Jones M., Fleming S.A. Organic Chemistry

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14.13 Special Topic: Stable Carbocations in “Superacid” 679

the presence of an acid chloride (here, SO

), in order to produce a good

leaving group. This reaction converts the oxidized pyridine into the substituted

nonoxidized pyridine.

Nature uses a similar reaction in biological oxidation, the transfer of hydride to

the para position of the pyridinium ion, NAD



, the oxidized form of nicotinamide

adenine dinucleotide. We will see more of this reaction in Chapter 16 (Fig. 14.108).

hydride

transfer

NAD

–

H H

FIGURE 14.108 Reduction of NAD



through hydride transfer.

Summary

There are similarities between nucleophilic aromatic substitution (S

Ar) and its

more usual counterpart, electrophilic aromatic substitution.Each involves the for-

mation of a resonance-stabilized intermediate,and each involves a temporary loss

of aromaticity that is regained in the final step of the reaction. But the similar-

ities are only so deep.The electrophilic reaction involves cationic intermediates;

the nucleophilic involves anionic intermediates. Use the differing effects of a

nitro group, strongly deactivating in the electrophilic substitution and strongly

activating in the nucleophilic substitution, to keep the two mechanisms distinct

in your mind.

SbF

–

SbF

nonnucleophilic

solvent, low

temperature

FIGURE 14.109 Cation formation

through reaction of an alkyl fluoride

with SbF

14.13 Special Topic: Stable Carbocations

in “Superacid”

The Friedel–Crafts reactions succeed because relatively stable carbocations can be

formed by the reaction of aluminum chloride (an exceptionally strong Lewis acid with

which an alkyl halide will react) in solvents that are nonnucleophilic. The strongest

nucleophile available in the reaction vessel is benzene, which attacks the cation to

start the Friedel–Crafts reaction. If stable carbocations cannot be formed, as with

methyl chloride or ethyl chloride, the AlCl

complexes can also alkylate benzenes.

How would we modify the reaction conditions so as to maximize the possibil-

ity of generating the carbocation and minimize the chances for its further reaction?

Antimony pentafluoride (SbF

) is an even stronger Lewis acid than AlCl

toward

halogens, particularly fluorine. When SbF

removes a fluoride ion from an alkyl

compound the anion antimony hexafluoride (SbF



) is produced (Fig.14.109).This

anion is nonnucleophilic, and fluoride ion is a poor nucleophile as well.

680 CHAPTER 14 Substitution Reactions of Aromatic Compounds

HC F

SbF

–

SbF

–

SbF

ClF, –60 ⬚C

SbF

ClF, –60 ⬚C

FIGURE 14.110 Stable secondary and tertiary carbocations can be formed in superacid.

Many carbocations,once thought to be impossible species even as reaction inter-

mediates, can be generated and studied under superacid conditions.This work was

pioneered by George A. Olah (b. 1927) and earned Olah the Nobel Prize in

Chemistry in 1994.In future chapters, we will see a number of examples of the util-

ity of looking at cations in superacid.

+ –

, –33 ⬚C

Aniline

(50%)

+ KCl

Chlorobenzene

FIGURE 14.111 Aniline can be

formed from chlorobenzene by

reaction with very strong base,

potassium amide in ammonia.

PROBLEM 14.39 Explain why SbF



is nonnucleophilic.

Treatment of alkyl fluorides with SbF

at very low temperatures in highly polar, non-

nucleophilic solvents such as SO

,SO

ClF,or FSO

H (so-called superacid conditions)

can generate many stable carbocations. This process is Friedel–Crafts chemistry

carried to extremes. Methyl and primary carbocations cannot be generated this way,

but tertiary and secondary cations can be as long as they are not prone to rearrange-

ment to more stable cations (Fig. 14.110).

PROBLEM 14.40 When 1-fluorobutane is dissolved in superacid at low tempera-

ture, it is possible to observe a stable species. However, the

C NMR spectrum

of that stable species shows only two signals. (a) Why doesn’t the “expected” spec-

trum appear? What is the species that is observed? (b) Although part (a) is quite

easy, this reaction actually poses a vexing problem. Try to write a mechanism for

formation of the stable species. What problems appear when you do?

14.14 Special Topic: Benzyne

In Section 14.12, we mentioned the lack of reactivity of halobenzenes with

nucleophiles. It took the activation of a nitro group on the ring for substitution

to occur (see Figs. 14.94–14.96). In truth, if the nucleophile is basic enough, even

chlorobenzene will react. In the very strongly basic medium potassium amide



) in liquid ammonia (NH

), chlorobenzene is converted into aniline

(Fig. 14.111).

14.14 Special Topic: Benzyne 681

–

FIGURE 14.112 A clue to the mechanism for this reaction is provided by Roberts’

observation of isotopic scrambling when labeled chlorobenzene is used.The red

dot represents

C, an isotopic label.

