Gasoline -
British Foreign Minister Ernest Bevin once said that "The Kingdom of Heaven runs on righteousness, but the Kingdom of Earth runs on alkanes." (Actually, he said "runs on oil" not "runs on alkanes," but they're essentially the same.) By far, the major sources of alkanes are the world's natural gas and petroleum deposits. Laid down eons ago, these deposits are thought to be derived primarily from the decomposition of tiny singlecelled marine organisms called foraminifera. Natural gas consists chiefly of methane but also contains ethane, propane, and butane. Petroleum is a complex mixture of hydrocarbons that must be separated into fractions and then further refined before it can be used.

FIGURE 3.12 Gasoline is a finite resource. It won't be around forever. (credit: "The first oil well" (https://www.loc.gov/item/ 2010649522/) by Unknown/Library of Congress)

The petroleum era began in August 1859, when the world's first oil well was drilled by Edwin Drake near Titusville, Pennsylvania. The petroleum was distilled into fractions according to boiling point, but it was highboiling kerosene, or lamp oil, rather than gasoline that was primarily sought. Literacy was becoming widespread at the time, and people wanted better light for reading than was available from candles. Gasoline was too volatile for use in lamps and was initially considered a waste by-product. The world has changed greatly since those early days, however, and it is now gasoline rather than lamp oil that is prized.

Petroleum refining begins by fractional distillation of crude oil into three principal cuts according to boiling point (bp): straight-run gasoline (bp 30-200 ${ }^{\circ} \mathrm{C}$ ), kerosene (bp $175-300^{\circ} \mathrm{C}$ ), and heating oil, or diesel fuel (bp $275-400{ }^{\circ} \mathrm{C}$ ). Further distillation under reduced pressure then yields lubricating oils and waxes and leaves a tarry residue of asphalt. The distillation of crude oil is only the first step in gasoline production, however. Straight-run gasoline turns out to be a poor fuel in automobiles because of engine knock, an uncontrolled combustion that can occur in a hot engine causing potentially serious damage.


Gasoline -
The octane number of a fuel is the measure by which its antiknock properties are judged. It was recognized long ago that straight-chain hydrocarbons are far more prone to inducing engine knock than highly branched compounds. Heptane, a particularly bad fuel, is assigned a base value of 0 octane number, and 2,2,4-trimethylpentane, commonly known as isooctane, has a rating of 100 .

Heptane
(octane number $=\mathbf{0}$ )

2,2,4-Trimethylpentane
(octane number $=100$ )

Because straight-run gasoline burns so poorly in engines, petroleum chemists have devised numerous methods for producing higher-quality fuels. One of these methods, catalytic cracking, involves taking the high-boiling kerosene cut ( $\mathrm{C}_{11}-\mathrm{C}_{14}$ ) and "cracking" it into smaller branched molecules suitable for use in gasoline. Another process, called reforming, is used to convert $\mathrm{C}_{6}-\mathrm{C}_{8}$ alkanes to aromatic compounds such as benzene and toluene, which have substantially higher octane numbers than alkanes. The final product that goes in your tank has an approximate composition of $15 \% \mathrm{C}_{4}-\mathrm{C}_{8}$ straight-chain alkanes, $25 \%$ to $40 \% \mathrm{C}_{4}-\mathrm{C}_{10}$ branched-chain alkanes, $10 \%$ cyclic alkanes, $10 \%$ straight-chain and cyclic alkenes, and $25 \%$ arenes (aromatics).


Key Terms -
```
- alcohol
ester
- aldehyde
- ether
- aliphatic
- functional group
- alkane
- gauche conformation
- alkene
- hydrocarbon
- alkyl group
- isomer
- alkyl halide
- ketone
- alkyne
- Newman projection
- amide
- nitrile
- amine
- R group
- anti conformation
- saturated
- arene
- sawhorse representation
- branched-chain alkane
- staggered conformation
- carbonyl group
- stereochemistry
- carboxylic acid
- steric strain
- conformation
- straight-chain alkane
- conformational isomer
- substituent
- conformer
- sulfide
- constitutional isomer
- thiol
- eclipsed conformation
- torsional strain
```


Summary -
Alkanes are relatively unreactive and rarely involved in chemical reactions, but they nevertheless provide a useful vehicle for introducing some important general ideas. In this chapter, we've used alkanes to introduce the basic approach to naming organic compounds and to take an initial look at some of the three-dimensional aspects of molecules.

A functional group is a group of atoms within a larger molecule that has a characteristic chemical reactivity. Because functional groups behave in approximately the same way in all molecules where they occur, the chemical reactions of an organic molecule are largely determined by its functional groups.

[^0]structures are called isomers. More specifically, compounds such as butane and isobutane, which differ in their connections between atoms, are called constitutional isomers.

Carbon-carbon single bonds in alkanes are formed by $\sigma$ overlap of carbon $s p^{3}$ hybrid orbitals. Rotation is possible around $\sigma$ bonds because of their cylindrical symmetry, and alkanes therefore exist in a large number of rapidly interconverting conformations. Newman projections make it possible to visualize the spatial consequences of bond rotation by sighting directly along a carbon-carbon bond axis. Not all alkane conformations are equally stable. The staggered conformation of ethane is $12 \mathrm{~kJ} / \mathrm{mol}(2.9 \mathrm{kcal} / \mathrm{mol}) \mathrm{more}$ stable than the eclipsed conformation because of torsional strain. In general, any alkane is most stable when all its bonds are staggered.


Visualizing Chemistry -
PROBLEM Identify the functional groups in the following substances, and convert each drawing into a 3-19 molecular formula (red $=0$, blue $=\mathrm{N}$ ).
(a)

Phenylalanine
(b)

Lidocaine
PROBLEM Give IUPAC names for the following alkanes, and convert each drawing into a skeletal structure.

3-20 (a)

(b)

(c)

(d)

PROBLEM Draw a Newman projection along the C2-C3 bond of the following conformation of 2-butanol.
3-21


Functional Groups -
PROBLEM Locate and identify the functional groups in the following molecules.

3-22 (a)

(e)

(f)

(c)
(d)

(d)

PROBLEM Propose structures that meet the following descriptions:
3-23 (a) A ketone with five carbons (b) A four-carbon amide (c) A five-carbon ester
(d) An aromatic aldehyde
(e) A keto ester
(f) An amino alcohol

PROBLEM Propose structures for the following:
3-24 (a) A ketone, $\mathrm{C}_{4} \mathrm{H}_{8} \mathrm{O}$ (b) A nitrile, $\mathrm{C}_{5} \mathrm{H}_{9} \mathrm{~N}$ (c) A dialdehyde, $\mathrm{C}_{4} \mathrm{H}_{6} \mathrm{O}_{2}$
(d) A bromoalkene, $\mathrm{C}_{6} \mathrm{H}_{11} \mathrm{Br}$ (e) An alkane, $\mathrm{C}_{6} \mathrm{H}_{14}$ (f) cyclic saturated hydrocarbon, $\mathrm{C}_{6} \mathrm{H}_{12}$
(g) A diene (dialkene), $\mathrm{C}_{5} \mathrm{H}_{8}$
(h) A keto alkene, $\mathrm{C}_{5} \mathrm{H}_{8} \mathrm{O}$

PROBLEM Predict the hybridization of the carbon atom in each of the following functional groups:
3-25 (a) Ketone
(b) Nitrile
(c) Carboxylic acid


Functional Groups -
PROBLEM Predict the hybridization of the carbon atom in each of the following functional groups:
3-25 (a) Ketone
(b) Nitrile
(c) Carboxylic acid

PROBLEM Draw the structures of the following molecules:
3-26 (a) Biacetyl, $\mathrm{C}_{4} \mathrm{H}_{6} \mathrm{O}_{2}$, a substance with the aroma of butter; it contains no rings or carbon-carbon multiple bonds.
(b) Ethylenimine, $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{~N}$, a substance used in the synthesis of melamine polymers; it contains no multiple bonds.
(c) Glycerol, $\mathrm{C}_{3} \mathrm{H}_{8} \mathrm{O}_{3}$, a substance isolated from fat and used in cosmetics; it has an-OH group on each carbon.


Isomers -
PROBLEM Draw structures that meet the following descriptions (there are many possibilities):
3-27 (a) Three isomers with the formula $\mathrm{C}_{8} \mathrm{H}_{18}$ (b) Two isomers with the formula $\mathrm{C}_{4} \mathrm{H}_{8} \mathrm{O}_{2}$
PROBLEM Draw structures of the nine isomers of $\mathrm{C}_{7} \mathrm{H}_{16}$.
3-28
PROBLEM In each of the following sets, which structures represent the same compound and which represent 3-29 different compounds?
(a)

(b)

(c)

PROBLEM Seven constitutional isomers have the formula $\mathrm{C}_{4} \mathrm{H}_{10}$ O. Draw as many as you can.
3-30
PROBLEM Draw as many compounds as you can that fit the following descriptions:
3-31 (a) Alcohols with formula $\mathrm{C}_{4} \mathrm{H}_{10} \mathrm{O}$
(b) Amines with formula $\mathrm{C}_{5} \mathrm{H}_{13} \mathrm{~N}$
(c) Ketones with formula $\mathrm{C}_{5} \mathrm{H}_{10} \mathrm{O}$
(d) Aldehydes with formula $\mathrm{C}_{5} \mathrm{H}_{10} \mathrm{O}$
(e) Esters with formula $\mathrm{C}_{4} \mathrm{H}_{8} \mathrm{O}_{2}$ (f) Ethers with formula $\mathrm{C}_{4} \mathrm{H}_{10} \mathrm{O}$

PROBLEM Draw compounds that contain the following:
3-32 (a) A primary alcohol (b) A tertiary nitrile (c) A secondary thiol
(d) Both primary and secondary alcohols (e) An isopropyl group (f) A quaternary carbon


Isomers -
Naming Compounds
PROBLEM Draw and name all monobromo derivatives of pentane, $\mathrm{C}_{5} \mathrm{H}_{11} \mathrm{Br}$.
3-33
PROBLEM Draw and name all monochloro derivatives of 2,5-dimethylhexane, $\mathrm{C}_{8} \mathrm{H}_{17} \mathrm{Cl}$.
3-34
PROBLEM Draw structures for the following:
3-35 (a) 2-Methylheptane (b) 4-Ethyl-2,2-dimethylhexane (c) 4-Ethyl-3,4-dimethyloctane
(d) 2,4,4-Trimethylheptane (e) 3,3-Diethyl-2,5-dimethylnonane
(f) 4-Isopropyl-3-methylheptane

PROBLEM Draw a compound that:
3-36 (a) Has only primary and tertiary carbons (b) Has no secondary or tertiary carbons
(c) Has no secondary or tertiary carbons

PROBLEM Draw a compound that:
3-37 (a) Has nine primary hydrogens (b) Has only primary hydrogens
PROBLEM Give IUPAC names for the following compounds:
3-38 (a)

(b)

(c)

(d)

(e)

(f)

PROBLEM Name the five isomers of $\mathrm{C}_{6} \mathrm{H}_{14}$.
3-39
PROBLEM Explain why each of the following names is incorrect:
3-40 (a) 2,2-Dimethyl-6-ethylheptane
(b) 4-Ethyl-5,5-dimethylpentane
(c) 3-Ethyl-4,4-dimethylhexane (d) 5,5,6-Trimethyloctane
(e) 2-Isopropyl-4-methylheptane

PROBLEM Propose structures and give IUPAC names for the following:
3-41 (a) A diethyldimethylhexane (b) A (3-methylbutyl)-substituted alkane


Conformations -
PROBLEM Consider 2-methylbutane (isopentane). Sighting along the C2-C3 bond:
3-42 (a) Draw a Newman projection of the most stable conformation.
(b) Draw a Newman projection of the least stable conformation.
(c) If a $\mathrm{CH}_{3} \longleftrightarrow \mathrm{CH}_{3}$ eclipsing interaction costs $11 \mathrm{~kJ} / \mathrm{mol}(2.5 \mathrm{kcal} / \mathrm{mol})$ and a $\mathrm{CH}_{3} \longleftrightarrow \mathrm{CH}_{3}$ gauche interaction costs $3.8 \mathrm{~kJ} / \mathrm{mol}(0.9 \mathrm{kcal} / \mathrm{mol})$, make a quantitative plot of energy versus rotation about the $\mathrm{C} 2-\mathrm{C} 3$ bond.

PROBLEM What are the relative energies of the three possible staggered conformations around the C2-C3
3-43 bond in 2,3-dimethylbutane? (See Problem 3-42.)
PROBLEM Construct a qualitative potential-energy diagram for rotation about the $\mathrm{C}-\mathrm{C}$ bond of 3-44 1,2-dibromoethane. Which conformation would you expect to be most stable? Label the anti and gauche conformations of 1,2-dibromoethane.

