Jones M., Fleming S.A. Organic Chemistry

19.4 Halogenation in the Position 949

␣

In NaOH/H

2

O,these trihalocarbonyl compounds react further because hydrox-

ide addition can lead to elimination of the stabilized trihalocarbanion. After these

addition–elimination steps, the carbanion can gain a proton from solvent or the

carboxylic acid to form a molecule of a haloform, which is a common name for a

trihalomethane (Fig. 19.38). Under these basic conditions the carboxylic acid will

remain deprotonated.

C

R

O

..

CI

3

C

R

O

–

..

–

HO

..

HO

..

–

OH

..

CI

3

CI

3

HCI

3

+

C

R

deprotonation

addition elimination

O

..

C

R

O

..

O

..

Tetrahedral

intermediate

Iodoform

(a yellow solid)

FIGURE 19.38 Addition of hydroxide to the carbonyl group leads to a tetrahedral intermediate that can lose

triiodomethide anion to generate the carboxylic acid.Transfer of a proton completes the reaction.

I

3

C I

3

C

HH H

pK

a

~ 14 pK

a

= 15.7

–

O

..

OH

..

+

FIGURE 19.39 The pK

a

of iodoform

is about 14. Iodoform is a relatively

strong acid, and the loss of



CI

3

as

a leaving group is a reasonable step.

Essentially all nucleophilic reagents add to carbonyl groups, often in a reversible

fashion. In this case, the addition reaction is strongly favored by the halogen substi-

tution (p. 780)! So, the addition of hydroxide to the carbonyl looks like a reaction

that is almost certain to occur. Once the addition reaction has taken place, there is

an opportunity to generate the acid and the haloform in an elimination step if the

triiodomethyl anion can be lost as the carbonyl group re-forms. Protonation of the

carbanion completes the reaction.

Is this mechanism reasonable? Is the triiodomethyl anion a good enough leav-

ing group to make this step a sensible one? The iodines are strongly electron with-

drawing, and that property will stabilize the anion. As a check we might look up

the pK

a

of iodoform to see how easily the molecule is deprotonated to form the anion

(Fig. 19.39). The value is about 14, and so iodoform is a relatively strong acid, at

least compared to water (pK

a

 15.7).

PROBLEM 19.9 Write out the mechanism for the reaction between acetophenone

and excess chlorine in aqueous sodium hydroxide.

PROBLEM 19.10 If we measure the rates of three reactions of the ketone shown on the

next page, exchange of the α hydrogen for deuterium in D

2

O/DO



, racemization in

H

2

O/HO



, and α bromination using Br

2

/H

2

O/HO



, we find that they are identical.

(continued)

So the loss of



CI

3

in Figure 19.38 does look reasonable. In fact, the iodoform

reaction serves as a diagnostic test for methyl ketones. Formation of iodoform, a

yellow solid, is a positive test for a molecule containing a methyl group attached to

a carbonyl carbon.

950 CHAPTER 19 Carbonyl Chemistry 2: Reactions at the Position␣

19.4b Halogenation of Carboxylic Acids Halogenation can occur at the

α position of carboxylic acids. Treatment of carboxylic acids with Br

2

and PBr

3

,

or the equivalent, a mixture of phosphorus and bromine, leads ultimately to

formation of the α-bromo acid (Fig. 19.40). The process is known as the

Hell–Volhard–Zelinsky (HVZ) reaction after Carl M. Hell (1849–1926), Jacob

35 ⬚C

D

2

O/NaOD

Br

2

NaOH/H

2

O

H

2

O/NaOH

C

*

* Means optically active

O

H

3

C

CH

2

CH

3

H

3

C

CH

2

CH

3

H

3

C

CH

2

CH

3

D

H

3

C

H

C

Deuterium exchange

O

Br

C

α-Bromination

O

A

B

H

C

CH

2

CH

3

C

Racemization

O

C

dioxane

35 ⬚C

..

Br

O

COOH

P/Br

2

100 ⬚C

Br

(~80%)

..

A SPECIFIC EXAMPLE

..

H

2

O

CH

2

RC

O

P

Br

2

OH

..

OH

..

CH

Br

..

RC

O

..

CH

Br

..

RC

O

THE GENERAL CASE

FIGURE 19.40 The

Hell–Volhard–Zelinsky reaction.

