04-chapter 2.pdf

AlkanesOrganic compounds all contain carbon, but they can also contain a wide variety of other ele-ments. Before we can appreciate such chemical diversity, however, we have to begin at the be-ginning, with the simplest organic compounds, the hydrocarbons. Hydrocarbons are com-pounds that contain only the elements carbon and hydrogen.

2.1 HYDROCARBONS

Methane, CHo, is the simplest hydrocarbon. As you learned in Sec. l 3B, all of the hydrogenatoms of methane are equivalent, occupying the corners of a regular tetrahedron. Imagine now,that instead ofbeing bound only to hydrogens, a carbon atom could be bound to a second car-bon with enough hydrogens to fulfill the octet rule. The resulting compound is ethane.

Lewis structures of ethane: HiC-CH:

space-filling model of ethane

tI

HHHH I I

H:C:e :H H-C-C-HHH I IHH

46

ball-and-stick model of ethane

2.1 HYDROCARBONS 47

'-tp'o bond

sp34sobond r

Figure 2.1 Hybrid-orbital description of the bonds in ethane. (The small rear lobes of the carbon sp3 orbitals areomitted for clarity.) The component sp3 and ls orbitals are shown with dashed lines for the two labeled bonds.

In ethane, the bond between the two carbon atoms is longer than a C-H bond, but, like theC-H bonds, it is a covalent bond in the Lewis sense. In terms of hybrid orbitals, thecarbon-carbon bond in ethane consists of two electrons in a bond formed by the overlap oftwo sp3 hybrid orbitals, one from each carbon. Thus, the carbon--carbon bond in ethane is ansp3-sp3 o bond (Fig. 2. 1). The C-H bonds in ethane are like those of methane. They consistof covalent bonds, each of which is formed by the overlap of a carbon sp3 orbital with a hy-drogen ls orbital; that is, they are sp3-1s o bonds. Both the H-C-C and H-C-H bondangles in ethane are approximately tetrahedral because each carbon bears four groups.

We can go on to envision other hydrocarbons in which any number of carbons are bondedin this way to form chains of carbons bearing their associated hydrogen atoms. Indeed, theability of a carbon to form stable bonds to other carbons is what gives rise to the tremendousnumber ofknown organic compounds. The idea ofcarbon chains, arevolutionary one in theearly days of chemistry, was developed independently by the German chemist August Kekuld(1829-1896) and the Scotsman Archibald Scott Couper (1831-1892) in about 1858. Kekul6'saccount of his inspiration for this idea is amusing.

During my stay in London I resided for a considerable time in Clapham Road in the neighborhoodof Clapham Common. . . . One fine summer evening I was retuming by the last bus, "outside" asusual, through the deserted streets of the city that are at other times so full of life. I fell into a reverie,and lo, the atoms were gamboling before my eyes. Whenever, hitherto, these diminutive beings hadappeared to me they had always been in motion. Now, however, I saw how, frequently, two smalleratoms united to form a pair. . . . I saw how the larger ones formed a chain, dragging the smaller onesafter them but only at the ends of the chain. . . . The cry of the conductor, "Clapham Road," awak-ened me from my dreaming, but I spent a part of the night putting on paper at least sketches of ttresedream forms. This was the origin of the "Structure Theory."

Hydrocarbons are divided into two broad classes: aliphatic hydrocarbons and aromatic hy-drocarbons. (The term aliphatic comes from the Greek aleiphntos, which means "fat." Fatscontain long carbon chains that, as you will learn, are aliphatic groups.) The aliphatic hydro-carbons consist of three hydrocarbon families: alkanes, alkcnes, and allqnes. We'1l begin ourstudy of aliphatic hydrocarbons with the alkanes, which are sometimes known as paraffins.Alkanes are hydrocarbons that contain only single bonds. Methane and ethane are the simplestalkanes. Later we'll consider the alkenes, or olefins, hydrocarbons that contain carbon--carbondouble bonds; and the alkynes, or acet5rlenes, hydrocarbons that contain carbon--carbon triple

spIlIII

II

48 CHAPTER 2 . ALKANES

an alkane

2.2

aliph ati c hy dr o c arb o n s

UNBRANCHED ALKANES

bonds. The last hydrocarbons we'll study are the aromatic hydrocarbons, which include ben-zene and its substituted derivatives.

HIH\ ^,rc\^ /'H(- -(-

lllCCH/" -Cy'" -H

I

H

benzene

an flromatic hydro carb on

HH\/

(1

-fU-U/\HH

an alkene

HHllH-C-C-HtlHHH-C-C-H

an a$rne

Carbon chains take many forms in the alkanes; they may be branched or unbranched, and theycan even exist as rings (cyclic alkanes). Alkanes with unbranched carbon chains are sometimescalled normal alkanes, or n-alkanes. A few of the unbranched alkanes are shown in Table 2. l,along with some of their physical properties. You should learn the names of the first twelve un-branched alkanes because they are the basis for naming many other organic compounds. Thenanes methane, ethane, propone, and butane have their origins in the early history of organicchemistry, but the names of the higher alkanes are derived from the conesponding Greek numer-ical names: pentane (pent : five); hexane (hex: six); and so on.

EEEE The unbranched Att€nes

Compoundname

Molecularformula

Condensed structuralformula

Melting Boiling Density*point ('C) point ('C) (g mt-l1

methane CHo

u'h:*"- !r!9propane crH,

butane CoH 'o

CHo

cH3cH3

- 182.5 -161.7

-183.3 -88.6cH3cH2cH3 - 187 .7

- 1 38.3cH3(cH2)rcH'

- 42.1 -- 0.5

pentane C,H,, iHt!cH:)icH,cH3(cH2)ocH'

- 129.8

-95.3

0.6262

0.6603

0.6837

0.7026

0.7177 _a.7299

0.7402

0.7 487

36.1

68.7

98.4

hexane

heptane_

octane

nona ne

CuH''o

C,H ,U c1!cH,)sCH.

cH3(cH2)ucH'

cH3(cH2)rcH'

CtH '*

- 56.8

- 53.512':]-150.8CnHro

deca ne c,oHz cH3(cH2)8cH3

undecane C, r Hro cH3(cH2)ncH.

dodeca ne

ercosane CroHo, cH3(cH2)'rcH'

c,rHru cH3(cH2),ocl

-29.7 17 4.0

-25.6 ',195.8

-9.6+ 36.8

216.3

*The densities tabulated in this text are of the liquids at 20 "C unless otherwise noted.

343.0 0.7886

2,2 UNBRANCHED ALKANES 49

Organic molecules are represented in different ways, which we'll illustrate using the alkanehexane. The molecular formula of a compound (for example, CuH,o for hexane) gives itsatomic composition. All noncyclic alkanes (alkanes without rings) have the general formulaC,H2,r,2, in which n is the number of carbon atoms. The structural formula of a molecule isits Lewis structure, which shows the connectivity of its atoms-that is, the order in which itsatoms are connected. For example, a structural formula for hexane is the following:

hexane

(Notice that this type of formula does not portray the molecular geometry.) Writing each hy-drogen atom in this way is very time-consuming, and a simpler representation of this mole-cule, called a condensed structural formula, conveys the same information.

H3C - CH2 - CH2 - CH2 - CH2 - CH3

hexane

In such a structure, the hydrogen atoms are understood to be connected to carbon atoms withsingle bonds, and the bonds shown explicitly are bonds betvveen carbon a/ons. Sometimeseven these bonds are omitted, so that hexane can also be written CH3CH2CH2CHTCHTCH,. Thestructural formula may be further abbreviated as shown in the third column of Thble 2.1 . In thistype of formula, for example, (9H)o means -cH 2cH2cH,cH2-, and hexane can thus bewritten CH3(CH2)4CH3.

cH3cH2cH2cH2cH2CH3 CH3(CH2)4CH3

two other representations of hexane

The family of unbranched alkanes forms a series in which successive members differ fromone another by one -CHr- group (methylene group) in the carbon chain. A series of com-pounds that differ by the addition of methylene groups is called a homologous series. Thus, theunbranched alkanes constitute one homologous series. Generally, physical properties within ahomologous series vary in a regular way. An examination of Table 2. 1 , for example, reveals thatthe boiling points and densities of the unbranched alkanes vary regularly with increasing num-ber of carbon atoms. This variation can be useful for quickly estimating the properties of amember of the series whose properties are not known.

The French chemist Charles Gerhardt (1816-1856) made an important chemical observa-tion in 1845 about members of homologous series. His observation still has significant impli-cations for learning organic chemistry. He wrote, "These (related) substances undergo reac-tions according to the same equations, and it is only necessary to know the reactions of one inorder to predict the reactions of the others." What Gerhardt was saying, for example, is thatwe can study the chemical reactions of propane with the confidence that ethane, butane, or do-decane will undergo analogous reactions.

2.r (a) How many hydrogen atoms are in the unbranched alkane with 18 carbon atoms?O) Is there an unbranched alkane containing 23 hydrogen atoms? Ifso, give its sructural for-

mula; if not, explain why not.Give the structural formula and estimate the boiling point of tridecane, C,rHrr.

HHHHHHtttltlH-C-C-C-C-C-C-HrttttlHHHHHH

)),hl.2

50 CHAPTER 2 O ALKANES

2.3 CONFORMATIONS OF ATKANESIn Section 1.3B, we learned that understanding the structures of many molecules requires thatwe specify not only their bond lengths and bond angles but also their dihedral angles. In this sec-tion, we'll use the simple alkanes ethane and butane to develop some widely applicable simpleprinciples that will allow us to predict the dihedral angles in more complex molecules.

A. Conformation of EthaneTo specify the dihedral angles in ethane, we must define the relationship between the C-Hbonds on one carbon and those on the other. A convenient way to do this is to view the mole-cule in a Newman projection. A Newman projection is a type of planar projection along onebond, which we'll call the projected bond.For example, suppose we wish to view the ethanemolecule in a Newman projection along the carbon-carbon bond, as shown in Fig. 2.2. In thisprojection, the carbon--carbon bond is the projected bond. To draw a Newman projection, startwith a circle. The remaining bonds to Ihe nearer atom in the projected bond are drawn to thecenter of the circle. The remaining bonds to the farther atom in the projected bond are drawnto the periphery of the circle. In the Newman projection of ethane (Fig.2.2c), the three C-H

b all - an d - stick m o d eI s :

Iine - and- w edge fo rmulqs :

//'Llprojectr-c1 botr

f+r-./

lJ -_

t,

,,,

H\ ri\\

j,'

*'"".,7b \ c--

H II

(a) viewing a model of ethane from one end

LIll

(b) end-on view (c) Newman projection( 0 = clitre-dral angle )

Figure 2.2 How to derive a Newman projection for ethane using ball-and-stick models (top) and line-and-wedge formulas (bottom). (The hydrogens and C-H bonds farthest from the observer are shown in blue.) Firstview the ethane molecule from the end of the bond you wish to project, as in part (a).The resulting end-on viewis shown in part (b).This is represented as a Newman projection (c) in the plane of the page. In the Newman pro-jection. the bonds drawn to the center of the circle are attached to the carbon closer to the observer; the bondsdrawn to the periphery of the circle (blue) are attached to the carbon farther from the observer.The projectedbond (the carbon-carbon bond) is hidden.

projected bond andrear carbon are hidden

1t

2,3 CONFORMATIONS OF ALKANES

bonds drawn to the center of the circle are bonds to the front carbon. The C-H bonds to theperiphery of the circle are the bonds to the rear carbon. The projected bond itself, which is thefourth bond to each carbon, is hidden.

