functional groupch5

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1 MODULE 5 ORGANIC 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 GROUPS Bromine 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 from addition across the double bond. This approach to understanding the chemistry of organic compounds presumes that certain atoms or groups of atoms known as functional groups give these compounds their characteristic properties. CH 3 CHPC CH 3 CH 2 Br 2 CH 3 C AA Br HC A Br CH 3 CH 3 A CH 3 CH 2 CH 3 CH P CHCH 3 Br 2 CH 3 C A Br HC A Br HCH 3

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Page 1: Functional GroupCH5

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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4 FUNCTIONAL GROUP CHEMISTRY

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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FUNCTIONAL GROUP CHEMISTRY 5

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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6 FUNCTIONAL GROUP CHEMISTRY

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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FUNCTIONAL GROUP CHEMISTRY 7

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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8 FUNCTIONAL GROUP CHEMISTRY

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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FUNCTIONAL GROUP CHEMISTRY 9

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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10 FUNCTIONAL GROUP CHEMISTRY

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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FUNCTIONAL GROUP CHEMISTRY 11

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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12 FUNCTIONAL GROUP CHEMISTRY

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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FUNCTIONAL GROUP CHEMISTRY 13

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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14 FUNCTIONAL GROUP CHEMISTRY

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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FUNCTIONAL GROUP CHEMISTRY 15

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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16 FUNCTIONAL GROUP CHEMISTRY

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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FUNCTIONAL GROUP CHEMISTRY 17

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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18 FUNCTIONAL GROUP CHEMISTRY

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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FUNCTIONAL GROUP CHEMISTRY 19

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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20 FUNCTIONAL GROUP CHEMISTRY

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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FUNCTIONAL GROUP CHEMISTRY 21

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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22 FUNCTIONAL GROUP CHEMISTRY

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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FUNCTIONAL GROUP CHEMISTRY 23

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

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24 FUNCTIONAL GROUP CHEMISTRY

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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FUNCTIONAL GROUP CHEMISTRY 25

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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26 FUNCTIONAL GROUP CHEMISTRY

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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FUNCTIONAL GROUP CHEMISTRY 27

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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28 FUNCTIONAL GROUP CHEMISTRY

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

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FUNCTIONAL GROUP CHEMISTRY 29

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

Chemistry in the World Around Us

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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30 FUNCTIONAL GROUP CHEMISTRY

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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FUNCTIONAL GROUP CHEMISTRY 31

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

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32 FUNCTIONAL GROUP CHEMISTRY

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

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FUNCTIONAL GROUP CHEMISTRY 33

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.

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34 FUNCTIONAL GROUP CHEMISTRY

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

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FUNCTIONAL GROUP CHEMISTRY 35

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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36 FUNCTIONAL GROUP CHEMISTRY

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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