How can the rates of three such different reactions be the same? Explain, using

an Energy versus Reaction progress diagram.

19.4 Halogenation in the Position 951

␣

Volhard (1834–1910), and Nicolai D. Zelinsky (1861–1953). The first step is for-

mation of the acid bromide through reaction with PBr

3

. The acid bromide is in

equilibrium with its enol form (Fig. 19.41). Bromination of the enol form with

Br

2

gives an isolable α-bromo acid bromide.

..

CH

2

RC

O

PBr

3

Br

2

OH

..

CH

2

Br

RC

Acid bromide

O

..

CH

Br

..

Br

RC

O

..

CH

Br

RC

Enol form

α-Bromo acid bromide

OH

FIGURE 19.41 An intermediate in the Hell–Volhard–Zelinsky reaction is the α-bromo acid bromide.

This compound can be isolated if a full equivalent of PBr

3

is used in the reaction.

..

CH

..

Br

CH

2

C

O

..

O

Cl

..

CH

..

Br

CH

2

C

Cl

..

CH

..

Br

CH

2

C OCH

2

CH

3

..

OCH

2

CH

3

..

O

..

O

CH

..

Br

CH

2

C

(>90%)

CH

3

CH

2

OH

(cooling)

A SPECIFIC EXAMPLE

..

H

2

O

..

CH

..

Br

..

Br

R

C

O

..

O

..

CH

..

Br

HOR

OR

..

CH

..

Br

R

C

OH

CH

..

Br

NH

2

NH

3

..

O

R

C

..

O

R

C

THE GENERAL REACTIONS

FIGURE 19.42 Compounds available

from reactions of an α-bromo acid

halide.

PROBLEM 19.11 Write a general mechanism that accounts for the reactions of

Figure 19.42.

The α-bromo acid halide reacts like any other acid halide, and it can be used to

generate the α-bromo acid, ester, and amide, for example (Fig. 19.42).

..

CH

2

Br

2

/PBr

3

RC

O

OH

..

OH

..

CH

..

Br

C

O

R

..

CH

2

Br

2

catalytic PBr

3

(see Fig. 19.41)

RC

O

OH

..

Br

..

CH

..

Br

C

O

R

Anhydride

–

..

Br

addition

Br

2

R

..

O

..

O

..

O

C C

..

CH

2

R

CH

..

Br

+

CH

2

R

HO

..

O

C

R

elimination

..

–

..

Br

..

O

..

O

CH

2

R

CH

..

Br

C

..

O

..

Br

C

..

O

R

–

..

O

CH

..

Br

C

CH

2

R

(a)

(b)

FIGURE 19.43 (a) The overall

reaction of a carboxylic acid using a

catalytic amount of PBr

3

. (b) The

complex mechanism of the HVZ

reaction using a catalytic amount of

PBr

3

.

The HVZ reaction is limited to the formation of α-bromo compounds and is,

in truth, sometimes awkward to carry out. The reagents, bromine and phosphorus,

are noxious; reaction times are often long,and reaction conditions are harsh.For these

reasons, methods have evolved to extend and replace the classic HVZ process. For

example, David Harpp (b. 1937) and his colleagues at McGill University use the

reaction of acid chlorides and N-bromosuccinimide (NBS, p. 613), in the presence

of a catalytic amount of HBr, to form α-bromo acid chlorides conveniently and in

excellent yields (Fig. 19.44).

952 CHAPTER 19 Carbonyl Chemistry 2: Reactions at the Position␣

If a catalytic amount of PBr

3

is used, and not the full equivalent as shown in the

reaction in Figure 19.41, the product of the reaction is the α-bromo acid, not the

α-bromo acid bromide (Fig. 19.43a). The α-bromo acid bromide is still an inter-

mediate, but only a small amount of it can be made at any one time, because its for-

mation depends on PBr

3

. The small amount of α-bromo acid bromide produced is

attacked by a molecule of the acid in an anhydride-forming reaction (Fig. 19.43b).

The anhydride undergoes addition–elimination by bromide to give the carboxylate

anion and a new molecule of α-bromo acid bromide that recycles. A final hydrolysis

gives the α-bromo acid itself.This reaction surely involves a complicated mechanism.

