Jones M., Fleming S.A. Organic Chemistry

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19.11 Condensation Reactions in Combination 999

PROBLEM SOLVING

Complex mechanistic problems absolutely require analysis. A good problem

solver always begins by analyzing the problem before writing any arrows. We

suggest first building a map. Ask yourself what atoms in the starting material

become what atoms in the product. You won’t always be able to identify the

fate of every atom, but you should usually be able to see what happens to some

of them. Use attached groups that are not changed in the reaction as markers.

Next, develop a set of goals. Your map should allow you to spot which bonds

in the product must be made and which bonds in the starting material must

be broken.

Finally, after building your map and establishing your goals, you can begin

to think about how to accomplish those goals. We’ll start you off on part (b) of

Problem 19.38, but please try the other problems this way. We guarantee that

you will get more answers right if you do. And a side benefit is that even if you

don’t get the problem completely right, the graders will “see” you thinking and

partial credit definitely flows to such answers.

Magid’s first rule is archaic, and we will shortly encounter Magid’s third rule.

PROBLEM 19.37 Although the mechanism shown in Figure 19.113 is the way the

Robinson annulation is always described, there is another (harder) way to write

the reaction. Write a mechanism for the general reaction in Figure 19.113 that

involves doing the aldol condensation first.

Problem 19.37 brings up a serious practical question. You may well be able

to analyze a problem by figuring out the different possible reactions, but choos-

ing among them is more difficult. How do you find the easiest route (there

are very often several) to the product you are after? Clearly, experience is invalu-

able and you are by definition short on that commodity. In such a situation,

it is wise to be receptive to good advice, and here is a practical hint from

one of the best problem solvers known, Ronald M. Magid (b. 1938), in

the form of Magid’s second rule:

When there is a choice, always try the

Michael first.

In fact, there are sound reasons for doing the Michael reaction first. Aldol con-

densation products of ketones are not generally favored at equilibrium, for exam-

ple, and doing the Michael reaction makes the subsequent aldol condensation

intramolecular, and thus more favorable.

There is no substitute for practice,though, and Problem 19.38 offers some good

opportunities to practice solving Fun in Base problems. Be sure to look at the

Problem Solving box first.

is followed by an intramolecular aldol condensation to give the new six-membered

ring (Fig. 19.113).

1000 CHAPTER 19 Carbonyl Chemistry 2: Reactions at the Position␣

C +

(a)

*(b)

(c)

(d)

(e)

COOEt

KOH

NaOEt

HOEt

NaOH

KOH

Ph Ph

BrCH

COEt

–

(f)

KOH

Hint: There is a 10-membered ring involved.

WORKED PROBLEM 19.38 Provide mechanisms for the reactions shown below.

(continued)

19.12 Special Topic: Alkylation of Dithianes 1001

COOEt

OEt

make

break

ANSWER (b) Those two Ph groups are just along for the ride—nothing happens

to them in this reaction. Use them as markers to identify where the carbon

attached to them in the starting material winds up in the product (red dot).

Similarly, that ester group is preserved. Use it as a marker to find the carbon

attached to it in the starting materials (green dot).

Goals:

1. Our map allows us to see that we must make the red dot–green dot bond

(arrow, above), so that’s goal number one.

2. We also must make a bond from oxygen to either the red or green dot—but

we can’t tell which (yet).

3. The bond to bromine must be broken—we know that because there is no Br

in the product.

Okay, now you are on your own. Before looking at the Study Guide, find a way to

make that red–green bond.Once you do that, it will be easy to see how to lose the

Br and how to make that oxygen–dot bond.

HSCH

FIGURE 19.114 Formation of a

dithiane (a thioacetal).

1,3-Dithiane

(1,3-dithiacyclohexane)

= 31.1

–

BuLi

= CH

BuH

THF

FIGURE 19.115 Deprotonation

of 1,3-dithiacyclohexane with

butyllithium.

PROBLEM 19.39 Explain how the sulfur atoms operate to increase the acidity of

the adjacent hydrogens in 1,3-dithiane (1,3-dithiacyclohexane).

19.12 Special Topic: Alkylation of Dithianes

Here is a clever way to alkylate carbonyl compounds at the carbonyl carbon atom.

Thioacetals, also called dithianes, can be made from carbonyl compounds and the

sulfur counterparts of diols (Fig. 19.114). Dithianes are more acidic than typical

alkanes and can be deprotonated with alkyllithium reagents (Fig. 19.115).

