Jones M., Fleming S.A. Organic Chemistry

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4.6 Properties of Enantiomers: Chemical Differences 159

There must be a net change in the plane of polarization, and this rotation is gen-

erally large enough to be observed by a device called a polarimeter (Fig. 4.21).

Because the amount of rotation varies with the wavelength of light, monochromat-

ic light, light of only one wavelength, must be used to produce results easily repro-

duced in other laboratories. Generally, the sodium D line is used, which is light that

Unpolarized light

from sodium lamp

Plane-polarized

light

Tube containing

sample of chiral

molecule

Emerging light

is rotated

Detector

Nicol prism

FIGURE 4.21 A highly schematic drawing of a polarimeter.

has a wavelength of 589 nm (5890 Å).The light is passed through a Nicol prism to

polarize it, and then through a tube containing a solution of the molecule in known

concentration. The amount of rotation is proportional to the number of molecules

in the path of the light, and so the length of the tube and the concentration of the

solution are important. For this reason, rotation of a sample is reported as the spe-

cific rotation,[α], which is the rotation induced by a solution of concentration

1 g/mL in a tube 10 cm long. The specific rotation [α] is related to the observed

rotation α in the following way:

[α]  α/cl, where c  concentration in grams per milliliter (g/mL),

and l  length of the tube in decimeters (dm)

Subscripts and superscripts are used to denote the wavelength and temperature of

the measurement.Thus, means the sodium D line was used and the measure-

ment was made at 20 °C.

[α]

PROBLEM 4.6 Create a scheme like those in Figures 4.19 and 4.20 to show what

happens when plane-polarized light is passed through a solution of racemic

3-methylhexane.

4.6 Properties of Enantiomers:

Chemical Differences

Now let’s look at how the two enantiomers of a chiral molecule differ from each

other chemically. We still can’t discuss specific chemical reactions, but we can make

some important general distinctions. Note that there is no difference in how the two

enantiomers of 3-methylhexane interact with a “Spaldeen,”an old-fashioned smooth

160 CHAPTER 4 Stereochemistry

pink rubber ball (Fig. 4.22).

Your left and right hands interact in exactly the same

way with the featureless Spaldeen (Fig. 4.23). In both the right and left hand, the

thumb and little finger are opposed by the same featureless pink rubber ball.

Chemically, the Spaldeen represents an achiral molecule.

aldeen

(

)

-3-Meth

lhexane

aldeen(S )-3-Methylhexane

FIGURE 4.22 (R)- and

(S)-3-methylhexane interact

identically with the achiral Spaldeen.

The inclusion of the term “Spaldeen” was the source of one of the few minor skirmishes between author and

editor of this book. MJ maintained that a Spaldeen was part of our cultural heritage, and that all educated

people would immediately know what one was. Perhaps he was overly optimistic. Here’s what the editors of

the New Yorker magazine replied when queried as to the meaning of the phrase “a kid bouncing a Spaldeen

off a wall of the Boston Museum of Fine Arts”: “Spaldeens are the small rubber balls beloved by generations

of New York stoopball and stickball players. In Boston, those balls are often known as ‘pinkies,’but what would

you have thought if we had written ‘bouncing a pinkie off a wall’?”

aldeen

Spaldeen

FIGURE 4.23 Mirror image left and

right hands also interact identically

with the achiral Spaldeen.

WORKED PROBLEM 4.7 Actually, the Spaldeen is achiral only if the name has been

worn off by too much stoopball playing. Show that the two enantiomers interact

differently with a Spaldeen with the name “Spalding” still visible.

ANSWER In the approach shown, (R)-3-methylhexane interacts with the new

Spaldeen to place the ethyl group near the “S”and the propyl group near the “G.”

These interactions are opposite in (S)-3-methylhexane; the propyl group

approaches the “S” and the ethyl group approaches the “G.”

(R )-3-Methylhexane

CCH

(S )-3-Methylhexane

CCH

You should be certain that this result is not caused by the particular approach cho-

sen for the two objects.Try interacting 3-methylhexane by bringing it toward the

Spaldeen “H and methyl first” or simply turn it over (H down, methyl up) in the

figure above.

4.6 Properties of Enantiomers: Chemical Differences 161

Figure 4.24 elaborates this subject—the interactions of chiral and achiral

objects—by showing the two enantiomers of 3-methylhexane approaching an achi-

ral molecule, propane.

(S )-3-Methylhexane Propane

ethyl–methyl

propyl–methyl

(R )-3-Methylhexane

Propane

propyl–methyl

ethyl–methyl

FIGURE 4.24 The two enantiomers of the chiral molecule 3-methylhexane interact in

exactly the same way with the achiral molecule propane.