PROBLEM Which conformation of 1,2-dibromoethane (Problem 3-44) would you expect to have the largest
3-45 dipole moment? The observed dipole moment of 1,2-dibromoethane is $\mu=1.0 \mathrm{D}$. What does this tell you about the actual conformation of the molecule?

PROBLEM Draw the most stable conformation of pentane, using wedges and dashes to represent bonds
3-46 coming out of the paper and going behind the paper, respectively.
PROBLEM Draw the most stable conformation of 1,4-dichlorobutane, using wedges and dashes to represent
3-47 bonds coming out of the paper and going behind the paper, respectively.


General Problems -
PROBLEM For each of the following compounds, draw an isomer that has the same functional groups.

3-48 (a)

(b)

(c) $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{C} \equiv \mathrm{N}$
(d)

(e) $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CHO}$
(f)

PROBLEM Malic acid, $\mathrm{C}_{4} \mathrm{H}_{6} \mathrm{O}_{5}$, has been isolated from apples. Because this compound reacts with 2 molar 3-49 equivalents of base, it is a dicarboxylic acid.
(a) Draw at least five possible structures.
(b) If malic acid is a secondary alcohol, what is its structure?

PROBLEM Formaldehyde, $\mathrm{H}_{2} \mathrm{C}=\mathrm{O}$, is known to all biologists because of its usefulness as a tissue preservative.
3-50 When pure, formaldehyde trimerizes to give trioxane, $\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}_{3}$, which, surprisingly enough, has no carbonyl groups. Only one monobromo derivative $\left(\mathrm{C}_{3} \mathrm{H}_{5} \mathrm{BrO}_{3}\right)$ of trioxane is possible. Propose a structure for trioxane.

PROBLEM The barrier to rotation about the $\mathrm{C}-\mathrm{C}$ bond in bromoethane is $15 \mathrm{~kJ} / \mathrm{mol}(3.6 \mathrm{kcal} / \mathrm{mol})$.
3-51 (a) What energy value can you assign to an $\mathrm{H} \leftrightarrow \mathrm{Br}$ eclipsing interaction?
(b) Construct a quantitative diagram of potential energy versus bond rotation for bromoethane.


General Problems -
PROBLEM Increased substitution around a bond leads to increased strain. Take the four substituted butanes
3-52 listed below, for example. For each compound, sight along the C2-C3 bond and draw Newman projections of the most stable and least stable conformations. Use the data in Table 3.5 to assign strain-energy values to each conformation. Which of the eight conformations is most strained? Which is least strained?
(a) 2-Methylbutane
(b) 2,2-Dimethylbutane
(c) 2,3-Dimethylbutane
(d) 2,2,3-Trimethylbutane

PROBLEM The cholesterol-lowering agents called statins, such as simvastatin (Zocor) and pravastatin
3-53 (Pravachol), are among the most widely prescribed drugs in the world, with annual sales estimated at approximately $\$ 25$ billion. Identify the functional groups in both, and tell how the two substances differ.

Simvastatin (Zocor)

Pravastatin
(Pravachol)

PROBLEM In the next chapter we'll look at cycloalkanes-saturated cyclic hydrocarbons-and we'll see that
3-54 the molecules generally adopt puckered, nonplanar conformations. Cyclohexane, for instance, has a puckered shape like a lounge chair rather than a flat shape. Why?

Nonplanar cyclohexane

Planar cyclohexane

PROBLEM We'll see in the next chapter that there are two isomeric substances, both named
3-55 1,2-dimethylcyclohexane. Explain.


1,2-Dimethylcyclohexane -
104 3 • Additional Problems


CHAPTER 4 -
Organic Compounds: Cycloalkanes and Their Stereochemistry

FIGURE 4.1 The musk gland of the male Himalayan musk deer secretes a substance once used in perfumery that contains cycloalkanes of 14 to $\mathbf{1 8}$ carbons. (credit: modification of work "Siberian musk deer in the tiaga" by ErikAdamsson/Wikimedia Commons, CCO 1.0)


CHAPTER CONTENTS -
4.1 Naming Cycloalkanes
4.2 Cis-Trans Isomerism in Cycloalkanes
4.3 Stability of Cycloalkanes: Ring Strain
4.4 Conformations of Cycloalkanes
4.5 Conformations of Cyclohexane
4.6 Axial and Equatorial Bonds in Cyclohexane
4.7 Conformations of Monosubstituted Cyclohexanes
4.8 Conformations of Disubstituted Cyclohexanes
4.9 Conformations of Polycyclic Molecules

WHY THIS CHAPTER? We'll see numerous instances in future chapters where the chemistry of a given functional group is affected by being in a ring rather than an open chain. Because cyclic molecules are encountered in most pharmaceuticals and in all classes of biomolecules, including proteins, lipids, carbohydrates, and nucleic acids, it's important to understand the behavior of cyclic structures.

Although we've only discussed open-chain compounds up to now, most organic compounds contain rings of carbon atoms. Chrysanthemic acid, for instance, whose esters occur naturally as the active insecticidal constituents of chrysanthemum flowers, contains a three-membered (cyclopropane) ring.


Chrysanthemic acid -
Prostaglandins, potent hormones that control an extraordinary variety of physiological functions in humans, contain a five-membered (cyclopentane) ring.

Prostaglandin $\mathrm{E}_{1}$

Steroids, such as cortisone, contain four rings joined together-three six-membered (cyclohexane) and one fivemembered. We'll discuss steroids and their properties in more detail in Sections 27.6 and 27.7.


Cortisone - 4.1 Naming Cycloalkanes
Saturated cyclic hydrocarbons are called cycloalkanes, or alicyclic compounds (aliphatic cyclic). Because cycloalkanes consist of rings of $-\mathrm{CH}_{2}$-units, they have the general formula $\left(\mathrm{CH}_{2}\right)_{n}$, or $\mathrm{C}_{n} \mathrm{H}_{2 n}$, and can be represented by polygons in skeletal drawings.

Substituted cycloalkanes are named by rules similar to those we saw in (Section 3.4) for open-chain alkanes. For most compounds, there are only two steps.


Find the parent. -
Count the number of carbon atoms in the ring and the number in the largest substituent. If the number of carbon atoms in the ring is equal to or greater than the number in the substituent, the compound is named as an alkyl-substituted cycloalkane. If the number of carbon atoms in the largest substituent is greater than the number in the ring, the compound is named as a cycloalkyl-substituted alkane. For example:

Methylcyclopentane

3 carbons 4 carbons


Number the substituents, and write the name. -
For an alkyl- or halo-substituted cycloalkane, choose a point of attachment as carbon 1 and number the substituents on the ring so that the second substituent has as low a number as possible. If ambiguity still exists, number so that the third or fourth substituent has as low a number as possible, until a point of difference is found.

NOT

1-Ethyl-2,6-dimethylcycloheptane

Higher

3-Ethyl-1,4-dimethylcycloheptane
$\uparrow_{\text {Higher }}^{\uparrow}$
(a) When two or more different alkyl groups are present that could potentially take the same numbers, number them by alphabetical priority, ignoring numerical prefixes such as di- and tri-.

1-Ethyl-2-methylcyclopentane

NOT

2-Ethyl-1-methylcyclopentane
(b) If halogens are present, treat them just like alkyl groups.

1-Bromo-2-methylcyclobutane

2-Bromo-1-methylcyclobutane

Some additional examples follow:

1-Bromo-3-ethyl-5-methylcyclohexane

(1-Methylpropyl)cyclobutane or sec-butylcyclobutane

1-Chloro-3-ethyl-2-methylcyclopentane

PROBLEM Give IUPAC names for the following cycloalkanes:
4-1 (a)

(b)

(c)

(d)

(e)

(f)

PROBLEM Draw structures corresponding to the following IUPAC names:
4-2 (a) 1,1-Dimethylcyclooctane
(b) 3-Cyclobutylhexane
(c) 1,2-Dichlorocyclopentane
(d) 1,3-Dibromo-5-methylcyclohexane

PROBLEM Name the following cycloalkane:


4-3 - 4.2 Cis-Trans Isomerism in Cycloalkanes
In many respects, the chemistry of cycloalkanes is like that of open-chain alkanes: both are nonpolar and fairly inert. There are, however, some important differences. One difference is that cycloalkanes are less flexible than open-chain alkanes. In contrast with the relatively free rotation around single bonds in open-chain alkanes (Section 3.6 and Section 3.7 ), there is much less freedom in cycloalkanes. Cyclopropane, for example, must be a rigid, planar molecule because three points (the carbon atoms) define a plane. No bond rotation can take place around a cyclopropane carbon-carbon bond without breaking open the ring (FIGURE 4.2).
(a)

(b)

FIGURE 4.2 Bond rotation in ethane and cyclopropane. (a) Rotation occurs around the carbon-carbon bond in ethane, but (b) no rotation is possible around the carbon-carbon bonds in cyclopropane without breaking open the ring.

Larger cycloalkanes have increasing rotational freedom, and very large rings ( $\mathrm{C}_{25}$ and up) are so floppy that they are nearly indistinguishable from open-chain alkanes. The common ring sizes ( $\mathrm{C}_{3}-\mathrm{C}_{7}$ ), however, are severely restricted in their molecular motions.

Because of their cyclic structures, cycloalkanes have two faces when viewed edge-on, a "top" face and a "bottom" face. As a result, isomerism is possible in substituted cycloalkanes. For example, there are two different 1,2-dimethylcyclopropane isomers, one with the two methyl groups on the same face of the ring and one with the methyl groups on opposite faces (FIGURE 4.3). Both isomers are stable compounds, and neither can be converted into the other without breaking and reforming chemical bonds.

cis-1,2-Dimethylcyclopropane

trans-1,2-Dimethylcyclopropane


4-3 - 4.2 Cis-Trans Isomerism in Cycloalkanes
cis-1,2-Dimethylcyclopropane

trans-1,2-Dimethylcyclopropane

FIGURE 4.3 There are two different 1,2-dimethylcyclopropane isomers, one with the methyl groups on the same face of the ring (cis) and the other with the methyl groups on opposite faces of the ring (trans). The two isomers do not interconvert.
Unlike the constitutional isomers butane and isobutane, which have their atoms connected in a different order (Section 3.2), the two 1,2-dimethylcyclopropanes have the same order of connections but differ in the spatial orientation of the atoms. Such compounds, with atoms connected in the same order but differing in threedimensional orientation, are called stereochemical isomers, or stereoisomers. As we saw in Section 3.6, the term stereochemistry is used generally to refer to the three-dimensional aspects of structure and reactivity.


Constitutional isomers (different connections between atoms) -
and $\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{CH}_{3}$

Stereoisomers (same connections but different threedimensional geometry)

and

The 1,2-dimethylcyclopropanes are members of a subclass of stereoisomers called cis-trans isomers. The prefixes cis- (Latin "on the same side") and trans- (Latin "across") are used to distinguish between them. Cis-trans isomerism is a common occurrence in substituted cycloalkanes and in many cyclic biological molecules.

cis-1,3-Dimethylcyclobutane

trans-1-Bromo-3-ethylcyclopentane


Naming Cycloalkanes -
Name the following substances, including the cis- or trans- prefix:
(a)

(b)


Strategy -
In these views, the ring is roughly in the plane of the page, a wedged bond protrudes out of the page, and a dashed bond recedes into the page. Two substituents are cis if they are both out of or both into the page, and they are trans if one is out of and one is into the page.


Solution -
(a) trans-1,3-Dimethylcyclopentane
(b) cis-1,2-Dichlorocyclohexane

PROBLEM Name the following substances, including the cis- or trans- prefix:
4-4 (a)

(b)

PROBLEM Draw the structures of the following molecules:
4-5 (a) trans-1-Bromo-3-methylcyclohexane (b) cis-1,2-Dimethylcyclobutane
(c) trans-1-tert-Butyl-2-ethylcyclohexane

PROBLEM Prostaglandin $\mathrm{F}_{2 \alpha}$, a hormone that causes uterine contraction during childbirth, has the following
4-6 structure. Are the two hydroxyl groups $(-\mathrm{OH})$ on the cyclopentane ring cis or trans to each other? What about the two carbon chains attached to the ring?

Prostaglandin $\mathrm{F}_{2 \alpha}$

PROBLEM Name the following substances, including the cis- or trans- prefix (red-brown $=\mathrm{Br}$ ):
4-7 (a)

(b)


Solution - 4.3 Stability of Cycloalkanes: Ring Strain
Chemists in the late 1800s knew that cyclic molecules existed, but the limitations on ring size were unclear. Although numerous compounds containing five-membered and six-membered rings were known, smaller and larger ring sizes had not been prepared, despite many attempts.

