Jones M., Fleming S.A. Organic Chemistry

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4.9 Physical Properties of Diastereomers 169

divides the molecule into halves that contribute equally and oppositely to the rota-

tion of plane-polarized light. Compound F  G provides an example of a mol-

ecule with a plane of symmetry in the eclipsed forms we have been using, and a

point of symmetry if we draw the molecule in its energy minimum, staggered

arrangement (Fig. 4.41).

PROBLEM 4.17 Draw all the stereoisomers of the following molecules, in which

there is free rotation about carbon–carbon single bonds. Some of the molecules

may be achiral:

(a) 2,2,3,3-tetrabromobutane

(b) 2,2-dibromo-3,3-dichlorobutane

(d) 2,3-dibromo-2-chloro-3-fluorobutane

Hint: It is easier to analyze these molecules in their eclipsed forms.

PROBLEM 4.18 How does your answer to Problem 4.17d change if there is no

free rotation about carbon–carbon single bonds for 2,3-dibromo-2-chloro-

3-fluorobutane?

PROBLEM 4.19 Identify each stereogenic carbon in the compound of Problem

4.17c as either (R) or (S).

Plane of symmetry

in eclipsed form

Point of symmetry

in staggered form

FIGURE 4.41 Two views of the meso

compound of Figure 4.40, F  G.

4.9 Physical Properties of Diastereomers:

Resolution, a Method of Separating Enantiomers

from Each Other

The formation of diastereomers allows the separation of enantiomers from each

other. Separation of enantiomers, called resolution, is a serious experimental diffi-

culty. So far we have ignored it.Enantiomers have identical physical properties except

for the ability to rotate the plane of plane-polarized light,and one might legitimate-

ly wonder how in the world we are ever going to get them apart. At several points,

we used a single enantiomer without giving any hint of how a pair of enantiomers

might be separated.The key to this puzzle is that diastereomers, unlike enantiomers,

have different physical properties—melting point, boiling point, and so on.

Dextro Levo

Crystals of () tartaric acid and

() tartaric acid are shown here.These

crystals are non-superimposable

mirror images. Although tedious, a

careful visual analysis would allow

separation of the two crystal forms.

This would be a “visual resolution.”

170 CHAPTER 4 Stereochemistry

One general procedure for separating enantiomers is to allow them to react with

a naturally occurring chiral molecule to form a pair of diastereomers.These can then

be separated from each other by taking advantage of one of their different physical

properties. For example, we can often separate such a pair by crystallization because

the members of the diastereomeric pair have different solubilities in a given solvent.

Then, if the original chemical reaction can be reversed, we have the pair of enan-

tiomers separated. Figure 4.42 outlines the general scheme and begins with a

schematic recapitulation of Figure 4.33, which first described the reaction of a sin-

gle enantiomer with a racemic mixture to give a pair of diastereomers. Be sure to

compare the two figures.

CX+

reverse

reaction

separate

physically

One

enantiomer

A racemic pair

of molecules

A pair of

diastereomeric

products

Now we have separated

the two enantiomers of the

original racemic mixture

Separated

racemic pair

CX+

FIGURE 4.42 Resolution is a general method for separating the constituent enantiomers of a racemic mixture. A single

enantiomer of a chiral molecule is used to form a pair of diastereomers, which can be separated physically. If the original

chemical reaction can be reversed, the enantiomers can be isolated.

It is not even necessary to form covalent bonds in order to separate enantiomers.

For example,in the traditional method for separating enantiomers of organic acids,opti-

cally active nitrogen-containing molecules, called alkaloids (see Section 6.11), are used

to form a pair of diastereomeric salts, which can then be separated by crystallization.

The term alkaloid refers to any nitrogen-containing compound extracted from

plants, although the word is used loosely and some compounds of nonplant origin

are also commonly known as alkaloids.These biologically active compounds are basic

like all amines, and it is this basicity that led to the name. Presumably, basicity was

also important in the relative ease of extraction of alkaloids from the myriad com-

pounds present in an organism as complex as a plant. Extraction of plant mass with

acid yields any amines present as water-soluble ammonium ion salts, from which

the amines can easily be regenerated. Figure 4.43 shows this process for a simple

alkaloid known as coniine. Coniine is the active ingredient in hemlock, and it is the

physiological activity of coniine that led to the demise of Socrates.

Plant (hemlock)

base

Ammonium ion

(water soluble)

(S)-Coniine

FIGURE 4.43 A scheme for isolation

of alkaloids from plant materials.

