. A commercial process has therefore been developed that usesan isomerase enzyme to convert about half of the glucose in corn syrup into fructose (seeExercise BIO.2). The resulting high-fructose corn sweetener is just as sweet as sucrose andhas found extensive use in soft drinks.

The monosaccharides and disaccharides represent only a small fraction of the totalamount of carbohydrates in the natural world. The great bulk of the carbohydrates in na-ture are present as polysaccharides, which have relatively large molecular weights. Thepolysaccharides serve two principal functions. They are used by both plants and animalsto store glucose as a source of future food energy, and they provide some of the mechan-ical structure of cells.

Very few forms of life receive a constant supply of energy from their environment. Inorder to survive, plant and animal cells have had to develop a way of storing energy dur-ing times of plenty in order to survive the times of shortage that follow. Plants store foodenergy as polysaccharides known as starch. There are two basic kinds of starch: amyloseand amylopectin. Amylose is found in algae and other lower forms of plants. It is a linearpolymer of glucose residues whose structure can be predicted by adding �-D-glucopyra-nose rings to the structure of maltose. Amylopectin is the dominant form of starch in thehigher plants. It is a branched polymer of about 6000 glucose residues with branches on 1in every 24 glucose rings. A small portion of the structure of amylose is shown in FigureBIO.16.

FIGURE BIO.16 The structure of amylose formed by linking �-D-glucopyranose rings.

The polysaccharide that animals use for the short-term storage of food energy is knownas glycogen. Glycogen has almost the same structure as amylopectin, with two minor dif-ferences. The glycogen molecule is roughly twice as large as amylopectin, and it has roughlytwice as many branches.

There is an advantage to branched polysaccharides such as amylopectin and glycogen.During times of shortage, enzymes attack one end of the polymer chain and cut off glu-cose molecules, one at a time. The more branches, the more points there are at which theenzyme attacks the polysaccharide. Thus, a highly branched polysaccharide is better suitedfor the rapid release of glucose than is a linear polymer.

OHO

HOHO

HOCH2 O

OHO

HO

O

HOCH2

HOCH2

OHOH

HO

On

Amylosen � 1000–6000

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BIOCHEMISTRY 23

Polysaccharides are also used to form the walls of plant and bacterial cells. Cells thatdo not have a cell wall often break open in solutions whose salt concentrations are eithertoo low (hypotonic) or too high (hypertonic). If the ionic strength of the solution is muchsmaller than that of the cell, osmotic pressure forces water into the cell to bring the systeminto balance, which causes the cell to burst. If the ionic strength of the solution is too high, os-motic pressure forces water out of the cell, and the cell breaks open as it shrinks. The cellwall provides the mechanical strength that helps protect plant cells that live in freshwaterponds (too little salt) or seawater (too much salt) from osmotic shock. The cell wall alsoprovides the mechanical strength that allows plant cells to support the weight of other cells.

The most abundant structural polysaccharide is cellulose. There is so much cellulose inthe cell walls of plants that it is the most abundant of all biological molecules. Cellulose isa linear polymer of glucose residues, with a structure that resembles amylose more closelythan amylopectin, as shown in Figure BIO.17. The difference between cellulose and amy-lose can be seen by comparing Figures BIO.16 and BIO.17. Cellulose is formed by linking�-glucopyranose rings, instead of the �-glucopyranose rings in starch and glycogen.

FIGURE BIO.17 The structure of cellulose formed by linking �-D-glucopyranose rings.

The OOH substituent that serves as the primary link between �-glucopyranose ringsin starch and glycogen is perpendicular to the plane of the six-membered ring. As a result,the glucopyranose rings in those carbohydrates form a structure that resembles the stairsof a staircase. The OOH substituent that links the �-glucopyranose rings in cellulose liesin the plane of the six-membered ring. The cellulose molecule therefore stretches out in alinear fashion. This makes it easier for strong hydrogen bonds to form between the OOHgroups of adjacent molecules. This, in turn, gives cellulose the rigidity required for it toserve as a source of the mechanical structure of plant cells.

Cellulose and starch provide an excellent example of the link between the structureand function of biomolecules. At the turn of the twentieth century, Emil Fischer suggestedthat the structure of an enzyme is matched to the substance on which it acts, in much thesame way that a lock and key are matched. Thus, the amylase enzymes in saliva that breakdown the � linkages between glucose molecules in starch cannot act on the � linkages incellulose.

Most animals cannot digest cellulose because they don’t have an enzyme that can cleave� linkages between glucose molecules. Cellulose in their diet therefore serves only as fiber,or roughage. The digestive tracts of some animals, such as cows, horses, sheep, and goats,contain bacteria that have enzymes that cleave the � linkages, so those animals can digestcellulose.

Exercise BIO.3

Termites provide an example of the symbiotic relationship between bacteria and higher or-ganisms. Termites cannot digest the cellulose in the wood they eat, but their digestive tractsare infested with bacteria that can. Propose a simple way of ridding a house from termites,without killing other insects that might be beneficial.

OO

HOCH2HO

HOOH

OO

HOCH2

HOOH

OOH

HOCH2

HO

Cellulosen = 5000–10,000

OH( (

n

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24 BIOCHEMISTRY

Solution

Killing termites isn’t difficult—there are a number of poisons that do an excellent job. Un-fortunately, the compounds can also poison other insects or even small animals or children.The simplest way of getting rid of termites is to treat the wood with a poison that kills thebacteria that digest the cellulose in wood, so that the termites starve to death.

