Jones M., Fleming S.A. Organic Chemistry

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4.13 Special Topic: Chirality without “Four Different Groups Attached to One Carbon” 179

The flat molecule, C

, called benzene, consists of a six-membered ring of car-

bons to which hydrogens are attached, one at each vertex (Fig. 4.60). The key point is

that benzene is flat as a pancake (for much more on benzene see Chapters 13 and 14).

Two benzenes can be attached to each other to make a molecule called “biphenyl”

(Fig. 4.61).

Top view

Side view

WEB 3D

FIGURE 4.60 Two views of benzene.

Biphenyl

COOH

HOOC

COOH

HOOC

Junction bond

Mirror

WEB 3D

FIGURE 4.61 Biphenyl and a substituted (chiral) biphenyl.

If we now substitute the four carbons directly adjacent to the junction between

the two rings in the correct way, a chiral molecule can be made! It is necessary that

the substituents be large enough so that rotation about the junction bond between

the two rings is restricted (Fig. 4.61).

PROBLEM 4.30 Show that rotation about the junction bond will racemize the two

enantiomers of Figure 4.61.

We can use benzene in another way to construct an odd chiral molecule. Two

benzenes can be attached in what is called a “fused” fashion to create the molecule

known as naphthalene (Fig. 4.62). We can imagine continuing this process in two

ways. We might simply continue to add benzene rings in a linear fashion, as in

Figure 4.62. Or, we might add the restriction that new rings be added so as to

produce a curve (Fig. 4.63).

Benzene

Naphthalene

Anthracene

FIGURE 4.62 Naphthalene and

anthracene: fused benzenes.

Benzene Naphthalene Phenanthrene

3,4-Benzphen-

anthrene

Dibenzo[c,g]phen-

anthrene

Phenanthro[3,4-c]phenanthrene

hexahelicene

WEB 3D WEB 3D

FIGURE 4.63 More fused benzenes. Hexahelicene is chiral. Why?

180 CHAPTER 4 Stereochemistry

If we follow the latter course, the addition of the sixth ring produces a chiral

molecule (hexahelicene) in which no carbon is attached to four different groups!

PROBLEM 4.31 Explain why hexahelicene is chiral. You may need models to do

this task.

4.14 Special Topic: Stereochemistry in the Real

World: Thalidomide, the Consequences of Being

Wrong-Handed

You should be able to spot the stereogenic carbon in thalidomide,which is the com-

mon name for the formidably named molecule (N-phthalidimido)glutarimide

shown in Figure 4.64. You should also be able to write structures for the (R) and

(S) enantiomeric forms.

(S)-Thalidomide(R)-Thalidomide

WEB 3D

FIGURE 4.64 The two enantiomers of

thalidomide.The stereogenic carbon

atom is highlighted.

Thalidomide was marketed as a sedative and as a palliative for morning sick-

ness in the 1950s, mostly in Europe. Let’s see why this compound, ironically now

used (carefully) in a number of other ways, including cancer therapy and as an

antileprosy agent, is so notorious. Much biological activity depends on fit—the

accommodation of a molecule in a receptor into which it fits exactly. That perfect

fit depends critically on the details of shape. As enantiomers have different shapes,

they are often bound very differently by receptor sites. Sometimes the difference is

not very important, as with carroway and spearmint, the enantiomers of carvone

(p. 163), but sometimes the difference is critical.The defining example for our time

is thalidomide. The (R) enantiomer is at least harmless, but the (S) enantiomer is

quite another story.This enantiomer is a most potent teratogen, causing a very high

incidence of fetal anomalies. At the same time, it operates as an antiabortive agent,

thus diminishing the body’s tendency to purge abnormal fetuses. Why should this

difference matter? Thalidomide was marketed as the racemic mixture, not the pure

(R) enantiomer, and those taking the drug were automatically ingesting the bad

along with the good.