1002 CHAPTER 19 Carbonyl Chemistry 2: Reactions at the Position␣

THE GENERAL CASE

SPECIFIC EXAMPLES

(Hg

)

(Hg

)

–

(84%)

I, THF

60 ⬚C

1. BuLi/THF

(70%)

I, THF

1. BuLi

3. repeat steps

(1) and (2)

1. BuLi

2. RX

FIGURE 19.116 A synthesis

of carbonyl compounds using

alkylation of 1,3-dithiane

(1,3-dithiacyclohexane).

–

BuLi repeat

Raney Ni

R I

RCH

WORKED PROBLEM 19.40 Suggest a synthetic route to alkanes starting from

1,3-dithiane.

ANSWER Easy (and effective in practice). First alkylate the dithiane, then desul-

furize with Raney nickel (p. 253).

The resulting anion is a powerful nucleophile and can undergo both displace-

ment reactions (subject, of course, to the usual restrictions of the S

2 reaction) and

additions to carbonyl compounds (Fig. 19.116).The products of these reactions are

dithianes and can be hydrolyzed, usually by mercury salts, to aldehydes or ketones.

The dithiane functions as a masked carbonyl group! Here is another synthesis of

aldehydes and ketones.

19.13 Special Topic: Amines in Condensation Reactions, the Mannich Reaction 1003

19.13 Special Topic: Amines in Condensation

Reactions, the Mannich Reaction

In the Mannich reaction, named for Carl Mannich (1877–1947), an aldehyde or

ketone is heated with an acid catalyst in the presence of formaldehyde and an amine.

The initially formed ammonium ion is then treated with base to liberate the final

condensation product, the free β-amino ketone (Fig. 19.117).

THE GENERAL CASE

A SPECIFIC EXAMPLE

N(CH

)

(70%)

/(CH

)

HCl/EtOH

NaOH

(CH

)

1. HCl/EtOH

80 ⬚C, 2 h

2. base

–

N(CH

)

O NaCl

N(CH

)

FIGURE 19.117 The Mannich reaction.

The Mannich reaction presents a nice mechanistic problem because two routes

to product seem possible, and it is the simpler (and incorrect) process that is the

easier to find. The difficulty is that we are now experienced in working out aldol

condensations and a reasonable person would probably start that way. An acid-cat-

alyzed crossed aldol condensation (p. 982) leads to a β-hydroxy ketone (Fig. 19.118).

Displacement of the hydroxyl group (after protonation, of course) would lead to

the initial product, the ammonium ion. Treatment with base would surely liberate

the free amine.

A β-hydroxy ketone

(from a crossed aldol

condensation)

HCl/EtOH HCl/ EtOH

(CH

)

NaOH/H

deprotonation

FIGURE 19.118 An attractive, but incorrect, mechanism for the Mannich reaction.

Sadly, this simple mechanism cannot be correct because, when the β-hydroxy

ketone is made in other ways, it is not converted into Mannich product under the

reaction conditions. What else can happen? Amines are nucleophiles and react rap-

idly with carbonyl compounds in acid to give iminium ions (p. 795). It is this imini-

um ion that is the real participant in the Mannich condensation reaction.

Formaldehyde is more reactive than the ketone, so the iminium ion is preferentially

1004 CHAPTER 19 Carbonyl Chemistry 2: Reactions at the Position␣

Iminium ion

Enol

(CH

)

NH/HCl

NaOH

NaCl

–

HCl

(CH

)

N(CH

)

(CH

)

(CH

)

FIGURE 19.119 The correct

mechanism of the Mannich reaction.

Benzaldehyde

KOH

Potassium benzoate

(oxidized)

Benzyl alcohol

(reduced)

–

FIGURE 19.120 Benzaldehyde gives

benzoate and benzyl alcohol on

treatment with hydroxide ion.

–

ONa

Na OD2

–

FIGURE 19.121 A labeling

experiment shows that no solvent

deuterium becomes attached to carbon

in the Cannizzaro reaction.The

hydrogen that reduces the aldehyde to

the alcohol must come from another

aldehyde, not the solvent.

formed from this carbonyl compound (Fig. 19.119). In an aldol-like process,the enol

of the ketone reacts with the iminium ion to give the penultimate product, the ammo-

nium ion. In the second step, the free amine is formed by deprotonation in base.

This process is known as the Cannizzaro reaction, for Stanislao Cannizzaro

(1826–1910). If the reaction is carried out in deuterated solvent, no carbon–deuterium

bonds appear in the alcohol (Fig. 19.121).