The Newman projection of ethane makes it very easy to see the dihedrat angles 0 betweenits C-H bonds. When we have specified all of the dihedral angles in a molecule, we havespecified its conformation. Thus, the conformation of a molecule is the spatial arrangementof its atoms when all of its dihedral angles are specified. We can also refer to conformationsof parts of molecules, for example, conformations about individual bonds.

Two limiting possibilities for the conformation of ethane can be seen from its Newman pro-jections; these are termed the staggered conformation and the eclipsed conformation.

f/ : 60"

51

HTil

staggered conformationof ethane

eclipsed conformationof ethane

In the staggered conformation, a C-H bond of one carbon bisects the angle between twoC-H bonds of the other. The smallest dihedral angle in the staggered conformation is g :60". (The other dihedral angles arc 0 : 180' and 0 : 300'.) In the eclipsed conformation,the C-H bonds on the respective carbons are superimposed in the Newman projection. Thesmallest dihedral angle is g : 0o. (The other dihedral angles are 0 : l20o and 0 : 240..) ofcourse, conformations intermediate between the staggered and eclipsed conformations arepossible, but these two conformations will prove to be of central importance.

Which is the preferred conformation of ethane? The ehergies of the ethane conformationscan be described by a plot of relative energy versus dihedral angle, which is shown in Fig.2.3on p. 52. In this figure, the dihedral angle is the angle between the bonds to the colored hydro-gens on the different carbons. To see the relationships inFig.2.3,build a model of ethane, holdeither carbon fixed, and turn the other carbon about the C-C bond. As the angle of rotationchanges, the model passes alternately through three identical staggered and three identicaleclipsed conformations. As short*n'by Fig. 2.3, identical conformations have identical ener-gies. The graph also shows that the eclipsed conformation is characterized by an energy max-imum, andthe staggered conformation is characterized by an energy minimum. The staggeredconformation is therefore the more stable conformation of ethane. The graph shows that thestaggered conformation is more stable than the eclipsed conformation by about, 12 kJ mol-l(about 2.9 kcal mol-r). This meaas that it would take about 12kJ of energy td convert onemole of staggered ethane into one mole of eclipsed ethane.

The reason for the relative stability of the staggered conformation has been debated foryears. One theory is that the staggered form is more stable because there is a favorable inter-action between the bonding and antibonding molecular orbitals associated with the C-Hbonds on the two carbons that lie at a dihedral angle of 180'. This dihedral angle, and hencethe stabilizing interaction, is only possible in the staggered form. A second theory holds thatthere is repulsion between the electrons in the C-H bonds on the two carbons. Because thebonds are closer when they have a dihedral angle of0o, this repulsion is greater in the eclipsedform. This repulsion is termed torsional strain. Notice that the repulsion is not between thehydrogens but between the electrons in the bonds themselves. An assessment in 2007 of thetwo effects indicated that torsional strain accounts for about 757o of the energy difference


60

HH>*n"olotH

r20

HHAHA\gHI1

180

dihedral angle, degrees

Hn)?rnH^tr-^H

H

300

Hff---r\rHH^-i H

H

240 360(=o)

(HH

HH/T\HIO\e

HH.

HHAiAH

Figufe 2.3 Variation of energy with dihedral angle about the carbon-carbon bond of ethane. In this diagram,the rear carbon is held fixed and the front carbon is rotated, as shown by the green arrows.The dihedral angleplotted is the angle between the bonds to the red and blue hydrogens. Note that the staggered conformationsare at the energy minima, and the eclipsed conformations are at the energy maxima.

:t-=EixHsE5x*sj

V

H

between the eclipsed and staggered forms, and that the favorable orbital interaction accountsfor abott25%o.

One staggered conformation of ethane can convert into another by rotation of either carbonrelative to the other about the carbon--carbon bond. Such a rotation about a bond is called aninternal rotation (to differentiate it from a rotation of the entire molecule). When an internalrotation occurs, an ethane molecule must briefly pass through the eclipsed conformation. Todo so, it must acquire the additional energy of the eclipsed conformation and then lose it again.What is the source of this energY?

At temperatures above absolute zero, molecules are in constant motion and therefore havekinetic energy. Heat is a manifestation of this kinetic energy. In a sample of ethane, the mole-cules move about in a random manner, much as people might mill about in a large crowd.These moving molecules frequently collide, and molecules can gain or lose energy in suchcollisions. (An analogy is the collision of a bat with a ball; some of the kinetic energy of thebat is lost to the ball.) When an ethane molecule gains sufficient energy from a collision, it canundergo internal rotation, passing through the more energetic eclipsed conformation into an-other staggered conformation. Whether a given ethane molecule acquires sufficient energy toundergo an internal rotation is strictly a matter of probability (random chance). However, aninternal rotation is more probable at higher temperature because wiumer molecules havegreater kinetic energy.

The probability that ethane undergoes internal rotation is reflected as its rate of rotation:how many times per second the molecule converts from one staggered conformation into an-other. This rate is determined by how much energy must be acquired for the rotation to occur:the more energy required, the smaller the rate. In the case of ethane, 12 kJ mol-' (2.9 kcalmol-l; is required. This amount of energy is small enough that the internal rotation of ethane

2.3 CONFORMATIONS OF ALKANES 53

is very rapid even uiu.ry low temperatures. At 25 "C, atypical ethane molecule undergoes arotation from one staggered conformation to another at a rate of about 10r1 times per second!This means that the interconversion between staggered conformations takes place about onceevery 10-rr second. Despite this short lifetime for any one staggered conformation, an ethanemolecule spends most of its time in its staggered conformations, passing only transientlythrough its eclipsed conformations. Thus, an internal rotation is best characterized not as acontinuous spinning but as a constant succession ofjumps from one staggered conformationto another.

B. Conformations of ButaneButane contains two distinguishable types of carbon--carbon bonds: the two terminal C-Cbonds (blue), and the central C-C bond (red).

H3C-CH2-CH2-CH3butane

two types of C-C bonds

We'll consider internal rotation about the central C-C bond. This rotation is a bit more com-plex than the ethane case, but examination of this rotation leads to important new insightsabout molecular conformation. As with ethane, we use Newman projections, as shown in Fig.2.4. Remember again that the projected bond-the central C-C bond in this case-is hiddenin the Newman projection.

b all- and- stick mo dels:

] t"'sroup

projected bon

\line-and-wedge formulas: , \1'\ ...-

/\H ?t,

CH:

(a) viewing a model of butane from one endof the central carbon-carbon bond

(b) end-on view (c) Newman projection

FiSure 2'4 How to derive the Newman projection ofthe central carbon-carbon bond in butane using ball-and-stick models (fop) and line-and-wedge formulas (botfom).The bonds and groups on the rear carbon of the pro-jected bond are shown in blue. (Only one of the butane conformations is shown.)

projected bond andcarbon- 3 are hidden


60 120 180dihedral angle, degrees

CH--anti

240

H{-,H1

300

Hgauche

360(:o)

H.C cH 1alil( /T\LHF\sA' ,H+-/

H CF{rAHp\fi,

T]II. H.C (,Fl 1A

HAffH

H

H

H

H

H

CH-.

HH

gauche

Figure 2.5 Variation of energy with dihedral angle about the central carbon-carbon bond of butane. In this di-agiam, the rear carbon is held fixed and the front carbon is rotated, as shown by the green arrows.The dihedralangle plotted is the one between the bonds to the two CH3 groups.

I

f-. )Y

+lYrn l4( 3.3

'1\kJ mol-r .

kcal mol*r )

IYz.skl mol- \t Q.6i kcal mol-r) \

cFI ,

The graph of energy as a function of dihedral angle in butane is given in Fig. 2.5. Note onceagain that the various rotational possibilities are generated with a model by holding either car-bon fixed (the carbon away from the observer in Fig. 2.5) and rotating the other one'

Figure 2.5 shows that the staggered conformations of butane, like those of ethane, are at en-ergy minima and are thus the stable conformations of butane. However, not all of the staggeredconformations (nor all of the eclipsed conformations) of butane are alike. The different stag-gered conformations have been given special names. The conformations with a dihedral angleof 60. and 300' in Fig.2.5 (or +60") between the two C-CH3 bonds are called gauche con-formations (from the French gauchi4 meaning "to turn aside"); the form in which the dihedralansle is 180" is called the anti conformation.

gauche conformation0=60"

The relationship betweenbonds that have a dihedral

anti conformation0 = 180o

bonds also can be describedrelationship of +60o are said

gauche conformation0 = 300o (= -60')

with the terms gauche and anti. Twoto be gauche bonds. Two bonds that

Further Exploratio n 2.1Atomic Radii and

van der WaalsRepulsion

2.3 CONFORMATIONS OF ALKANES 55

have a dihedral relationship of 180' are said to be anti bonds. Notice that these terms refer tobonds on adjacent carbons.

Figure 2.5 shows that the gauche and anti conformations of butane have different energies.The anti conformation is the more stable of the two by 2.g0 kJ mol-r (0.67 kcal mol r;. Thegauche conformation is the less stable of the two because the CH, groups are very close to-gether-so close that the hydrogens on the two groups occupy each other's space. You can seethis with the aid of the space-filling model inFrg.2.6a.

This problem can be discussed more precisely in terms of atomic size. One measure of anatom's effective size is its van der Waals radius. Energy is required to force two nonbondedatoms together more closely than the sum of their van der Waals radii. Because the van derWaals radius of a hydrogen atom is about |.2 A, forcing the centers of two nonbonded hydro-gens to be closer than twice this distance requires energy. Furthermore, the more the two hy-drogens are pushed together, the more energy is required. The extra energy required to forcetwo nonbonded atoms within the sum of their van der Waals radii is called a van der Waalsrepulsion. Thus, to attain the gauche conformation, butane must acquire more energy. In otherwords, gauche-butane is destabilized by van der Waals repulsions between nonbonded hydro-Sens on the two CH, groups. Such van der Waals repulsions are absent in anti-butane (see Fig.2.6b). Thus, anti-butane is more stable than gauche-butane.

As with ethane, the eclipsed conformations of butane are destabilized by torsional strain.But, in the conformation in which the two C-CHr bonds are eclipsed, the major source of in-stability is van der Waals repulsions between the methyl hydrogens (Fig. 2.6c), which areforced to be even closer than they are in the gauche conformation. Notice that this is the mostunstable of the eclipsed conformations (0 : 0' in Fig. 2.5).

_It is important to understand the relative energies of the butane conformations because,when different stable conformations are in equilibrium, the most stable conformation-theconformation of lowest energy-is present in greatest amount. Thus, the anti conformation ofbutane is the predominant conformation of butane. At room temperature, there are about twiceas many molecules of butane in the anti conformation as there are in the gauche conformation.

H-H distanr-e is less than tiresultt of r.l1n rler \\Iirals rarlii

H-H clistiinres ilre k:ss thrrn theslun cf van ct-r \\rirals rircliino van der Waals repulsions

(a) gauche-butane (b) anfi-butane (c) butane with C-CH:bonds eclipsed

Figure 2.6 Space-filling models of different butane conformations with the methyl hydrogens shown in color.(a) Gouche-butane.A hydrogen atom from one CH, group is so close to a hydrogen atom ofthe other CH. groupthat these hydrogens, shown in pink, violate each other's van der Waals radii.The resulting van der Waals repul-sions cause gouche-butane to have a higher energy than dntl-butane, in which this interaction is absent. (b) Anti-butane.This conformation is.most stable because it contains no van der Waals repulsions. (c) Butane with theC-CH3 bonds eclipsed. In this conformation, van der Waals repulsions between the hydrogens of the two CHrgroups (pink) are even greater than they are in gauche-butane.