For many years, biochemists considered carbohydrates to be dull, inert compounds thatfilled the space between the exciting molecules in the cell—the proteins. Carbohydrateswere impurities to be removed when “purifying” a protein. Biochemists now recognize thatmost proteins are actually glycoproteins, in which carbohydrates are covalently linked tothe protein chain. Glycoproteins play a particularly important role in the formation of therigid cell walls that surround bacterial cells.

BIO.9 LIPIDSAny biomolecule that dissolves in relatively nonpolar solvents—such as chloroform(CHCl3), benzene (C6H6), and diethyl ether (CH3CH2OCH2CH3)—is classified as a lipid(from the Greek word lipos, “fat”). Because they are soluble in nonpolar solvents, lipidsare often insoluble—or only marginally soluble—in water, and they often feel oily or greasyto the touch.

Neutral Fats and Oils

Long-chain carboxylic acids such as stearic acid [CH3(CH2)16CO2H] are called fatty acidsbecause they can be isolated from animal fats. The fatty acids are subdivided into two cat-egories on the basis of whether they contain CPC double bonds: saturated fatty acids andunsaturated fatty acids.

The common names of carboxylic acids trace back to Latin or Greek stems that indi-cate a natural source of the acid. The destructive distillation of ants, for example, producesformic acid (from the Latin word formica, “ant”). Vinegar is a 5–6% solution of acetic acidin water. Acetic acid therefore takes its common name from the Latin term for vinegar:acetum. The next acid, as we build up the hydrocarbon chain, is propionic acid, which takesits name from the Greek stems protos and pion. The name literally means “first fat,” be-cause it is the simplest carboxylic acid that can be isolated from fat. The next member ofthe family is butyric acid, from the Latin word butyrum, or “butter,” because it can be ob-tained from rancid butter. The fifth carboxylic acid is known as valeric acid, because it canbe obtained from plants in the genus Valeriana, which are perennial herbs.

HCO2H Formic acid

CH3CO2H Acetic acid

CH3CH2CO2H Propionic acid

CH3CH2CH2CO2H Butyric acid

CH3CH2CH2CH2CO2H Valeric acid

From this point on, common carboxylic acids tend to have an even number of carbon atoms.The next three derivatives are all given names from the Latin term for goat, caper. Thecarboxylic acids with 12, 14, 16 and 18 carbon atoms are named from the Latin stem forthe bay tree, laurel; the genus for the spice nutmeg, Myristica; the Latin stem for the palmtree, palma; and the Greek stem for the tallow used to make candles, stear.

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BIOCHEMISTRY 25

CH3(CH2)4CO2H Caproic acid

CH3(CH2)6CO2H Caprylic acid

CH3(CH2)8CO2H Capric acid

CH3(CH2)10CO2H Lauric acid

CH3(CH2)12CO2H Myristic acid

CH3(CH2)14CO2H Palmitic acid

CH3(CH2)16CO2H Stearic acid

The very small carboxylic acids have a sharp odor. (Formic acid has an odor that iseven sharper than acetic acid.) By the time the hydrocarbon chain has grown to a total offour carbon atoms, the odor of the compounds has taken a significant turn for the worse.(Butyric acid is the source of the characteristic odor of rancid butter and spoiled meat.)As the length of the hydrocarbon chain increases further, the odor of the acid changes onceagain—this time, becoming more pleasant.

There are four important unsaturated fatty acids. One of them, a derivative of palmiticacid, is known as palmitoleic acid.

The other three are derivatives of stearic acid. The first has a single CPC double bond inthe center of the fatty acid chain, and is known as oleic acid.

The second, which is known as linoleic acid, has another CPC double bond in the non-polar half of the fatty acid chain.

The third—linolenic acid—has one more CPC double bond in the same half of the fattyacid chain.

There are several regularities in the chemistry of the unsaturated fatty acids. First, theycontain cis double bonds. Second, the double bonds are always isolated from each otherby a CH2 group.

So much attention is paid to the structures of the fatty acids in discussions of the chem-istry of lipids that it is easy to miss an important point: Free fatty acids are seldom foundin nature. They are usually tied up with alcohols or amines to form esters (RCO2R) or

CH3CH2GC

DH

PCG

H

DCH2G

CD

HPC

GH

D(CH2)7CO2H

Linolenic acidC CPCG

H

DCH2G

DH

CH3(CH2)4GC

DH

PCG

H

DCH2G

CD

HPC

GH

D(CH2)7CO2H

Linoleic acid

CG

CH3(CH2)7

DH

PCD

(CH2)7CO2H

GH

Oleic acid

CG

CH3(CH2)5

DH

PCD

(CH2)7CO2H

GH

Palmitoleic acid

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26 BIOCHEMISTRY

amides (RCONHR). The most abundant lipids are the triesters formed when a glycerol mol-ecule reacts with three fatty acids, as shown in Figure BIO.18. These lipids have been knownby a variety of names, including fat, neutral fat, glyceride, triglyceride, and triacylglyceride.

FIGURE BIO.18 The triester trimyristin can beisolated from nutmeg.