4.9 Physical Properties of Diastereomers 171

Other alkaloids have wondrously complex structures. Two further examples,

brucine and strychnine, are shown in Figure 4.44 along with the general procedure

for resolution.

COOH

Brucine

Strychnine

An optically

active alkaloid

COO

–

A racemic

pair of

acids

Separation

COOH

Pure (S)-acid

Pure (R)-acid

A pair of

diastereomeric

ammonium ion

salts; these will

have different

physical properties

COOH

COO

–

COO

–

COO

–

FIGURE 4.44 Two alkaloids, brucine and strychnine, are commonly used to separate the enantiomers of chiral organic

acids. Diastereomeric salts are first formed, separated by crystallization, and the individual enantiomeric acids are

regenerated.

WORKED PROBLEM 4.20 Identify with an asterisk (*) all the stereogenic carbons

in brucine.

ANSWER

PROBLEM 4.21 Identify each stereogenic carbon in brucine as (R) or (S).

172 CHAPTER 4 Stereochemistry

These days, the general procedure outlined in Figure 4.42 has been extended so

that all manner of enantiomeric pairs can be separated by chromatography. In chro-

matography, as in the salt formation in Figure 4.44, covalent chemical bonds are not

formed. Rather, advantage is taken of the formation of complexes with partial bonds

as the pair of enantiomers passes over an optically active substrate (Fig. 4.45). The

optically active substrate (X*) forms a complex with the enantiomers A and A′.These

complexes are diastereomers and so have different physical properties,including bond

strengths of AX* and A′ X*. Both AX* and A′X* are in equilibrium with

the free enantiomers, and these equilibria will be different for the two diastereomeric

complexes. Therefore, A and A′ move through the column at different rates and

emerge at different times.

4.10 Determination of Absolute Configuration

(R or S)

Now that we have achieved the separation of our racemic mixture of enantiomers

into a pair of optically active stereoisomers, we face the difficult task of finding out

which enantiomer is (R) and which is (S). This problem is not trivial! Indeed, in

------

AAX*

A' A'X*

A chromatography

column packed with

X*, a chiral material

A racemic

mixture of

enantiomers

A and A'

FIGURE 4.45 Separation of

enantiomers through

chromatography.

Strychnine

STRYCHNINE

The notorious poison strychnine was first isolated from

the beans of strychnos ignati Berg by Pelletier and Caventou

in 1818. It constitutes about one-half of the alkaloids pres-

ent in the beans and makes up 5–6% of their weight. Its

structure is obviously complicated and was only determined

correctly by Sir Robert Robinson in 1946. It was a mere two

years more before a physical determination of the structure

by X-ray diffraction was reported by Bijvoet, confirming

Robinson’s deductions, and presaging the demise of chemi-

cal, as opposed to physical, structure determination as a

viable enterprise. Robert B. Woodward provided the first

synthesis of strychnine in 1954 in a landmark paper that

begins with the exclamation “Strychnine!” (hardly the usual

dispassionate scientific writing). The introduction to

Woodward’s paper makes good reading, and it provides an

interesting defense of the art of synthesis in the face of

critics who thought that the profession would surely be

rendered obsolete by the increasingly powerful physical

methods. The ensuing years have shown Woodward’s

defense to be correct. (See Tetrahedron, 1963, 19, 247;

your chemistry library probably has it.)

Of course, much interest in strychnine centers on its

pharmacological properties. It is a powerful convulsant, lethal

to an adult human in a dose as small as 30 mg. Death comes

from central respiratory failure and is preceded by violent

convulsions. Strychnine is the deadly agent in many a murder

story, real and imagined. One example is Sir Arthur Conan

Doyle’s Sherlock Holmes mystery, “The Sign of the Four,” in

which Dr. Watson suggests the lethal agent to be a “powerful

vegetable alkaloid ...some strychnine-like substance.”

4.11 Stereochemical Analysis of Ring Compounds (a Beginning) 173

Chapter 23, when we deal with sugars,we’ll find that until the mid 1900s, there was

no way to be certain, and one just had to guess (correctly in the case of the sugars,

it turns out). One would like to peer directly at the structures, of course, and under

some circumstances this kind of inspection is possible.