A theoretical interpretation of this observation was proposed in 1885 by Adolf von Baeyer, who suggested that small and large rings might be unstable due to angle strain-the strain induced in a molecule when bond angles are forced to deviate from the ideal $109^{\circ}$ tetrahedral value. Baeyer based his suggestion on the simple geometric notion that a three-membered ring (cyclopropane) should be an equilateral triangle with bond angles of $60^{\circ}$ rather than $109^{\circ}$, a four-membered ring (cyclobutane) should be a square with bond angles of $90^{\circ}$, a fivemembered ring should be a regular pentagon with bond angles of $108^{\circ}$, and so on. Continuing this argument, large rings should be strained by having bond angles that are much greater than $109^{\circ}$.

What are the facts? To measure the amount of strain in a compound, we have to measure the total energy of the compound and then subtract the energy of a strain-free reference compound. The difference between the two values should represent the amount of extra energy in the molecule due to strain. The simplest experimental way to do this for a cycloalkane is to measure its heat of combustion, the amount of heat released when the compound burns completely with oxygen. The more energy (strain) the compound contains, the more energy (heat) is released by combustion.

$$
\left(\mathrm{CH}_{2}\right)_{n}+3 n / 2 \mathrm{O}_{2} \longrightarrow n \mathrm{CO}_{2}+n \mathrm{H}_{2} \mathrm{O}+\text { Heat (add) }
$$


Solution - 4.3 Stability of Cycloalkanes: Ring Strain
Because the heat of combustion of a cycloalkane depends on size, we need to look at heats of combustion per $\mathrm{CH}_{2}$ unit. Subtracting a reference value derived from a strain-free acyclic alkane and then multiplying by the number of $\mathrm{CH}_{2}$ units in the ring gives the overall strain energy. FIGURE 4.4 shows the results.

FIGURE 4.4 Cycloalkane strain energies, as calculated by taking the difference between cycloalkane heat of combustion per $\mathbf{C H}_{2}$ and acyclic alkane heat of combustion per $\mathbf{C H}_{2}$, and multiplying by the number of $\mathbf{C H}_{2}$ units in a ring. Small and medium rings are strained, but cyclohexane rings and very large rings are strain-free.

The data in FIGURE 4.4 show that Baeyer's theory is only partially correct. Cyclopropane and cyclobutane are indeed strained, just as predicted, but cyclopentane is more strained than predicted, and cyclohexane is strainfree. Cycloalkanes of intermediate size have only modest strain, and rings of 14 carbons or more are strain-free. Why is Baeyer's theory wrong?

Baeyer's theory is wrong for the simple reason that he assumed all cycloalkanes to be flat. In fact, as we'll see in the next section, most cycloalkanes are not flat; instead, they adopt puckered three-dimensional conformations that allow bond angles to be nearly tetrahedral. As a result, angle strain occurs only in three- and fourmembered rings, which have little flexibility. For most ring sizes, particularly the medium-ring ( $\mathrm{C}_{7}-\mathrm{C}_{11}$ ) cycloalkanes, torsional strain caused by $\mathrm{H} \leftrightarrows \mathrm{H}$ eclipsing interactions at adjacent carbons (Section 3.6) and steric strain caused by the repulsion between nonbonded atoms that approach too closely (Section 3.7) are the most important factors. Thus, three kinds of strain contribute to the overall energy of a cycloalkane.


Solution - 4.3 Stability of Cycloalkanes: Ring Strain
- Angle strain-the strain due to expansion or compression of bond angles
- Torsional strain-the strain due to eclipsing of bonds between neighboring atoms
- Steric strain-the strain due to repulsive interactions when atoms approach each other too closely

PROBLEM Each $\mathrm{H} \longleftrightarrow \mathrm{H}$ eclipsing interaction in ethane costs about $4.0 \mathrm{~kJ} / \mathrm{mol}$. How many such interactions $\mathbf{4 - 8}$ are present in cyclopropane? What fraction of the overall $115 \mathrm{~kJ} / \mathrm{mol}(27.5 \mathrm{kcal} / \mathrm{mol})$ strain energy of cyclopropane is due to torsional strain?

PROBLEM cis-1,2-Dimethylcyclopropane has more strain than trans-1,2-dimethylcyclopropane. How can you 4-9 account for this difference? Which of the two compounds is more stable?


Cyclopropane -
Cyclopropane is the most strained of all rings, primarily due to the angle strain caused by its $60^{\circ} \mathrm{C}-\mathrm{C}-\mathrm{C}$ bond angles. In addition, cyclopropane has considerable torsional strain because the $\mathrm{C}-\mathrm{H}$ bonds on neighboring carbon atoms are eclipsed (FIGURE 4.5).

FIGURE 4.5 The structure of cyclopropane, showing the eclipsing of neighboring C-H bonds that gives rise to torsional strain. Part (b) is a Newman projection along a $\mathrm{C}-\mathrm{C}$ bond.

How can the hybrid-orbital model of bonding account for the large distortion of bond angles from the normal $109^{\circ}$ tetrahedral value to $60^{\circ}$ in cyclopropane? The answer is that cyclopropane has bent bonds. In an unstrained alkane, maximum bonding is achieved when two atoms have their overlapping orbitals pointing directly toward each other. In cyclopropane, though, the orbitals can't point directly toward each other; instead, they overlap at a slight angle. The result is that cyclopropane bonds are weaker and more reactive than typical alkane bonds- $255 \mathrm{~kJ} / \mathrm{mol}$ ( $61 \mathrm{kcal} / \mathrm{mol}$ ) for a C-C bond in cyclopropane versus $370 \mathrm{~kJ} / \mathrm{mol}(88 \mathrm{kcal} / \mathrm{mol}$ ) for a $\mathrm{C}-\mathrm{C}$ bond in open-chain propane.

Typical alkane $\mathrm{C}-\mathrm{C}$ bonds

Typical bent cyclopropane $\mathrm{C}-\mathrm{C}$ bonds


Cyclobutane -
Cyclobutane has less angle strain than cyclopropane but has more torsional strain because of its larger number of ring hydrogens. As a result, the total strain for the two compounds is nearly the same $-110 \mathrm{~kJ} / \mathrm{mol}(26.4 \mathrm{kcal} /$ $\mathrm{mol})$ for cyclobutane versus $115 \mathrm{~kJ} / \mathrm{mol}(27.5 \mathrm{kcal} / \mathrm{mol})$ for cyclopropane. Cyclobutane is not quite flat but is slightly bent so that one carbon atom lies about $25^{\circ}$ above the plane of the other three (FIGURE 4.6). The effect of this slight bend is to increase angle strain but to decrease torsional strain, until a minimum-energy balance between the two opposing effects is achieved.
(a)

(b)

(c) Not quite

FIGURE 4.6 The conformation of cyclobutane. Part (c) is a Newman projection along a $\mathrm{C}-\mathrm{C}$ bond, showing that neighboring $\mathrm{C}-\mathrm{H}$ bonds are
not quite eclipsed.


Cyclopentane -
Cyclopentane was predicted by Baeyer to be nearly strain-free, but it actually has a total strain energy of 26 $\mathrm{kJ} / \mathrm{mol}(6.2 \mathrm{kcal} / \mathrm{mol})$. Although planar cyclopentane has practically no angle strain, it has a large torsional strain. Cyclopentane therefore twists to adopt a puckered, nonplanar conformation that strikes a balance between increased angle strain and decreased torsional strain. Four of the cyclopentane carbon atoms are in approximately the same plane, with the fifth carbon atom bent out of the plane. Most of the hydrogens are nearly staggered with respect to their neighbors (FIGURE 4.7).
(a)

(b)

Observer
(c)

FIGURE 4.7 The conformation of cyclopentane. Carbons $1,2,3$, and 4 are nearly coplanar, but carbon 5 is out of the plane. Part (c) is a Newman projection along the C1-C2 bond, showing that neighboring C-H bonds are nearly staggered.

PROBLEM How many $\mathrm{H} \longleftrightarrow \mathrm{H}$ eclipsing interactions would be present if cyclopentane were planar? Assuming
4-10 an energy cost of $4.0 \mathrm{~kJ} / \mathrm{mol}$ for each eclipsing interaction, how much torsional strain would planar cyclopentane have? Since the measured total strain of cyclopentane is $26 \mathrm{~kJ} / \mathrm{mol}$, how much of the torsional strain is relieved by puckering?

PROBLEM Two conformations of cis-1,3-dimethylcyclobutane are shown. What is the difference between
4-11 them, and which do you think is likely to be more stable?
(a)

(b)


Cyclopentane - 4.5 Conformations of Cyclohexane
Substituted cyclohexanes are the most common cycloalkanes and occur widely in nature. A large number of compounds, including steroids and many pharmaceutical agents, have cyclohexane rings. The flavoring agent menthol, for instance, has three substituents on a six-membered ring.

Menthol

Cyclohexane adopts a strain-free, three-dimensional shape that is called a chair conformation because of its similarity to a lounge chair, with a back, seat, and footrest (FIGURE 4.8). Chair cyclohexane has neither angle strain nor torsional strain-all C-C-C bond angles are near the $109^{\circ}$ tetrahedral value, and all neighboring $\mathrm{C}-\mathrm{H}$ bonds are staggered.
(a)

(b)

Observer
(c)

FIGURE 4.8 The strain-free chair conformation of cyclohexane. All C-C-C bond angles are $111.5^{\circ}$, close to the ideal $109^{\circ}$ tetrahedral angle, and all neighboring $\mathrm{C}-\mathrm{H}$ bonds are staggered.

The easiest way to visualize chair cyclohexane is to build a molecular model if you have access to a model kit, or alternatively to explore with one of the many computer-based modeling programs you may have access to.

The chair conformation of cyclohexane can be drawn in three steps.


STEP 1 -
Draw two parallel lines, slanted downward and slightly offset from each other. This means that four of the cyclohexane carbons lie in a plane.


STEP 2 -
Place the topmost carbon atom above and to the right of the plane of the other four, and connect the bonds.


STEP 3 -
Place the bottommost carbon atom below and to the left of the plane of the middle four, and connect the bonds. Note that the bonds to the bottommost carbon atom are parallel to the bonds to the topmost carbon.

When viewing cyclohexane, it's helpful to remember that the lower bond is in front and the upper bond is in back. If this convention isn't defined, it can appear that the reverse is true. For clarity, all cyclohexane rings drawn in this book will have the front (lower) bond heavily shaded to indicate nearness to the viewer.

In addition to the chair conformation of cyclohexane, there is an alternative conformation of cyclohexane that bears a slight resemblance to a twisted boat. Called the twist-boat conformation, it is nearly free of angle strain. It does, however, have both steric strain and torsional strain and is about $23 \mathrm{~kJ} / \mathrm{mol}(5.5 \mathrm{kcal} / \mathrm{mol})$ higher in
energy than the chair conformation. As a result, molecules adopt the twist-boat geometry only rarely.

Twist-boat cyclohexane ( $23 \mathrm{~kJ} / \mathrm{mol}$ strain)


STEP 3 - 4.6 Axial and Equatorial Bonds in Cyclohexane
The chair conformation of cyclohexane has many consequences. We'll see in Section 11.9, for instance, that the chemical behavior of many substituted cyclohexanes is influenced by their conformation. In addition, we'll see in Section 25.5 that simple carbohydrates, such as glucose, adopt a conformation based on the cyclohexane chair and that their chemistry is directly affected as a result.

Cyclohexane (chair conformation)

Glucose
(chair conformation)

Another trait of the chair conformation is that there are two kinds of positions for substituents on the cyclohexane ring: axial positions and equatorial positions (as shown in FIGURE 4.9). The six axial positions are parallel to the ring axis, while the six equatorial positions are in the rough plane of the ring, around the ring equator.

FIGURE 4.9 Axial and equatorial positions in chair cyclohexane. The six axial hydrogens are parallel to the ring axis, and the six equatorial hydrogens are in a band around the ring equator.

As shown in FIGURE 4.9, each carbon atom in chair cyclohexane has one axial and one equatorial hydrogen. Furthermore, each side of the ring has three axial and three equatorial hydrogens in an alternating arrangement. For example, if the top side of the ring has axial hydrogens on carbons 1, 3, and 5, then it has equatorial hydrogens on carbons 2,4 , and 6 . The reverse is true for the bottom side: carbons 1,3 , and 5 have equatorial hydrogens, but carbons 2, 4, and 6 have axial hydrogens (FIGURE 4.10).

FIGURE 4.10 Alternating axial and equatorial positions in chair cyclohexane, looking directly down the ring axis. Each carbon atom has one axial and one equatorial position, and each face has alternating axial and equatorial positions.

Note that we haven't used the words cis and trans in this discussion of cyclohexane conformation. Two hydrogens on the same side of the ring are always cis, regardless of whether they're axial or equatorial and regardless of whether they're adjacent. Similarly, two hydrogens on opposite sides of the ring are always trans.