In each instance, the 3-methyl group of 3-methylhexane approaches the hydrogen

of propane and both the ethyl and propyl groups are opposite methyl groups of

propane. Because of the symmetry of propane, one can always find exactly equiva-

lent interactions for the two enantiomers and propane,no matter what approach you

pick. Try it with models.

Now imagine your hands, or the two enantiomers of 3-methylhexane, interact-

ing with a chiral object. The Spaldeen with the label will do, but let’s use a more

common object, such as the shell in Figure 4.25.

Right handShell

Left handShell

FIGURE 4.25 All sorts of common

objects are chiral. A shell is one

example.The interactions between a

chiral object and each of a pair of

enantiomers—here represented by a

shell and two hands—are different.

The enantiomeric right and left

hands interact differently with the

chiral shell.

PROBLEM 4.8 Convince yourself that the shell in Figure 4.25 is chiral.

As your right hand approaches the shell (Fig. 4.25), it is the little finger that oppos-

es the opening in the shell,and the thumb that is directed away from it.For the left hand

the situation is reversed: It is the thumb that opposes the opening and the little finger

that does not.The interactions of the two hands with the shell are distinctly different.

PROBLEM 4.9 Convince yourself that turning the shell upside down doesn’t

change the situation.

162 CHAPTER 4 Stereochemistry

For the molecular counterpart of what Figure 4.25 shows, look at the approach

of the two enantiomers of 3-methylhexane to another chiral molecule,(R)-sec-butyl

alcohol (Fig. 4.26). As the (S) enantiomer approaches the alcohol, propyl is opposite

methyl and ethyl is opposite H. For the (R) enantiomer, the interactions are propyl

with H and ethyl with methyl. The two approaches are very different. Indeed, there

is no approach to the chiral alcohol that can give identical interactions for the two

enantiomeric 3-methylhexanes.Try some. Contrast this situation with the approaches

to the achiral propane shown in Figure 4.24. The two enantiomers interact identi-

cally with achiral propane but differently with the chiral molecule (R)-sec-butyl

alcohol. This situation is general for any chiral molecule: The enantiomers have

identical chemistries with achiral reagents but different chemistries with chiral ones.

WORKED PROBLEM 4.10 A reader of an early version of this chapter suggested

that students would ask, quite correctly, Why can’t I turn the picture of

(R)-3-methylhexane upside down,and then the ethyl–hydrogen and propyl–methyl

interactions with (R)-sec-butyl alcohol would be just like the interactions with

(S)-3-methylhexane, propyl–methyl and ethyl–hydrogen? Explain carefully why

turning the molecule upside down doesn’t produce a situation identical to that in

Figure 4.24.

ANSWER It is true that turning the molecule upside down will make two of the

interactions the same. Propyl will be opposite methyl and ethyl will be opposite

hydrogen. However, everything else is wrong! No longer are methyl and OH

being opposed, for example (red bonds). Simply turning (R)-3-methylhexane

over does not produce the same interactions—or even interactions similar to

those between (S)-3-methylhexane and (R)-sec-butyl alcohol.

(S )-3-Methylhexane (R)-sec-Butyl alcohol

ethyl–hydrogen

propyl–methyl

(R )-3-Methylhexane (R)-sec-Butyl alcohol

propyl–hydrogen

ethyl–methyl

FIGURE 4.26 The approach of (S )- and (R )-3-methylhexane to (R )-sec-butyl alcohol.The two approaches are

very different and in sharp contrast to the interaction shown in Figure 4.24 with the achiral molecule propane.

ethyl–hydrogen

propyl–methyl

(S )-3 -Methylhexane (R )-sec -Butyl alcohol

ethyl–hydrogen

(R)-3-Methylhexane

(R )-sec -Butyl alcohol

propyl–methyl

4.7 Interconversion of Enantiomers by Rotation about a Single Bond: gauche-Butane 163

A practical example of the consequences of chiral interactions is the difference

in smells of enantiomeric compounds. One smells a molecule by binding it in a decid-

edly chiral receptor in the nose.There are many examples in which one enantiomer

binds either differently from the other or not at all. For example,(R)-carvone smells

like spearmint and (S)-carvone like caraway (Fig. 4.27).

PROBLEM 4.11 Convince yourself that the assignment of (R) and (S) in Figure

4.27 is correct.