OH

Br

1. SOCl

2

65 ⬚C

2. NBS, HBr

85 ⬚C

N BrNBS =

O

Not isolated (75%)

O

Cl

O

Cl

O

FIGURE 19.44 The Harpp modification of the HVZ reaction.

19.4 Halogenation in the Position 953

␣

Cl

O

H

Br N

O

HO

+

O

Cl

O

Cl

+

O

Br

Cl

H

O

H

Cl

N Br

H Br

N

–

Br

–

Br

O

OH

O

H

..

FIGURE 19.45 The mechanism of the Harpp modification of the HVZ reaction.

The mechanism of Harpp’s modification is an ionic one, and uses many of the

steps we have seen already. The HBr catalyst protonates the carbonyl oxygen of

the acid halide (Fig. 19.45). Then NBS deprotonates the α carbon to form an enol.

Cl

..

Nu

KCN

K

2

CO

3

/H

2

O

95–100 ⬚C

1. NH

3

25 ⬚C, 1 week

2. H

2

O/H

3

O

+

S

N

2

..

–

..

–

Br

COOH

..

NH

2

COOH

CH

X

= Br or Cl

R

X

CH

2

COOH CH

2

COOH

(~90%)

An

α-amino

acid, valine

(48%)

CN

COOH

CH

R

Nu

COOH

+ X

THE GENERAL CASE

SPECIFIC EXAMPLES

FIGURE 19.46 Some reactions of α-halo acids.

The enol now functions as a base to remove a Br from NBS. Finally deproto-

nation of the protonated carbonyl gives the product and a new molecule of acid

catalyst.

Like the α-bromo aldehydes and ketones, the α-bromo acids are very reactive

in displacement reactions and therefore serve as sources of many α-substituted

carboxylic acids (Fig. 19.46). A particularly important example is the formation of

α-amino acids through reaction with ammonia.

THE GENERAL CASE

A SPECIFIC EXAMPLE

1. LDA

THF

2. Br

2

OCH

3

O

OCH

3

Br

(87%)

O

OR

1. Strong base

2. Br

2

R

C

CH

2

O

OR

C

O

R

Br

CH

FIGURE 19.47 Formation of an α-bromo ester.

954 CHAPTER 19 Carbonyl Chemistry 2: Reactions at the Position␣

CH

2

–

..

C

R

O

..

CH

2

C

The methylated

enolate, the

product ketone

Enolate

R

O

..

CH

2

CH

3

CH

3

+

S

N

2

–

..

C

R

O

..

I

..

–

I

..

FIGURE 19.48 If the enolate could

act as a nucleophile in the S

N

2

reaction, we would have a way of

alkylating at the α position, thus

forming a new carbon–carbon bond.

Summary

We have now discussed two reactions that take place at the α position of car-

bonyl compounds.We will use these as prototypes on which to base further, more

complicated processes. The enolate can react with an electrophilic source of H



or the electrophile X

2

.The unifying theme is formation of an enolate ion (in base)

or an enol (in acid) that can act as a nucleophile. With what other electrophiles

might such a nucleophile react?

We can see some potential difficulties and limitations right away. Alkylation of

an enolate is an S

N

2 reaction,and that means we cannot use a tertiary halide, a species

too hindered for participation in the S

N

2 reaction.

19.4c Halogenation of Esters In theory, one should be able to halogenate an

ester in the α position.However, there are very few examples of this reaction. Here’s

one (Fig. 19.47).

19.5 Alkylation in the ␣ Position

In the previous sections,we saw several examples of enolate anions acting as nucleo-

philes. A logical extension would be to attempt to use enolates as nucleophiles in

S

N

2 displacements. If we could do this reaction,we would have a way to alkylate the

α position of carbonyl compounds,and a new and most useful carbon–carbon bond-

forming reaction would result (Fig. 19.48).

It turns out that most enolates are better nucleophiles at carbon. So, alkylation

takes place faster at carbon than at oxygen (Fig.19.50). One explanation for this fact

is that the highest occupied molecular orbital of the nucleophile has more electron

density at the carbon than at oxygen. Selectivity might also be a result of counter

ion (Li



, for example) coordination between the enolate and the electrophile. If the

counterion is complexed with the site of overall higher electron density, the oxygen

in this case, it sterically hinders alkylation at that position.

19.5 Alkylation in the Position 955

␣

CH

2

–

..

C

R

–

..

