Organic Chemistry

204 CHAPTER 5 Stereochemistry

SOLUTION To be chiral, a molecule must not have plane of symmetry. Therefore, only

the following compounds are chiral.

In the top row of compounds, only the third compound is chiral. The first, second, and

fourth compounds each have a plane of symmetry. In the bottom row of compounds, the

first and third compounds are chiral. The second and fourth compounds each have a plane

of symmetry.

PROBLEM 28

Draw all the stereoisomers for each of the following compounds:

a. 1-bromo-2-methylbutane h. 2,4-dichloropentane

b. 1-chloro-3-methylpentane i. 2,4-dichloroheptane

c. 2-methyl-1-propanol j. 1,2-dichlorocyclobutane

d. 2-bromo-1-butanol k. 1,3-dichlorocyclohexane

e. 3-chloro-3-methylpentane l. 1,4-dichlorocyclohexane

f. 3-bromo-2-butanol m. 1-bromo-2-chlorocyclobutane

g. 3,4-dichlorohexane n. 1-bromo-3-chlorocyclobutane

5.11 The R,SSystem of Nomenclature for Isomers

with More than One Asymmetric Carbon

If a compound has more than one asymmetric carbon, the steps used to determine

whether an asymmetric carbon has the Ror the Sconfiguration must be applied to each

of the asymmetric carbons individually. As an example, let’s name one of the

stereoisomers of 3-bromo-2-butanol.

First, we will determine the configuration at C-2. The OH group has the highest prior-

ity, the C-3 carbon (the C attached to Br, C, H) has the next highest priority, is

next, and H has the lowest priority. Because the group with the lowest priority is bond-

ed by a hatched wedge, we can immediately draw an arrow from the group with the

highest priority to the group with the next highest priority. Because that arrow points

counterclockwise, the configuration at C-2 is S.

Now we need to determine the configuration at C-3. Because the group with the low-

est priority (H) is not bonded by a hatched wedge, we must put it there by temporarily

switching two groups.

H 3 C

H H

HO

Br

CH 3

C C

S

1

4

2

3

CH 3

a stereoisomer of 3-bromo-2-butanol

H 3 C

H H

HO

Br

CH 3

C C

H 3 C CH 3 CH 3

H 3 C Cl

Cl CH 3