19.14 Special Topic: Carbonyl Compounds

without ␣ Hydrogens

Throughout this chapter we have concentrated on molecules capable of forming either

enolates or enols, and the subsequent chemistry has been dependent on those inter-

mediates. It is a legitimate question to ask if there is a base-induced chemistry of

carbonyl compounds that do not have α hydrogens, and cannot form enolates.

19.14a The Cannizzaro Reaction Hydroxide ion will give products from

nonenolizable carbonyl compounds. If we examine a sample of benzaldehyde that

has been allowed to stand in the presence of strong base, we find that a redox reac-

tion has occurred. Some of the aldehyde has been oxidized to benzoate and some

reduced to benzyl alcohol (Fig. 19.120).

Benzaldehyde is especially prone to hydrate formation, and the first step of the

Cannizzaro reaction is addition of hydroxide to benzaldehyde. Much of the time this

reaction simply reverses to re-form hydroxide and benzaldehyde, but some of the

19.14 Special Topic: Carbonyl Compounds without ␣ Hydrogens 1005

time the intermediate finds a molecule of benzaldehyde in a position to accept a

transferred hydride ion (Fig. 19.122).

–

Hydride transfer

–

FIGURE 19.122 A hydride shift is at the heart of the

mechanism of the Cannizzaro reaction.

Do not confuse this hydride transfer step with the expulsion of hydride as a leaving

group. Hydride is not a decent leaving group and cannot be lost in this way. However,

if there is an electrophile positioned in exactly the right place to accept the hydride,

it can be transferred in a step that simultaneously reduces the receptor aldehyde and

oxidizes the donor aldehyde. It is very important to be aware of the requirement for

the Lewis acid acceptor of hydride in the Cannizzaro reaction. The hydride must

be transferred, and in the transition state for the reaction there is a partial bond

between the hydride and the accepting carbonyl group.A subsequent deprotonation

of the carboxylic acid and protonation of the alkoxide formed from benzaldehyde

completes the reaction. Notice how critical the observation about the nonparticipa-

tion of the deuterated solvent is. It identifies the source of the reducing hydrogen

atom. Because it cannot come from solvent (all the “hydrogens” in the solvent are

deuteriums) it must come from the aldehyde.

PROBLEM 19.41 Draw the transition state for the hydride shift reaction.

If there were a way to hold the hydride acceptor in the right position, the Cannizzaro

reaction would be much easier, because the requirements for ordering the reactants in

just the correct way (entropy) would be automatically satisfed.The two partners in the

reaction would not have to find each other. This notion leads to the next reaction in

this section, which is largely devoted to hydride shift reactions with wonderful names.

19.14b The Meerwein–Ponndorf–Verley–Oppenauer Equilibration In

the Meerwein–Ponndorf–Verley–Oppenauer (MPVO) equilibration an alu-

minum atom is used to clamp together the two halves (oxidation and reduction) of

the Cannizzaro reaction. An aluminum alkoxide, typically aluminum triisopropox-

ide, is used as a hydride source to reduce the carbonyl compound (Fig. 19.123).

1. Al[OCH(CH

)

]

HOCH(CH

)

2. H

(89%)

FIGURE 19.123 The reduction

of benzaldehyde using the

Meerwein–Ponndorf–Verley–

Oppenauer reaction.

1006 CHAPTER 19 Carbonyl Chemistry 2: Reactions at the Position␣

+ +

–

Reduced

Oxidized

FIGURE 19.124 The formation of a bond between

the nucleophilic carbonyl oxygen and the Lewis

acid aluminum alkoxide to give complex A,

which undergoes an intramolecular hydride

transfer facilitated by the aluminum atom

chemical clamp.

The first step in this reaction is the formation of a bond between the carbonyl

oxygen and aluminum. Aluminum is a very strong Lewis acid and the bond to the

carbonyl oxygen forms easily (Fig. 19.124).This new molecule (A) has the two react-

ing groups clamped together by the aluminum and is nicely set up for an intramol-

ecular transfer of hydride, which simultaneously reduces a molecule of the carbonyl

compound and oxidizes an isopropoxide group to acetone.

PROBLEM 19.43 Write a mechanism for the second stage in the reaction, the

formation of .

PROBLEM 19.44 Write a mechanism for the final hydrolysis reaction of Al(OR)

to Al(OH)

CH(CH

)

3 R

CHOHAl(OH)

CHR

FIGURE 19.125 Repetition and

hydrolysis lead to more acetone

and the new alcohol.