Most animal fats are complex mixtures of different triglycerides. As the percentage ofunsaturated fatty acids in the fats increases, the melting point of the triglyceride decreasesuntil it eventually becomes an oil at room temperature. Beef fat, which is one-third un-saturated fatty acids, is a solid. Olive oil, which is roughly 80% unsaturated, is a liquid.

The effect of unsaturated fatty acids on the melting point of a triglyceride can be un-derstood by recognizing that the cis CPC double bond introduces a rigid 30° bend in thehydrocarbon chain, as shown in Figure BIO.19.

TABLE BIO.2 Relative Abundance ofFatty Acids in a Typical Fat and aTypical Oil

Fatty Acid Butter Olive Oil

Butyric 3–4%Caproic 1–2%Caprylic �1%Capric 2–3%Lauric 2–5%Myristic 8–15% �1%Palmitic 25–29% 5–15%Stearic 9–12% 1–4%Palmitoleic 4–6%Oleic 18–33% 67–84%Linoleic 2–4% 8–12%Linolenic �1%

FIGURE BIO.19 A phospholipid containing one saturated (bottom)and one unsaturated (top) fatty acid.

This bend or “kink” increases the average distance between triglyceride molecules,which decreases the van der Waals interactions between neighboring molecules. Thus, theintroduction of unsaturated fatty acids into a triglyceride increases the fluidity of the lipid.Table BIO.2 compares the relative abundances of the common fatty acids in a typical an-imal fat (butter) and a vegetable oil (olive oil).

CH3(CH2)12COCH

CH2OC(CH2)12CH3

CH2OC(CH2)12CH3

Trimyristin

OP

O

PO

P

AA

AA

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BIOCHEMISTRY 27

Fats and oils are used by living cells for only one purpose—to store energy. They area far more efficient storage system than glycogen or starch because they give off betweentwo and three times as much energy when they are burned. (The metabolism of glycogenreleases 15.7 kilojoules per gram of carbohydrate consumed, whereas the metabolism oflipids gives approximately 40 kJ/g.) This explains why the seeds of many plants are rela-tively rich in oils, which provide the energy the seed needs to grow until the leaves can be-gin to produce energy by photosynthesis.

The average human contains enough fat (21% of the body weight for men, 26% forwomen) to provide the energy they need to survive for up to 3 months. But there is onlyenough glycogen stored in the human body at any time to provide enough energy for oneday. Thus, glycogen is only used for the short-term storage of food energy. In “times ofplenty,” the body stores energy in the form of fat to compensate for “times of shortage”to come.

CheckpointSeveral hundred people were poisoned in Spain a few years ago when an unscrupu-lous merchant diluted olive oil with diesel oil (see Section O1.11). Explain both thesimilarities and differences between the structures of the components of the two“oils.”

Polar Lipids

Fats and oils are neutral compounds. When one of the fatty acids in a triglyceride is re-placed by a phosphate group, a phospholipid is obtained that has two nonpolar hydrophobictails and a charged hydrophilic head.

A variety of biochemically important molecules can be obtained by forming a second es-ter linkage to the phosphate group. These phosphate diesters are often called phosphatides,and they contain an alcohol at the position labeled with an X in the following structure.

CH3(CH2)14CO2CH2A

CH3(CH2)14CO2CHACH2OOOP

A

O

O�

BOX

CH2CH2CH3CH2CH2 CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2CBO

OOOCH2

BO

A

A

A

O

�

B

O

CH2OOOPOO�

CH3CH2CH2CH2CH2CH2CH2CH2CH CHCH2CH2CH2CH2CH2CH2CH2COOOCHP

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28 BIOCHEMISTRY

The most important phosphatides contain the following X groups.

X � OCH2CH2N(CH3)3�

Phosphatidylcholine

X � OCH2CH2NH3�

Phosphatidylethanolamine

Note that the phosphatidylcholines and phosphatidylethanolamines are zwitterions becausethey simultaneously carry both positive and negative charges within a molecule that hasno net charge. The phosphatidylcholines are also known as lecithins, while the phos-phatidylethanolamines are known as cephalins.

The phosphatidylcholines and phosphatidylethanolamines are described as amphipathic(literally, “both paths”) because they contain a polar, hydrophilic head and a pair of non-polar, hydrophobic tails. The compounds can therefore spontaneously associate in aque-ous solutions to form a bilayer in which the molecules are oriented so that the nonpolartails of adjacent phospholipids form a hydrophobic pocket and the polar heads point to-ward the water that surrounds both sides of the bilayer. If we symbolize the hydrophilichead by a small circle and the hydrophobic tails by a pair of wavy lines, we can see thatthe bilayer is the first step toward the formation of a membrane (see Figure BIO.20).

FIGURE BIO.20 Schematic drawing of the phospholipid bilayer that forms the membranes that separatecells from each other and compartmentalize each cell.

The best model for the structure of cell membranes involves a bilayer of amphipathiclipids approximately 8 nm wide into which various proteins are embedded. It is a commonmistake to assume that the membrane is static. It is not. There is a considerable amountof mobility or flexibility in the hydrocarbon tails of the lipid molecules. However, the stronghydrophobic character of the region between the inner and outer surface of the membraneresists the passage of highly charged or polar intermediates across the membrane.


