Organic Chemistry

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228 CHAPTER 5 Stereochemistry


Table 5.2 Stereochemistry of Alkene Addition Reactions

Reaction Type of addition Stereoisomers formed

Addition reactions that create one 1. If the reactant does not have an asymmetric
asymmetric carbon in the product carbon, a pair of enantiomers will be obtained
(equal amounts of the Rand Sisomers).


  1. If the reactant has an asymmetric carbon, unequal
    amounts of a pair of diastereomers will be obtained.
    Addition reactions that create two
    asymmetric carbons in the product


Addition of reagents that form a syn and anti Four stereoisomers can be obtaineda
carbocation or radical intermediate (the cis and trans isomers form the same products)

Addition of syn cis erythro enantiomersa
Addition of borane trans threo enantiomers

Addition of anti cis threo enantiomers
trans erythro enantiomersa
aIf the two asymmetric carbons have the same substituents, a meso compound will be obtained instead of the pair of erythro enantiomers.

¡

Br 2 ¡

¡

H 2 ¡

Start by drawing the product without regard to its configuration, to check whether the
reaction has created any asymmetric carbons. Then determine the configuration of the
products, paying attention to the configuration (if any) of the reactant, how many asym-
metric carbons are formed, and the mechanism of the reaction. Let’s start with a.

a.

The product has one asymmetric carbon, so equal amounts of the Rand Senantiomers
will be formed.

b.
The product does not have an asymmetric carbon, so it has no stereoisomers.

c.

Two asymmetric carbons have been created in the product. Because the reaction forms
a radical intermediate, two pairs of enantiomers are formed.

CH 3 CH 2 CHCHCH 2 CH 3

CH 3

Br

CH 3 CH 2 CH 2 CH 2 Br

C
CH 3 CH 2

Cl

H
CH 3

C
CH 2 CH 3

Cl

H
CH 3

CH 3 CH 2 CHCH 3

Cl

Br
H

C C
CH 2 CH 3

CH 3 CH 2
CH 3

H

H
Br

C C
CH 2 CH 3

CH 3 CH 2
CH 3

H

Br

CH 3 CC
CH 3 CH 2

H CH 2 CH 3

H

H
Br

CC
CH 3 CH 2

CH 2 CH 3
CH 3

H

or

CH 2 CH 3
H
H CH 3

Br

CH 2 CH 3

CH 2 CH 3

CH 3

Br H
H
CH 2 CH 3

CH 2 CH 3

CH 3

H
H

Br

CH 2 CH 3

CH 2 CH 3
H
Br

CH 3
H
CH 2 CH 3
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