56 CHAPTER 2 O ALKANES

The gauche and anti conformations of butane interconvert rapidly at room temperature-almost as rapidly as the staggered forms of ethane. Because the eclipsed conformations lie at

energy maxima and are unstable, they do not exist to any measurable extent.The investigation of molecular conformations and their relative energies is called confor-

mational analysis. In this section, we've learned some important principles of conformationalanalysis that we'll be able to apply to more complex molecules. Here is a summary of these

principles:

l. Staggered conformations about single bonds are favored.2. Yan der Waals repulsions (repulsions between nonbonded atoms) occur when atoms are

"squeezed" closer together than the sum of their van der Waals radii'3. Conformations containing van der Waals repulsions are less stable than conformations

in which such repulsions are absent.4. Rotation about C-C single bonds is so rapid that it is hard to imagine separating con-

formations except at very low temperature.

Draw a Newman projection for the anti conformation about the C3-C4 bond of 2-methylhexane,viewing the bond so that C3 is nearest the observer.

H3C-CH- CH2-CH2-CH2-CH3lt{CH.II'C3C42-methylhexane

Sglgtion First draw a "blank" Newman projection to represent the projected bond. Rememberthat the projected bond itself (the C3{4 bond) is invisible in the projection. Either templatebelow can be used.

We arbitrarily pick the template on the left. In the view dictated by the problem, the front carbonis C3. Identify the three groups attached to C3 with bonds other than the projected bond. These

groups ,ue H, H, and the H3C-C, H- group. Put these on the front carbon of the Newman

CH:projection. It doesn\ mnner which bonds go to which Sroups as long as all groups are on the

front carbon.It's important to understand that, because we are not examining the bonds within the

large group, we can condense this group to (CHr)rCH- or even CrHr-.

cH(cH3)2

We then identify the groups attached to the back carbon (C4) by bonds other than the projected

bond. These groups are H, H, and -CH2CH3. Now it does matter where we put these groups, be-

cause we are asked for the anti conformation. The -CHTCH: glouP must be placed anti to the

(CH,),CH- group. Remember that "anti" means a dihedral angle of 180".

H

2.4 CONSTITUTIONAL ISOMERS AND NOMENCLATURE 57

HI

cH3cH'z tVHuf\cH1cHry,

H

2-nethy'hexaneanti conformation about C3-C4 bond

Rerrcmber that Newman projections are used to examine oonfsrmations abovt a p,rticular bondIf we want to examine the confounations about several differelt bonds, we must draw a differentset of Newman projections for each bond.

@ 2.3 (a)Drawtn"r".*p*iecrionforeachsonfonnation".n "n-!! **;r*o*o*",

a compound containing a branched carbon chein23

H3C-CH-CH2-CH3J''r. ':l 'i',

icopentane

Show both staggered and eclipsed conformations.

", ;Iff : ;"il:;J ffi Iffi ffiNLTlffi ffi *H"trilH :iTJ#*::tions you drcw in part (a).

(c) Which conforrnations "* ,n"r, to be present in gr€abst amowt i1 n 5nmFle of isopentane?Explain.

24,,, Repeat, tho analysis in ttoblem Z.l fot either one of the terminal bo+ds oJ buane.

2,4 CONSTITUTIONAL ISOMERS AND NOMENCTATURE

A. lsomensWhen a carbon atom in an alkane is bound to more than two other carbon atoms, a branch in thecarbon chain occurs at that position. The smallest branched alkane has four carbon atoms. As aresult, there are two four-carbon alkanes; one is butane, and the other is isobutane.

H,C\H3C-CH2-CH2-CH3 /CH-CH3/butane HtC

bp -0.5'isobutanebp -11.7'

These are different compounds with different properties. For example, the boiling point of bu-tane is - 0.5 "C, whereas that of isobutane is - I | .7 'C. Yet both have the same molecular for-


mula, CoH,o. Different compounds that have the same molecular formula are said to be iso'mers or isomeric compounds.

There are different types of isomers. Isomers that differ in the connectivity of their atoms,such as butane and isobutane, are called constitutional isomers or structural isomers. Recall(Sec. 1.3) that connectiulry is the order in which the atoms of the molecule are bonded.Theatomic connectivities of butane and isobutane differ because in isobutane a carbon is attachedto three other carbons, whereas in butane no carbon is attached to more than two other carbons.

Which of the following four structures represent constitutional isomers, which represent the same

molecule, and which one is neither isomeric nor identical to the others? Explain your answers.

CH.I

CH3CHCHCH3I

CHr

CHaI

cH3cHcHCH2CHrI

CH:

cH3cH2cHCH3I

"CH"HrCt -- - -CH:C

cHjcHCH2CH2CH2CH3'r1CH:

Solution Compounds must have the same molecular formula to be either identical or isomeric.Structure A has a different molecular formula (CuH,.) from the other struchrres (C7H16), and hencestructure A represents a molecule that is neither identical nor isomeric to the others. To solve therest of the problem, we must understand that Izwis structures show connectivity only. They do notrepresent the actual shapes of molecules unless we start adding spatial elements such as wedges

and dashed wedges. This means thatwe can draw a given structure many dffirent ways. Haveyou ever heard the old spiritual, "The foot-bone's connected to the ankle-bone . . . "? That's a

song about connectivity of the typical human body. If the description fits you, its validity doesn'tchange whether you are sifting down, standing up, standing on your head, or doing yoga. Simi-larly, the connectivity of a molecule doesn't change whether it is drawn forwards, backwards, orupside-down. With that in mind, let's trace the connectivity of each structure above. Considersffuctures B and C. Each has two CH, groups connected to a CH, and that CH is connected to an-

other CH, which in turn is connected to both a CH, group and a CHTCH, group. In B, this connec-tivity pattern starts on the left in C it starts on the bottom. But it's the same in both. Because bothsffuctures have identical connectivities, they represent the same molecule.

Structures D and B (or D and C) have the same molecular formula CrH'u, but, as you should ver-ify, their connectivities are different, so they are constitutional isomers.

Butane and isobutane are the only constitutional isomers with the formula CoH'g. However,more constitutional isomers are possible for alkanes with more carbon atoms. There are nineconstitutional isomers of the heptanes (CrH16), 75 constitutional isomers of the decanes(Cr0H22) and366,319 constitutional isomers of the eicosanes (CroHor)! From these few exam-ples, it is easy to understand that millions of organic compounds are known and millions moreare conceivable. It follows that organizing the body of chemical knowledge requires a systemof nomenclature that can provide an unambiguous name for each compound.

Organic NomenclatureAn organized effort to standardize organic nomenclature dates from proposals made atGeneva in 1892. From those proposals the International Union of Pure and Applied Chemistry

B.


(IUPAC), a professional association of chemists, developed and sanctioned several acceptedsystems of nomenclature. The most widely applied system in use today is called substitutivenomenclature.

The IUPAC rules for the nomenclature of alkanes form the basis for the substitutivenomenclature of most other compound classes. Hence, it is important to learn these rules andbe able to apply them.

C. Substitutive Nomenclature of AlkanesAlkanes are named by applying the following 10 rules in order. This means that if one ruledoesn't unambiguously determine the name of a compound of interest, we proceed down thelist in order until we find a rule that does.

l. The unbranched alkanes are named according to the number of carbons, as shown inTable 2.1.

2. For alkanes containing branched carbon chains, determine the principal chain.

The principal chain is the longest continuous carbon chain in the molecule. To illustrate:

H3C-CH2-CH2-CH-CH2-CH3 principal charn

3r,When identifying the principal chain, take into account thatthe condensed structure of a givenmolecule may be drawn in several dffirent ways (Study Problem 2.2). Thus, the followingstructures represent the same molecule, with the principal chain shown in red:

H3C-CH2-CH2-CH-CH2-CH3 H3C-CH-CH2-CH3J", J",-.",-.",(Be sure to veriS that these structures have identical connectivities and thus represent thesame molecule.)

3. If two or more chains within a structure have the same length, choose as the principalchain the one with the greater number of branches.

The following structure is an example of such a situation:

cH, --- CH.rl

H:C- CH- CH-- L-Hr - CH, -- CHIIl--

cHz-cH:IH3C- CH- CH- CH 2- CHZ- CH'-=-_l

CH: six-carbon chain\\,ith tlr.o branches

(this is the proper choicefor principal chain)

CH: six-r-arl-ron L-hain\\'ith olte Lrritnrlt

The correct choice ofprincipal chain is the one on the right, because it has two branches; thechoice on the left has only one. (It makes no difference that the branch on the left is larger orthat it has additional branching within itself.)

4. Number the carbons of the principal chain consecutively from one end to the other in thedirection that gives the lower number to the first branching point.

60 cHAPTER2. ALKANES

In the following structure, the carbons of the principal chain are numbered consecutively fromone end to give the lower number to the carbon at the -CH, branch.

6 5 4 3 2 |

-oroPernumheringH]C-CH2-CH2-CH-CH2-CH3 , .-l )' J--' +1 - . 6 i impropernttmberitrg

CHr

5. Name each branch and identifu the carbon number of the principal chain at which itoccurs.

In the previous example, the branching group is a -CH, group. This group, called a methylgroup, is located at carbon-3 ofthe principal chain.

Branching groups are in general called substituents, and substituents derived from alkanesare called alkyl groups. An alkyl group may contain any number of carbons. The name of anunbranched alkyl group is derived from the name of the unbranched alkane with the samenumber of carbons by dropping the final ane and adding y/.

-CHr methyl (- methtld + yl)

-CHzCH: or -CzHs ethyl (: ethttd + yl)

-cHrcH2cH3 propyl

Alkyl substituents themselves may be branched. The most common branched alkyl groups

ual have special names, given in Table2.2. These should be learned because they will be encoun-sruDycutDEulr2.l tered frequently. Notice that the "iso" prefix is used for substituents containing two methylN.omenclatllr€ of sroups at the end of a carbon chain. Also notice carefully the difference between an isobutylslmple Braiched- cbmpourioi group and a sec-butyl group; these two groups are frequently confused.

@Nomenc|atureofSomeShortBranched.ChainA|ky|Groups

Group structure Condensed stru(ture Written name Pronounced name

H.C

CH-H:C

(cH3)2cH- isopropyl isopropyl

H.C

iHCH2-H:C

(CH3)2CH CHr- isobutyl isobutyl

CH.CH,CH --'lCH:

sec-butyl secondary butyl or"sec-butyl"

fH,IH.C-C--lCH:

(cH3)3c- ferf-butyl tertiary butyl or(or r-butyl) "tert-butyl"

CH.l-H.C-C-CH?-"lCH:

(cH3)3ccH 2- neopentyl neopentyl


6. Construct the name by writing the carbon number of the principal chain at which thesubstituent occurs, a hyphen, the name of the branch, and the name of the alknne corre-sponding to the principal chain.

H3C- CH2 - CH2 - CH- CH2.- CH3

CH:

name:3-methylhexanellI nu-. of principal chainnumber and nameofalkyl substituent

Notice that the name of the branch and the name of the principal chain are written togetheras one word. Notice also that the name itself has no relationship to the name of the isomericunbranched alkale; that is, {he preceding compound is a constitutional isomer ofheptanebe-cause it hds seven carbon atoms, but it is named as a derivative of hexane, because its princi-pal chain contains six carbon atoms.