O

..

CH

2

C

R

O

..

R

X

CH

2

C

C-Alkylation

O-Alkylation

R

O

..

O

..

CH

2

C

R

RX

FIGURE 19.49 In principle, alkylation

of the enolate could take place at

either carbon or oxygen.

(a) (b)

FIGURE 19.50 (a) The enolate of acetone. (b) The highest occupied molecular orbital of the

enolate of acetone is represented. Note the greater contribution of electron density on the

carbon of the enolate. In practice, alkylation generally takes place at carbon because of the

greater electron density of the orbital of the nucleophile involved in the reaction.

19.5a Alkylation of Ketone and Aldehyde Enolates Even though

alkylation at oxygen is not common, there are other problems with the alkylation

reaction of simple ketones using hydroxide base. Consider the alkylation of

2-methylcyclohexanone in H

2

O/HO



. Both α positions are active, and two enolates

The resonance formulation of enolate anions clearly shows that the negative

charge is shared between an oxygen and a carbon atom (Fig. 19.49). At which atom

will alkylation be faster? If there is little or no selectivity in the alkylation reaction,

it will surely be of limited use.

..

–

..

–

O

..

O

..

H

Two enolates

H

3

C

O

..

H

3

C

H

3

C

H

3

C

H

..

–

O

..

–

O

..

H

3

C

Both α positions

are active

+

H

FIGURE 19.51 For many ketones there

are at least two possible enolates, and

therefore mixtures are obtained in the

alkylation reaction.

H

3

C

H

3

C

H

3

C

H

3

C

H

3

C

H

3

C

..

O

1. (Ph)

3

C K

+

2. CH

3

I

H

3

C

O

(9%) (41%) (21%)(6%)

O

H

3

C

CH

3

CH

3

OO

+

++

–

CH

3

CH

3

FIGURE 19.52 For ketones, there can

often be more than one alkylation.

Mixtures of products can be formed.

LDA

–78 °C

+

O

..

CH

2

CH

3

S

N

2

H

C

O

..

C

CH

3

RX

X

H

C

..

–

..

–

(–)

–

=

–

O

..

CH

3

H

R

C

O

..

C

CH

3

H

C

..

O

..

C

CH

3

H

C

–

..

Li

+

N

FIGURE 19.53 Strong base (LDA)

and low temperatures are effective at

forming enolates.

PROBLEM 19.12 LDA is a nonnucleophilic base. It does not add to the carbonyl

carbon. Explain why LDA is a poor nucleophile.

956 CHAPTER 19 Carbonyl Chemistry 2: Reactions at the Position␣

CONVENTION ALERT

will be produced (Fig. 19.51). Moreover, once alkylated, the compounds can undergo

further alkylation. Reactions of the initial products compete with the desired single

alkylation reaction. Mixtures of products can be formed, and that is not a very useful

situation (Fig. 19.52). Most aldehydes encounter similar problems.

However, if a strong, poorly nucleophilic base such as LDA is used to form the

enolate at very low temperatures (Fig. 19.53), then one can usually control the reac-

tion.LDA is strong enough to convert all available carbonyl compound into an eno-

late.This technique has become a useful tool for the synthesis of organic molecules.

Note again the use in Figure 19.53 of the shorthand notation for resonance

forms. Instead of always drawing out each important resonance form, one summa-

ry form is drawn. The positions sharing the charge or electron are indicated with

parentheses, (), (), or ( ).

.

19.5 Alkylation in the Position 957

␣

C

O

..

LDA

Anion

1. NaH, 55 ⬚C, THF

2. LDA, 10 –35 ⬚C

heptane

0–35 ⬚C

1.

2. H

2

O

(63%)

CH

2

R

H

O

OH

..

O

Ph

Br

O

..

C

O

..

CH

2

R

..

–

Li

+

Li

+

–

Li

+

Na

+

O

..

–

Na

+

C

O

..

CH

Br

2

R

–

C

O

..

CH

R

–

C

O

..

CH

R

–

N

LDA

Dianion

α-bromination

α-alkylation

(R primary)

R I

Ph

R

Br

THE GENERAL CASE

A SPECIFIC EXAMPLE

FIGURE 19.54 Dianions of carboxylic

acids can function as enolates.They

can be alkylated in the α position

with primary or other reactive

halides.