There is much more to the story of thalidomide, and it is not a nice tale at all. If you would like to read more

about the extraordinarily shabby science that went into the “testing” of this drug, and the even shabbier soci-

ology, see Roald Hoffmann, The Same and Not the Same, Chapter 20 (Columbia University Press, 1995). Here

you will also encounter Dr. Frances Kelsey, the FDA physician who played a major role in preventing thalido-

mide from being licensed in the United States. She is a real hero of our time.

4.15 Summary 181

This chapter deals almost exclusively with the concept and con-

sequences of the “handedness” of molecules—chirality. Some

molecules are related to each other as are your left and right

hands. These are nonsuperimposable mirror images, or enan-

tiomers.

An absolutely safe method of determining whether a mol-

ecule is chiral or not is to examine the mirror image to see if it

is superimposable on the original. Finding a carbon attached

to four different groups is neither a necessary nor a sufficient

condition for chirality, although it is a good way to begin a

search for chirality. A meso compound is an example of an

achiral molecule containing carbons attached to four different

groups.

Enantiomers have identical physical properties except that

one stereoisomer rotates the plane of plane-polarized light by

some amount to the right, whereas the other rotates the plane

by the same amount in the opposite direction. Enantiomers

have identical chemistries with achiral molecules, but interact

differently with other chiral molecules.

An examination of the consequences of the presence of more

than one stereogenic carbon in a single molecule leads to the dis-

covery of diastereomers; stereoisomers that are not mirror images.

4.15 Summary

New Concepts

absolute configuration (p. 152)

achiral (p. 150)

alkaloid (p. 170)

allene (p. 178)

chiral (p. 148)

chirality (p. 148)

conformational enantiomers (p. 164)

constitutional isomers (p. 174)

dextrorotatory (p. 156)

diastereomers (p. 165)

enantiomers (p. 151)

levorotatory (p. 156)

meso compound (p. 168)

optical activity (p. 155)

plane-polarized light (p. 155)

polarimeter (p. 159)

racemic mixture (racemate) (p. 156)

resolution (p. 169)

specific rotation (p. 159)

stereochemistry (p. 148)

stereogenic atom (p. 152)

Key Terms

The arbitrary but important (R/S) convention is based upon the

Cahn–Ingold–Prelog priority system and allows us to specify

absolute configuration of molecules.

Resolution, or separation of enantiomers, is generally

accomplished by allowing a racemic mixture of enantiomers to

react with a single enantiomer of a chiral agent to form a pair

of diastereomers. Diastereomers, unlike enantiomers, have dif-

ferent physical properties and can be separated by crystallization

or other techniques. If the original chemical reaction that

formed the diastereomers can be reversed, the pure enantiomers

can be regenerated.

Reactions, Mechanisms, and Tools

This chapter continues the journey into three dimensions

begun in Chapter 3. Most commonly encountered problems

have to do with learning to see molecules in three

dimensions and in particular, with the difficulty of

translating from the two-dimensional page into the three-

dimensional world. There can be no hiding the difficulty of

this endeavor for many people, but it will yield to careful

work and practice.

There are a few small errors you should be careful

to avoid. For example, there are many arbitrary parts to

the (R/S) convention, and there is no way to figure them

out. The (R/S) convention is something that must be

memorized.

When a single enantiomer is drawn, be sure to ask yourself

whether this is done deliberately, in order to specify that partic-

ular enantiomer, or simply as a matter of convenience to avoid

the work of drawing both enantiomers of the racemic pair.

Usually some indication will be made if it is indeed a single

enantiomer that is meant.

There is no connection between the absolute configuration

of a molecule, (R or S) and the direction of rotation of plane-

polarized light.

Remember: Finding a carbon with four different groups

attached is a fine way to start looking for chirality, but it is nei-

ther a sufficient nor necessary condition. Problem-makers love

to find exotic molecules that prove this point!

Common Errors

182 CHAPTER 4 Stereochemistry

PROBLEM 4.32 Find all the chiral isomers of the heptanes

(pp. 82–83). Designate the stereogenic atoms with an asterisk (*).

