Organic Chemistry

796 CHAPTER 19 Carbonyl Compounds III

In the first step of this base-promoted reaction, hydroxide ion removes a proton

from the The enolate ion then reacts with the electrophilic bromine. These

two steps are repeated until all the are replaced by bromine.

Each successive halogenation is more rapidthan the previous one because the electron-

withdrawing bromine increases the acidity of the remaining This is why

allthe are replaced by bromines. Under acidic conditions, on the other

hand, each successive halogenation is slowerthan the previous one because the electron-

withdrawing bromine decreases the basicity of the carbonyl oxygen, thereby making

protonation of the carbonyl oxygen less favorable.

The Haloform Reaction

In the presence of excess base and excess halogen, a methyl ketone is first converted

into a trihalo-substituted ketone. Then hydroxide ion attacks the carbonyl carbon of the

trihalo-substituted ketone. Because the trihalomethyl ion is a weaker base than hy-

droxide ion (the of is 14; the of is 15.7), the trihalomethyl ion is the

group more easily expelled from the tetrahedral intermediate, so the final product is a

carboxylic acid. The conversion of a methyl ketone to a carboxylic acid is called

a haloform reactionbecause one of the products is haloform— (chloroform),

(bromoform), or (iodoform). Before spectroscopy became a routine ana-

lytical tool, the haloform reaction served as a test for methyl ketones. The presence of a

methyl ketone was indicated by the formation of iodoform, a bright yellow compound.

PROBLEM 7

Why do only methyl ketones undergo the haloform reaction?

PROBLEM 8

A ketone undergoes acid-catalyzed bromination, acid-catalyzed chlorination, and acid-cat-

alyzed deuterium exchange at the all at about the same rate. What does this tell

you about the mechanism of the reactions?

19.5 Halogenation of the -Carbon of Carboxylic

Acids: The Hell–Volhard–Zelinski Reaction

Carboxylic acids do not undergo substitution reactions at the because a base will

remove a proton from the OH group rather than from the since the OH group is

more acidic. If, however, a carboxylic acid is treated with and then the

can be brominated. (Red phosphorus can be used in place of since P and excess

react to form PBr 3 .) This halogenation reaction is called the Hell–Volhard–Zelinski

PBr 3 , Br 2

PBr 3 Br 2 , a-carbon

a-carbon,

a-carbon

A

a-carbon,

CHBr 3 CHI 3

CHCl 3

pKa CHI 3 pKa H 2 O

a-hydrogens

a-hydrogens.

a-hydrogens

a-carbon.

RC

O

CH 3 RC

O

CI 3 RC

O

OH + −CI 3 R

HO

C

O

RC O− + CHI 3

O

OH

CI 3

− −

HO−

I 2

the haloform reaction

a trihalo-

substituted

ketone

excess

excess

RCH C

O

R

H Br Br

HO− + H 2 O + Br−

RCH C

O

R

Br

RC C

O

R

Br

repeat first Br

two steps

base-promoted halogenation

RCH C

O−

CR R

O

RCH

−