The Search for New Drugs

In 1763, the Reverend Edmund Stone took the first step toward the discovery of one ofthe most commonly used medicines when he noted that the bark of the English willowwas an effective treatment for patients suffering from a fever. Stone explained the effect

Phospholipidbilayer

Hydrophobictails

Integralprotein

Hydrophilicareas

Hydrophilicareas

Hydrophobicareas

Hydrophilicheads

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BIOCHEMISTRY 29

of willow bark by noting that “many natural maladies carry their cures along with them,or their remedies lie not far from their causes.” Thus, he argued, the English willowgrows in the same moist regions where one was likely to catch the fever treated with itsbark.

It took 50 years before the active ingredient in willow bark was isolated and namedsalicin, from the Latin name for the willow (Salix alba). Another 50 years elapsed beforea large-scale synthesis for the compound was available. By that time, the compound wasknown as salicylic acid because saturated solutions in water are highly acidic (pH � 2.4).

By the end of the nineteenth century, salicylic acid was being used to treat rheumaticfever, gout, and arthritis. Many patients treated with the drug complained of chronicstomach irritation because of its acidity and the large doses required (6–8 g/day). Be-cause his father was one of these patients, Felix Hoffman searched the chemical litera-ture for a less acidic derivative of salicylic acid. In 1898, Hoffman reported that theacetyl ester of salicylic acid was simultaneously more effective and easier to toleratethan the parent compound. He named the compound aspirin, taking the prefix a- fromthe name of the acetyl group and spirin from the German name of the parent compoundspirsäure.

The existence of a drug that reduced both pain and fever initiated a search for othercompounds that could achieve the same result. Although based on trial and error, thissearch inevitably produced a variety of substances, such as those in Figure BIO.21, thatare analgesics, antipyretics, and/or anti-inflammatory agents. Analgesics relieve painwithout decreasing sensibility or consciousness. Antipyretics reduce the body tempera-ture when it is elevated. Anti-inflammatory agents counteract swelling or inflammationof the joints, skin, and eyes.

FIGURE BIO.21 The structures of some common analgesics, antipyretics, and anti-inflammatory agents.

Although the use of aspirin has been widespread since the nineteenth century, themechanism for its action was first described in 1971 [J. R. Vane, Nature, 231(25),232–235 (1971)]. Vane noted that injury to tissue was often followed by the release of agroup of hormones known as the prostaglandins, which have widespread physiologicaleffects at very low concentrations. The prostaglandins regulate blood pressure, mediatethe inflammatory response of the joints, induce the process by which blood clots, regu-late the sleep/wake cycle, and, when appropriate, induce labor.

Vane suggested that aspirin and other nonsteroidal anti-inflammatory drugs (orNSAIDs) inhibit the enzyme that starts the process by which prostaglandins such as PGE2

and PGF2� are synthesized from the 20-carbon unsaturated fatty acid known as arachidonic

CO2H

OH

Salicylic acid

P

PC O

NH

OH

Acetaminophen(Tylenol)

CH3

CH

CH2

CH

Ibuprofen(Motrin, Advil)

CH3

CH3

CH3

CO2H

CO2H

OCCH3

O

Acetylsalicylic acid(aspirin)

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30 BIOCHEMISTRY

acid shown in Figure BIO.22. The steroidal anti-inflammatory drugs (such as hydrocorti-sone) achieve a similar effect by inhibiting the enzyme that releases arachidonic acid intothe cell.

FIGURE BIO.22 The structures of the PGE2 and PGF2� prostaglandins synthesized from arachidonic acid.

Now that they are beginning to understand the mechanism by which drugs operate,medicinal chemists can approach the design of drugs by a rational process. One paperdescribed progress toward the design of a drug to treat the debilitating diseases causedby protozoan parasites that afflict millions of people in Latin America, Africa, and Asia[W. N. Hunter et al., Journal of Molecular Biology, 227, 322–333 (1992)]. The potentialtarget for the drug is an enzyme—trypanothione reductase (TR)—that protects the para-site from oxidative damage from the immune system of its mammalian host. Mammaliancells use a similar enzyme, known as glutathione reductase (GR), to protect againstdamage from oxidation reactions.

Hunter and co-workers found that the human GR enzyme has a smaller, more posi-tively charged active site compared to the TR enzyme in the parasite. The structural in-formation in this study can now be used to rationally modify a substrate of the enzymesuntil it possesses the following characteristics.

• The substrate must be too large to bind to the GR enzyme in humans.• The substrate must have a high affinity for binding to the TR enzyme in the parasite.• The substrate must inhibit the activity of the TR enzyme, thereby allowing the im-

mune system of the mammalian host to attack and eventually destroy the parasite.

BIO.10 NUCLEIC ACIDSNucleic acids, which are relatively strong acids found in the nuclei of cells, were first iso-lated in 1869. The nucleic acids include polymers with molecular weights as high as100,000,000 grams per mole. They can be broken down, or digested, to form monomersknown as nucleotides. Each nucleotide contains three units: a sugar, an amine, and a phos-phate, as shown in Figure BIO.23.

Nucleic acids are divided into classes on the basis of the sugar used to form the nu-cleotides. Ribonucleic acid (RNA) is built on a �-D-ribofuranose ring. Deoxyribonucleicacid (DNA) contains a modified ribofuranose in which the OOH group on the second car-bon atom has been removed, as shown in Figure BIO.24.