It is not at all easy to show (R)- and (S)-carvone fitting into different (and not

yet exactly known) nasal receptors, but it is easy to show in principle how fit and

non-fit works. Imagine a chiral binding site as the blue tetrahedron in Figure 4.28,

with the smaller red and green tetrahedra representing two enantiomeric molecules.

The red enantiomer fits perfectly inside,with the “proper”attachments, shown as 1–1,

2–2,3–3,and 4–4. If we try the green tetrahedron,the enantiomer of the red one, there

is no possible way for the proper attachments to be achieved. The green tetrahedron

cannot bind to the blue one with the proper 1–1, 2–2, 3–3, and 4–4 interactions.

(R )- (–)-Carvone

(spearmint)

(S )- (+)-Carvone

(caraway)

WEB 3D

FIGURE 4.27 (R)- and (S)-Carvone.

Good fit

No fit possible

FIGURE 4.28 The red tetrahedron fits

inside the blue one with the proper

interactions (1–1, 2–2, 3–3, and 4–4),

but the green one does not.

4.7 Interconversion of Enantiomers by Rotation

about a Single Bond: gauche-Butane

In Chapter 2 (p. 74), we analyzed the conformations of butane,finding an anti form

lower in energy than either of the two enantiomeric gauche forms.Figure 4.29 shows

the anti and gauche forms of butane as both three-dimensional drawings and

Newman projections.

anti Form gauche Form

120⬚

FIGURE 4.29 anti- and

gauche-Butane.

164 CHAPTER 4 Stereochemistry

The Newman projections clearly reveal the high-energy, destabilizing gauche

methyl–methyl interaction. Moreover, they show clearly that the gauche form of

butane is chiral. It cannot be superimposed on its mirror image (Fig. 4.30).

Mirror

rotation

120⬚

FIGURE 4.30 The gauche form of

butane is chiral! The mirror images

are not superimposable. However,

rapid rotation around carbon–carbon

single bonds interconverts these two

“conformational enantiomers.”

Why then is butane achiral? No matter how hard we try, no optical rotation can

be observed for butane. The nearly free rotation about the carbon–carbon bond in

butane rapidly interconverts the two enantiomers of the gauche form. If we were able

to work at the extraordinarily low temperatures required to “freeze out” the equilib-

rium between the two gauche forms we could separate butane into enantiomers

(Fig. 4.30).The two gauche forms are conformational enantiomers, enantiomers that

are interconverted through rotation about carbon-carbon single bonds.We will encounter

this phenomenon again when we study the stereochemistry of cyclic compounds.

In this and many succeeding figures, the molecules are drawn in eclipsed forms for clarity. As you know,

these arrangements are not stable and will be converted into staggered forms by rotation about the central

carbon–carbon bond. None of the arguments presented is changed by this rotation, and in the staggered

arrangements it is more difficult to visualize the relationships involved.

some

process

X + X

FIGURE 4.31 A process that removes

the two X substituents and joins the

carbons. The new connection (bond)

is shown in blue.

4.8 Diastereomers and Molecules Containing More

than One Stereogenic Atom

We have just looked at the different interactions between a single enantiomer of a

chiral molecule and an enantiomeric pair of molecules (Fig. 4.26). Let’s take this

interaction business to its limit and see what happens if we actually form bonds to

produce new molecules. We can’t worry yet about real chemical reactions—we are

still in the tool-building stages—but that doesn’t matter. We can imagine some

process that replaces the two X substituents in Figure 4.31 with a chemical bond,

shown in all succeeding figures with a blue screen.

If we follow this hypothetical process (which mimics some very real chemical

reactions), we generate a new molecule containing two potentially stereogenic car-

bons.First,let’s look again at what happens when a single chiral molecule reacts with

each isomer of a racemic (50:50) pair of enantiomers. It’s worth taking a moment

to see if we can figure out what’s going to happen in general before writing real mol-

4.8 Diastereomers and Molecules Containing More than One Stereogenic Atom 165

ecules. Let’s assume we are using the (R) form of our single enantiomer. It will react

with the (R) form of the enantiomeric pair to generate a molecule we might call

(R–R), and with the (S) form to generate (R–S) (Fig. 4.32).

PROBLEM 4.12 What is the mirror image of (R–S)? This question sounds insult-

ingly simple, but most people get it wrong!

We can tell a lot even from the schematic coded picture of Figure 4.32. First of all,

(R–R) and (R–S) are clearly neither identical nor enantiomers (nonsuperimposable mir-

ror images), because the mirror image of (R–R) is (S–S), not (R–S). Yet these molecules

are stereoisomers—they differ only in the arrangement of their constituent parts in space.