H3C - CH2-CH2 - CH-CH2 -CH2-CH1-lcH-cH3

I.iCHl

SOlUtiOn Becauro the,principal chain has seven carbons, ttre compound is narnEd as asubstituted heptane. The branch is at carbon4, and the substituerrt group at this biranch is

I

gH_cH3I

CH,Table 2.2 shows that this gfoup is an isopropyl gLtoup. Thus, the name of the compound is4-isopropylheptane:

H3C -CH2-CH2 -CH - CH2 -CH2-CHq-lcH-cH3J*.

4-isopropylheptane\--l

Ialkyl group name from Table 2.2

Because this compound has the molecular formula C,oHrr, it is a constihrtional isom€f of theunbranched akane de c arw.

Name the following compcxrnd, and grve the name of the unb,ranched alkane of which it is aconstitutional isomer.

62 CHA|TER 2 . ALKANES

7. If the principal chain contains multiple substituent groups, each substituent receives itsown numben The prefixes di, tri, tetra, and so on, are used to indicate the number ofidentical substituents.

CH:

H3C-C-CH2CH2CH3

CH:

2l-dimethylpentanelfI l*" methvl substituentsI

shows that both methyl branches areat carbon-2 ofthe principal chain

Which trro of tlre following sftuctlrr€s represnt the same compound? Name the compound"

Hrc-ll'-cHr

IHrC-CHz*C-HICHr

HrC -?H - CH, - T" - cHz - cHr Hgc-?**?"-.H2*cH3

?", cHa

CHr

CHs CHr

Solutlon Tb comectivities of both A ald C are the samq [CHr, CH2, (CH connected b CIlr),(CH conncst€d to C[tr), C&, Cfis]. The compound rep'resented by these stnictures has sir car$onsin its Fiqcipal chain and is tberefore named as a hexane. There are methyl bmnches at carbons 3ad 4. Hence tlp nane is 3,44imotbyllpxane. (You should name cornpound B after you study thenext nde.)

8. If substituent groups occur at more than one carbon of the pincipal chain, alternativenumbering schemes are compared number by numbe4 and the one is chosen that givesthe smaller number at the first point of difference.

This is one of the trickiest nomenclature rules, but it is easy to handle if we are systematic.To apply this rule, write the two possible numbering schemes derived by numbering from ei-ther end of the chain. In the following example, the two schemes are 3,3,5- and 3,5,5-.

9Ht1 2 3l 4 5 6 7H3C-CH2- Q-CHz- QH- CH2- CH37 6 sl 4 31 2 1t", i",

possible names:3,3,5-trimethylheptane ( correct)3,5,5-trimethylheptane (incorrect)

A decision between the two numbering schemes is made by a pairwise comparison of thenumber sets (3,3,5) and (3,5,5).


How to do a pairwise comparison:

(3,3,5 )

Because the first point of dffirence in these sets occurs at the second pair-3 versus 5-thedecision is made at this point, and the first scheme is chosen, because 3 is smaller than 5. Ifthere are differences in the remaining numbers, they are ignored. The sum of the numbers isalso irrelevant. Finally, it makes no difference whether the names of the substituents are thesame or different; only the numerical locations are used.

The next rule deals with the order in which substituents are listed, or "cited," in the name.Don't confuse the citation order of a substituent with its numerical prefix; they aren't neces-sarily the same.

9. Substituent groups are cited in alphabetical order in the name regardless of theirIocation in the principal chain. The numerical prefaes di, tri, and so on, as well as thehyphenated prefixes tert- and sec-, are ignored in alphabetizing, but the prefixes iso,neo, and cyclo are considered in alphabetizing substituent groups.

The following compounds illustrate the application of this rule:

s,t - 3u,-.6" - 6n, - 3", - 3r - i",I

QH, CH:

CHr

5-ethyl-2-methylheptane(ethylis cited before methyl eventhough it has a higher number)

CH.t-cH3cH2cH2 - Q - CHz - CH - CH2CH3

cHr cH2cH3

3-ethyl-5,5 -dimethyloctane(note that dimethyl begins

with the letter rr forpurposes ofcitation)

10. If the numbering of different groups is not resolved by the other rules, the first-citedgroup receives the lowest number

In the following compound, rules 1-9 do not dictate a choice between the names 3-ethyl-5-methylheptane and 5-ethyl-3-methylheptane. Because the ethyl group is cited first in thename (rule 9), it receives the lower numbeq by rule 10.

compere these firstC o fttp sr e t'h e s t: st'c:rl Fld

(3,5,5 )

compare these last

64 CHA TER 2 . ALKANES

H3C - CH2- CH-CH2 - CH - CH2 - CH3IIcHr czHs

3 -ethyl-5-methylheptane

Some situations of greater complexity are not covered by these 10 rules; however, theserules will suffice for most cases.

2.5 Name the follornring compounds.(n) T*,

CH3CHCHCH2CHCH3I

CHr

(c) CHzCHaCHT (d)l-

9H, (b) compound B in Study Problem 2.4 on p.62I

?"' ?"'cH3cHzcHzl,YCHzCHzCFIs CH3CH2CCH2CH2CHCH3

cHZcfls CHg

D. Hlghly condensed structuresWhen space is at a premium, parentheses are sometimes used to form highly condensed struc-tures that can be written on one line, as in the following example.

QHr

(CH3)4C or C(CH3)4 means H3C-C-CH3CH:

When such structures are complex, it is sometimes not immediately obvious, particularlyto the beginner, which atom inside the parentheses is connected to the atom outside the paren-theses, but a little analysis will generally solve the problem. Usually the structure is drawn sothat one of the parentheses intervenes between the atoms that are connected (except for at-tached hydrogens). However, if in doubt, look for the atom within the parentheses that is miss-ing its usual number of bonds. When the group inside the parentheses is CH3, as in the previ-ous example, the carbon has only three bonds (to the H's). Hence, it must be bound to the atomoutside the parentheses. Consider as another example the CH,OH groups in the followingstrucnue.

(cH:)2cH-cH(cH2oH)zH.C CH?OH'\ /MEANS CH-CH/\H:C CH2OH

Because the oxygen is bound to a carbon and to a hydrogen, it has its full complement of twobonds (the two unshared pairs are understood). The carbon of each CH2OH group, however, isbound to only three groups (two H's and the O); hence, it is the atom that is connected to thecarbon outside the parentheses.


If the meaning of a condensed structure is not immediately clear, write it out in less con-densed form. If you will take the time to do this in a few cases, it should not be long before theinterpretation of condensed structures becomes more routine.

Research in student learning strategies has shown that student success in organic chemistry is highlycorrelated with whether a student takes the time to write out intermediate steps in a problem. Suchsteps in many cases involve writing structures and partial structures. Students may be tempted to skipsuch steps because they see their professors working things out quickly in their heads and perhapsfeel that they are expected to do the same. Professors can do this because they have years of experi-ence. Most of them probably gained their expertise through step-by-step problem solving. In somecases, the temptation to skip steps may be a consequence of time pressure. If you are tempted in thisdirection, remember that a step-by-step approach applied to relatively few problems is a better expen-diture of time than rushing through many problems. Study Problem 2.5 illustrates a step-by-step ap-proach to a nomenclature problem.

Write the Lewis structure of 4-ser-butyl-5-ethyl-3-methyloctane. Then write the structure in acondensed form.

SOlution To this point, we've been giving names to structures. This problem now requires thatwe work "in reverse" and construct a structure from a name. Don't try to write out the structureimmediately; rather, take a systematic, stepwise approach involving intermediate structures. First,write the principal chain. Because the name ends in octane, tlte principal chain contains eight car-bons. Draw the principal chain without its hydrogen atoms:

c-c-c-c-c-c-c-cNext, number the chain from either end and attach the branches indicated in the name at the ap-propriate positions: a sec-butyl group at carbon-4, an ethyl group at carbon-5, and a methyl groupat carbon-3. (Use Thble 2.2 to learn or releam the structure of a sac-butyl group, if necessary.)

H3C-CH2-9H-CH3 =- sec-butyl group

c-c-c-C-c-c-c-c1 2 31 4 sl 6 7 8

cH: cH2-cH3Finally, fill in the proper number of hydrogens at each carbon of the principal chain so that eachcarbon has a total of four bonds:

H:C-CHz-CH-CH:-lH:C - CHz - CH -CH - CH - CHz - CHz - CHrtl

CH: cHz-cH:4-sec-butyl-5-ethyl-3-rnethyloctane

To write the structure in condensed form, put like groups attached to the same carbon withinparentheses. Notice that the structure contains within it two sec-butyl groups (red in the followingstructure), even though only one is mentioned in the name; the other consists of a methyl branchand part of the principal chain.

H_.C_-*CHo-tlH -CH-,H:C -. CH: - CH - CH - CH - CHz - CHz -CH:rlcH- cHz-cHr

+ {cI-IrcHrcH) -cHCHCH2cHzcH:"ltLlcH3 cH2cH3


E.

Nomenclature and chemical IndexingThe world's chemical knowledge is housed in the chemical llterature, which is the collection ofbook,journals,patents,technical reporttand reviews that constitute the published record of chem-ical research.To find out what, if anything, is known about an organic compound of interest, we have

to search the entire chemical literature.To carry out such a search, organic chemists rely on twomajor indexes.One is ChernicalAbstracts,published by the Chemical Abstracts 5ervice of the Ameri-can Chemical Society, which has been the major index of the entire chemical literature since 1 907.The second index is Beitstein's Handbaok of Organic Chemistry,known to all chemists simply as Betl-

stein,which has published detailed information on organic compounds since 1881.Initially,a search

of these indexes was a laborious manual process that could require hours or days in the library'Today, however,both Chemical Abstracts and Beilstein have efficient search engines that enable

chemists to search for chemical information from a personal computer. Nomenclature plays a key

role in focating chemical compounds, particularly in Chemical Absfrdcfs, but it is also possible tosearch for a compound of interest by submitting its structure.A search of Chem icalAbstractsyields a

short summary called an abstract,of every article that references the compound of interest, alongwith a detailed reference to each article. Beilsfein yields not only the appropriate references but also

detailed summaries of compound properties.

2.6 Draw structures for all isomers of (a) heptane and (b) hexane. Give their systematic names.

2.? Name the following compounds. Be sure to designate the principal chain properly before con-structing the name.(a) CH3CHT- CH2CH2CH3

I9H, CHr

iHr-cHr-cH,(b) 9H, PH'tl

cH3cH2cH2GH - C- CH - CH2CH2CH3

3t, lt,2.8 Draw a structue for (CHTCHTCI!)2CHCH(CH2CHr), in which all carbon--carbon bonds are

shown explicitly; then name the compound.

2.9 Draw the sfiucture of 4-isopropyl-2,4,5-trimethylheptane.

Classification of Carbon SubstitutionWhen we begin our study of chemical reactions, it will be important to recognize differenttypes of carbon substitution in branched compounds. A carbon is said to be primary'secondary, tertiary, or quaternary when it is bonded to one, two, three, or four othercarbons, respectively.

3 CH:| -/ quaternary carbonty'C-CH:licfl: I\l'bon \pri,-,',i1r'\'i,-irrb.rrs

frritntrn' ctlrl)olls

2.5 CYCLOALKANES AND SKELETAL STRUCTURES 67

Likewise, the hydrogens bonded to each type of carbon are called primary, secondary, or ter-tiary hydrogens, respectively.