HO

O

OHPGE2

CO2H

CO2H

HO

HO

OHPGF2

Arachidonic acid

CO2H

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BIOCHEMISTRY 31

FIGURE BIO.23 The structure of the nucleotideknown as cytidine monophosphate.

FIGURE BIO.24 The structuresof the carbohydrates on whichDNA and RNA are built.

The amines that form nucleic acids fall into two categories: purines and pyrimidines.There are three pyrimidines—cytosine, thymine, and uracil—and two purines—adenineand guanine—as shown in Figure BIO.25. DNA and RNA each contain four nucleotides.Both contain the same purines—adenine and guanine—and both also contain the pyrimi-dine cytosine. But the fourth nucleotide in DNA is thymine, whereas RNA uses uracil tocomplete its quartet of nucleotides.

FIGURE BIO.25 The two classes of aminesubstituents in nucleic acids.

The carbon atoms in the sugar at the center of a nucleotide are numbered from 1′to 5′. The OOH group on the 3′ carbon of one nucleotide can react with the phosphate

AOO

B

OO

OH

P O

OPhosphate

O�

�O OCH2

NH2

Amine

Sugar

N

N

O

C

HC C

C

OH3�

4�

5�

2�

1�

-D-Ribofuranosefound in RNA

H

OH

HOOCH2 Amine

HH

O

C

HC C

C

OH3�

4�

5�

2�

1�

-D-Deoxyribofuranosefound in DNA

H

H

HOOCH2 Amine

HH

NH2

CH3

O

N

HCytosine

N

O

NH2

N

HAdenine Guanine

N

H

N

HThymine

PYRIMIDINES

NH

N N

N

H

O

O

N

N

H

O

N

HUracil

NH

O

NH

H2N

PURINES

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32 BIOCHEMISTRY

FIGURE BIO.26 Short segment of a single strand of DNA.

attached to the 5′ carbon of another to form a dinucleotide held together by phosphate es-ter bonds. As the chain continues to grow, it becomes a polynucleotide. A short segmentof a DNA chain is shown in Figure BIO.26. Reading from the 5′ end of the chain to the3′ end, this DNA segment contains the following sequence of amine substituents: adenine(A), cytosine (C), guanine (G), and thymine (T).

For many years, the role of nucleic acids in living systems was unknown. In 1944Oswald Avery presented evidence that nucleic acids were involved in the storage and trans-fer of the genetic information needed for the synthesis of proteins. This suggestion was ac-tively opposed by many of his contemporaries, who believed that the structure of the nu-cleic acids was too regular—and therefore too dull—to carry the information that codesfor the thousands of different proteins a cell needs to survive.

In retrospect, the first clue about how nucleic acids function was obtained by Er-win Chargaff, who found that DNA always contains the same amounts of certain pairs ofbases. There is always just as much adenine as thymine, for example, and just as much gua-nine as cytosine.

In 1954, James Watson and Francis Crick proposed a structure for DNA that ex-plained how DNA could be used to store genetic information. Their structure consisted oftwo polynucleotide chains running in opposite directions that were linked by hydrogenbonds between a specific purine (A or G) on one strand and a specific pyrimidine (C or T)

A

AOO

O

O

P O

O�

O�

O

A

AOOP

O

O�O

OCH2

OCH2

O

A

AOOP

O

O�O

OCH2

OH

A

AOOP

O

O�O

OCH2

CH3

NH2

N

N N

N

H

N

O

N

N

H

O

N

H2N

O

NH2

N

N

O

N

NT

G

C

A

5� end

3� end

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BIOCHEMISTRY 33

on the other, as shown in Figure BIO.27. The strands form a helix that is not quite as tightlycoiled as the � helix Pauling and Corey proposed for proteins.

The helix structure must be able to explain two processes. There must be some way tomake perfect copies of the DNA that can be handed down to future generations (replica-tion). There also must be some way to decode the information on the DNA chain (tran-scription) and to translate the information into a sequence of amino acids in a protein(translation).

Replication is easy to understand. According to Watson and Crick, an adenine on onestrand of DNA is always paired with a guanine on the other, and a cytosine is always pairedwith a thymine. The two strands of DNA therefore complement each other perfectly; thesequence of nucleotides on one strand can always be predicted from the sequence on theother. Replication occurs when the two strands of the parent DNA molecule separate andboth strands are copied simultaneously. Thus, one strand from the parent DNA is presentin each of the daughter molecules produced when a cell divides.

BIO.11 PROTEIN BIOSYNTHESISThe information that tells a cell how to build the proteins it needs to survive is coded inthe structure of the DNA in the nucleus of that cell. The code can’t be based on a one-to-one match between nucleotides and amino acids because there are only four nucleotidesbut there are 20 amino acids that must be coded. If the nucleotides are grouped in threes,however, there are 64 possible triplets, or codons, which is more than enough combina-tions to code for the 20 amino acids.

To understand how proteins are made, we have to divide the decoding process into twosteps. DNA only stores the genetic information; it isn’t involved in the process by whichthe information is used. The first step in protein biosynthesis therefore has to involve tran-scribing the information in the DNA structure into a useful form. In a separate step, thisinformation can be translated into a sequence of amino acids.