PROBLEM 4.33 Draw the (S) enantiomers of the chiral heptanes.

PROBLEM 4.34 Find all the chiral isomers of the octanes (see

Problem 2.28).

PROBLEM 4.35 Draw the (R) enantiomers of all of the chiral

isomers of octane whose name must end in “pentane.”

PROBLEM 4.36 Now find all the chiral isomers of the

nonanes (see Problem 2.59). Indicate those with more than one

stereogenic carbon.

PROBLEM 4.37 Draw three-dimensional pictures of all the

stereoisomers of 3,4-dimethylheptane and 3,5-dimethylheptane.

It is probably easiest to draw them in the eclipsed arrangement,

even though this is not a low-energy conformation. Determine

the absolute configuration (R or S) of each stereogenic carbon.

Designate the stereogenic carbons with an asterisk.

PROBLEM 4.38 You are given a bottle labeled 3,4-dimethyl-

heptane. Presumably, all of the 3,4-dimethylheptane isomers

you found in your answer to Problem 4.37 are in that bottle.

How many signals will appear in the

C NMR spectrum of the

contents of the bottle?

PROBLEM 4.39 You are given a bottle labeled 3,5-dimethyl-

heptane. Presumably, all of the 3,5-dimethylheptane isomers

you found in your answer to Problem 4.37 are in that bottle.

How many signals will appear in the

C NMR spectrum of the

contents of the bottle?

PROBLEM 4.40 Find all the chiral isomers of the hexenes (see

Problem 3.5) and heptenes (see Problem 3.40).

PROBLEM 4.41 Find all the chiral isomers of the hexynes, hep-

tynes (see Problem 3.23), and octynes (see Problem 3.41).

PROBLEM 4.42 For each of the following compounds deter-

mine the number of stereogenic carbons and draw all possible

stereoisomers. You can treat the cyclohexanes as if they were flat.

PROBLEM 4.43 Which of the following compounds are chi-

ral? Determine the configuration (R or S) for all stereogenic

atoms.

4.16 Additional Problems

(a) (b)

Br Br

(a) (b) (c)

OH F

(d)

HO OH

(e)

Estradiol

Narbomycin

N(CH

)

PROBLEM 4.44 Estradiol (shown below) is the active ingredi-

ent in one birth control method. Indicate the stereogenic car-

bons in the molecule and determine the configuration for each.

PROBLEM 4.45 Narbomycin is an antibiotic substance pro-

duced by Streptomyces narbonensis. It is shown below. Put an

asterisk by each stereogenic carbon.

4.16 Additional Problems 183

PROBLEM 4.48

In Problem 2.45, you drew all the isomers of

“chlorohexane” C

Cl. Which of these isomers contain

stereogenic carbons? Be sure to be alert for molecules containing

more than one stereogenic carbon. Designate the stereogenic

carbons with an asterisk.

PROBLEM 4.49 Draw three-dimensional pictures of the

stereoisomers of the molecules in your answer to Problem 4.48

that contain more than one stereogenic carbon.

PROBLEM 4.46 Which of the following molecules are chiral?

Specify each stereogenic atom as (R) or (S).

PROBLEM 4.47 Taxol is a natural product with antitumor

activity. It is a component of the Yew tree. Put an asterisk by

each stereogenic atom.

(a) (b)

BrH

Taxol

(a)

(CH

)

C CHCH

CH(CH

)CH

(b)

(

)

C CHCH

C(CH

) CHCH

PROBLEM 4.50 Which of the following molecules can exist as

cis/trans (Z/E) isomers? Draw the (E) and (Z) forms. Which

molecules possess stereogenic carbons? Designate the stereogenic

carbons with an asterisk.

PROBLEM 4.51 Draw the (R) enantiomer for each molecule

containing a stereogenic carbon in Problem 4.50.

PROBLEM 4.52 Draw all the stereoisomers of the following

molecule. Identify all pairs of enantiomers and diastereomers.