H:C-CH2-CHr-CH C-CH:t\CHo \ primary hydrogens

secoIldary hydrogenstertiary hydrogen

@ 2.10 In tk sm* of 4isqrogyl-2,a5-trinetryrhepae (probbm 29)(a) ldeutify ftc pimwy, s€dtry, b-dart, ad quatsn"ty [email protected]) Iddify tbe pimmy, mdery, and tertiary hy&ogeffi.(c) Circb om exary$ of erch of fu following group$: a mefhyl grurp; u ethyl g[anp; an

ieopropyl groug a sec{utyl group; an isobutyl group.2.ll Identify the etlryl groups and the rethyl groups in the structurc of 4-rec-hrtJ/l-5dttl-3-

methyloctane, the conpound discussed in Study hoblem 2.5. Note thn thee€ grurps are ncneceesa4ily eouffim 6ose qpecifica[y rnentioned in the name"

iriirlll,1 ,I 1'l

2.6 CYCTOALKANES AND SKEIETAI STRUCTURES

primary hydrogens

Some alkane contain carbon chains in closed loops, or rings; these are called cycloalkanes.Cycloalkanes are named by adding the prefix cyclo to the name of the alkane. Thus, the six-membered cycloalkane is called cyclohexane. -

cH"Hrc/ -cH,

,lH2c\,,'cH2i. , CHz

cyclohbxane

The names and some physical properties of the simple cycloalkanes are given in Thble 2.3.The general formula for an alkane containing a single ring has two fewer hydrogens than thatof the open-chain alkane with the same number of carbon atoms. For example, cyclohexanehas the formula CuH,r, whereas hexane has the formula CuH,o. The general formula for the cy-cloalkanes with one ring is C,Hrn.

Because ofthe tetrahedral configuration ofcarbon in the cycloalkanes, the carbon skeletonsof the cycloalkanes (except for cyclopropane) are not planar. We'll study the conformations ofcycloalkanes in Chapter 7. For now, remember only that planar condensed structures for thecycloalkanes convey no information about their conformations.

Skeletal Structureg An important structure-drawing convention is the use of skeletalstructures, which are structures that show only the carbon-carbon bonds. In this notation, acycloalkane is drawn as a closed geometric figure. In a skeletal structure, it is understood that

ALKAN ES

@Compound

68 CHAPTER 2 o

Physical Prspertles of $ome Cycloafkanes

Boiling point ('C) Melting point ('C) Density (g mL-t)

cyclopropane - 32.7 - 127 .6

cyclobutane 12.5

cyclopentane 49,3 - 93.9 0.7 457

cyclohexane 80.7 6.6 0.7786

cycloheptane 1 18.5 - 12.0 0.8098

cyclooctane 150.0 14.3 0.8340

a carbon is located at each vertex of the figure, and that enough hydrogens are present oneach carbon to fulfill its tetravalence. Thus. the skeletal structure ofcyclohexane is drawn asfollows:

r a carbon and two hydrogensare at each vertex

Skeletal structures may also be drawn for open-chain alkanes. For example, hexane can beindicated this wav:

* for Hrc 'zcHz-cH/cHz-c"/cH:

When drawing a skeletal structure for an open-chain compound, don't forget that carbons arenot only at each vertex, but also atthe ends ofthe structure. Thus, the six carbons ofhexane inthe preceding structure are indicated by the four vertices and two ends of the skeletal structure.Here are two other examples of skeletal structures:

isop

3,3,4-triethylhexane

tlomenclaErre of GYcloalkanes The nomenclature of cycloalkanes follows essentiallythe same rules used for open-chain alkanes.

H3CHr

tl\-AczHsI - ethyl- 2-methylcyclohexane

(Note alphabe!:it citation,

/| \=Jl-/ \ropylcyclopentane

CJCJtl,-

H3

H:C

l, 3 - dimethylcyclob utanemethylryclobutane

2,5 CYCLOALKANES AND SKELETAL STRUCTURES 69

The numerical prefix l- is not necessary for monosubstituted cycloalkanes. Thus, the firstcompound is methylcyclobutane, not 1-methylcyclobutane. Two or more substituents, how-ever, must be numbered to indicate their relative positions. The lowest number is assigned inaccordance with the usual rules.

Most of the cyclic compounds in this text, like those in the preceding examples, involverings with small alkyl branches. In such cases, the ring is treated as the principal chain. How-ever, when a noncyclic carbon chain contains more carbons than an attached ring, the ring istreated as the substituent.

cHrcH2cH2cH2cH2 -<1l -ryclopropylpentane

( not penrylcyclopropane )

Name the following compound.

SOlutiOn This problem, in addition to illustrating the nomenclature of cyclic alkanes, is a goodillustration of rule 8 for nomenclature, the "first point of difference" rule (p.62). The compound isa cyclopentane with two methyl substituents and one ethyl substituent. If we number the ring car-bons consecutively, the following numbering schemes (and corresponding names) are possible,depending on which carbon is designated as carbon- I :

1 ,2,4- 4-ethyl- I ,2-dimethylcyclopentane1,3,4- 1-ethyl-3,4-dimethylcyclopentane

1,3,5- 3-ethyl-1,5-dimethylcyclopentane

,The correct name is decided by nomenclature rule 8 using the numbering schemes (zof the namesthemselves). Because all numbering schemes begin with l, the second number must be used todecide on the correct numbering. The scheme 1,2,4- has the lowest number at this point.Consequently, the correct name is 4-ethyl- 1,2-dimethylcyclopentane.

Drawaskeletalstructureottert-butylcyclohexane.

SOfuti0n The real question in this problem is how to represent a tert-butyl group with a skeletalstructure. The branched carbon in this group has four other bonds, three of which go to CH,groups. Hence:

CH.t

9-cH3 _I

CH..

l---\/ \ | \ ()nc Lilrbort is itl llte\ en(l trl c.lch hrirrrr'lr

l.- .r,

skeletal structure oftert-butylcyclohexane


@2.12Representeachofthefollowingcompoundswithaskeletalstructufe.(a) 9H,I

cH3cH2cH2cH - cH - c(cHr )r

lrn,(b) ethylcyclopentane

2.13 Name the following compounds.(a) (b) f^'

ft*2cH3H:C

2.14 How many hydrogens are in an alkane of n carbons containing (a) two rings? (b) three rings?(c) n rings?

2. 15 How many rings does an alkane have if its formula is (a) CrH,o? (b) CrH,r? Explain how youknow.

2.6 PHYSICAT PROPERTIES OF ALKANES

A.

Each time we come to a new family of organic compounds, we'll consider the trends in theirmelting points, boiling points, densities, and solubilities, collectively referred to as theirpftys-ical properties. The physical properties of an organic compound are important because theydetermine the conditions under which the compound is handled and used. For example, theform in which a drug is manufactured and dispensed is affected by its physical properties. Incommercial agriculture, ammonia (a gas at ordinary temperatures) and urea (a crystallinesolid) are both very important sources of nitrogen, but their physical properties dictate thatthey are handled and dispensed in very different ways.

Your goal should not be to memorize physical properties of individual compounds, butrather to learn to predict trends in how physical properties vary with structure.

Boiling PointsThe boiling point is the temperature at which the vapor pressure of a substance equals atmos-pheric pressure (which is typically 760 mm Hg). Table 2.1 shows that there is a regular changein the boiling points of the unbranched alkanes with increasing number of carbons. This trendof boiling point within the series of unbranched alkanes is particularly apparent in a plot ofboiling point against carbon number (Fig. 2.1). The regular increase in boiling point of20-30 "C per carbon atom within a series is a general trend observed for many types of or-ganic compounds.

What is the reason for this increase? The key point for understanding this trend is that boil-ing points are a crude measure of the attractive forces among molecules-intermolecular at-tractions-in the liquid state. The greater are these intermolecular attractions, the more energy(heat, higher temperature) it takes to overcome them so that the molecules escape into the gasphase, in which such attractions do not exist. The greater are the intermolecular attractionswithin a liquid, the greater is the boiling point. Now, it is important to understand that thereare no covalent bonds between molecules, and furthermore, that intermolecular attractions

2.6 PHYSICAL PROPERTIES OF ALKANES 71

250

200

rs0

100U

.E so

botrn

-50

- 100

- 1s0

- 200 46810t2number of carbon atoms

Figure 2.7 Boiling points of some unbranched alkanes plotted against number of carbon atoms. Notice thesteady increase with the size of the alkane, which is in the range of 20-30 "C per carbon atom.

have nothing to do with the strengths of the covalent bonds within the molecules themselves.What, then, is the origin of these intermolecular attractions?

In Chapter 1, we learned that electrons in bonds are not confined between the nuclei butrather reside in bonding molecular orbitals that surround the nuclei. We can think of the totalelectron distribution as an "electron cloud." Electron clouds are rather "squishy" and can un-dergo distortions. Such distortions occur rapidly and at random, and when they occur, they re-sult in the temporary formation of regions of local positive and negative charge; that is, thesedistortions cause a temporary dipole moment within the molecule (Fig. 2.8, p. 72).When asecond molecule is located nearby, its electron cloud distorts to form a complementary dipole,called an induced dipole. The positive charge in one molecule is attracted to the negativecharge in the other. The attraction between temporary dipoles, called a van der Waals attrac-tion or a dispersion interaction, is the cohesive interaction that must be overcome to vapor-ize a liquid. Alkanes do not have significant permanenr dipole moments. The dipoles dis-cussed here are temporary, and the presence of a temporary dipole in one molecule induces atemporary dipole in another. We might say, "Nearness makes the molecules grow fonder."

Now we are ready to understand why larger molecules have higher boiling points. Van derWaals attractions increase with the surface areas of the interacting electron clouds. That is, thelarger the interacting surfaces, the greater the magnitude of the induced dipoles. A larger mole-cule has a greater surface area ofelectron clouds and therefore greater van der Waals interactionswith other molecules. It follows, then, that large molecules have higher boiling points.

Tlre shape of a molecule is also important in determining its boiling point. For example, acomparison of the boiling point of the highly branched alkane neopentane (9.4 "C) and its un-branched isomer pentane (36.1 'C) is particularly striking. Neopentane has four methylgroups disposed in a tetrahedral arrangement about a central carbon. As the following space-filling models show, the molecule almost resembles a compact ball, and could fit readily

72 CHAPTER 2 o ALKANES

tzitf molecules moving at random

approach each other

0

Ots

O

the electron cloud of onemolecule distorts randomlyto form a temporary dipole

+

0tq: temporary dipoles dissipate

t{ the dipole in one molecule inducesa complementary dipole in the other

.F $:0weak attractions developbetween opposite charg6s

(van der Waals attractions)

Figure 2.8 A stop-frame cartoon showing the origin of van derWaals attraction.The frames are labeled tl, t2, andso on, for successive points in time.The time scale is about 1O-10 s.The colors represent electrostatic potentialmaps (EPMs).The green color of the isolated molecules (tr and tJ shows the absence of a permanent dipole mo-ment.As the molecules approach (t,),the electron cloud of one molecule undergoes a random distortion (t2) thatproduces a temporary dipole, indicated by the red and blue colors.This dipole induces a complementary chargeseparation (induced dipole) in the second molecule (t3), and attractions between the two dipoles result.Throughrandom fluctuations of the electron clouds (to), the temporary dipoles vanish. Averaged over time, this phenome-non results in a small net attraction.This is the van derWaals attraction.

within a sphere. On the other hand, pentane is rather extended, is ellipsoidal in shape, andwould not fit within the same sphere.

neopentane:compact, nearly spherical

pentane:extended, ellipsoidal

The more a molecule approaches spherical proportions, the less surface area it presents toother molecules, because a sphere is the three-dimensional object with the minimum surface-to-volume ratio. Because neopentane has less surface area at which van der Waals interactionswith other neopentane molecules can occur, it has fewer cohesive interactions than pentane,and thus, a lower boiling point.