X-ray crystallography can determine the relative positions of atoms in a crystal,

and a special kind of X-ray diffraction called anomalous dispersion can tell the

absolute configuration of the molecule. But this technique is not generally applicable—

to do X-ray analysis one needs a crystalline compound, for example. It does

serve to give us some benchmarks, though. If we know the absolute configuration

(R or S) of some compounds, we may be able to determine the absolute configura-

tions of other molecules by relating them to those compounds of known absolute

configuration. We must be very careful, however.The chemical reactions that inter-

convert the molecules of known and unknown absolute configuration must not alter

the stereochemical arrangement at the stereogenic atoms, or if they do, it must be

in a known fashion. How do we know whether a given chemical reaction will or will

not change the stereochemistry? We need to know the reaction mechanism—to

know how the chemical changes occur—in order to answer this question. This rea-

son is just one of many for the study of reaction mechanisms, and we’ll take great

pains to examine carefully the stereochemical consequences of the various chemical

reactions we encounter. Figure 4.46 shows the general case, as well as examples of

two chemical reactions that do not alter stereochemistry at the stereogenic carbon.

In both reactions, the absolute configuration (R or S)

of the carbon marked with the asterisk is unchanged

If we know the absolute configuration (R or S)

of the starting material and the mechanism of

the reaction, we can know the absolute

configuration of the product

1. LiAlH

2. H

/PtO

FIGURE 4.46 Two reactions that do not change

the configuration at an adjacent carbon (*).

Summary

To know the absolute configuration of the product in the general reaction in

Figure 4.46 you must know two things: the absolute configuration of the starting

material and the mechanism of the reaction with Y. Only if these two things are

known will the absolute configuration of the product be known with certainty.

4.11 Stereochemical Analysis of Ring Compounds

(a Beginning)

Most of the stereochemical principles discussed in the previous sections apply quite direct-

ly to ring compounds. We’ll start with a simple question. How many isomers of chloro-

cyclopropane exist? That looks like an easy question,and it is.There is only one (Fig.4.47).

WEB 3D

FIGURE 4.47 Different

representations of the single isomer

of chlorocyclopropane.The bold

(thick) bond is toward you.

174 CHAPTER 4 Stereochemistry

To be certain there is only one chlorocyclopropane, however, we should exam-

ine this compound for chirality. Again this process is simple. As Figure 4.48 shows,

the mirror image is easily superimposed on the original, through a 120° rotation.

Chlorocyclopropane is an achiral molecule.

Mirror

Rotate 120

FIGURE 4.48 Chlorocyclopropane is

superimposable on its mirror image

and is, therefore, an achiral molecule.

A rotation of 120° about the dotted

axis passing through the center of the

ring easily shows that the mirror

image and the original are identical.

How many isomers of dichlorocyclopropane exist? This question is much

tougher than the last one. There are four isomers of dichlorocyclopropane. Most

people don’t get them all at this point, so it’s worth going carefully through an analy-

sis.Where can the second chlorine be placed in the single monochlorocyclopropane?

As Figure 4.49 shows, there are only two places, either on the same carbon as the

first chlorine, or on one of the two equivalent adjacent carbons.This analysis yields

two compounds called structural isomers, molecules with the same formula, but

differing atom-to-atom connectivity. Structural isomers are sometimes called

constitutional isomers.

1,1-Dichloro-

cyclopropane

1,2-Dichloro-

cyclopropane

Structural isomers

replace a

hydrogen

with a

chlorine

Cl Cl Cl

FIGURE 4.49 Replacement of one

hydrogen of chlorocyclopropane

with another chlorine can occur in

only two places, on the carbon to

which the first chlorine is attached

or on one of the other two equivalent

carbons. This procedure generates

the structural isomers

1,1-dichlorocyclopropane and

1,2-dichlorocyclopropane.

Figure 4.49 brings us only two isomers; we’re still two short of reality. As first

mentioned in Chapter 2 (p. 84), however, rings have sides. We can find one more

isomer if we notice that in 1,2-dichlorocyclopropane the second chlorine can be on

the same side (the cis isomer) or on the opposite side (the trans isomer) of the ring

(Fig. 4.50). These compounds are stereoisomers, more precisely, diastereomers

(stereoisomers that are not mirror images). Now we are just one isomer short.

1,2-Dichloro-

cyclopropane

Diastereomers

cis-1,2-Dichloro-

cyclopropane

trans-1,2-Dichloro-

cyclopropane

Cl Cl

C C

FIGURE 4.50 1,2-Dichlorocyclopropane

can exist in cis and trans forms. In the

figure, all the ring hydrogens have

been drawn in to help you see the

difference between these two

molecules. Be absolutely certain you

see why these two molecules cannot

be interconverted, no matter how

they are rotated or translated in space.

Use models.