FIGURE BIO.27 The two strands in DNAare held together by hydrogen bonds be-tween specific pairs of purine and pyrimi-dine bases. The hydrogen bonds between Aand T and between G and C are shown.

CH3

N

N

N

N

NN

H

H

HH bonds

Sugar–phosphatechain

Thymine(T) Adenine

(A)

O

O

N

H

H

N

,

††

N

N

N

O

NN

H

H

H

H

H bonds

Cytosine(C) Guanine

(G)

N

O

N

H

,

†

Sugar–phosphatechain

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34 BIOCHEMISTRY

Transcription

Before the information in DNA can be decoded, a small portion of the DNA double he-lix must be uncoiled. A strand of RNA is then synthesized that is a complementary copyof one strand of the DNA.

Assume that the section of the DNA that is copied has the following sequence ofnucleotides, starting from the 3 end.

When we predict the sequence of nucleotides in the RNA complement, we have to re-member that RNA uses U where T would be found in DNA. We also have to rememberthat base pairing occurs between two chains that run in opposite directions. The RNA com-plement of the DNA should therefore be written as follows.

Since the RNA strand contains the message that was coded in the DNA, it is called mes-senger RNA, or mRNA.

Translation

The messenger RNA now binds to a ribosome, where the message is translated into a se-quence of amino acids. The amino acids that are incorporated into the protein being syn-thesized are carried by relatively small RNA molecules known as transfer RNA, or tRNA.There are at least 60 tRNAs, which differ slightly in their structures, in each cell. At oneend of each tRNA is a specific sequence of three nucleotides that can bind to the mes-senger RNA. At the other end is a specific amino acid. Thus, each three-nucleotide seg-ment of the messenger RNA molecule codes for the incorporation of a particular aminoacid. The relationship between the triplets, or codons, on the mRNA and the amino acidsis shown in Table BIO.3.

Exercise BIO.4

Assume that the DNA chain that codes for the synthesis of a particular protein containsthe triplet A-G-T (reading from the 3′ to the 5′ end). Predict the sequence of nucleotidesin the triplet, or codon, that would be built in the messenger RNA constructed on theDNA template. Then predict the amino acid that would be incorporated at this point inthe protein.

Solution

The messenger RNA synthesized when the DNA chain is read is a complement of theDNA. Every time a G appears, a C is placed on the complementary chain, and vice versa.Each time a T appears in the DNA message, an A is incorporated into the mRNA chain.When the DNA contains an A, the complementary RNA chain contains a U.

DNA 3� ...A-G-T... 5�

mRNA 5� ...U-C-A... 3�

3� T-A-C-A-A-G-C-A-G-T-T-G-G-T-C-G-T-G... 5� DNA

5� A-U-G-U-U-C-G-U-C-A-A-C-C-A-G-C-A-C... 3� m-RNA

3� T-A-C-A-A-G-C-A-G-T-T-G-G-T-C-G-T-G... 5� DNA

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BIOCHEMISTRY 35

According to Table BIO.3, the mRNA chain is read from the 5′ to the 3′ end. We thereforelook up the U-C-A codon in Table BIO.3 and find that the triplet codes for the amino acidserine.

The signal to start making a polypeptide chain in simple, prokaryotic cells4 is the tripletAUG, which codes for the amino acid methionine (Met). The synthesis of every proteinin these cells therefore starts with a Met residue at the N-terminal end of the polypeptidechain. After the tRNA that carries Met binds to the start signal on the messenger RNA,a tRNA carrying the second amino acid binds to the next codon. A dipeptide is synthe-sized when the Met residue is transferred from the first tRNA to the amino acid on thesecond tRNA. If the DNA described in this section were translated, the dipeptide wouldbe Met-Phe (reading from the N-terminal to the C-terminal amino acid).

The messenger RNA now moves through the ribosome, and a tRNA carrying the thirdamino acid (Val) binds to the next codon. The dipeptide is then transferred to the aminoacid on the third tRNA to form a tripeptide. This sequence of steps continues until one ofthree codons is encountered: UAA, UGA, or UAG. These codons give the signal for ter-minating the synthesis of the polypeptide chain, and the chain is cleaved from the last tRNAresidue.

The sequence of DNA described in this section would produce the following sequenceof amino acids.

TABLE BIO.3 The Genetic Code

First PositionSecond Position

Third Position(5′ End) U C A G (3′ End)

U Phe Ser Tyr Cys UPhe Ser Tyr Cys CLeu Ser —a —a ALeu Ser —a Trp G

C Leu Pro His Arg ULeu Pro His Arg CLeu Pro Gln Arg ALeu Pro Gln Arg G

A Ile Thr Asn Ser UIle Thr Asn Ser CIle Thr Lys Arg AMet Thr Lys Arg G

G Val Ala Asp Gly UVal Ala Asp Gly CVal Ala Glu Gly AVal Ala Glu Gly G

aThere are three triplets that code for termination of the polypeptide chain:UAA, UGA, and UAG.

4Prokaryotic cells do not have a cell nucleus. Cells that have a nucleus are called eukaryotic.

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36 BIOCHEMISTRY

Met-Phe-Val-Asn-Gln-His- ...

This polypeptide is not necessarily an active protein. All proteins in prokaryotic cells startwith Met when synthesized, but not all proteins have Met first in their active form. It is of-ten necessary to clip off the Met after the polypeptide has been synthesized to give a pro-tein with a different N-terminal amino acid.

