Organic Chemistry

166 CHAPTER 4 Reactions of Alkenes

Mechanistic Tutorial:

Hydroboration–oxidation

Synthetic Tutorial:

Hydroboration–oxidation

The mechanism of the oxidation reaction shows that a hydroperoxide ion (a Lewis

base) reacts with (a Lewis acid). Then, a 1,2-alkyl shift displaces a hydroxide ion.

These two steps are repeated two more times. Then, hydroxide ion (a Lewis base) re-

acts with (a Lewis acid), and an alkoxide ion is eliminated. Protonation of the

alkoxide ion forms the alcohol. These three steps are repeated two more times.

We have seen that, in the overall hydroboration–oxidation reaction, 1 mole of

reacts with 3 moles of alkene to form 3 moles of alcohol. The OH ends up on the

carbon that was bonded to the greater number of hydrogens because it replaces

boron, which was the original electrophile in the reaction.

Because carbocation intermediates are not formed in the hydroboration reaction, car-

bocation rearrangements do not occur.

PROBLEM 23

How many moles of are needed to react with 2 moles of 1-pentene?

PROBLEM 24

What product would be obtained from hydroboration–oxidation of the following alkenes?

a. 2-methyl-2-butene b. 1-methylcyclohexene

BH 3

CH 3 CHCH

3-methyl-1-butene 3-methyl-1-butanol

CH 2 CH 3 CHCH 2 CH 2 OH

1. BH 3 /THF


2. HO−, H 2 O 2 , H 2 O

CH 3 CH 3

CH 3 CCH

3,3-dimethyl-1-butene 3,3-dimethyl-1-butanol

CH 2 CH 3 CCH 2 CH 2 OH

1. BH 3 /THF


2. HO−, H 2 O 2 , H 2 O

CH 3

CH 3

CH 3

CH 3

sp^2

BH 3

(RO) 3 B

R 3 B

3 CH 3 CH CH 2 + BH 3 (CH 3 CH 2 CH 2 ) 3 B3 CH 3 CH 2 CH 2 OH + BO

THF HO−, H 2 O 2

H 2 O^3

3 −

+ −

R

ROHB

R

repeat the two

preceding steps

two times

repeat the three

preceding steps

two times

O −

R

ROHB

R

O

R

RB

+ HO−

OR

OR

RO B

−OH

OR

OR

OH

RO B− OR

+

OH

OR

RO− RO B

O−

OR

3 ROH + BO 33 − ROH + RO B