2.6 PHYSICAL PROPERTIES OF ALKANES 7 3

In summary, by analysis of the boiling points of alkanes, we have learned two generaltrends in the variation of boiling point with structure:

1. Boiling points increase with increasing molecular weight within a homologous series-typically 20-30 "C per carbon atom. This increase is due to the greater van der Waals at-tractions between larger molecules.

2. Boiling points tend to be lower for highly branched molecules that approach spherical pro-portions because they have less molecular surface available for van der Waals attractions.

B. Melting PointsThe melting point of a substance is the temperature above which it is transformed sponta-neously and completely from the solid to the liquid state. The melting point is an especiallyimportant physical property in organic chemistry because it is used both to identify organiccompounds and to assess their purity. Melting points are usually depressed, or lowered, by im-purities. Moreover, the melting range (the range of temperature over which a substance melts),usually quite narrow for a pure substance, is substantially broadened by impurities. The melt-ing point largely reflects the stabilizing intermolecular interactions between molecules in thecrystal as well as the molecular symmetry, which determines the number of indistinguishableways in which the molecule fits into the crystal. The higher the melting point, the more stableis the crystal structure relative to the liquid state. Although most alkanes are liquids or gasesat room temperature and have relatively low melting points, their melting points neverthelessillustrate trends that are observed in the melting points of other types of organic compounds.

One such trend is that melting points tend to increase with the number of carbons (Fig. 2.9).Another trend is that the melting points of unbranched alkanes with an even number of carbonatoms lie on a separate, higher curve from those of the alkanes with an odd number of carbons.This reflects the more effective packing of the even-carbon alkanes in the crystalline solidstate. In other words, the odd-carbon alkane molecules do not "fit together" as well in thecrystal as the even-carbon alkanes. Similar alternation of melting points is observed in otherseries of comDounds. such as the cycloalkanes in Table 2.3.

0

-24-40-60-80

- 100

- 120

- 140

- 160

- 180

- 200 4 6 B 10

number of carbons

Fig$re2"9 Aplotof meltingpointsoftheunbranchedalkanesagainstnumberofcarbonatoms.Noticethegen-eral increase of melting point with molecular size. Also notice that the alkanes with an even number of carbons(red) lie on a different curve from the alkanes with an odd number of carbons (b/ue).This trend is observed in anumber of different types of organic compounds.

Qlja-

i4

bo

UE

t2

74 CHAPTER 2 I ALKANES

Branched-chain hydrocarbons tend to have lower melting points than linear ones becausethe branching interferes with regular packing in the crystal. When a branched molecule has asubstantial symmetry however, its melting point is typically relatively high because of theease with which symmetrical molecules fit together within the crystal. For example, the melt-ing point of the very symmetrical molecule neopentane, - 16.8 oC, is considerably higher thanthat of the less symmetrical pentane, -129.8 oC. (See models onp.72.) Compare also themelting points of the compact and symmetrical molecule cyclohexane, 6.6 oC, and the ex-tended and less symmetrical hexane, -95.3 "C.

In summary, melting points show the following general trends:

l. Melting points tend to increase with increasing molecular mass within a series.2. Many highly symmetrical molecules have unusually high melting points.3. A sawtooth pattern of melting point behavior (see Fig. 2.9) is observed within many ho-

mologous series.

E@2.l6Matcheachofthefol1owingisomerswiththecorrectboilingPointsandmeltingpoints.Ex-plain your choices.Compounds: 2,2,3,3-tetxa&ethylbutane and octaneBoiling points: 106.5 oC, 125.7 "CMelting points: -56.8 oC, +100.7 "C

2.17 Which compound has (a) the greater boiling point? (b) the greater melting point? Explain.(HintiWhat is the geometr5r of benzene?)

HHbenzene

HHtoluene

C. Other Physical PropertiegAmong the other significant physical properties of organic compounds are dipole moments,solubilities, and densities. A molecule's dipole moment (Sec. l.2D) determines its polarity,which, in turn, affects its physical properties. Because carbon and hydrogen differ little in theirelectronegativities, alkanes have negligible dipole moments and are therefore nonpolar mole-cules.We can see this graphically by comparing the EPMs of ethane, with a dipole moment ofzero, and fluoromethane, a polar molecule with a dipole moment of 1.82 D.

EPM of ethane

Solubilities are important inmost reactions are carried out in

EPM of fluoromethane (H3C-F)

determining which solvents can be used to form solutions;solution. Water solubility is particularly important for several

2.6 PHYSICAL PROPERTIES OF ALKANES 75

Fi$ure 2.1O The lower density of hydrocarbons and their insolubility in water allows an oil spill in flood watersto be contained by plastic tubes at a Texas refinery in the aftermath of Hurricane Rita in 2005.

reasons. For one thing, watei is the solvent in biological systems. For this reason, water solu-bility is a crucial factor in the activity of drugs and other biologically important compounds.There has also been an increasing interest in the use of water as a solvent for large-scale chem-ical processes as part of an effort to control environmental pollution by organic solvents. Thewater solubility of the compounds to be used in a water-based chemical process is crucial.(We'11 deal in greater depth with the important question of solubility and solvents in Chapter8.) The alkanes are, for all practical pulposes, insoluble in water-thus the saying, "Oil andwater don't mix." (Alkanes are a major constituent of crude oil')

The density of a compound is another property, like boiling point or melting point, that de-termines how the compound is handled. For example, whether a water-insoluble compound ismore or less dense than water determines whether it will appear as a lower or upper layer whenmixed with water. Alkanes have considerably lower densities than water. For this reason, amixture of an alkane and water will separate into two distinct layers with the less dense alkanelayer on top. An oil slick is an example of this behavior (Fig. 2.10).

2.18 Gasoline consists mostly of alkanes. Explain why water is not usually very effective in extin-guishing a gasoline fire.

2.19 (g,llnto a separatory funnel is poured 50 mL of CHTCHTBT (bromoethane), a water-insolublecompound with a density of 1.460 glrrr,l-,and 50 mL of water. The funnel is stoppered andthe mixture is shaken vigorously. After standing, two layers separate. Which substance isin which layer? Explain.

(b) Into the same funnel is poured carefully 50 mL of hexane (density : 0.660 g/ml) so thatthe other two layers are not disturbed. The hexane forms a third layer. The funnel is stop-pered and the mixture is shaken vigorously. After standing, two layets separate. Whichcompound(s) are in which layer? Explain.

76 cHAPTER2. ALKANES

2.7 GOMBUSTIONAlkanes are among the least reactive types of organic compounds. They do not react withcommon acids or bases, nor do they react with common oxidizing or reducing agents.

Alkanes do, however, share one type of reactivity with many other types of organic com-pounds: they are flammable. This means that they react rapidly with oxygen to give carbondioxide and water, provided that the reaction is initiated by a suitable heat source, such as aflame or the spark from a spark plug. This reaction is called combustion. For example, thecombustion of methane, the major alkane in natural gas, is written as follows.

CH++2Oz + COz+2H2O (2. 1)

This reaction is an example of complete combustion: combustion in which carbon dioxide andwater are the only products. Under conditions of oxygen deficiency, incomplete combustionmay also occur with the formation of such byproducts as carbon monoxide, CO. Carbonmonoxide is a deadly poison because it bonds to, and displaces oxygen from, hemoglobin, theprotein in red blood cells that transports oxygen to tissues. It is also colorless and odorless, andis therefore difficult to detect without special equipment.

The fact that we can carry a container of gasoline in the open air without its going up inflames shows that simple mixing of alkanes and oxygen does not initiate combustion. However,once a small amount of heat is applied (in the form of a flame or a spark from a spark plug), thecombustion reaction proceeds vigorously with the liberation of large amounts of energy.

Combustion is of tremendous commercial importance because it liberates energy that canbe used to keep us warm, generate electricity, or move motor vehicles. However, combustionalso liberates larse amounts of carbon dioxide and water.

CoHr* + Oz =-+octane or

its isomers

8CO2 + 9H,Ocarbondioxide (2.2)

As Eq. 2.2 illustrates, every carbon atom of a hydrocarbon combines with two atoms of oxy-gen to generate a molar equivalent of carbon dioxide, and every pair of hydrogens combineswith one oxygen atom to generate a molar equivalent of water. The atmosphere can hold a rel-atively small amount of water, and when that is exceeded, water returns to the Earth as rain orsnow. However, natural processes of removing carbon dioxide from the atmosphere are lim-ited. After eons in which the CO, content of the atmosphere remained relatively constant atabout 290 parts per million (ppm), the amount of CO, in the Earth's atmosphere began to risedramatically with the advent of the industrial age. The CO, level now approaches 400 ppm, anincrease of more than one-third (Fig. 2. I I ). Most of this increase has taken place in the last ]Oyears. Because so much of it is produced, carbon dioxide is the most significant of severalcompounds known to be greenhouse gases, atmospheric compounds that act as a heat-reflec-tive blanket over the Earth. Most scientists are now convinced that the temperature of theEarth is being increased by the effect of greenhouse gases; this phenomenon is known asglobal warming. These scientists believe that global warming is beginning to have significantadverse environmental consequences, such as an increase in the intensity of hurricanes, therapid receding ofglaciers, and the loss ofanimal and plant species at an increased rate. Globalwarming predicts that ultimately the ocean levels will rise as polar ice melts, and coastal flood-ing that will displace hundreds of millions of people will likely occur. As a result of these con-cerns, along with concerns about the political instability of the oil-producing regions of theworld, the development of alternative fuels has become increasingly urgent. ldeally, the goalis to produce cheap and abundant fuels that will not, on combustion, increase the net CO, con-tent of the atmosphere. (This issue is considered further in Sec. 8.9C.)

2.7 COMBUSTION 77

a

t\\\\390

a-t

U5 3so

l-r()O.s 300L{cd

400

375

275

250I 000 I 200 1400 1 600 1 800 2000

year

Fiture 2.11 Atmospheric CO, levels for the past 1000 years.The data prior to 1958 were obtained from air bub-bles trapped in dated ice core samples. More recent data were obtained from air sampling towers on Mauna Loa,Hawaii, by the Scripps Institute of Oceanography (1958-1974) and the National Oceanic and Atmospheric Admin-istration (NOAA, 1 974-present).The inset shows data obtained since I 975 in more detail.These data show the sea-sonal fluctuations normally observed in CO, levels. Notice the continuous rise in CO, levels since the nineteenthcentury.

Combustion finds a minor but important use as an analytical tool for the determination ofmolecular formulas. In this type of analysis, the mass of CO2 produced in the combustion of aknown mass of an organic compound is used to calculate the amount of carbon in the sample.Similarly, the mass of HrO produced is used to calculate the amount of hydrogen in the sam-ple. (Procedures have been developed for the combustion analysis of other elements.) Com-bustion analysis is illustrated in Problems 2.43 and 2.44 on p. 86.

2.20

z2l

Give agenoral bdarced reaction for(e) the complete combustion of an alkane (formula C,Hr"*r).O) tlro.cpqpleto combirtio! of a cycloalkane containing one ring fiormula Cull2j.Calculare the number of pounds of CO, rcleased into the atmospherc when a lS-gallon tankof gpsoline in bumed in an autorebileengine. Assffte €o4lplete oombrtion AIso assumethat gasoline is a mixture of octane isomers and that fte. @n*ityrof:ggsolim is 0 .692 gmr '-t .(lhis assumption ignores about l0 volumspqrg@to{o qdditiv€s.) Useful conver-sion factors: 1 gallon : 3.785 L; I kg * 2..fu- ..Carv and Di Oxhide drive their family car about |Z,OOO mitrc per year. Their car gets about 25miles per gallon of gasoline. What is th ' carbon foorydnfl (pornds of CO, rcleased into theatnosphere) of the Oxhide family car ovor <xro lq:ffs&e assu4ptions and conversionfactors in Problem 2.21.)