PROBLEM 4.53 In Problem 3.48, you worked out all the iso-

mers of the formula C

Cl. Find all the chiral isomers of the

ring compounds with this formula. Indicate the stereogenic car-

bons with an asterisk.

PROBLEM 4.54 Draw the four chiral isomers found in

Problem 4.53, show the mirror images, and indicate the

absolute configuration (R or S) of the stereogenic carbons.

PROBLEM 4.55 In Problem 3.46, you found 10 cyclopropanes

of the formula C

. Find the chiral ones. Indicate the

stereogenic carbons with an asterisk.

PROBLEM 4.56 One of the isomers in the answer to Problem

4.55 has a pair of bromines on one carbon. Draw the stereoiso-

mers of this molecule, and use the (R/S) convention to show

the absolute configuration of the stereogenic carbons.

184 CHAPTER 4 Stereochemistry

PROBLEM 4.58 In Dorothy Sayers’ and Robert Eustace’s mys-

tery novel, The Documents in the Case (Harper Paperbacks,

1995), mushroom expert George Harrison is murdered by

Harwood Latham. Latham doses Harrison’s stew of Amanita

rubescens with muscarine, the toxin of the related, but deadly,

Amanita muscaria. Lathom hopes to make it look like an acci-

PROBLEM 4.57 Write the IUPAC name for each of the fol-

lowing compounds. Don’t forget the (R/S) or (E/Z) designation

if relevant.

Cl Br

Cl Cl

(b)(a)

(d)(c)

(f) (g)(e)

dental poisoning through a mistaken identification by Harrison.

See if you can guess how Latham is caught. Here is a clue:

Latham obtains the poison muscarine not from the mushroom

itself, but from a laboratory in which it has been synthesized by

an unwitting accomplice from simple starting materials. Thanks

go to Ms. Dana Guyer, a former student of organic chemistry at

Princeton University, for finding this book.

Use Organic Reaction Animations (ORA) to answer the fol-

lowing questions:

PROBLEM 4.59 Observe the reactions titled “Acetylide

addition” and “S

2 with cyanide.” What stereochemical event

happens at the carbon being attacked in each of these

reactions?

PROBLEM 4.60 Observe the reactions “Unimolecular nucleo-

philic substitution” and “Unimolecular elimination.”The first

intermediate in each reaction has a carbon that is planar. What

is the hybridization of that carbon in each case? What would be

the outcome if a stereogenic carbon were involved in the reac-

tion? Would chirality be retained? Why or why not?

PROBLEM 4.61 Observe the “Bimolecular elimination” reac-

tion. When the base reacts with the hydrogen in the first step of

the reaction, what is the spatial relationship between that

hydrogen and the bromide leaving group? Do you suppose this

relationship is important? Why or why not?

PROBLEM 4.62 Is the starting material shown in the reaction

“Carbocation rearrangement” chiral? If yes, what is the configu-

ration of the molecule shown in the animation? Is the product

chiral? Is this a case of: (a) a chiral molecule becoming achiral,

(b) a chiral molecule retaining its chirality, (c) an achiral mole-

cule staying achiral, or (d) an achiral molecule becoming chiral?

Rings

185

5.1 Preview

5.2 Rings and Strain

5.3 Quantitative Evaluation of

Strain Energy

5.4 Stereochemistry of

Cyclohexane: Conformational

Analysis

5.5 Monosubstituted Cyclohexanes

5.6 Disubstituted Ring

Compounds

5.7 Bicyclic Compounds

5.8 Special Topic: Polycyclic

Systems

5.9 Special Topic: Adamantanes in

Materials and Biology

5.10 Summary

5.11 Additional Problems

RING FORMATION This chapter is about rings, from the smallest to the largest.

We’ll learn about their properties here, but their formation will remain a mystery

for later chapters to solve.

186 CHAPTER 5 Rings

Two schematic views

of cyclopentane

A complicated, polycyclic compound,

bicyclo[2.2.1]heptane

A really complicated, polycyclic

compound that nobody tries to name

systematically—it’s called

“dodecahedrane”

FIGURE 5.1 Some simple and complex ring structures.