STEP 3 - 4.6 Axial and Equatorial Bonds in Cyclohexane
Axial and equatorial bonds can be drawn following the procedure shown in FIGURE 4.11. If possible, look at a molecular model as you practice.

Axial bonds: The six axial bonds, one on each carbon, are parallel and alternate up-down.

Equatorial bonds: The six equatorial bonds, one on each carbon, come in three sets of two parallel lines. Each set is also parallel to two ring bonds. Equatorial bonds

alternate between sides around the ring.

FIGURE 4.11 A procedure for drawing axial and equatorial bonds in chair cyclohexane.
Because chair cyclohexane has two kinds of positions-axial and equatorial-we might expect to find two isomeric forms of a monosubstituted cyclohexane. In fact, we don't. There is only one methylcyclohexane, one bromocyclohexane, one cyclohexanol (hydroxycyclohexane), and so on, because cyclohexane rings are conformationally mobile at room temperature. Different chair conformations readily interconvert, exchanging axial and equatorial positions. This interconversion, called a ring-flip, is shown in FIGURE 4.12.

FIGURE 4.12 A ring-flip in chair cyclohexane interconverts axial and equatorial positions. What is axial in the starting structure becomes equatorial in the ring-flipped structure, and what is equatorial in the starting structure is axial after ring-flip.

As shown in FIGURE 4.12, a chair cyclohexane can be ring-flipped by keeping the middle four carbon atoms in place while folding the two end carbons in opposite directions. In so doing, an axial substituent in one chair form becomes an equatorial substituent in the ring-flipped chair form and vice versa. For example, axial bromocyclohexane becomes equatorial bromocyclohexane after a ring-flip. Since the energy barrier to chair-chair interconversion is only about $45 \mathrm{~kJ} / \mathrm{mol}(10.8 \mathrm{kcal} / \mathrm{mol})$, the process is rapid at room temperature and we see what appears to be a single structure rather than distinct axial and equatorial isomers.

Axial bromocyclohexane

Ring-flip


STEP 3 - 4.6 Axial and Equatorial Bonds in Cyclohexane
Axial bromocyclohexane

Ring-flip

Equatorial bromocyclohexane


Drawing the Chair Conformation of a Substituted Cyclohexane -
Draw 1,1-dimethylcyclohexane in a chair conformation, indicating which methyl group in your drawing is axial and which is equatorial.


Strategy -
Draw a chair cyclohexane ring using the procedure in FIGURE 4.11, and then put two methyl groups on the same carbon. The methyl group in the rough plane of the ring is equatorial, and the one directly above or below the ring is axial.


Solution -
PROBLEM Draw two different chair conformations of cyclohexanol (hydroxycyclohexane), showing all 4-12 hydrogen atoms. Identify each position as axial or equatorial.

PROBLEM Draw two different chair conformations of trans-1,4-dimethylcyclohexane, and label all positions 4-13 as axial or equatorial.

PROBLEM Identify each of the colored positions-red, blue, and green-as axial or equatorial. Then carry out a 4-14 ring-flip, and show the new positions occupied by each color.


Solution - 4.7 Conformations of Monosubstituted Cyclohexanes
Even though cyclohexane rings flip rapidly between chair conformations at room temperature, the two conformations of a monosubstituted cyclohexane aren't equally stable. In methylcyclohexane, for instance, the equatorial conformation is more stable than the axial conformation by $7.6 \mathrm{~kJ} / \mathrm{mol}(1.8 \mathrm{kcal} / \mathrm{mol})$. The same is true of other monosubstituted cyclohexanes: a substituent is almost always more stable in an equatorial position than in an axial position.

You might recall from your general chemistry course that it's possible to calculate the percentages of two isomers at equilibrium using the equation $\Delta E=-R T \ln K$, where $\Delta E$ is the energy difference between isomers, $R$ is the gas constant [8.315 J/(K•mol)], $T$ is the Kelvin temperature, and $K$ is the equilibrium constant between isomers. For example, an energy difference of $7.6 \mathrm{~kJ} / \mathrm{mol}$ means that about $95 \%$ of methylcyclohexane molecules have an equatorial methyl group at any given instant while only $5 \%$ have an axial methyl group. FIGURE 4.13 plots the relationship between energy and isomer percentages.

FIGURE 4.13 A plot of the percentages of two isomers at equilibrium versus the energy difference between them. The curves are calculated using the equation $\Delta E=-R T \ln K$.

The energy difference between axial and equatorial conformations is due to steric strain caused by $\mathbf{1 , 3} \mathbf{3}$ diaxial interactions. The axial methyl group on C1 is too close to the axial hydrogens three carbons away on C3 and C5, resulting in $7.6 \mathrm{~kJ} / \mathrm{mol}$ of steric strain (FIGURE 4.14).

FIGURE 4.14 Interconversion of axial and equatorial methylcyclohexane, represented in several formats. The equatorial conformation is more stable than the axial conformation by $7.6 \mathrm{~kJ} / \mathrm{mol}$.


Solution - 4.7 Conformations of Monosubstituted Cyclohexanes
The 1,3-diaxial steric strain in substituted methylcyclohexane is already familiar-we saw it previously as the steric strain between methyl groups in gauche in Section 3.7. Gauche butane is less stable than anti butane by $3.8 \mathrm{~kJ} / \mathrm{mol}(0.9 \mathrm{kcal} / \mathrm{mol})$ because of steric interference between hydrogen atoms on the two methyl groups. Comparing a four-carbon fragment of axial methylcyclohexane with gauche butane shows that the steric interaction is the same in both (FIGURE 4.15). Because axial methylcyclohexane has two such interactions, it has $2 \times 3.8=7.6 \mathrm{~kJ} / \mathrm{mol}$ of steric strain. Equatorial methylcyclohexane has no such interactions and is therefore more stable.

FIGURE 4.15 The origin of 1,3-diaxial interactions in methylcyclohexane. The steric strain between an axial methyl group and an axial hydrogen atom three carbons away is identical to the steric strain in gauche butane. (To display clearly the diaxial interactions in methylcyclohexane, two of the equatorial hydrogens are not shown.)

The exact amount of 1,3-diaxial steric strain in a given substituted cyclohexane depends on the nature and size of the substituent, as indicated in TABLE 4.1. Not surprisingly, the amount of steric strain increases through the series $\mathrm{H}_{3} \mathrm{C}-<\mathrm{CH}_{3} \mathrm{CH}_{2}-<\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CH}-\ll\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}-$, paralleling the increasing size of the alkyl groups. Note that the values given in TABLE 4.1 refer to 1,3-diaxial interactions of the substituent with a single hydrogen atom. These values must be doubled to arrive at the amount of strain in a monosubstituted cyclohexane.

TABLE 4.1 Steric Strain in Monosubstituted Cyclohexanes


Solution - 4.7 Conformations of Monosubstituted Cyclohexanes
TABLE 4.1 Steric Strain in Monosubstituted Cyclohexanes

| 1,3-Diaxial strain
 (kJ/mol)
 (kcal/mol) |  |  |  |
| :--- | :--- | :--- | :--- |
| Y |  |  |  |
| F | 0.5 | 0.12 |  |

TABLE 4.1 Steric Strain in Monosubstituted Cyclohexanes

| 1,3-Diaxial strain |  |  |  |
| :---: | :---: | :---: | :---: |
| Y | ( $\mathrm{kJ} / \mathrm{mol}$ ) | ( $\mathrm{kcal} / \mathrm{mol}$ ) | $\sim$ |
| $\mathrm{Cl}, \mathrm{Br}$ | 1.0 | 0.25 |  |
| OH | 2.1 | 0.5 |  |
| $\mathrm{CH}_{3}$ | 3.8 | 0.9 |  |
| $\mathrm{CH}_{2} \mathrm{CH}_{3}$ | 4.0 | 0.95 |  |
| $\mathrm{CH}\left(\mathrm{CH}_{3}\right)_{2}$ | 4.6 | 1.1 |  |
| $\mathrm{C}\left(\mathrm{CH}_{3}\right)_{3}$ | 11.4 | 2.7 |  |
| $\mathrm{C}_{6} \mathrm{H}_{5}$ | 6.3 | 1.5 |  |
| $\mathrm{CO}_{2} \mathrm{H}$ | 2.9 | 0.7 |  |
| CN | 0.4 | 0.1 |  |

PROBLEM What is the energy difference between the axial and equatorial conformations of cyclohexanol 4-15 (hydroxycyclohexane)?

PROBLEM Why do you suppose an axial cyano (-CN) substituent causes practically no 1,3-diaxial steric strain 4-16 ( $0.4 \mathrm{~kJ} / \mathrm{mol}$ )?


Solution - 4.7 Conformations of Monosubstituted Cyclohexanes
PROBLEM Look back at Figure 4.13 and estimate the percentages of axial and equatorial conformations 4-17 present at equilibrium in bromocyclohexane.


Solution - 4.8 Conformations of Disubstituted Cyclohexanes
Monosubstituted cyclohexanes are always more stable with their substituent in an equatorial position, but the situation with disubstituted cyclohexanes is more complex because the steric effects of both substituents must be taken into account. All steric interactions for both possible chair conformations must be analyzed before deciding which conformation is favored.

Let's look at 1,2-dimethylcyclohexane as an example. There are two isomers, cis-1,2-dimethylcyclohexane and trans-1,2-dimethylcyclohexane, which must be considered separately. In the cis isomer, both methyl groups are on the same face of the ring and the compound can exist in either of the two chair conformations shown in FIGURE 4.16. (It may be easier for you to see whether a compound is cis- or trans-disubstituted by first drawing the ring as a flat representation and then converting it to a chair conformation.)
cis-1,2-Dimethylcyclohexane
One gauche
interaction ( $3.8 \mathrm{~kJ} / \mathrm{mol}$ ) Two $\mathrm{CH}_{3} \leftrightarrow \mathrm{H}$ diaxial interactions ( $7.6 \mathrm{~kJ} / \mathrm{mol}$ )
Total strain: $\mathbf{3 . 8}+\mathbf{7 . 6}=\mathbf{1 1 . 4} \mathbf{~ k J} / \mathbf{m o l}$

One gauche interaction ( $3.8 \mathrm{~kJ} / \mathrm{mol}$ ) Two $\mathrm{CH}_{3} \leftrightarrow \mathrm{H}$ diaxial interactions ( $7.6 \mathrm{~kJ} / \mathrm{mol}$ )
Total strain: $3.8+7.6=11.4 \mathrm{~kJ} / \mathbf{m o l}$


Solution - 4.8 Conformations of Disubstituted Cyclohexanes
FIGURE 4.16 Conformations of cis-1,2-dimethylcyclohexane. The two chair conformations are equal in energy because each has one axial methyl group and one equatorial methyl group.
Both chair conformations of cis-1,2-dimethylcyclohexane have one axial methyl group and one equatorial methyl group. The top conformation in FIGURE 4.16 has an axial methyl group at C2, which has 1,3-diaxial interactions with hydrogens on C4 and C6. The ring-flipped conformation has an axial methyl group at C1, which has 1,3-diaxial interactions with hydrogens on C3 and C5. In addition, both conformations have gauche butane interactions between the two methyl groups. The two conformations are equal in energy, with a total steric strain of $3 \times 3.8 \mathrm{~kJ} / \mathrm{mol}=11.4 \mathrm{~kJ} / \mathrm{mol}(2.7 \mathrm{kcal} / \mathrm{mol})$.


Solution - 4.8 Conformations of Disubstituted Cyclohexanes
In trans-1,2-dimethylcyclohexane, the two methyl groups are on opposite sides of the ring and the compound can exist in either of the two chair conformations shown in FIGURE 4.17. The situation here is quite different from that of the cis isomer. The top conformation in FIGURE 4.17 has both methyl groups equatorial with only a gauche butane interaction between them ( $3.8 \mathrm{~kJ} / \mathrm{mol}$ ) but no 1,3-diaxial interactions. The ring-flipped conformation, however, has both methyl groups axial. The axial methyl group at C1 interacts with axial hydrogens at C3 and C5, and the axial methyl group at C2 interacts with axial hydrogens at C4 and C6. These four 1,3-diaxial interactions produce a steric strain of $4 \times 3.8 \mathrm{~kJ} / \mathrm{mol}=15.2 \mathrm{~kJ} / \mathrm{mol}$ and make the diaxial conformation $15.2-3.8=11.4 \mathrm{~kJ} / \mathrm{mol}$ less favorable than the diequatorial conformation. We therefore predict that trans-1,2-dimethylcyclohexane will exist almost exclusively in the diequatorial conformation.


trans-1,2-Dimethylcyclohexane -
One gauche interaction ( $3.8 \mathrm{~kJ} / \mathrm{mol}$ )

Four $\mathrm{CH}_{3} \leftrightarrow \mathrm{H}$ diaxial interactions ( $15.2 \mathrm{~kJ} / \mathrm{mol}$ )

FIGURE 4.17 Conformations of trans-1,2-dimethylcyclohexane. The conformation with both methyl groups equatorial (top) is favored by $11.4 \mathrm{~kJ} / \mathrm{mol}(2.7 \mathrm{kcal} / \mathrm{mol})$ over the conformation with both methyl groups axial (bottom).
The same kind of conformational analysis just carried out for cis- and trans-1,2-dimethylcyclohexane can be done for any substituted cyclohexane, such as cis-1-tert-butyl-4-chlorocyclohexane (see Worked Example 4.3).