There is a new word for such molecules—they are diastereomers, “stereoisomers that

are not mirror images of each other—in other words,stereoisomers but not enantiomers.”

To sum up,there are two kinds of stereoisomers: enantiomers (nonsuperimposable

mirror images) and diastereomers.

The (R) form

of a chiral

molecule

These molecules, called

diastereomers are

stereoisomers, but not

enantiomers!

The products are

different molecules,

diastereomers

A racemic mixture

of another molecule

(R)(R)

(S)(S )

(R)

(R) (S)(R)and

FIGURE 4.32 A schematic analysis of

the reaction of one enantiomer with a

racemic mixture of another molecule.

The products of this reaction are

stereoisomers but not enantiomers.

They are called diastereomers.

WORKED PROBLEM 4.13 If the two gauche forms of butane shown in Figure 4.30 are

enantiomers, what is the relationship between one gauche form and the anti form?

ANSWER The two gauche forms are mirror images and, therefore, enantiomers.

Each gauche form is a stereoisomer of the anti form, but not its mirror image.

Stereoisomers that are not mirror images are diastereomers.

PROBLEM 4.14 We have seen diastereomers twice before. Show two different kinds

of diastereomeric molecules. Hint: Start with a good definition of diastereomer.

(S)-2-Chlorobutane

2-Chloro-1,1,1-trifluoro

ane

(S) Enantiomer

(R) Enantiomer

CCl

Diastereomers

FIGURE 4.33 The hypothetical

reaction of (S)-2-chlorobutane with

the two enantiomers of 2-chloro-

1,1,1-trifluorochloropropane.Two

stereoisomeric products, A and B,are

formed.They are not mirror images

(enantiomers) so they must be

diastereomers.

Now let’s make all this theory a little more real, and look at some molecules.Imagine

a process (Figure 4.31,X  Cl) in which the (S) enantiomer of 2-chlorobutane (sec-butyl

chloride) reacts with the two enantiomers of 2-chloro-1,1,1-trifluoropropane (Fig.4.33).

166 CHAPTER 4 Stereochemistry

Look carefully at the two new products, A and B, produced from our reaction. Are

they identical? Certainly not.Are they enantiomers? No,again.They are stereoisomers,

though, and so they must be diastereomers.

PROBLEM 4.15 Verify the claims of the preceding paragraph. Are molecules A and

B diastereomers?

The (R) form of 2-chlorobutane would also produce two new products, C and

D (Fig. 4.34) that are neither identical nor enantiomeric and thus must also be

diastereomers.

(R)-2-Chlorobutane

(S) Enantiomer

(R) Enantiomer

A pair of

diastereomers

2-Chloro-1,1,1-trifluoropropane

FIGURE 4.34 The hypothetical

reaction of (R)-2-chlorobutane with

the two enantiomers of 2-chloro-

1,1,1-trifluoropropanes gives another

pair of diastereomers, C and D.

Now the question is, What are the relationships among molecules A, B, C, and

D? As Figure 4.35 shows, molecules A and D are a pair of enantiomers and B and

C are another pair of enantiomers. So all possible combinations of 2-chlorobutane

with the two enantiomers of 2-chloro-1,1,1-trifluoropropanes have produced two

pairs of enantiomers, or four stereoisomers in all. The (R) and (S) relationships are

also shown in Figure 4.35.

Mirror

(R)

(S)

(R)

(S)

(R)

(S)

(R)

180⬚

(R)

180⬚

FIGURE 4.35 Two pairs of

enantiomers (A and D, B and C)

are formed from this hypothetical

reaction. Combinations A and B, A

and C, B and D, and C and D are

diastereomeric pairs. Each of these

pairs is composed of two

stereoisomers that are not mirror

images.

4.8 Diastereomers and Molecules Containing More than One Stereogenic Atom 167

As a simple theoretical analysis would predict (Fig. 4.36), we have formed all pos-

sible isomers in this combination [(S–R), A;(S–S), B;(R–S), D; and (R–R), C].

(S)

C(R)

(R)

Mirror

FIGURE 4.36 (R)- and (S)-2-

chlorobutane reacting with (R)- and

(S)-2-chloro-1,1,1-trifluoropropane

should give four stereoisomers: two

pairs of enantiomers.

PROBLEM 4.16 (S)-2-Chlorobutane and (S)-2-chloro-1,1,1-trifluoropropane

(Fig. 4.33) are used to produce isomer A, which is properly designated as the

(S,R) isomer (Fig. 4.35). Check that there is no mistake in this analysis, and then

explain how the two (S) compounds can produce the (S,R) isomer, A.