Clearly the ring had an unwholesome power...

—J. R. R. TOLKIEN,

THE LORD OF THE RINGS

John Ronald Reuel Tolkien (1892–1973) was Merton Professor of English language at Oxford University.

5.1 Preview

In these early chapters, we have used our powers of imagination to picture the three

dimensionality of organic molecules. But nowhere is thinking in three dimensions

more important than in depicting the myriad structures formed from chains of atoms

tied into rings. It’s not always easy to see these structures clearly, so do not be reluc-

tant to work with models,especially at the beginning. All organic chemists use them.

Polycyclic compounds are especially complicated. No one can see all the subtlety of

these compounds without models.Much of chemistry involves interactions of groups

in proximity, and the two-dimensional page is often quite ineffective in showing which

atoms are close to others. As you go on in this subject, your ability to use the two-

dimensional surface to depict the three-dimensional molecular world will increase,

but none of us will ever outgrow the need for models. Rings were first encountered

in Section 2.12, and Figure 5.1 recalls some simple and complex ring compounds.

ESSENTIAL SKILLS AND DETAILS

1. Craftsmanship.This chapter is entirely about the structures of rings. Drawing those

rings accurately is an essential skill for any chemist. Ring compounds abound in

Nature, and the positions of groups attached to rings can be known quite well. If you

know for certain the shape of the ring compound, you also know where the groups

attached to it are located.The distance between groups and the orientation of groups

can be controlled in this way if you know the ring structure. It is really more than a

detail, but drawing the chair form of a cyclohexane perfectly (well, at least very

accurately) is a necessary skill. Fortunately, this skill may look hard but, in fact, it is easy.

Just follow the directions in Section 5.2.

2. Flexibility. Many rings are highly mobile. One example of this phenomenon is the ring

“flip” of cyclohexane.You must be able to “flip”cyclohexanes and draw the two

interconverting chair forms well.

3. Strain. You have to be able to identify two sources of instability in rings. Eclipsing

(torsional) strain and angle strain both act to make rings less stable and thus more

reactive.

4. Intermediate versus transition state.The distinction between an intermediate and a

transition state reappears in this chapter. An intermediate is a potentially isolable

compound in an energy well. However, often it is high in energy and thus not easy to

isolate. On the other hand, a transition state is the top of an energy hill—the top of the

barrier separating two compounds in energy wells—and cannot be isolated.

5.2 Rings and Strain 187

5.2 Rings and Strain

Much can be learned from the simplest of all rings, cyclopropane, (CH

)

. Indeed,

even at this early stage we can make some preliminary judgments about this com-

pound. In doing so, we will be recapitulating the thoughts of Adolf von Baeyer

(1835–1917), who in 1885 appreciated the importance of the deviation of the inter-

nal angles in most cycloalkanes from the ideal tetrahedral angle,109.5° (Fig. 5.2). He

109.5⬚

normal tetrahedron

with a bond angle of

109.5⬚

A small ring will

constrict this angle

to less than 109.5⬚

<109.5⬚

A large ring will

expand this angle

to more than 109.5⬚

>109.5⬚

FIGURE 5.2 Ring compounds have

angle strain if angles are forced to be

significantly smaller or larger than

the ideal 109.5°.

suggested that the instabilities of ring compounds should parallel the deviations of

their internal angles from the ideal; the further the angle from tetrahedral, the more

angle strain the molecule has, and the less stable (higher energy) it is.The idea that

angle strain is an important factor in ring stability has continued to this day.

Cyclopropane represents an extreme example of the effects of angle strain.The strain

induced by the reduction of the ideal tetrahedral angle of 109.5° to 60° is so great as

to make other bonding arrangements possible. One way of looking at cyclopropane

views the bonds between the carbons as “bent.”The internuclear angle in an equilat-

eral triangle is, of course, 60° and cannot be otherwise.However, the interorbital angle

is apparently quite a bit larger (Fig. 5.3) and is estimated at 104°.Thus, the internu-

clear angle (60°) and the interorbital angle ( 104°) are different for cyclopropane.