2.22

78 cHAPTER2 o ALKANES

2,8 OCGURRENCE AND USE OF ALKANESMost alkanes come from petroleum, or crude oil. (The word petroleum comes from the Latinwords for "rock" and "oil": thus, "oil from rocks.") Petroleum is a dark, viscous mixture com-posed mostly of alkanes and aromatic hydrocarbons (benzene and its derivatives) that are sep-arated by a technique called fractional distillation. In fractional distillation, a mixture ofcompounds is slowly boiled; the vapor is then collected, cooled, and recondensed to a liquid.Because the compounds with the lowest boiling points vaporize most readily, the condensatefrom a fractional distillation is enriched in the more volatile components of the mixture. Asdistillation continues, components ofprogressively higher boiling point appear in the conden-sate. A student who takes an organic chemistry laboratory course will almost certainly becomeacquainted with this technique on a laboratory scale. Industrial fractional distillations are car-ried out on a large scale in fractionating towers that are several stories tall (Fig.2.l2). The typ-ical fractions obtained from distillation of petroleum are shown in Fig.2.l3.

Another important alkane source is natural gas, which is mostly methane. Natural gascomes from gas wells of various types. Significant biological sources of methane also existthat could someday be exploited commercially. For example, methane is produced by the ac-tion of certain anaerobic bacteria (bacteria that function without oxygen) on decaying organicmatter (Fig.2.14, p. 80). This type of process, for example, produces "marsh gas," as methanewas known before it was characteized by organic chemists. This same biological process canbe used for the production of methane from animal and human waste. Methane produced thisway is becoming practical as a local source of power (Fig. 2.14). The methane is burned toproduce heat that is converted into electricity. Although this process generates carbon dioxide,the source of the carbon is the food eaten by the humans and animals, and the carbon in thatfood comes from atmospheric carbon dioxide by photosynthesis. In other words, the CO, pro-duced in this process is "recycled COr" and does not conffibute to a net increase in atmos-pheric COr.

FigUfe 2.12 Fractionating towers such as these are used in the chemical industry to separate mixtures of com-pounds on the basis oftheir boiling points.

2.8 OCCURRENCE AND USE OF ALKANES 79

Figure2.13 Schematicviewof anindustrial fractionatingcolumn.Thetemperatureofthecolumndecreasesfrom bottom to top. Crude oil is introduced at the bottom and heated. As the vapors rise, they cool and condenseto liquids. Fractions of progressively lower boiling points are collected from bottom to top of the column.The fig-ure shows typical fractions, their boiling points, the nu mber of ca rbons in the compou nds collected, and the typ-ical uses of each fraction after further processing.

tttt||tlIllheat

Fractionnarne

Frirctiontemprerature

lrlumber 0fcarhons

Tl pir-aluse

petroleum gas up to 20 "C Ct-Cq bottled gas

gasoline 20-70 "c Ct-C t o Gutomobile fuel

naphtha 70-t20 "c Ca-Ctt chemical feedstock

kerosene t20-240 "c Cro-Ctu jet fuel, paraffin

light tuel oil 240-320 "C Crr-Cro diesel fuel

heaw oil 320-500 'c Cz r-Cz* lubricants, heating oil

asphalt, tar above 500 oC )Ces road surfacing

Alkanes of low molecular mass are in great demand for a variety of purposes----especiallyas motor fuels-and alkanes available directly from wells do not satisfy the demand. The pe-troleum industry has developed methods (called catalytic cracking) for converting alkanes ofhigh molecular mass into alkanes and alkenes of lower molecular mass (Sec. 5.8). The petro-leum industry has also developed processes (called reforming) for converting unbranchedalkanes into branched-chain ones, which have superior ignition properties as motor fuels.

Typically, motor fuels, fuel oils, and aviation fuels account for most of the world's hydro-carbon consumption. An Arabian oil minister once remarked, "Oil is too precious to burn." Hewas undoubtedly referring to the important uses for petroleum other than as fuels. Petroleumwill remain for the foreseeable future the principal source of carbon, from which organic start-ing materials are made for such diverse products as plastics and pharmaceuticals. Petroleum isthus the basis for organic chemical feedstocks-the basic organic compounds from whichmore comolex chemical substances are fabricated.


Figure 2.14 Manure fermenters at Fair Oaks Farms, a large commercial dairy farm in northwestern Indiana, areused to produce methane from cow manure. Electricity produced from burning the methane is fed into the localpower grid.The power produced is frequently sufficient to support a large fraction of the farm's power require-ments. Such fermentations are carried out by methanogens (methane-producing bacteria). The inset showsMethanococcus voltae, one of many known methanogen strains.This bacterium was named for Alessandro Volta,who,in l776,collected tombustible air" (methane) that was liberated while he was exploring a marshy section ofLake Maggiore in northern ltaly.

Alkanes as Motor FuelsAlkanes vary significantly in their quality as motor fuels. Branched-chain alkanes are better motorfuels than unbranched ones.The quality of a motor fuel relates to its rate of ignition in an internalcombustion engine. Premature ignition results in'engine knock,"a condition that indicates poor en-gine performance. Severe engine knock can result in significant engine damage.The octane numberis a measure of the quality of a motor fuel:the higher the octane number.the better the fuel.The oc-tane number is the number you see associated with each grade of gasoline on the gasoline pump.Octane numbers of 100 and 0 are assigned to 2,2,4-trimethylpentane and heptane, respectively. Mix-tures of the tvvo compounds are used to define octane numbers between 0 and 100. For example, a

fuel that performs as well as a | :1 mixture of 2,2,4-trimethylpentane and heptane has an octane num-ber of 50.The motor fuels used in modern automobiles have octane numbers in the 87-95 range.

Various additives can be used to improve the octane number of motor fuels. In the past,tetraethyllead, (CHrCHr)rPb, was used extensively for this purpose, but concerns over atmosphericlead pollution and the use of catalytic converters (which are adversely affected by lead) resulted ina phase-out of tetraethyllead over the period 1 976-1 986 in the United States and in the EuropeanUnion by 2000.This was followed by the use of methyl tert-butyl ether IMTBE, (CH')'C-O-CH,J asthe major gasoline additive. After a meteoric rise in MTBE production, this compound became anobject of environmental concern when its leakage from storage vessels into groundwater was dis-covered in several communities in the mid-l990s. Because MTBE has shown some carcinogenic(cancer-causing) activity in laboratory animals, many cities and states have enacted a phase-out ofMTBE usage as a gasoline additive. Ethanol (ethyl alcohol) can be used as a substitute for MTBE, andethanol is produced by the fermentation of sugars in corn. Farming interests in the United Stateshave advocated the use of ethanol for fuel, and it appears that the MTBE in gasoline will soon be re-placed by ethanol in the United States.The demand for ethanol is so great that the price of corn hasescalated sharply.This increase, in turn, has had a noticeable impact on the price of foods that de-pend on corn as an animal food source (for example, milk, chicken, and beef). MTBE. ethanol, andother oxygen-containing additives are collectively referredto as oxygenafes within the fuel industry.

2.9

2.9 FUNCTIONAL GROUPS, COMPOUND CLASSES, AND THE "R" NOTATION 81

In the 1970s, the world experienced a period in which a relative scarcity of petroleum prod-ucts was caused largely by political forces. The resulting dramatic effects on energy prices andthe consequences in all sectors of the economy afforded a tiny foretaste of a chaotic world inwhich energy is in truly short supply. In 2005, oil prices and natural gas prices (the latter as a re-sult ofHurricane Katrina) again reached record highs. There is no doubt that eventually the worldwill exhaust its natural petroleum reserves. It is thus important that scientists develop new sourcesof energy, which may include new ways of producing petroleum from renewable sources.

FUNCTIONAT GROUPS, COMPOUND CIASSES, ANDTHE "R",NOTATION

A. Functional Groups and Compound ClassesAlkanes are the conceptual "rootstock" of organic chemistry. Replacing C-H bonds of alka-nes gives the many functional groups of organic chemistry. A functional group is a charac-teristically bonded group of atoms that has about the same chemical reactivity whenever it oc-curs in a variety of compounds. Compounds that contain the same functional group comprisea compound class. Consider the following examples:

H:C

H:C

HIH:C-F--0Hil

H

H\ ,/'(-:()\, \,/ \\\

H

ethyl alcoholI

f r-rnr-tional gruup: - [- - []Hil

cornpoLlncl class: alcohol

t),{

H-,C-(. ,\\'

OH

acetic acid

ft-rnu:tional qrr)rlLr: ( -()lI I

compounci class: carboxTlic acicl

isobutylene

tuuctionr1l grllLtp:

compoLrnd clarss: alkene

For example, the functional group that is characteristic of the alkene compound class is thecarbon--carbon double bond. Most alkenes undergo the same types of reactions, and these re-actions occur at or near the double bond. Similarly, all compounds in the alcohol compoundclass contain an -OH group bound to the carbon atom of an alkyl group. The characteristicreactions of alcohols occur at the -OH group or the directly attached carbon, and this func-tional group undergoes the same general chemical transformations regardless of the structureof the remainder of the molecule. Needless to say, some compounds can contain more than onefunctional group. Such compounds belong to more than one compound class.

iH2C:CH-C

\"acrylic acid

:ontains both C:C and CO2H functional groupsand is thus both an alkene and a carboxylic acid

The organization of this text is centered for the most part on the common functional groupsand corresponding compound classes. Although you will study in detail each major functionalgroup in subsequent chapters, you should leam to recognize the common functional groupsand compound classes now. These are shown on the inside front cover.

,,(- : ( -\, \,,/' "*,

82 CHA TER 2 . ALKANES

B. "R" NotationSometimes we'll want to use a general structure to represent an entire class of compounds. Insuch a case, we can use the R notation, in which an R is used to represent all all<yl groups(Sec. 2.4C). For example, R-Cl can be used to represent an alkyl chloride.

R-Cl could represent H:C-CI

R-: H:C- P-: (CHr)2CH- P-: rycloheryl-

Just as alkyl groups such as methyl, ethyl, and isopropyl are substituent groups derivedfrom alkanes, aryl groups are substituent groups derived from benzene and its derivatives.The simplest aryl group is the phenyl grcupr abbreviated Ph-, which is derived from the hy-drocarbon benzene. Notice that each ring carbon of an aryl group notjoined to another groupbears a hydrogen atom that is not shown. (This is the usual convention for skeletal structures;see Sec. 2.5.)

skeletal structure ofbenzene

H.C\CH-CI/H:C

G"

-CH-CH2 can be written ,_ Ph-CH-CHz\ phenylgroup

Other aryl groups are designated by Ar-. Thus, AI-OH could refer to any one of the fol-lowing compounds, or to many others.

Ar-oH could represent H:C- '/=1t\r/-oH

whereAr-- H:CnV Ar- -

Although you will not study benzene and its derivatives until Chapter 15, before then you willsee many examples in which phenyl and aryl groups are used as substituent goups.

l' i,A Oorr s strucfiuel formula fq crch of thc folibi '@slpoun&. ($evcral formulas may beposrible in cach cese.)(s) acafroxylic acid with thc molceulsfftffithqrHlo,(b) rn eldo,hol with tilo &rlie''f@C5l{,6o

ADDITIONAL PROBLEMS 83

2.r+ A certain compound was found to have.the molecular formula CrH,rOr. To which of the fol-lowing compound classet could the compound belong? Give one example for each positiveanswer, and explain any negative responses.

an amide an ether a carboxylic acid a phenol an alcohsl an ester

r

t

I

I

t

Alkanes are hydrocarbons that contain onlycarbon-carbon single bonds; alkanes may containbranched chains, unbranched chains, or rings.