As you might imagine, though, the situation becomes more complex as the number of substituents increases. For instance, compare glucose with mannose, a carbohydrate present in seaweed. Which do you think is more strained? In glucose, all substituents on the six-membered ring are equatorial, while in mannose, one of the -OH groups is axial, making it more strained.

Glucose

Mannose

A summary of the various axial and equatorial relationships among substituent groups in the different possible cis and trans substitution patterns for disubstituted cyclohexanes is given in TABLE 4.2.

TABLE 4.2 Axial and Equatorial Relationships in Cis- and Trans-Disubstituted Cyclohexanes


trans-1,2-Dimethylcyclohexane -
TABLE 4.2 Axial and Equatorial Relationships in Cis- and Trans-Disubstituted Cyclohexanes

| Cis/trans substitution pattern |  | Axial/equatorial relationships |  |
| :--- | :--- | :--- | :--- |
| 1,2-Cis disubstituted | a,e | or | e,a |
| 1,2-Trans disubstituted | a,a | or | e,e |
| 1,3-Cis disubstituted | a,a | or | e,e |
| 1,3-Trans disubstituted | a,e | or | e,a |
| 1,4-Cis disubstituted | a,e | or | e,a |
| 1,4-Trans disubstituted | a,a | or | e,e |


Drawing the Most Stable Conformation of a Substituted Cyclohexane -
Draw the more stable chair conformation of cis-1-tert-butyl-4-chlorocyclohexane. By how much is it favored?


Strategy -
Draw the two possible chair conformations, and calculate the strain energy in each. Remember that equatorial substituents cause less strain than axial substituents.


Solution -
First draw the two chair conformations of the molecule:

In the conformation on the left, the tert-butyl group is equatorial and the chlorine is axial. In the conformation
on the right, the tert-butyl group is axial and the chlorine is equatorial. These conformations aren't of equal energy because an axial tert-butyl substituent and an axial chloro substituent produce different amounts of steric strain. TABLE 4.1 shows that the 1,3-diaxial interaction between a hydrogen and a tert-butyl group costs $11.4 \mathrm{~kJ} / \mathrm{mol}(2.7 \mathrm{kcal} / \mathrm{mol})$, whereas the interaction between a hydrogen and a chlorine costs only $1.0 \mathrm{~kJ} /$ $\mathrm{mol}(0.25 \mathrm{kcal} / \mathrm{mol})$. An axial tert-butyl group therefore produces $(2 \times 11.4 \mathrm{~kJ} / \mathrm{mol})-(2 \times 1.0 \mathrm{~kJ} / \mathrm{mol})=20.8$ $\mathrm{kJ} / \mathrm{mol}(4.9 \mathrm{kcal} / \mathrm{mol})$ more steric strain than an axial chlorine, and the compound preferentially adopts the conformation with the chlorine axial and the tert-butyl equatorial.

PROBLEM Draw the more stable chair conformation of the following molecules, and estimate the amount of
4-18 strain in each:
(a) trans-1-Chloro-3-methylcyclohexane
(b) cis-1-Ethyl-2-methylcyclohexane
(c) cis-1-Bromo-4-ethylcyclohexane
(d) cis-1-tert-Butyl-4-ethylcyclohexane

PROBLEM Identify each substituent in the following compound as axial or equatorial, and tell whether the 4-19 conformation shown is the more stable or less stable chair form (green $=\mathrm{Cl}$ ):


Solution - 4.9 Conformations of Polycyclic Molecules
The final point we'll consider about cycloalkane stereochemistry is to see what happens when two or more cycloalkane rings are fused together along a common bond to construct a polycyclic molecule-for example, decalin.


Decalin-two fused cyclohexane rings -
Decalin consists of two cyclohexane rings joined to share two carbon atoms (the bridgehead carbons, C1 and C6) and a common bond. Decalin can exist in either of two isomeric forms, depending on whether the rings are trans fused or cis fused. In cis-decalin, the hydrogen atoms at the bridgehead carbons are on the same side of the rings; in trans-decalin, the bridgehead hydrogens are on opposite sides. FIGURE 4.18 shows how both compounds can be represented using chair cyclohexane conformations. Note that the two decalin isomers are not interconvertible by ring-flips or other rotations. They are cis-trans stereoisomers and have the same relationship to each other that cis- and trans-1,2-dimethylcyclohexane have.

cis-Decalin

trans-Decalin

FIGURE 4.18 Representations of cis- and trans-decalin. Hydrogen atoms at the bridgehead carbons are on the same face of the rings in the cis isomer but on opposite faces in the trans isomer.

Polycyclic compounds are common in nature, and many valuable substances have fused-ring structures. For example, steroids, such as testosterone, the primary sex hormone in males, have three six-membered rings and one five-membered ring fused together. Although steroids look complicated compared with cyclohexane or decalin, the same principles that apply to the conformational analysis of simple cyclohexane rings apply equally well (and often better) to steroids.


Testosterone (a steroid) -
Another common ring system is the norbornane, or bicyclo[2.2.1]heptane, structure. Like decalin, norbornane is a bicycloalkane, so called because two rings would have to be broken open to generate an acyclic structure. Its systematic name, bicyclo[2.2.1]heptane, reflects the fact that the molecule has seven carbons, is bicyclic, and has three "bridges" of 2, 2, and 1 carbon atoms connecting the two bridgehead carbons.

Norbornane (bicyclo[2.2.1]heptane)
Norbornane has a conformationally locked boat cyclohexane ring (Section 4.5) in which carbons 1 and 4 are joined by an additional $\mathrm{CH}_{2}$ group. In drawing this structure, a break in the rear bond indicates that the vertical bond crosses in front of it. Making a molecular model is particularly helpful when trying to see the three-
dimensionality of norbornane.
Substituted norbornanes, such as camphor, are found widely in nature, and many have been important historically in developing organic structural theories.

PROBLEM Which isomer is more stable, cis-decalin or trans-decalin (Figure 4.18)? Explain. 4-20

PROBLEM Look at the following structure of estrone, the primary sex hormone in females, and tell whether 4-21 each of the two indicated (red) ring-fusions is cis or trans.

Estrone


Molecular Mechanics -
All the structural models in this book are computer-drawn. To make sure they accurately represent bond angles, bond lengths, torsional interactions, and steric interactions, the most stable geometry of each molecule has been calculated on a desktop computer using a commercially available molecular mechanics program based on work by Norman Allinger at the University of Georgia.

The idea behind molecular mechanics is to begin with a rough geometry for a molecule and then calculate a total strain energy for that starting geometry, using mathematical equations that assign values to specific kinds of molecular interactions. Bond angles that are too large or too small cause angle strain; bond lengths that are too short or too long cause stretching or compressing strain; unfavorable eclipsing interactions around single bonds cause torsional strain; and nonbonded atoms that approach each other too closely cause steric, or van der Waals, strain.

$$
E_{\text {total }}=E_{\mathrm{bond} \text { stretching }}+E_{\text {angle strain }}+E_{\text {torsional strain }}+E_{\text {van der Waals }}
$$

FIGURE 4.19 Computer programs make it possible to accurately represent molecular geometry. (credit: "Molecular geometry" by Jane Whitney/Flickr, CC BY 4.0)

After calculating a total strain energy for the starting geometry, the program automatically changes the geometry slightly in an attempt to lower strain-perhaps by lengthening a bond that is too short or decreasing an angle that is too large. Strain is recalculated for the new geometry, more changes are made, and more calculations are done. After dozens or hundreds of iterations, the calculation ultimately converges on a minimum energy that corresponds to the most favorable, least strained conformation of the molecule.

Similar calculations have proven to be particularly useful in pharmaceutical research, where a complementary fit between a drug molecule and a receptor molecule in the body is often the key to designing new pharmaceutical agents (FIGURE 4.20).

FIGURE 4.20 The structure of Tamiflu (oseltamivir), an antiviral agent active against type A influenza, along with a molecular model of its minimum-energy conformation as calculated by molecular mechanics.


Key Terms -
- alicyclic
- angle strain
- axial position (cyclohexane)
- boat cyclohexane
- Bridgehead atom
- chair conformation
- cis-trans isomers
- conformational analysis
- cycloalkanes


- 1,3-diaxial interaction -
- Equatorial position (cyclohexane)
- polycyclic molecule
- ring-flip (cyclohexane)
- stereoisomer
- steric strain
- torsional strain
- twist-boat conformation


Summary -
Cyclic molecules are so commonly encountered throughout organic and biological chemistry that it's important to understand the consequences of their cyclic structures. Thus, we've taken a close look at cyclic structures in this chapter.

Cycloalkanes are saturated cyclic hydrocarbons with the general formula $\mathrm{C}_{n} \mathrm{H}_{2 n}$. In contrast to open-chain alkanes, where nearly free rotation occurs around $\mathrm{C}-\mathrm{C}$ bonds, rotation is greatly reduced in cycloalkanes. Disubstituted cycloalkanes can therefore exist as cis-trans isomers. The cis isomer has both substituents on the same side of the ring; the trans isomer has substituents on opposite sides. Cis-trans isomers are just one kind of stereoisomer-compounds that have the same connections between atoms but different threedimensional arrangements.

Not all cycloalkanes are equally stable. Three kinds of strain contribute to the overall energy of a cycloalkane: (1) angle strain is the resistance of a bond angle to compression or expansion from the normal $109^{\circ}$ tetrahedral value, (2) torsional strain is the energy cost of having neighboring $\mathrm{C}-\mathrm{H}$ bonds eclipsed rather than staggered, and (3) steric strain is the repulsive interaction that arises when two groups attempt to occupy the same space.

Cyclopropane ( $115 \mathrm{~kJ} / \mathrm{mol}$ strain) and cyclobutane ( $110.4 \mathrm{~kJ} / \mathrm{mol}$ strain) have both angle strain and torsional strain. Cyclopentane is free of angle strain but has a substantial torsional strain due to its large number of eclipsing interactions. Both cyclobutane and cyclopentane pucker slightly away from planarity to relieve torsional strain.


Summary -
Cyclohexane is strain-free because it adopts a puckered chair conformation, in which all bond angles are near $109^{\circ}$ and all neighboring $\mathrm{C}-\mathrm{H}$ bonds are staggered. Chair cyclohexane has two kinds of positions: axial and equatorial. Axial positions are oriented up and down, parallel to the ring axis, while equatorial positions lie in a belt around the equator of the ring. Each carbon atom has one axial and one equatorial position.

Chair cyclohexanes are conformationally mobile and can undergo a ring-flip, which interconverts axial and equatorial positions. Substituents on the ring are more stable in the equatorial position because axial substituents cause 1,3-diaxial interactions. The amount of 1,3-diaxial steric strain caused by an axial substituent depends on its size.


Visualizing Chemistry -
PROBLEM Name the following cycloalkanes:

4-22 (a)

(b)

PROBLEM Name the following compound, identify each substituent as axial or equatorial, and tell whether the
4-23 conformation shown is the more stable or less stable chair form (green $=\mathrm{Cl}$ ):

PROBLEM A trisubstituted cyclohexane with three substituents-red, green, and blue-undergoes a ring-flip
4-24 to its alternate chair conformation. Identify each substituent as axial or equatorial, and show the positions occupied by the three substituents in the ring-flipped form.

PROBLEM The following cyclohexane derivative has three substituents-red, green, and blue. Identify each
4-25 substituent as axial or equatorial, and identify each pair of relationships (red-blue, red-green, and blue-green) as cis or trans.

PROBLEM Glucose exists in two forms having a 36:64 ratio at equilibrium. Draw a skeletal structure of each,
4-26 describe the difference between them, and tell which of the two you think is more stable (red = 0).

Cycloalkane Isomers
PROBLEM Draw the five cycloalkanes with the formula $\mathrm{C}_{5} \mathrm{H}_{10}$.
4-27
PROBLEM Give IUPAC names for the following compounds.
4-28 (a)

(b)

(c)

(d)

(e)

PROBLEM Draw a stereoisomer of trans-1,3-dimethylcyclobutane.
4-29
PROBLEM Tell whether the following pairs of compounds are identical, constitutional isomers, stereoisomers, 4-30 or unrelated.
(a) cis-1,3-Dibromocyclohexane and trans-1,4-dibromocyclohexane
(b) 2,3-Dimethylhexane and 2,3,3-trimethylpentane
(c)
and

PROBLEM Draw three isomers of trans-1,2-dichlorocyclobutane, and label them as either constitutional 4-31 isomers or stereoisomers.