4.11 Stereochemical Analysis of Ring Compounds (a Beginning) 175

Once we are sure we have found all the different positions on which to put

the two chlorines, it’s time to look for chirality. Figure 4.51 shows that neither

1,1-dichlorocyclopropane nor cis-1,2-dichlorocyclopropane is chiral.The mirror images

are easily superimposable on the originals. Note that cis-1,2-dichlorocyclopropane

is a meso compound. It contains stereogenic carbons but is superimposable on its

mirror image, and therefore is achiral.

1,1-Dichloro-

cyclopropane

Mirror

cis-1,2-Dichloro-

cyclopropane

(a meso compound)

Mirror

WEB 3D

FIGURE 4.51 Both

1,1-dichlorocyclopropane and

cis-1,2-dichlorocyclopropane are

achiral.

trans-1,2-Dichlorocyclopropane is different from the cis molecule we just exam-

ined for chirality. In the trans diastereomer, the mirror image is not superimposable

on the original and this stereoisomer is chiral (Fig. 4.52).

trans-1,2-Dichloro-

clo

ane

Mirror

WEB 3D

FIGURE 4.52 trans-1,2-Dichlorocyclopropane is not

superimposable on its mirror image. It is a chiral

molecule and therefore two enantiomers exist. This

stereochemical relationship is probably the most

difficult you’ve encountered so far, so be sure you can

see why no manipulation of one enantiomer will

make it superimposable on the other.This is a point

at which models are very useful.

Now we have the four isomers: 1,1-dichlorocyclopropane,cis-1,2-dichlorocyclopropane,

trans-(1S,2S)-dichlorocyclopropane,and trans-(1R,2R)-dichlorocyclopropane (Fig.4.53).

Notice that we have to designate each stereogenic carbon—two in each trans

stereoisomer—as either R or S.

1,1-Dichloro-

cyclopropane

cis-1,2-Dichloro-

cyclopropane

The two enantiomers of

trans-1,2-dichlorocyclopropane

(S)(R)(R)(S)

C C

Cl Cl

FIGURE 4.53 The four isomers of

dichlorocyclopropane.

PROBLEM 4.22 Verify the assignment of absolute configuration in the two enan-

tiomers of trans-1,2-dichlorocyclopropane in Figure 4.53.

PROBLEM 4.23 Identify the stereochemical relationships among the four isomers of

Figure 4.53. In other words, find all the pairs of enantiomers and diastereomers.

176 CHAPTER 4 Stereochemistry

This section illustrates one procedure for working out an isomer problem, but

any rational technique will work. The important thing is to devise some systemat-

ic way of searching. We happen to like the one described above in which one first

finds all the structural isomers (in this case 1,1- and 1,2-dichlorocyclopropane), then

searches for stereoisomerism of the cis/trans type, and finally examines each isomer

for chirality. But it doesn’t matter what procedure you use as long as you are system-

atic. What will not work is a nonrational scheme. So don’t start a “find the isomers”

problem by just writing out compounds without thinking first. No one can do it

that way.

PROBLEM 4.24 There are six isomers of dichlorocyclobutane. Find them all and

name them carefully.

We started this discussion of stereochemistry in ring compounds with small rings

for a reason—the inflexibility of small rings makes them flat, or very nearly so.Three

points determine a plane, so cyclopropane must be planar. Cyclobutane need not be

absolutely flat, and we can see that it is not flat if we take the time to make a model.

At the same time, it cannot be far from planar. Larger rings are more complicated

though, and we’ll defer an examination of their stereochemistries until Chapter 5,

which looks at a number of structural questions about rings.

4.12 Summary of Isomerism

The words isomer and stereoisomer are often tossed around quite loosely, and the

subject of isomerism is worth a bit of review. Structural isomers (often called con-

stitutional isomers) are molecules of the same formula, but differing atom-to-atom

connectivity.Their constituent parts may well be different (but need not be).Butane

(two methyl groups and two methylene groups) and isobutane (three methyl groups

and one methine group) are typical examples of structural isomers (Fig. 4.54).

PROBLEM 4.25 Find a pair of structural isomers whose constituent parts are not

different from one another.

FIGURE 4.54 Two structural isomers.

Stereoisomers have the same connectivity, but differ in the arrangement of their

parts in space. Two kinds of stereoisomers exist: enantiomers and diastereomers.

Enantiomeric molecules are nonsuperimposable mirror images of each other. Simple

examples are (R)- and (S)-3-methylhexane. Slightly more complicated are (2S,3S)-

dichlorobutane and its enantiomer, (2R,3R)-dichlorobutane. These pairs are not

structural isomers because they have the same connectivity (Fig. 4.55).