Alkanes have sp3-hybridized carbon atoms with tetra-hedral geometry.They exist in various staggered con-formations that rapidly interconvert at room tempera-ture. The conformation that minimizes van der Waalsrepulsions has the lowest energy and is the predomi-nant one. In butane, the major conformation is the anticonformation; the gauche conformations exist to a

lesser extent.

lsomers are different compounds with the same mol-ecular formula. Compounds that have the same mole-cular formula but differ in their atomic connectivitiesare called constitutional isomers.

Alkanes are named systematically according to thesubstitutive nomenclature rules of the IUPAC. Thename of a compound is based on its principal chain,which, for an alkane, is the longest continuous carbonchain in the molecule.

The boiling point of an alkane is determined by vander Waals attractions between molecules, which inturn depend on molecular size and shape. Large mol-

ecules have relatively high boiling points; highlybranched molecules have relatively low boiling points.The boiling points of compounds within a homolo-gous series increase by 20-30 'C per carbon atom.

Melting points of alkanes increase with molecularmass. Highly symmetrical molecules have particularlyhigh melting points.

Combustion is the most important reaction of alka-nes. lt finds practical application in the generation ofmuch of the world's energy,Alkanes are derived from petroleum and are usedmostly as fuels; however, they are also important asraw materials for the industrial preparation of otherorganic compounds.

Organic compounds are classified by their functionalgroups. Different compounds containing the samefunctional groups undergo the same types ofreaction s.

The "R" notation is used as a general abbreviation foralkyl groups; Ph is the abbreviation for a phenyl group,and Ar is the abbreviation for an aryl (substitutedphenyl) group.

I

I

t

t

I

2.25 Given the boiling point of the flrst compound in eachset, estimate the boiling point of the second.{a} CH,CH2CH2CH2CH2CH2BT (bp 155'C)

CH3CH2CH2CH2CH2CH2CH2BT

(b) oil

CH3CCH2CH2CH2CH3 (bp 128'C)otl

cH3ccH2cH2cHZcH2cHzcH:

{c} Otl

CH3CCH2CH2CH2CH2CH3 (bp 152 "C)otl

cH-3cH2ccHzcHZcH2cH3

2.26 Draw the structures and give the names of all isomersof octane with{a} five carbons (b) six carbonsin their principal chains.

84

2.27

2.28

2.29

CHAPTER 2 o ALKAN

Label edch carbon in themary, secondary, tertiary,(a) 9H, (b)a-h'. /tlw

ules as pri-

ES

following molec, or quaternary.

Draw ttfe structure of an alkane or cycloalkane thatmeets e[ch of the following criteria.(a) a cofnpound that has more than three carbons and

only primary hydrogens.(b) a cofnpound that has five carbons and only sec-

ond{ry hydrogens.(c) a colrnpound that has only tertiary hydrogens.(d) a compound that has a molecular mass of 84.2.

Name fach of the following compounds using IUPACsubstitultive nomenclature.(a) 9H, fHr-CHz-CHrlrHr9 9-CH-CHz-CH:

IcHz-cHz-cH:

the correct name for any compounds that are not namedcorrectly.(a) 2-eth yl-2,4,6-trimethy lheptane(b) S-neopentyldecane(c) 1 -cyclopropyl-3,4-dimethylcyclohexane(d) 3 -bu tyl-2,2-dimethylhexane

2.32 Although compounds are indexed by their IUPAC sub-stitutive names, sometimes chemists give whimsicalnames to compounds that they discover. Assist thesetwo chemists by providing substitutive names for theircompounds.(a) Chemist Val Losipede isolated an alkane with the

following skeletal structure from asphalt scrapingsfollowing a bicycle race and named it "Tourde-francane."

(b) Chemist Slim Pickins isolated a compound with thefollowing structure from the floor of a henhouse anddubbed it "pullane" (ytlillus, Latin for chick).

2.33 Within each set, which two structures represent thesame compound?(a)

CHrI

"YtYcHrL,A.JHrC CzHs

CH:

(b) cH2cHrcHr

CHbCHCHCH2CHE

cH2cHzcH:

(c)YI

(d)

Hrc-cH-ffi'cHz-cH:

230 Draw s[ructures that corespond to the following names.(a) a-idobutyl-2, 5 -dimethylheptane(b) 2,3,5 -trimethyl-4-propylheptang ( skeletal structure )( c ) 5 - sec-buty I - 6 + e rt -butyl-Z,Z -di methy I n onane

2.31 The foflowing labels were found on bottles of liquid hy-drocarbons in the laboratory of Dr. Ima Tirrkey follow-ing his disappearance under mysterious circumstances.Althou[h each name defines a structure unambiguously,some alre not correct IUPAC substitutive names. Give

CHrcH2cH3

(b)Yr234 {a) Draw a skeletal structure of the compound in part

(a) of Problem 2.33 that is different from the othertwo compounds, and name the compound.

(b) Draw a Newman projection for the most stable con-formation of the compound in part (b) of Problem2.33 that is different from the other two compounds.Draw your Newman projection about the bond be-tween carbons 3 and 4 in the IUPAC standard num-bering system, with the projection viewed fiom thedirection of carbon-3. Describe any ambiguity youencounter in drawing this structure. Name the com-pound.

2.-15 Sketch a diagram of potential energy versus angle of ro-tation about the carbon-carbon bond of chloroethane,H3C-CH'-C1. The magnitude of the energy barrierto internal rotation is 15.5 kJ mol-t (3.7 kcal mol-r).Label this barrier on your diagram.

2.36 Explain how you would expect the diagram of potentialenergy versus dihedral angle about the C2-C3 (central)carbon-carbon bond of 2,2,3,3-tetramethylbutane todiff-er from that for ethane (Fig. 2.3), if at all.

2.37 The anti confbrmation of 1,2-dichloroethane,C1-CHr-CHr-Cl, is 4.81 kJ mol-' ( 1.15 kcalmol-r; more stable than the gauche conformation. Thetwo energy barriers (measured relative to the energy ofthe gauche conformation) for carbon-carbon bond rota-tion are 21.5 kJ mol-' (5.15 kcal mol-r) and 38.9 kJmol-t (9.3 kcal mol-r).(a) Sketch a graph of potential energy versus dihedral

angle about the carbon-carbon bond. Show the en-ergy differences on your graph and label each mini-mum and maximum with the appropriate confbrma-tion of 1 .2-dichloroethane.

(b) Which conformation of this compound is present ingreatest amount? Explain.

2.38 Predict the most stable conformation of hexane. Builda model of this conformation. (Hint: Consider the con-formation about each carbon-carbon bond separately.)

2.39 When the structure of compound A was determined in1912, it was found to have an unusually long C-C

ADDITIONAL PROBLEMS 8 5

bond and unusually large C-C-C bond angles, com-pared with the similar parameters for compound B(isobutane).

HI _1.61 I A

..C ,/(CH:)rC"" / -c(CH.r).,f

c(cH3)_3

4 c-c-c : 116o

H| _l.s3s A

.C. r/H.,C ""*/" -an.,/

CH:

x c-c-C - lto.8"+v\J

ABExplain why the indicated bond length and bond angleare larger for compound A.

2.1{l Which of the following compounds should have thelarger energy barrier to internal rotation about the indi-cated bond? Explain your reasoning carefully.

(cH3)3c-c(cH:): (cH3)3si -si(cH3)3AB

2.11 From what you learned in Sec. 1.38 about the relativelengths of C-C ancl C-O bonds, predict which of thefollowing compounds should have the larger energy d{'-ference between gauche and anti conformations aboutthe indicated bond. Explain.

cH3o-cHzcH-, CH3CH?-CH2CH3

AB

2.12 (a) What value is expected for the dipole moment of theanti conformation of 1,2-dibromoethane,Br- CHz -CH, -Br? Explain.

(b) The dipole moment g, of any compound that under-goes internal rotation can be expressed as aweighted average of the dipole moments of each ofits conformations by the following equation:

p: ltrNr * pzlr{z + p:l/:in which pi is the dipole moment of conformation i,and { is the mole fraction of conformation i. (Themole fraction of any conformation i is the numberof moles of i divided by the total moles of all con-formations.) There are about 82 mole percent of anticonformation and about 9 mole percent of eachgauche conformation present at equilibrium in 1,2-dibromoethane, and the observed dipole moment ;.r,

of 1,2-dibromoethane is 1.0 D. Using the precedingequation and the answer to part (a), calculate the di-pole moment of a gauche conformation of 1,2-dibro-moethane.


2.4-1 This problem illustrates how combustion can be used todetermine the molecular formula of an unknown com-pound. A compound X (8.00 mg) undergoes combustionin a str{am of oxygen to give ?4.60 mg of CO, andI 1.51 rng of HrO.(a) Calculate the mass of carbon and hydrogen in X.(b) How many moles of H are present in X per mole of

C? Express this as a formula CrH,.(c) Multiply this formula by successive integers until

the amount of H is also an integer. This is the molec-ular formula of X.

2.44 A hydrocarbon )'is found by combustion analysis tocontain 87.l7Vo carbon and 12.83Vo hydrogen by mass.(a) Use the procedure in Problem 2.43(b) and (c) to de-

termine the molecular formula of I'. (Hint: Remem-ber fhat all alkanes and cycloalkanes must containeverl numbers of hydrogens.)

(b) Draw the structure of an alkane (which may containone or more rings) consistent with the analysis givenin part (a) that has two tertiary carbons and all othercarbons secondary. (More than one coffect answer ispossible.)

(c ) Drafv the structure of an alkane (which may containone or more rings) consistent with the analysis givenin part (a) that has no primary hydrogens, no tertiarycarbon atoms, and one quaternary carbon atom.(More than one correct answer is possible.)

2.45 Imaginp a reaction that can replace one hydrogen atomof an alkane at random with a chlorine atom.

tlc-H -* -c-cltt(a) If pentane were subjected to such a reaction, how

many different compounds with the formulaCrH,,Cl would be obtained? Give their Lewis struc-

tures. Then build models (or draw wedge-dashedwedge formulas) for each compound. Does this alteryour answer in any way? Explain.

(b) Provide the same analysis as in part (a) for the samereaction carried out on 2.2-dimethvlbutane.

2.46 To which compound class does each of the followingcompounds belong?(a) gg.CH, (b)

|- \" | .c-Nb:oCH.CH,

''' ryoHtu'r-o)]2.47 Organic compounds can contain many different func-

tional groups. Identify the functional groups (aside fromthe alkane carbons) present in acebutolol (Fig. P2.47), a

drug that blocks a certain part of the nervous system.Name the compound class to which each groupbelongs.

2.48 (a) Two amides are constitutional isomers and have theformula C4Hel'{O, and each contains an isopropylgroup as part of its structure. Give structures forthese two isomeric amides.

(b) Draw the strucJure of two other amides with the for-mula C4HeI{O that do not contain isopropyl groups.

(c) Draw the structure of a compound X that is a consti-tutional isomer of the amides in parts (a) and (b), butis not an amide, and contains both an amine and analcohol functional group.

(d) Could a compound with the formula CoHnNO con-tain a nitrile functional group? Explain.

:O:tlC-CH:

:OH

acebutolol

Figure P2.47

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