Visualizing Chemistry -
PROBLEM Draw three isomers of trans-1,2-dichlorocyclobutane, and label them as either constitutional 4-31 isomers or stereoisomers.

PROBLEM Identify each pair of relationships among the -OH groups in glucose (red-blue, red-green, 4-32 red-black, blue-green, blue-black, green-black) as cis or trans.

Glucose

PROBLEM Draw 1,3,5-trimethylcyclohexane using a hexagon to represent the ring. How many cis-trans 4-33 stereoisomers are possible?


Cycloalkane Conformation and Stability -
PROBLEM Hydrocortisone, a naturally occurring hormone produced in the adrenal glands, is often used to
4-34 treat inflammation, severe allergies, and numerous other conditions. Is the indicated -OH group axial or equatorial?

Hydrocortisone

PROBLEM A 1,2-cis disubstituted cyclohexane, such as cis-1,2-dichlorocyclohexane, must have one group 4-35 axial and one group equatorial. Explain.

PROBLEM A 1,2-trans disubstituted cyclohexane must have either both groups axial or both groups equatorial. 4-36 Explain.

PROBLEM Why is a 1,3-cis disubstituted cyclohexane more stable than its trans isomer? 4-37

PROBLEM Which is more stable, a 1,4-trans disubstituted cyclohexane or its cis isomer?
4-38
PROBLEM cis-1,2-Dimethylcyclobutane is less stable than its trans isomer, but cis-1,3-dimethylcyclobutane is
4-39 more stable than its trans isomer. Draw the most stable conformations of both, and explain.
PROBLEM From the data in Figure 4.13 and Table 4.1, estimate the percentages of molecules that have their
4-40 substituents in an axial orientation for the following compounds:
(a) Isopropylcyclohexane
(b) Fluorocyclohexane
(c) Cyclohexanecarbonitrile, $\mathrm{C}_{6} \mathrm{H}_{11} \mathrm{CN}$


Cycloalkane Conformation and Stability -
PROBLEM Assume that you have a variety of cyclohexanes substituted in the positions indicated. Identify
4-41 the substituents as either axial or equatorial. For example, a 1,2-cis relationship means that one substituent must be axial and one equatorial, whereas a 1,2-trans relationship means that both substituents are axial or both are equatorial.
(a) 1,3-Trans disubstituted (b) 1,4-Cis disubstituted (c) 1,3-Cis disubstituted
(d) 1,5-Trans disubstituted
(e) 1,5-Cis disubstituted (f) 1,6-Trans disubstituted

Cyclohexane Conformational Analysis
PROBLEM Draw the two chair conformations of cis-1-chloro-2-methylcyclohexane. Which is more stable, and
4-42 by how much?
PROBLEM Draw the two chair conformations of trans-1-chloro-2-methylcyclohexane. Which is more stable? 4-43

PROBLEM Galactose, a sugar related to glucose, contains a six-membered ring in which all the substituents
4-44 except the -OH group, indicated below in red, are equatorial. Draw galactose in its more stable chair conformation.

Galactose

PROBLEM Draw the two chair conformations of menthol, and tell which is more stable.
4-45


Menthol -
PROBLEM There are four cis-trans isomers of menthol (Problem 4-45), including the one shown. Draw the
4-46 other three.
PROBLEM The diaxial conformation of cis-1,3-dimethylcyclohexane is approximately $23 \mathrm{~kJ} / \mathrm{mol}(5.4 \mathrm{kcal} / \mathrm{mol})$
4-47 less stable than the diequatorial conformation. Draw the two possible chair conformations, and suggest a reason for the large energy difference.

PROBLEM Approximately how much steric strain does the 1,3-diaxial interaction between the two methyl
4-48 groups introduce into the diaxial conformation of cis-1,3-dimethylcyclohexane? (See Problem 4-47.)

PROBLEM In light of your answer to Problem 4-48, draw the two chair conformations of
4-49 1,1,3-trimethylcyclohexane and estimate the amount of strain energy in each. Which conformation is favored?

PROBLEM One of the two chair structures of cis-1-chloro-3-methylcyclohexane is more stable than the other
4-50 by $15.5 \mathrm{~kJ} / \mathrm{mol}(3.7 \mathrm{kcal} / \mathrm{mol})$. Which is it? What is the energy cost of a 1,3 -diaxial interaction between a chlorine and a methyl group?


General Problems -
PROBLEM We saw in Problem 4-20 that cis-decalin is less stable than trans-decalin. Assume that the
4-51 1,3-diaxial interactions in cis-decalin are similar to those in axial methylcyclohexane [that is, one $\mathrm{CH} 2 \leftrightarrow \mathrm{H}$ interaction costs $3.8 \mathrm{~kJ} / \mathrm{mol}(0.9 \mathrm{kcal} / \mathrm{mol})]$, and calculate the magnitude of the energy difference between cis- and trans-decalin.

PROBLEM Using molecular models as well as structural drawings, explain why trans-decalin is rigid and
4-52 cannot ring-flip whereas cis-decalin can easily ring-flip.
PROBLEM trans-Decalin is more stable than its cis isomer, but cis-bicyclo[4.1.0]heptane is more stable than its
4-53 trans isomer. Explain.

trans-Decalin

cis-Bicyclo[4.1.0]heptane

PROBLEM As mentioned in Problem 3-53, the statin drugs, such as simvastatin (Zocor), pravastatin (Pravachol),
4-54 and atorvastatin (Lipitor) are the most widely prescribed drugs in the world.

(a) Are the two indicated bonds on simvastatin cis or trans?
(b) What are the cis/trans relationships among the three indicated bonds on pravastatin?
(c) Why can't the three indicated bonds on atorvastatin be identified as cis or trans?

PROBLEM myo-Inositol, one of the isomers of 1,2,3,4,5,6-hexahydroxycyclohexane, acts as a growth factor in
4-55 both animals and microorganisms. Draw the most stable chair conformation of myo-inositol.

myo-Inositol


General Problems -
myo-Inositol

PROBLEM How many cis-trans stereoisomers of myo-inositol (Problem 4-55) are there? Draw the structure of
4-56 the most stable isomer.
PROBLEM Julius Bredt, discoverer of the structure of camphor, proposed in 1935 that bicycloalkenes such as
4-57 1-norbornene, which have a double bond to a bridgehead carbon, are too strained to exist. Explain. (Making a molecular model will be helpful.

1-Norbornene

PROBLEM Tell whether each of the following substituents on a steroid is axial or equatorial. (A substituent that
4-58 is "up" is on the top side of the molecule as drawn, and a substituent that is "down" is on the bottom side.
(a) Substituent up at C3 (b) Substituent down at C7
(c) Substituent down at C11

PROBLEM Amantadine is an antiviral agent that is active against influenza type A infection. Draw a three-
4-59 dimensional representation of amantadine, showing the chair cyclohexane rings.

Amantadine

PROBLEM There are two different isomers named trans-1,2-dimethylcyclopentane. Similarly, you have two
4-60 different appendages called hands. What is the relationship between them? (We'll explore this kind of isomerism in the next chapter.)

and

PROBLEM Ketones react with alcohols to yield products called ketals. Why does the all-cis isomer of
4-61 4-tert-butyl-1,3-cyclohexanediol react readily with acetone and an acid catalyst to form a ketal, but other stereoisomers do not react? In formulating your answer, draw the more stable chair conformations of all four stereoisomers and the product ketal for each one.


General Problems -
A ketal
PROBLEM Alcohols undergo an oxidation reaction to yield carbonyl compounds when treatment with $\mathrm{CrO}_{3}$.
4-62 For example, 2-tert-butylcyclohexanol gives 2-tert-butylcyclohexanone. If axial -OH groups are generally more reactive than their equatorial isomers, which do you think reacts faster, the cis isomer of 2-tert-butylcyclohexanol or the trans isomer? Explain.


CHAPTER 5 <br> Stereochemistry at Tetrahedral Centers -
FIGURE 5.1 Like the mountain whose image is reflected in a lake, many organic molecules also have mirror-image counterparts. (credit: modification of work "Crystal Lake sunrise reflection" by Sandy Horvath-Dori/Wikimedia Commons, CC BY 2.0)


CHAPTER CONTENTS - 5.1 Enantiomers and the Tetrahedral Carbon
5.2 The Reason for Handedness in Molecules: Chirality


CHAPTER CONTENTS - 5.3 Optical Activity
5.4 Pasteur's Discovery of Enantiomers
5.5 Sequence Rules for Specifying Configuration
5.6 Diastereomers
5.7 Meso Compounds


CHAPTER CONTENTS - 5.9 A Review of Isomerism
5.10 Chirality at Nitrogen, Phosphorus, and Sulfur


CHAPTER CONTENTS - 5.11 Prochirality
5.12 Chirality in Nature and Chiral Environments

WHY THIS CHAPTER? Understanding the causes and consequences of molecular handedness is crucial to understanding organic and biological chemistry. The subject can be a bit complex at first, but the material covered in this chapter nevertheless forms the basis for much of the remainder of the book.

Are you right-handed or left-handed? You may not spend much time thinking about it, but handedness plays a surprisingly large role in your daily activities. Many musical instruments, such as oboes and clarinets, have a handedness to them; the last available softball glove always fits the wrong hand. The reason for these difficulties is that our hands aren't identical; rather, they're mirror images. When you hold a left hand up to a mirror, the image you see looks like a right hand. Try it.

Left hand

Right hand

Handedness is also important in organic and biological chemistry, where it arises primarily as a consequence of the tetrahedral stereochemistry of $s p^{3}$-hybridized carbon atoms. Many drugs and almost all the molecules in our bodies-amino acids, carbohydrates, nucleic acids, and many more-have a handedness. Furthermore, molecular handedness enables the precise interactions between enzymes and their substrates that are involved in the hundreds of thousands of chemical reactions on which life is based.


CHAPTER CONTENTS - 5.1 Enantiomers and the Tetrahedral Carbon
What causes molecular handedness? Look at generalized molecules of the type $\mathrm{CH}_{3} \mathrm{X}, \mathrm{CH}_{2} \mathrm{XY}$, and CHXYZ shown in FIGURE 5.2. On the left are three molecules, and on the right are their images reflected in a mirror. The $\mathrm{CH}_{3} \mathrm{X}$ and $\mathrm{CH}_{2} \mathrm{XY}$ molecules are identical to their mirror images and thus are not handed. If you make a molecular model of each molecule and its mirror image, you find that you can superimpose one on the other so that all atoms coincide. The CHXYZ molecule, by contrast, is not identical to its mirror image. You can't superimpose a model of this molecule on a model of its mirror image for the same reason that you can't superimpose a left hand on a right hand: they simply aren't the same.

FIGURE 5.2 Tetrahedral carbon atoms and their mirror images. Molecules of the type $\mathrm{CH}_{3} \mathrm{X}$ and $\mathrm{CH}_{2} \mathrm{XY}$ are identical to their mirror images, but a molecule of the type CHXYZ is not. $\mathrm{A} C H X Y Z$ molecule is related to its mirror image in the same way a right hand is related to a left hand.


CHAPTER CONTENTS - 5.1 Enantiomers and the Tetrahedral Carbon
A molecule that is not identical to its mirror image is a kind of stereoisomer (Section 4.2) called an enantiomer (e-nan-tee-oh-mer, from the Greek enantio, meaning "opposite"). Enantiomers are related to each other as a right hand is related to a left hand and result whenever a tetrahedral carbon is bonded to four different substituents (one need not be H). For example, lactic acid (2-hydroxypropanoic acid) exists as a pair of enantiomers because there are four different groups ( $-\mathrm{H},-\mathrm{OH},-\mathrm{CH}_{3},-\mathrm{CO}_{2} \mathrm{H}$ ) bonded to the central carbon atom. The enantiomers are called ( + )-lactic acid and ( - )-lactic acid. Both are found in sour milk, but only the ( + ) enantiomer occurs in muscle tissue.


Lactic acid: a molecule of the general formula CHXYZ -
(+)-Lactic acid

No matter how hard you try, you can't superimpose a molecule of (+)-lactic acid on a molecule of (-)-lactic acid. If any two groups match up, say -H and $-\mathrm{CO}_{2} \mathrm{H}$, the remaining two groups don't match (FIGURE 5.3).