Modifications to the polypeptide often have to be made before an active protein isformed. Insulin, for example, consists of two polypeptide chains connected by disulfidelinkages. In theory, it would be possible to make the chains one at a time and then try toassemble them to make the final protein. Nature, however, has been more subtle. Thepolypeptide chain that is synthesized contains a total of 81 amino acids. All of the disul-fide bonds that will be present in insulin are present in that chain. The protein is madewhen a sequence of 30 amino acids is clipped out of the middle of the polypeptide chain.

KEY TERMS

Amino acidAmphipathicAnomerBiochemistryBreathing motionCarbohydrateCodonCooperative interactionDenaturationDeoxyribonucleic acid

(DNA)DisaccharideDisulfideFatty acidGel electrophoresisGlycoproteinHeme

HemoglobinHenderson–Hasselbalch

equationHydrophobic interactionIsoelectric pointLipidMembraneMessenger RNA

(mRNA)MonosaccharideMyoglobinNucleic acidNucleotidePeptidePhospholipidPolynucleotidePolysaccharide

PorphyrinPrimary structureProteinProtein biosynthesisPurinePyrimidineQuaternary structureReplicationRibonucleic acid (RNA)Secondary structureTertiary structureTranscriptionTransfer RNA (tRNA)TranslationTriglycerideZwitterion

PROBLEMSThe Amino Acids

1. Explain why amino acids such as glycine (H2NCH2CO2H) exist in aqueous solution aszwitterions (H3N�CH2CO2

�).2. Which of the following forms of the amino acid glycine is present in strongly acidic so-

lutions? In strongly basic solutions?(a) H3N�CH2CO2H (b) H3N�CH2CO2

� (c) H2NCH2CO2�

3. Write the formula for lysine as it would exist in aqueous solution at pH � 1.4. Write the formula for serine as it would exist at pH � 1 and pH � 13.

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BIOCHEMISTRY 37

5. Which of the following amino acids contain side chains that cannot contribute to form-ing a hydrophobic pocket?(a) alanine (b) isoleucine (c) methionine (d) serine (e) lysine

6. The common amino acids in proteins have the generic formula H2NCHRCO2H. Clas-sify the following amino acids as either hydrophilic or hydrophobic.(a) alanine, R � OCH3

(b) glutamic acid, R � OCH2CH2CO2�

(c) arginine, R � OCH2CH2CH2NHC(NH2)PNH2�

(d) methionine, R � OCH2CH2SCH3

(e) threonine, R � OCH(CH3)OH7. Use examples to explain why amino acids with an R group that is either positively

charged or negatively charged at neutral pH are hydrophilic.8. The three titratable groups in the following fully protonated amino acid are labeled I,

II, and III. Arrange the groups in order of increasing pKa.

(a) I � II � III (b) I � III � II (c) II � I � III(d) II � III � I (e) III � I � II (f) III � II � I

9. The pI for tyrosine is 5.7. What would be the charge on the average tyrosine moleculewhen the pH of an aqueous solution was 3?

10. The pKa values for the �-carboxylic acid and the �-amino groups in arginine are 1.82and 8.99, respectively. The pKa for the titratable CPNH2

� side chain in the amino acidis 12.48. Calculate the pI for arginine.

11. The pKa values for the �-carboxylic acid and the �-amino groups in cysteine are 1.92and 10.78, respectively. The pKa for the titratable OSH side chain in the amino acidis 8.33. Calculate the pI for cysteine.

12. Which of the following amino acids would be the most likely to have an isoelectricpoint near pH 3?(a) Ala (b) Asp (c) Lys (d) Ser (e) Gln

Peptides and Proteins

13. How do amino acids, peptides, and proteins differ?14. How many dipeptides can be formed when alanine, R � OCH3, is condensed with cys-

teine, R � OCH2SH? What is the difference between the dipeptides?15. Describe the tripeptides that can form when alanine, R � OCH3, is condensed with

cysteine, R � OCH2SH.16. We can describe peptide chains by listing the amino acids from the N-terminal to the

C-terminal residue. Write the amino acid sequences for all the tetrapeptides contain-ing four different amino acids that can be formed from the amino acids Ala, Lys, Ser,and Tyr.

17. Write the Lewis structure for the following tripeptide: Met-Thr-Glu.

H3N�CACACO2H

I

H2

HCO2HII III

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38 BIOCHEMISTRY

The Structure of Proteins

18. Explain why proteins, polysaccharides, and nucleic acids can all be classified as con-densation polymers.

19. Describe how the primary, secondary, tertiary, and quaternary structures of a proteindiffer.

20. Describe the four factors that give rise to the tertiary structure of a protein. Give anexample of each.

21. Which of the following amino acids has the fewest interactions that contribute to theformation of the tertiary structure of a protein?(a) Gly (b) Glu (c) Cys (d) Met (e) Pro

22. List the amino acids that can form hydrogen bonds in the tertiary structure of aprotein.

23. List the amino acids that can form ionic bonds in the tertiary structure of a protein.24. List the amino acids that can form the hydrophobic pockets in the tertiary structure of

a protein.25. Explain how each of the following can denature a protein.