PROBLEM 19.13 Draw the resonance forms for the dianion formed from butanoic

acid. Why do you suppose the electrophile adds to the α carbon?

PROBLEM 19.14 Provide mechanisms for the following reactions that show related

electrophiles that can react at the α position of carboxylic acids:

O

COOH

O

OH

COOHHO

1. 2 equiv. LDA

2.

3. H

2

O/H

3

O

+

1. 2 equiv. LDA

2.

3. H

2

O/H

3

O

+

..

HO

..

19.5b Alkylation of Carboxylic Acids Carboxylic acids can also be alkylated, but

two equivalents of strong base are necessary, because this reaction proceeds through a

dianion.The first equivalent of base removes the carboxyl hydrogen to give the carboxy-

late salt. If the base used is a strong base and a weak nucleophile, a second hydrogen can

be removed, this time from the α position.The typical base used is LDA. The dianion

can now be brominated or alkylated at the α position, as long as the alkylating agent is

reactive (Fig. 19.54). After the alkylation step the resulting carboxylate anion can be

reprotonated to obtain the neutral product. Primary halides work well in these S

N

2

alkylation reactions, but more substituted halides lead mostly to the products of an E2

reaction.Steric hindrance around the carbon–halogen antibond blocks the S

N

2 path.As

a result, the enolate reacts as a base to give the alkene product in the E2 elimination.

958 CHAPTER 19 Carbonyl Chemistry 2: Reactions at the Position␣

19.5c Alkylation of ␤-Dicarbonyl

Compounds

β-Dicarbonyl com-

pounds are quite strong acids. The

pK

a

of diethyl malonate is about 13

and other β-dicarbonyl compounds,

β-cyano carbonyl compounds, and

dicyano compounds are similarly

acidic (Table 19.3). The anions

formed from these compounds can

be alkylated in S

N

2 reactions (Fig.

19.55). The once-alkylated com-

pounds still contain one doubly α

hydrogen, and therefore remain

acidic. A second alkylation reaction

is often possible, as long as the

requirements of the S

N

2 reaction are

kept in mind. These β-dicarbonyl

compounds are sufficiently acidic so

that alkylation through the enolate

using an alkoxide base is easy. Both

mono- and dialkylation reactions of

β-keto esters and related diesters are

common.

GENERAL CASES

A SPECIFIC EXAMPLE

(–) (–)

–

..

C

O

..

O

..

C

HH

CH

3

..

RO

C

O

..

O

..

C

Once alkylated

H

CH

3

..

RO

C

O

..

O

..

C

H

CH

3

..

RO

..

RO

..

HOR

RX

(S

N

2)

R

(–) (–)

–

..

C

O

..

O

..

C

H

CH

3

..

RO

C

O

..

O

..

C

Twice alkylated

CH

3

..

RO

C

O

..

O

..

C

CH

3

..

RO

..

RO

..

HOR

R

X

(S

N

2)

R

(75%)

Not isolated

CH

2

Cl, 25 ⬚C

(NC)

2

CH

2

(NC)

2

CH

1. NaH, dimethyl sulfoxide

repeat

2.

CH

2

CH

2

(NC)

2

C

C C C

CCC

–

..

–

..

FIGURE 19.55 Alkylations of some β-dicarbonyl compounds.

TABLE 19.3 pK

a

Values for Some Dicarbonyl Compounds

Compound Conjugate Base (Enolate) pK

a

13.3

11

10.7

8.9

8.5

5

EtO

C

O

CH

2

C

O

OEt

NC

CH

2

CN

C

O

CH

2

H

3

C

O

OEt

C

O

CH

2

H

3

CCH

3

C

O

C

O

CH

2

Ph CH

3

C

O

C

O

CH

2

HH

C

O

EtO

C

O

–

C

O

NC

CH

CN

OEt

H

3

C

O

C

O

OEt

H

3

C

CH

3

C

O

C

O

Ph CH

3

C

O

C

O

HH

C

O

C

O

..

–

CH

..

–

CH

..

–

CH

..

–

CH

..

–

CH

..

PROBLEM 19.15 Write a mechanism for the base-induced alkylation of

1, 1-dicyanomethane (malononitrile) with ethyl iodide.

Jones M., Fleming S.A. Organic Chemistry

Подождите немного. Документ загружается.