FIGURE 5.3 Attempts at superimposing the mirror-image forms of lactic acid. (a) When the -H and -OH substituents match up, the $-\mathrm{CO}_{2} \mathrm{H}$ and $-\mathrm{CH}_{3}$ substituents don't; (b) when $-\mathrm{CO}_{2} \mathrm{H}$ and $-\mathrm{CH}_{3}$ match up, -H and -OH don't. Regardless of how the molecules are oriented, they aren't identical.


Lactic acid: a molecule of the general formula CHXYZ - 5.2 The Reason for Handedness in Molecules: Chirality
A molecule that is not identical to its mirror image is said to be chiral (ky-ral, from the Greek cheir, meaning "hand"). You can't take a chiral molecule and its enantiomer and place one on the other so that all atoms coincide.

How can you predict whether a given molecule is or is not chiral? A molecule is not chiral if it has a plane of symmetry. A plane of symmetry is a plane that cuts through the middle of a molecule (or any object) in such a way that one half of the molecule or object is a mirror image of the other half. A coffee mug, for example, has a plane of symmetry. If you were to cut the coffee mug in half from top to bottom, one half would be a mirror image of the other half. A hand, however, does not have a plane of symmetry. One "half" of a hand is not a mirror image of the other half (FIGURE 5.4).

FIGURE 5.4 The meaning of symmetry plane. (a) An object like the coffee mug has a symmetry plane cutting through it so that right and left halves are mirror images. (b) An object like a hand has no symmetry plane; the right "half" of a hand is not a mirror image of the left half.

A molecule that has a plane of symmetry in any conformation must be identical to its mirror image and must be nonchiral, or achiral. Thus, propanoic acid, $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CO}_{2} \mathrm{H}$, has a plane of symmetry when lined up as shown in FIGURE 5.5 and is achiral, while lactic acid, $\mathrm{CH}_{3} \mathrm{CH}(\mathrm{OH}) \mathrm{CO}_{2} \mathrm{H}$, has no plane of symmetry in any conformation
and is chiral.

FIGURE 5.5 The achiral propanoic acid molecule versus the chiral lactic acid molecule. Propanoic acid has a plane of symmetry that makes one side of the molecule a mirror image of the other. Lactic acid has no such symmetry plane.


Lactic acid: a molecule of the general formula CHXYZ - 5.2 The Reason for Handedness in Molecules: Chirality
The most common, although not the only, cause of chirality in organic molecules is the presence of a tetrahedral carbon atom bonded to four different groups-for example, the central carbon atom in lactic acid. Such carbons are referred to as chirality centers, although other terms such as stereocenter, asymmetric center, and stereogenic center have also been used. Note that chirality is a property of the entire molecule, whereas a chirality center is the cause of chirality.

Detecting a chirality center in a complex molecule takes practice because it's not always immediately apparent whether four different groups are bonded to a given carbon. The differences don't necessarily appear right next to the chirality center. For example, 5 -bromodecane is a chiral molecule because four different groups are bonded to C 5 , the chirality center (marked with an asterisk). A butyl substituent is similar to a pentyl substituent, but it isn't identical. The difference isn't apparent until looking four carbon atoms away from the chirality center, but there's still a difference.

As other possible examples, look at methylcyclohexane and 2-methylcyclohexanone. Methylcyclohexane is achiral because no carbon atom in the molecule is bonded to four different groups. You can immediately eliminate all $-\mathrm{CH}_{2}$ - carbons and the $-\mathrm{CH}_{3}$ carbon from consideration, but what about C 1 on the ring? The C 1 carbon atom is bonded to $\mathrm{a}-\mathrm{CH}_{3}$ group, to an -H atom, and to C 2 and C 6 of the ring. Carbons 2 and 6 are equivalent, however, as are carbons 3 and 5 . Thus, the C6-C5-C4 "substituent" is equivalent to the C2-C3-C4 substituent, and methylcyclohexane is achiral. Another way of reaching the same conclusion is to realize that methylcyclohexane has a symmetry plane, which passes through the methyl group and through C1 and C4 of the ring.


Lactic acid: a molecule of the general formula CHXYZ - 5.2 The Reason for Handedness in Molecules: Chirality
The situation is different for 2-methylcyclohexanone. 2-Methylcyclohexanone has no symmetry plane and is chiral because its C 2 is bonded to four different groups: $-\mathrm{CH}_{3}$ group, an -H atom, $\mathrm{a}-\mathrm{COCH}_{2}-$ ring bond (C1), and a $-\mathrm{CH}_{2} \mathrm{CH}_{2}-$ ring bond (C3).

Several more examples of chiral molecules are shown below. Check for yourself that the labeled carbons are chirality centers. You might note that carbons in $-\mathrm{CH}_{2}-,-\mathrm{CH}_{3}, \mathrm{C}=\mathrm{O}, \mathrm{C}=\mathrm{C}$, and $\mathrm{C} \equiv \mathrm{C}$ groups can't be chirality centers. (Why not?)

Carvone (spearmint oil)

Nootkatone (grapefruit oil)


Drawing the Three-Dimensional Structure of a Chiral Molecule -
Draw the structure of a chiral alcohol.


Strategy -
An alcohol is a compound that contains the -OH functional group. To make an alcohol chiral, we need to have four different groups bonded to a single carbon atom, say $-\mathrm{H},-\mathrm{OH},-\mathrm{CH}_{3}$, and $-\mathrm{CH}_{2} \mathrm{CH}_{3}$

Solution

2-Butanol
(chiral)

PROBLEM Which of the following objects are chiral?
5-1
(a) Soda can
(b) Screwdriver
(c) Screw
(d) Shoe

PROBLEM Which of the following molecules are chiral? Identify the chirality center(s) in each.
(a)

Coniine (poison hemlock)

Menthol (flavoring agent)
(c)

Dextromethorphan (cough suppressant)

PROBLEM Alanine, an amino acid found in proteins, is chiral. Draw the two enantiomers of alanine using the 5-3 standard convention of solid, wedged, and dashed lines.

Alanine
PROBLEM Identify the chirality centers in the following molecules (gray $=\mathrm{H}$, black $=\mathrm{C}$, red $=\mathrm{O}$, green $=\mathrm{Cl}$, 5-4 yellow-green = F):

Threose
(a sugar)
(b)

Enflurane (an anesthetic)


Strategy - 5.3 Optical Activity
The study of chirality originated in the early 19th century during investigations by the French physicist JeanBaptiste Biot into the nature of plane-polarized light. A beam of ordinary light consists of electromagnetic waves that oscillate in an infinite number of planes at right angles to its direction of travel. When a beam of ordinary light passes through a device called a polarizer, however, only the light waves oscillating in a single plane pass through and the light is said to be plane-polarized. Light waves in all other planes are blocked out.

Biot made the remarkable observation that when a beam of plane-polarized light passes through a solution of certain organic molecules, such as sugar or camphor, the plane of polarization is rotated through an angle, $\alpha$. Not all organic substances exhibit this property, but those that do are said to be optically active.

The angle of rotation can be measured with an instrument called a polarimeter, represented in FIGURE 5.6. A solution of optically active organic molecules is placed in a sample tube, plane-polarized light is passed through the tube, and rotation of the polarization plane occurs. The light then goes through a second polarizer called the analyzer. By rotating the analyzer until the light passes through it, we can find the new plane of polarization and can tell to what extent rotation has occurred.

FIGURE 5.6 Schematic representation of a polarimeter. Plane-polarized light passes through a solution of optically active molecules, which rotate the plane of polarization.

In addition to determining the extent of rotation, we can also find the direction. From the vantage point of the observer looking directly at the analyzer, some optically active molecules rotate polarized light to the left (counterclockwise) and are said to be levorotatory, whereas others rotate polarized light to the right (clockwise) and are said to be dextrorotatory. By convention, rotation to the left is given a minus sign (-) and rotation to the right is given a plus sign (+). (-)-Morphine, for example, is levorotatory, and (+)-sucrose is dextrorotatory.


Strategy - 5.3 Optical Activity
The extent of rotation observed in a polarimetry experiment depends on the number of optically active molecules encountered by the light beam. This number, in turn, depends on sample concentration and sample pathlength. If the concentration of the sample is doubled, the observed rotation doubles. If the concentration is kept constant but the length of the sample tube is doubled, the observed rotation doubles. It also happens that the angle of rotation depends on the wavelength of the light used.

To express optical rotations in a meaningful way so that comparisons can be made, we have to choose standard conditions. The specific rotation, $[\boldsymbol{\alpha}]_{\mathbf{D}}$, of a compound is defined as the observed rotation when light of 589.6 nanometer ( $\mathrm{nm} ; 1 \mathrm{~nm}=10^{-9} \mathrm{~m}$ ) wavelength is used with a sample pathlength 1 of 1 decimeter ( $\mathrm{dm} ; 1 \mathrm{dm}=10$ cm ) and a sample concentration $c$ of $1 \mathrm{~g} / \mathrm{cm}^{3}$. (Light of 589.6 nm , the so-called sodium D line, is the yellow light emitted from common sodium street lamps.)

$$
[\alpha]_{\mathrm{D}}=\frac{\text { Observed rotation (degrees) }}{\text { Pathlength, } l(\mathrm{dm}) \times \text { Concentration, } c\left(\mathrm{~g} / \mathrm{cm}^{3}\right)}=\frac{\alpha}{l \times c}
$$


Strategy - 5.3 Optical Activity
When optical rotations are expressed in this standard way, the specific rotation, $[\alpha]_{\mathrm{D}}$, is a physical constant characteristic of a given optically active compound. For example, (+)-lactic acid has $[\alpha]_{D}=+3.82$, and (-)-lactic acid has $[\alpha]_{D}=-3.82$. That is, the two enantiomers rotate plane-polarized light to exactly the same extent but in opposite directions. Note that the units of specific rotation are [ $\left.\left(\mathrm{deg} \cdot \mathrm{cm}^{2}\right) / \mathrm{g}\right]$ but the values are usually expressed without units. Some additional examples are listed in TABLE 5.1.

| TABLE 5.1 Specific Rotations of Some Organic Molecules |  |  |  |
| :--- | :--- | :--- | :---: |
| Compound | $[\boldsymbol{\alpha}]_{\mathrm{D}}$ | Compound | $[\boldsymbol{\alpha}]_{\mathrm{D}}$ |
| Penicillin V | +233 | Cholesterol | -31.5 |
| Sucrose | +66.47 | Morphine | -132 |
| Camphor | +44.26 | Cocaine | -16 |
| Chloroform | 0 | Acetic acid | 0 |


Calculating an Optical Rotation -
A 1.20 g sample of cocaine, $[\alpha]_{D}=-16$, was dissolved in 7.50 mL of chloroform and placed in a sample tube having a pathlength of 5.00 cm . What was the observed rotation?

Cocaine


Strategy -
Since $[\alpha]_{\mathrm{D}}=\frac{\alpha}{l \times c}$
Then $\alpha=l \times c \times[\alpha]_{\mathrm{D}}$
where $[\alpha]_{\mathrm{D}}=-16 ; 1=5.00 \mathrm{~cm}=0.500 \mathrm{dm} ; c=1.20 \mathrm{~g} / 7.50 \mathrm{~cm}^{3}=0.160 \mathrm{~g} / \mathrm{cm}^{3}$


Solution -
$\alpha=(-16)(0.500)(0.160)=-1.3^{\circ}$.

PROBLEM Is cocaine (Worked Example 5.2) dextrorotatory or levorotatory?
5-5
PROBLEM A 1.50 g sample of coniine, the toxic extract of poison hemlock, was dissolved in 10.0 mL of ethanol
5-6 and placed in a sample cell with a 5.00 cm pathlength. The observed rotation at the sodium D line was $+1.21^{\circ}$. Calculate $[\alpha]_{D}$ for coniine.


Solution - 5.4 Pasteur's Discovery of Enantiomers
Little was done to build on Biot's discovery of optical activity until 1848, when Louis Pasteur began work on a study of crystalline tartaric acid salts derived from wine. On crystallizing a concentrated solution of sodium ammonium tartrate below $28^{\circ} \mathrm{C}$, Pasteur made the surprising observation that two distinct kinds of crystals were obtained. Furthermore, the two kinds of crystals were nonsuperimposable mirror images and were related in the same way that a right hand is related to a left hand.

Working carefully with tweezers, Pasteur was able to separate the crystals into two piles, one of "right-handed" crystals and one of "left-handed" crystals, like those shown in FIGURE 5.7. Although the original sample, a 50 : 50 mixture of right and left, was optically inactive, solutions of the crystals from each of the sorted piles were optically active and their specific rotations were equal in magnitude but opposite in sign.