We can generalize even at this early stage. A molecule with one stereogenic

center has two possible stereoisomers, a single pair of (R) and (S) enantiomers. A

molecule containing two stereogenic atoms has two pairs of enantiomers, or four

stereoisomers in all. It seems that for n stereogenic carbons, we get 2

stereoisomers.

We can immediately predict that a molecule with n  3 stereogenic atoms has 2

8 stereoisomers. This surmise is easy to check—what stereoisomers are possible?

The easiest way to tell is not to examine some complicated molecule and thus incur

the task of drawing a large number of stereoisomers, but to analyze the problem in

a simpler fashion. A stereogenic carbon can be only (R) or (S).Figure 4.37 shows the

possibilities for any molecule containing three stereogenic atoms.

The eight possible isomers appear easily in this coded form. Note that there is

nothing in this analysis that requires the stereogenic atoms to be adjacent—they need

only be in the same molecule.

This procedure reliably gives you the maximum number of possible stereoiso-

mers. There are some interesting (and important) special cases, however, and this

very general method won’t spot them. Let’s imagine using our chemical coupling

reaction again, but this time let’s combine two molecules of 2-chlorobutane. This

procedure is exactly the same as the one we used before to generate the four stereoiso-

mers of Figures 4.33 and 4.34, A, B, C, and D. We allow our (R) and (S) starting

materials to combine in all possible ways (Fig. 4.38).

(RRR)

All (R)

Two (R), one (S)

One (R), two (S)

(RSS)

(SSR)

(SRS)

All (S)

(SSS)

(RRS)(RSR )(SRR)

FIGURE 4.37 The eight possible

stereoisomers for a molecule

containing three stereogenic atoms.

(S)-2-Chlorobutane

(R)-2-Chlorobutane

(S)(S)

(R)-2-Chlorobutane

(S)-2-Chlorobutane

(R)-2-Chlorobutane

(S)(R)

(R)(S)

(R)(R)

FIGURE 4.38 All possible

combinations of (R)- and

(S)-2-chlorobutane (sec-butyl

chloride) produce four isomers,

E, F, G, and H.

168 CHAPTER 4 Stereochemistry

In the previous example in Figure 4.35, we found that we had made two pairs of

enantiomers, (S–R) and (R–S), A and D, and (S–S) and (R–R), B and C. When we

combine two molecules of 2-chlorobutane, one pair of enantiomers, E and H,does

appear (Fig. 4.39).

However, the second potentially enantiomeric pair of molecules,F and G, are in fact

identical species (Fig. 4.40)! So in this example of a molecule containing two

stereogenic atoms, we get only three stereoisomers, not the maximum number of

 4. Clearly, we are one stereoisomer short because of the identity of F and G.

This molecule, F  G, contains stereogenic carbons but is not chiral. Such a

molecule is called a meso compound.

Twice before we have seen molecules that for different reasons give no observ-

able rotation of plane-polarized light. Neither achiral molecules nor racemic

mixtures of enantiomers can induce optical rotation. A racemic mixture of chiral

Mirror

(R)(R)

180

(S)(S)

FIGURE 4.39 Molecules E and H in

Figure 4.38 are enantiomers.

(S)(R)

(R)(S)

180

FIGURE 4.40 The F and G molecules

are identical. You can see this identity

if you rotate F 180° around the axis

shown.This compound F  G is

meso.

molecules contains equal numbers of right- and left-handed compounds, which

rotate the light equally in clockwise and counterclockwise directions, respectively,

for a net zero rotation. Our third example, a meso compound, does contain stereo-

genic atoms, but it is achiral and can induce no optical rotation.

On page 151 we asked if finding four different groups attached to a stereogenic

carbon was a sufficient condition for chirality. We just answered this question! No,

“four different groups attached to a stereogenic carbon” is not a sufficient condition

for optical activity. Our meso compound F  G certainly has carbons to which four

different groups are bonded, but it is equally clearly not chiral. Even though search-

ing for carbons attached to four different groups is definitely a good way to begin

a hunt for chirality, you must be careful. Finding such a carbon does not guarantee

that you have found a chiral molecule, unless you are certain that the molecule con-

tains one and only one such carbon. Alas, it is also the case that not finding such a

carbon does not guarantee that there will be no chirality. The ultimate test remains

the superimposability of mirror images. If the molecule and its mirror image are

superimposable, the molecule is not chiral; if the mirror images are different, non-

superimposable, the molecule is chiral.

When will meso compounds appear? Meso compounds occur when there is a

plane or point of symmetry in the molecule. In effect, the symmetry element