PROBLEM 19.42 Aluminum alkoxides may look very strange to you. Compare the

structure of Al(OR)

to that of a trivalent boron compound such as B(OH)

Three repetitions of this process leads to a new aluminum oxide that is hydrolyzed

at the end of the reaction to give the corresponding alcohol (Fig. 19.125).

19.15 The Aldol Condensation in the Real World, an Introduction to Modern Synthesis 1007

33+

AlCl

3 equiv.

FIGURE 19.126 The Meerwein–

Ponndorf–Verley–Oppenauer

equilibration can be used to

oxidize an alcohol.

The effect of the aluminum clamp on the rate of the reaction is profound. This

reaction,here shown as a reduction, takes place under very mild conditions and with

few side reactions.

The reaction can also be used to oxidize alcohols to carbonyl compounds.In this

case, acetone is used as the acceptor, and the alcohol to be oxidized is used to form

the aluminium oxide (Fig. 19.126).

–

(a)

KOH/H

(b)

O/H

(c)

O/KOD

Hints: Draw a good three dimensional picture of this molecule first! The other

stereoisomer exchanges only five hydrogens for deuteriums.

O/KOD

Problems involving hydride shifts tend to be hard enough to thwart even excel-

lent problem solvers. That observation leads directly to Magid’s third rule, which

states, When all else fails and desperation is setting in, look for the hydride shift.

Problem 19.45 involves some practice hydride shifts.

19.15 Special Topic: The Aldol Condensation in the

Real World, an Introduction to Modern Synthesis

In this section,we introduce some of the difficulties of real-world chemistry.The detail

isn’t so important here, although it will be if you go on to become a synthetic chemist!

It is important to get an idea of the magnitude of the problems, and of the ways in

which chemists try to solve them.The general principles and relatively simple reactions

PROBLEM 19.45 Write mechanisms for the following reactions.

1008 CHAPTER 19 Carbonyl Chemistry 2: Reactions at the Position␣

base

Two enolates… …lead to two products

–

(–)

–

(–)

2. H

base

–

2. H

FIGURE 19.127 In principle, an unsymmetrical ketone leads to two enolates, and therefore two products.

The two enolates differ in stability, depending on the number of alkyl groups attached to the carbon-

carbon double bond.

The two enolates are quite different in stability. Consider the enolate resonance

structure that contributes most to the overall structure for the two enolates.One con-

tains a disubstituted carbon–carbon double bond and will be more stable than the

other, which has only a monosubstituted carbon–carbon double bond.

Effective methods have evolved to form the less stable enolate by taking advan-

tage of the relative ease of access to the less hindered α hydrogen (Fig. 19.128a). As

we have learned,LDA is especially effective at forming these less stable, kinetic eno-

lates.The key point is that this strong but large base has difficulty in gaining access

to the more hindered parts of the carbonyl compound. Removal of the less hindered

proton leads to the less stable, or kinetic enolate.

Other methods allow formation of the more

stable, thermodynamic enolate (Fig. 19.128b).

In one of these, boron enolates are formed.

Another option is use of a slightly weaker base

and higher temperatures (thermodynamic condi-

tions) in making the enolate.

We have described only one of the difficul-

ties encountered in just one important syn-

thetic reaction,the aldol condensation.One goal

of all synthetic chemists is selectivity, ideally,

specificity. How do we do only one reaction?

For example, how do we form one, and only

one, stereoisomer of the many often possible?

FIGURE 19.128 Use of the boron

enolate or thermodynamic conditions

to make the more stable enolate.

we have studied (e.g., the straightforward aldol condensation) set the stage for a glance

at the difficulties encountered when chemists try to do something with these reactions.

In the real world of practical organic synthesis, one rarely needs to do a simple aldol

condensation between two identical aldehydes or two identical ketones. Far more com-

mon is the necessity to do a crossed aldol between two different aldehydes, two differ-

ent ketones, or an aldehyde and a ketone. As noted earlier, there are difficulties in doing

crossed aldol reactions.Suppose,for example,that we want to condense 2-pentanone with

benzaldehyde. Benzaldehyde has no α hydrogen, so no enolate can be formed from it.

Some version of the Claisen–Schmidt reaction (p. 984) seems feasible. But 2-pentanone

can form two enolates, and the first problem to solve is the specific formation of one or

the other enolate (Fig. 19.127).

DME

(dimethoxyethane)

25 ⬚C

CLi

–

BEt

25 ⬚C

OBEt

–

(b)

THF

–78 ⬚C

LDA

–

(a)