Sodium ammonium tartrate
FIGURE 5.7 Drawings of sodium ammonium tartrate crystals taken from Pasteur's original sketches. One of the crystals is dextrorotatory
in solution, and the other is levorotatory.
Pasteur was far ahead of his time. Although the structural theory of Kekulé had not yet been proposed, Pasteur explained his results by speaking of the molecules themselves, saying, "There is no doubt that [in the dextro tartaric acid] there exists an asymmetric arrangement having a nonsuperimposable image. It is no less certain that the atoms of the levo acid have precisely the inverse asymmetric arrangement." Pasteur's vision was extraordinary, for it was not until 25 years later that his ideas regarding asymmetric carbon atoms were confirmed.

Today, we would describe Pasteur's work by saying that he had discovered enantiomers. Enantiomers, also called optical isomers, have identical physical properties, such as melting point and boiling point, but differ in the direction in which their solutions rotate plane-polarized light.


Solution - 5.5 Sequence Rules for Specifying Configuration
Structural drawings provide a visual representation of stereochemistry, but a written method for indicating the three-dimensional arrangement, or configuration, of substituents at a chirality center is also needed. The method used a set of sequence rules to rank the four groups attached to the chirality center and then looks at the handedness with which those groups are attached. Called the Cahn-Ingold-Prelog rules after the chemists who proposed them, the sequence rules are as follows:


RULE 1 -
Look at the four atoms directly attached to the chirality center, and rank them according to atomic number. The atom with the highest atomic number has the highest ranking (first), and the atom with the lowest atomic number (usually hydrogen) has the lowest ranking (fourth). When different isotopes of the same element are compared, such as deuterium $\left({ }^{2} \mathrm{H}\right)$ and protium $\left({ }^{1} \mathrm{H}\right)$, the heavier isotope ranks higher than the lighter isotope. Thus, atoms commonly found in organic compounds have the following order.

| Atomic number | 35 | 17 | 16 | 15 | 8 | 7 | 6 | $(2)$ | $(1)$ |
| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |
| Higher rank | $\mathrm{Br}>\mathrm{Cl}>\mathrm{S}$ | $>\mathrm{P}$ | $>\mathrm{O}$ | $>\mathrm{N}$ | $>\mathrm{C}$ | $>{ }^{2} \mathrm{H}>{ }^{1} \mathrm{H} \quad$ Lower rank |  |  |  |

RULE 2
If a decision can't be reached by ranking the first atoms in the substituent, look at the second, third, or fourth atoms away from the chirality center until the first difference is found. $\mathrm{A}-\mathrm{CH}_{2} \mathrm{CH}_{3}$ substituent and a $-\mathrm{CH}_{3}$ substituent are equivalent by rule 1 because both have carbon as the first atom. By rule 2, however, ethyl ranks higher than methyl because ethyl has a carbon as its highest second atom, while methyl has only hydrogen as its second atom. Look at the following pairs of examples to see how the rule works:

Higher

Lower

Lower

Higher

Lower

Higher


RULE 1 -
Higher

Lower

Lower

Higher

Lower

Higher

RULE 3
Multiple-bonded atoms are equivalent to the same number of single-bonded atoms. For example, an aldehyde substituent $(-\mathrm{CH}=\mathrm{O})$, which has a carbon atom doubly bonded to one oxygen, is equivalent to a substituent having a carbon atom singly bonded to two oxygens:

As further examples, the following pairs are equivalent:

Having ranked the four groups attached to a chiral carbon, we describe the stereochemical configuration around the carbon by orienting the molecule so that the group with the lowest ranking (4) points directly away from us. We then look at the three remaining substituents, which now appear to radiate toward us like the spokes on a steering wheel (FIGURE 5.8). If a curved arrow drawn from the highest to second-highest to thirdhighest ranked substituent $(1 \rightarrow 2 \rightarrow 3)$ is clockwise, we say that the chirality center has the $\boldsymbol{R}$ configuration ( $S$ for the Latin rectus, meaning "right"). If an arrow from $1 \rightarrow 2 \rightarrow 3$ is counterclockwise, the chirality center has the $\boldsymbol{S}$ configuration (Latin sinister, meaning "left"). To remember these assignments, think of a car's steering wheel when making a Right (clockwise) turn.

FIGURE 5.8 Assigning $\boldsymbol{R}$ and $\boldsymbol{S}$ configurations to chirality centers. When the molecule is oriented so that the lowest-ranked group (4) is toward the rear, the remaining three groups radiate toward the viewer like the spokes of a steering wheel. If the direction of travel $1 \rightarrow 2 \rightarrow 3$ is clockwise (right turn), the center has the $R$ configuration. If the direction of travel $1 \rightarrow 2 \rightarrow 3$ is counterclockwise (left turn), the center is $S$.


RULE 1 -
Look at (-)-lactic acid in FIGURE 5.9 for an example of how to assign configuration. Sequence rule 1 says that -OH is ranked 1 and -H is ranked 4 , but it doesn't allow us to distinguish between $-\mathrm{CH}_{3}$ and $-\mathrm{CO}_{2} \mathrm{H}$ because both groups have carbon as their first atom. Sequence rule 2, however, says that $-\mathrm{CO}_{2} \mathrm{H}$ ranks higher than $-\mathrm{CH}_{3}$ because O (the highest second atom in $-\mathrm{CO}_{2} \mathrm{H}$ ) outranks H (the highest second atom in $-\mathrm{CH}_{3}$ ). Now, turn the
molecule so that the fourth-ranked group ( -H ) is oriented toward the rear, away from the observer. Since a curved arrow from $1(-\mathrm{OH})$ to $2\left(-\mathrm{CO}_{2} \mathrm{H}\right)$ to $3\left(-\mathrm{CH}_{3}\right)$ is clockwise (right turn of the steering wheel), ( - )-lactic acid has the $R$ configuration. Applying the same procedure to (+)-lactic acid leads to the opposite assignment.
(a)

$R$ configuration
(-)-Lactic acid

(b)

$S$ configuration
(+)-Lactic acid
FIGURE 5.9 Assigning configuration to (a)(R)-(-)-lactic acid and (b) (S)-(+)-lactic acid.

Further examples are provided by naturally occurring (-)-glyceraldehyde and (+)-alanine, which both have the $S$ configuration as shown in FIGURE 5.10. Note that the sign of optical rotation, $(+)$ or $(-)$, is not related to the $R, S$ designation. (S)-Glyceraldehyde happens to be levorotatory (-), and (S)-alanine happens to be dextrorotatory $(+)$. There is no simple correlation between $R, S$ configuration and direction or magnitude of optical rotation.
(a)

(S)-Glyceraldehyde
[(S)-(-)-2,3-Dihydroxypropanal]
$[\alpha]_{D}=-8.7$
(b)


RULE 1 -
(S)-Glyceraldehyde
[(S)-(-)-2,3-Dihydroxypropanal]
$[\alpha]_{D}=-8.7$
(b)

(S)-Alanine
[(S)-(+)-2-Aminopropanoic acid]
$[\alpha]_{D}=+8.5$

FIGURE 5.10 Assigning configuration to (a) (-)-glyceraldehyde. (b) (+)-alanine. Both happen to have the Sconfiguration, although one is levorotatory and the other is dextrorotatory.

One additional point needs to be mentioned-the matter of absolute configuration. How do we know that the assignments of $R$ and $S$ configuration are correct in an absolute sense, rather than a relative, sense? Since
there is no correlation between the $R, S$ configuration and the direction or magnitude of optical rotation, how do we know that the $R$ configuration belongs to the levorotatory enantiomer of lactic acid? This difficult question was finally solved in 1951, when an X-ray diffraction method was found for determining the absolute spatial arrangement of atoms in a molecule. Based on those results, we can say with certainty that the $R, S$ conventions are correct.


Assigning Configuration to Chirality Centers -
Orient each of the following drawings so that the lowest-ranked group is toward the rear, and then assign $R$ or $S$ configuration to each:
(a)

(b)


Strategy -
It takes practice to be able to visualize and orient a chirality center in three dimensions. You might start by indicating where the observer must be located $-180^{\circ}$ opposite the lowest-ranked group. Then imagine yourself in the position of the observer, and redraw what you would see.


Solution -
In (a), you would be located in front of the page toward the top right of the molecule, and you would see group 2 to your left, group 3 to your right, and group 1 below you. This corresponds to an $R$ configuration.

In (b), you would be located behind the page toward the top left of the molecule from your point of view, and you would see group 3 to your left, group 1 to your right, and group 2 below you. This also corresponds to an $R$ configuration.


Drawing the Three-Dimensional Structure of a Specific Enantiomer -
Draw a tetrahedral representation of ( $R$ )-2-chlorobutane.


Strategy -
Begin by ranking the four substituents bonded to the chirality center: (1) -Cl , (2) $-\mathrm{CH}_{2} \mathrm{CH}_{3}$, (3) $-\mathrm{CH}_{3}$, (4) -H . To draw a tetrahedral representation of the molecule, orient the lowest-ranked group ( -H ) away from you and imagine that the other three groups are coming out of the page toward you. Then, place the remaining three substituents such that the direction of travel $1 \rightarrow 2 \rightarrow 3$ is clockwise (right turn), and tilt the molecule toward you to bring the rear hydrogen into view. Using molecular models is a real help in working problems of this sort.


Solution -
(R)-2-Chlorobutane

PROBLEM Which member in each of the following sets ranks higher?
5-7 (a) -H or -Br
(b) -Cl or -Br
(c) $-\mathrm{CH}_{3}$ or $-\mathrm{CH}_{2} \mathrm{CH}_{3}$
(d) $-\mathrm{NH}_{2}$ or -OH
(e) $-\mathrm{CH}_{2} \mathrm{OH}$ or $-\mathrm{CH}_{3}$
(f) $-\mathrm{CH}_{2} \mathrm{OH}$ or $-\mathrm{CH}=\mathrm{O}$

PROBLEM Rank each of the following sets of substituents:
5-8 (a) $-\mathrm{H},-\mathrm{OH},-\mathrm{CH}_{2} \mathrm{CH}_{3},-\mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH}$
(b) $-\mathrm{CO}_{2} \mathrm{H},-\mathrm{CO}_{2} \mathrm{CH}_{3},-\mathrm{CH}_{2} \mathrm{OH},-\mathrm{OH}$
(c) $-\mathrm{CN},-\mathrm{CH}_{2} \mathrm{NH}_{2},-\mathrm{CH}_{2} \mathrm{NHCH}_{3},-\mathrm{NH}_{2}$ (d) $-\mathrm{SH},-\mathrm{CH}_{2} \mathrm{SCH}_{3},-\mathrm{CH}_{3},-\mathrm{SSCH}_{3}$

PROBLEM Orient each of the following drawings so that the lowest-ranked group is toward the rear, and then
5-9 assign $R$ or $S$ configuration:
(a)

(b)

(c)

PROBLEM Assign $R$ or $S$ configuration to the chirality center in each of the following molecules:
5-10 (a)

(b)

(c)

PROBLEM Draw a tetrahedral representation of ( $S$ )-2-pentanol (2-hydroxypentane).
5-11
PROBLEM Assign $R$ or $S$ configuration to the chirality center in the following molecular model of the amino 5-12 acid methionine (blue $=\mathrm{N}$, yellow $=\mathrm{S}$ ):


Solution - 5.6 Diastereomers
Molecules like lactic acid, alanine, and glyceraldehyde are relatively simple because each has only one chirality center and thus only two stereoisomers. The situation becomes more complex, however, with molecules that have more than one chirality center. As a general rule, a molecule with $n$ chirality centers can have up to
$2^{n}$ stereoisomers (although it may have fewer, as we'll see below). Take the amino acid threonine (2-amino-3-hydroxybutanoic acid), for example. Since threonine has two chirality centers (C2 and C3), there are four possible stereoisomers, as shown in FIGURE 5.11. Check for yourself that the $R, S$ configurations of all stereoisomers are correct.
Enantiomers

Enantiomers

FIGURE 5.11 The four stereoisomers of 2-amino-3-hydroxybutanoic acid.
The four stereoisomers of 2-amino-3-hydroxybutanoic acid can be grouped into two pairs of enantiomers. The $2 R, 3 R$ stereoisomer is the mirror image of $2 S, 3 S$, and the $2 R, 3 S$ stereoisomer is the mirror image of $2 S, 3 R$. But what is the relationship between any two molecules that are not mirror images? What, for instance, is the relationship between the $2 R, 3 R$ isomer and the $2 R, 3 S$ isomer? They are stereoisomers, yet they aren't enantiomers. To describe such a relationship, we need a new term-diastereomer.

Diastereomers (dia-stair-e-oh-mers) are stereoisomers that are not mirror images. Since we used the right-hand/left-hand analogy to describe the relationship between two enantiomers, we might extend the analogy by saying that the relationship between diastereomers is like that of hands from different people. Your hand and your friend's hand look similar, but they aren't identical and they aren't mirror images. The same is true of diastereomers: they're similar, but they aren't identical and they aren't mirror images.