Cyclopropane

(interorbital angle is 104⬚)

104⬚

Cyclopropane

(internuclear angle is 60⬚)

60⬚

WEB 3D

FIGURE 5.3 Bent bonds in cyclopropane.

Of course, the carbon skeleton of cyclopropane is flat—three points determine

a plane—and planarity causes still other problems for cyclopropane.Figure 5.4 shows

a Newman projection looking down one of the three equivalent carbon–carbon

bonds of cyclopropane. In cyclopropane, all carbon–hydrogen bonds are eclipsed.

Remember: The solid wedges are coming toward you, and the dashed wedges are

retreating from you.

We have seen eclipsing before in the eclipsed form of ethane (and other alkanes).

Ethane has a simple way to avoid this destabilizing opposition of atoms and elec-

trons; it adopts the staggered conformation and minimizes the problem.There is no

way for cyclopropane to do likewise. Its hydrogens are stuck in the eclipsed arrange-

ment, and there is no possible release. We can even make an estimate of how much

damage this torsional strain (eclipsing strain) might cause in cyclopropane, of how

This Newman projection attempts

to show the eclipsed C

H bonds;

in cyclopropane all the C

H bonds

are ecli

sed

FIGURE 5.4 Eclipsing of

bonds in cyclopropane.

188 CHAPTER 5 Rings

rotate 60⬚

Eclipsed conformation

Staggered conformation

(~3 kcal/mol lower in energy)

rotate 60⬚

Torsional strain is 3 kcal/mol in the eclipsed form of ethane;

this destabilization is about 1 kcal/mol per C

H bond pair

FIGURE 5.5 The destabilization

introduced by a pair of eclipsed

carbon–hydrogen bonds is about

1 kcal/mol.

much torsional strain raises the energy of cyclopropane. The rotational barrier in

ethane is the result of three pairs of eclipsed hydrogens (Chapter 2, p. 67), and

amounts to 3 kcal/mol (Fig. 5.5). Thus, a first guess would put the energy cost of

each pair of eclipsed hydrogens at about 1 kcal/mol. That estimate is a maximum

value,however,because the carbon–hydrogen bonds in cyclopropane point away from

each other more than do the ones in ethane.There is some debate about the amount

of eclipsing strain in cyclopropane, but the consensus now is that it is substantially

less than that 6 kcal/mol maximum. The total strain of cyclopropane is composed

of torsional strain plus angle strain, which is even more difficult to estimate with-

out the help of calculations or experiments (Fig. 5.6).

The effects of high strain in cyclopropane show up in a number of ways,

including the unusual reactivity of the carbon–carbon bonds. Breaking the cen-

tral carbon–carbon bond in butane requires 88 kcal/mol, but the carbon–carbon

bond in cyclopropane needs only 65 kcal/mol (Fig. 5.7). This difference allows

a first quantitative estimate of the strain of cyclopropane (on p. 193 we will

see two more quantitative estimates). Strain destabilizes cyclopropane by about

(88  65) kcal/mol  23 kcal/mol.

The six C

H bonds are all

eclipsed in cyclopropane;

there should be about

6 kcal/mol of torsional strain

FIGURE 5.6 There are six pairs of

eclipsed bonds in

cyclopropanes.

Butane

Bond dissociation energy about 88 kcal/mol

Energy

Cyclopropane

Bond dissociation energy about 65 kcal/mol

88 kcal/mol 65 kcal/mol

23 kcal/mol

FIGURE 5.7 The bond dissociation energy of a carbon–carbon bond in cyclopropane is only

65 kcal/mol. Cyclopropane is strained by about 23 kcal/mol.

At first glance, the other cycloalkanes would seem to have similar difficulties. In

planar cyclobutane, for example, angle strain is less of a problem because cyclobu-

tane’s 90° bond angle is a smaller deviation from the ideal 109.5° thanC