(a) increasing the temperature (b) changing the pH (c) adding a detergent(d) adding an oxidizing or reducing agent(e) adding compounds, such as urea, (H2N)2CPO, that form strong hydrogen bonds

Carbohydrates

26. Which of the following compounds satisfy the literal definition of the term carbohy-drate?(a) glucose, C6H12O6 (b) sucrose, C12H22O11 (c) cellulose, (C6H10O5)n

(d) glyceraldehyde, HOCH2CHOHCHO (e) dihydroxyacetone, (HOCH2)2CPO(f) ribose, C5H10O5

27. Use examples to explain the difference between an aldose and a hexose and betweena triose and a tetrose.

28. Describe the difference between �-D-glucopyranose and �-D-glucopyranose. Describethe difference between �-D-glucopyranose and �-D-glucofuranose.

29. Which of the following compounds is not an aldose?

30. How do maltose, lactose, and sucrose differ?31. How do glycogen, amylose, amylopectin, and cellulose differ?32. Explain why humans are able to digest glycogen, amylose, and amylopectin but not

cellulose. What would enable us to digest cellulose?

HOCA

HOCA

HOOCA

HOCACH2OH

OOH

OH

OOH

ACH2OH

OOHHOCA

OCA

HO

HO

OCA

HOCACHO

OOH

OH

HO

ACH2OH

OOHHOCA

OCA

HOOCA

HOCACHO

OOH

OH

OH

H O

ACH2OH

O

OH

Erythrose

Glucose Galactose Glucitol

HOCA

OCA

OH

H O

ACH2OH

CH2OH

O

OH

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BIOCHEMISTRY 39

33. The following formulas are the Fischer projections for D-erythrose and D-threose. Drawthe Fischer projections for the L isomers of these carbohydrates.

Lipids

34. Explain why lipids, such as fats and oils, are not soluble in water but are soluble in sol-vents such as chloroform, CHCl3, benzene, C6H6, and diethyl ether, (CH3CH2)2O.

35. In what ways are corn oil and olive oil similar to petroleum oil? What is the differencebetween these oils that allows us to digest the first two but not the third?

36. Draw the structure of the triglyceride formed when stearic acid combines withglycerol.

37. What is the difference between the structures of the fatty acids in fats and oils? Howdoes this difference cause one of these to be a solid and the other a liquid at roomtemperature?

38. Calculate the average oxidation states of the carbon atoms in a typical carbohydrate,such as glucose, C6H12O6, and a typical fatty acid, such as palmitic acid, C16H32O2. Usethe oxidation states to explain why fatty acids give off much more energy than carbo-hydrates when burned.

39. Glucose, C6H12O6, gives off 2870 kJ/mol when burned, and palmitic acid, C16H32O2,gives off 9790 kJ/mol when burned. Use the data to estimate the energy released pergram when carbohydrates and lipids are burned.

40. Which of the following carboxylic acids is least likely to be found in olive oil?(a) butyric acid (b) palmitic acid (c) oleic acid

41. The starting materials for the synthesis of phospholipids are the phosphatidic acids. Drawthe structure of the phosphatidic acid that would be produced by esterifying glycerol withtwo molecules of stearic acid (C17H35CO2H) and one molecule of phosphoric acid.

42. Explain why phospholipids, such as the phosphatidylcholines, can be described asphosphodiesters.

43. Explain why phospholipids, such as the phosphatidylcholines (lecithins), are such im-portant components of the structure of biological membranes.

44. Imagine a lipid bilayer formed by phosphatidylcholine. Which part(s) of the moleculewould form the hydrophilic surface of the bilayer?

Nucleic Acids

45. Draw the structure of one of the nucleotides found in nucleic acids. Show how thestructures of the monomers that form DNA differ from those that form RNA.

46. Classify the five common nucleic acids—adenine, cytosine, guanine, thymine, anduracil—as either purines or pyrimidines.

47. Explain why studies of nucleic acids invariably find roughly equivalent amounts ofpurine and pyrimidine nucleotides. Explain why nucleic acids that contain more ade-nine than guanine also contain more thymine than cytosine.

CA

HOCA

HOCA

OOH

OOH

H2OH CA

HOCA

HOOCA

OH

OOH

H2OH

CH2OH CH2OH

D-Erythrose D-Threose

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40 BIOCHEMISTRY

48. Which of the following equations are correct statements of the results of the work ofEdwin Chargaff, which provided the basis for Watson and Crick’s discovery of the struc-ture of DNA?(a) (%G � %A) � (%C � %T) (b) %G � %A (c) %C � %T(d) %G � %C (e) %A � %T

49. Draw the structures of the side chains on nucleic acids and show how hydrogen bondsform between adenine and thymine and between cytosine and guanine.

50. Describe the difference between replication, transcription, and translation.

Protein Biosynthesis

51. Describe the difference between the roles played by mRNA and tRNA in proteinbiosynthesis.

52. Assume that the following sequence of nucleotides on one strand of a DNA moleculeis used as the template to code for a protein.

3′ T-A-C-A-A-G-C-A-G-T-T-G-G-T-C-G-T-G- 5′

Write the sequence of nucleotides, reading from the 5′ to the 3′ end, in the messengerRNA produced when this strand of DNA is transcribed. Describe the polypeptide,starting from the N-terminal end, that would be produced during protein biosynthesis.

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