Section 5.19 Stereochemistry of Electrophilic Addition Reactions of Alkenes 229
d.
Two asymmetric carbons have been created in the product. Because the reactant is cis
and addition of is anti, the threo enantiomers are formed.
e.
Two asymmetric carbons have been created in the product. Because the reactant is trans
and addition of is anti, the erythro enantiomers are formed.
f.
Two asymmetric carbons have been created in the product. Because the reactant is trans
and addition of is anti, one would expect the erythro enantiomers. However, the two
asymmetric carbons are bonded to the same four groups, so the erythro product is a
meso compound. Thus, only one stereoisomer is formed.
Now continue on to Problem 47.
PROBLEM 47
Give the configuration of the products obtained from the following reactions:
a. d.
b. e.
c. f.
PROBLEM 48
When adds to an alkene with different substituents on each of the two carbons,
such as cis-2-heptene, identical amounts of the two threo enantiomers are obtained even
though is more likely to attack the less sterically hindered carbon atom of the bromo-
nium ion. Explain why identical amounts of the stereoisomers are obtained.
Br-
Br 2 sp^2
(Z)-3-methyl-2-pentene+HBr+peroxide 1-hexene+Br 2
(Z)-3-methyl-2-pentene+HBr cis-2-pentene+Br 2
trans-2-butene+HBr+peroxide cis-3-hexene+HBr
CH 2 CH 3
HBr
HBr
CH 2 CH 3
CH 3 CH 2 CH 2 CH 3
H
C C
Br
H
Br
or
Br 2
CH 3 CH 2 CHCHCH 2 CH 3
BrBr
CH 3 CH 2 CH 2 CH 2 CH 3
H
C C
Br
CH 3 CH 2 CH 2 CH 2 CH 3
H
C C
Br
H
Br
H
Br
or
CH 2 CH 3
HBr
HBr
CH 2 CH 2 CH 3
CH 2 CH 3
Br H
Br H
CH 2 CH 2 CH 3
Br 2
CH 3 CH 2 CHCHCH 2 CH 2 CH 3
BrBr
CH 3 CH 2 CH 2 CH 2 CH 3
H
H
C C
Br
Br
CH 3 CH 2 CH 2 CH 2 CH 3
H
H
C C
Br
Br
or
CH 2 CH 3
HBr
Br H
CH 2 CH 2 CH 3
CH 2 CH 3
Br H
HBr
CH 2 CH 2 CH 3
Br 2
CH 3 CH 2 CHCHCH 2 CH 2 CH 3
BrBr