Mirror

(R)-3-Methylhexane

(2R,3R)-Dichlorobutane (2S,3S)-Dichlorobutane

(S)-3-Methylhexane

CCH

FIGURE 4.55 Two pairs of enantiomers.

4.13 Special Topic: Chirality without “Four Different Groups Attached to One Carbon” 177

Diastereomers are stereoisomers that are not mirror images; cis- and trans-2-

butene are examples,as are meso-2,3-dichlorobutane and either the (R,R) or the (S,S)

isomer of 2,3-dichlorobutane in Figure 4.55 (Fig. 4.56).

trans-2-Butene

HCH

cis-2-Butene

CCH

(2R,3R)-Dichlorobutane

meso-2,3-Dichlorobutane

(2S,3R)-dichlorobutane

FIGURE 4.56 Two pairs of

diastereomers.

Stereoisomers also include conformational isomers, in which different isomers are

generated through rotations about bonds.Conformational isomers are often called “con-

formers.” Eclipsed and staggered ethane are typical examples. Note that a conforma-

tional isomer need not be an energy minimum—the eclipsed conformation of ethane

is an energy maximum,for example.Conformational isomers can be either enantiomer-

ic or diastereomeric.The two gauche forms of butane are conformational enantiomers,

but the gauche and the anti forms of butane are conformational diastereomers (Fig.4.57).

60⬚

Enantiomers

Diastereomers

120⬚

FIGURE 4.57 Examples of

conformational isomers.

WORKED PROBLEM 4.26 Show that the two conformational isomers labeled

“enantiomers” in Figure 4.57 really are mirror images.

ANSWER The trick is to find a way to make an easy comparison. In this case, it

takes only a simple 180° rotation to put the molecules into the proper orientation.

Now it is easy to see that the two molecules from Figure 4.57 (labeled A and B

below) are indeed mirror images.

4.13 Special Topic: Chirality without “Four Different

Groups Attached to One Carbon”

We already know from Section 4.8 that finding four different groups attached

to one carbon is not a sufficient condition for chirality. There are achiral meso

compounds that satisfy this condition.But is the presence of a carbon atom attached

Mirror

180⬚

AAB

178 CHAPTER 4 Stereochemistry

to four different groups a necessary condition for chirality? No. We asked this

question first on page 151.This section answers it by giving a number of examples

of chiral molecules without this kind of atom.

trans-Cyclooctene is chiral, and yet it contains no carbon atom attached to four

other groups.Figure 4.58 gives one representation of this molecule alongside its non-

superimposable mirror image, but you may need to use models to convince yourself

that two enantiomers really do exist. Be careful! This molecule is racemized by

rotation around carbon–carbon bonds, and your models are likely to be eager to do

exactly that.

In some molecules, two double bonds can be connected directly to each other

(Fig. 4.59). We will see these molecules, called allenes, again in detail in Chapter

12, but you are well equipped after this chapter to understand their bonding and

geometry, which we’ll work through in Problems 4.27–4.29. In summary, trans-

cyclooctene and substituted allenes can be chiral, although these molecules have no

“carbons attached to four different groups.”

trans-Cyclooctene

Mirror

WEB 3D

FIGURE 4.58 trans-Cyclooctene is

chiral, but it contains no carbon to

which four different groups are

attached.

Allene

CCH

FIGURE 4.59 Allene contains a

central carbon attached to two double

bonds.

WORKED PROBLEM 4.27 What is the hybridization of each carbon of allene (see

Fig. 4.59)?

ANSWER Each end carbon is attached to three groups (two hydrogens and the

other carbon), and is therefore hybridized sp

. The central carbon is attached to

two other groups (the two methylene groups), and is therefore hybridized sp.

WORKED PROBLEM 4.28 Make a careful,three-dimensional drawing of 2,3-butadiene

(1,3-dimethylallene). Pay special attention to the geometry at the ends of the

molecule.

ANSWER The key to this problem is to notice that overlap of the 2p orbitals on

the end sp

–hybridized carbons with the 2p orbitals on the central, sp-hybridized

carbon requires that the groups on the end carbons lie in perpendicular planes.

C CC

CCC

Mirror

CCC C CC

WORKED PROBLEM 4.29 Draw the mirror image of 1,3-dimethylallene and veri-

fy that the two enantiomers are not superimposable.

ANSWER The mirror images of 1,3-dimethylallene are not superimposable,

although it may take some work with models before you are certain!