The mechanism is surely not an S

2 displacement,and there is no reason to think

Ar addition–elimination is possible with an unstabilized benzene ring.This dif-

ficult mechanistic problem begins to unravel with the observation by John D.Roberts

(b. 1918) and his co-workers that labeled chlorobenzene produces aniline labeled

equally in two positions (Fig. 14.112).

–

NaCl

–

(a)

(b)

(a)

(b)

–

Benzyne

–

WEB 3D

FIGURE 14.113 The elimination–addition variation of the S

Ar reaction begins with

a halobenzene.The symmetrical intermediate benzyne is formed. Addition of the

amide ion to benzyne produces the observed labeling results.

PROBLEM SOLVING

Almost all “magical migration” reactions, such as the mysterious movement of the

red label in Figure 14.112, involve the formation of a symmetrical intermediate—

here benzyne (Fig. 14.113). The Special Topic in Chapter 21 is full of such

problems, but they have come up elsewhere as well. Problems 14.67, 20.16, and

20.42b are nice examples.

Benzyne formation

What reactions of halides do you know besides displacement? The E1 and E2

reactions compete with displacement reactions (Chapter 7). If HCl were lost

from chlorobenzene, a cyclic alkyne, called benzyne (or dehydrobenzene), would

be formed. This symmetrical bent alkyne might be reactive enough to undergo an

addition reaction with the amide ion.If this were the case, the labeled material must

produce two differently labeled products of addition (Fig. 14.113), because the



can attack either carbon of the alkyne.This mechanism is the elimination–addition

version of the S

Ar reaction.

682 CHAPTER 14 Substitution Reactions of Aromatic Compounds

NaCH

–

OCH

(c)

(d)

Benzyne

(a)

(b)

One would therefore not be too surprised to find that this molecule is able to

add strong bases such as the amide ion. In fact, one would expect very high reactiv-

ity for this strained alkyne. Problem 14.41 gives you a chance to see some of the

reactions of benzyne.

PROBLEM 14.41 Provide arrow formalisms for the formation of these four products

from benzyne:

14.15 Special Topic: Biological Synthesis

of Aromatic Rings; Phenylalanine

Some of the amino acids from which our tissues are constructed contain benzene rings,

and so nature must have a way of making them. You might actually set yourself that

problem right now; you have at your disposal the means to generate benzene from

simple hydrocarbon precursors.

Benzyne is surely an unusual species and deserves a close look.Although this inter-

mediate retains the aromatic sextet, the triple bond is badly bent, and there must be

very severe angle strain indeed.Remember, the optimal angle in triple bonds is 180°.

Moreover, the “third” bond is not composed of 2p/2p overlap like a normal alkyne,

but rather of overlap of two hybrid orbitals (Fig. 14.114).

Benzyne

These π bonds are

the remainder of

the

π system

These orbitals are

part of the

π system

These orbitals are not in the

π system—they make up

the “extra” bond in benzyne

FIGURE 14.114 The structure of

benzyne.

PROBLEM 14.42 Design a synthesis of benzene starting from inorganic com-

pounds, and organic compounds containing only hydrogen and no more than four

carbon atoms. Hint: Your primary goal is to construct the six-membered ring. You

have only one easy way to do that (so far).

14.15 Special Topic: Biological Synthesis of Aromatic Rings; Phenylalanine 683

Nature can’t use the method you designed in Problem 14.42, because too much

energy is needed to do the crucial reaction that constructs the six-membered ring.

Nature starts from simple sugars, which contain the carbons necessary for the

ring, along with many hydroxyl groups. (For much more on carbohydrates, wait for

Chapter 22.) The overall strategy, if one may speak of Nature having a strategy, is

to build a cyclohexane ring, then use the OH groups in various ways to do elimi-

nation reactions and put in the necessary double bonds.

Here is how the process works for phenylalanine, one of your amino acids.

A sugar is transformed enzymatically into dehydroquinate (A), which undergoes an

elimination reaction,also enzyme mediated,to give dehydroshikimate (B) (Fig.14.115).

COO

–

COO

–

COO

–

A sugar

Dehydroshikimate

(B)

Dehydroquinate

(A)

OPO

2–

FIGURE 14.115 In the biosynthesis of phenylalanine, enzymes transform a sugar into

dehydroquinate (A) and then dehydroshikimate (B).

The carbon–oxygen double bond in B is reduced and phosphorylated to give

shikimate-3P (C) (Fig. 14.116). Another enzyme adds a three-carbon fragment to

one of the hydroxyl groups (you’ll see why this is critical in a moment) to give D,

which undergoes a second elimination to place the second double bond in the ring

in E, chorismate.

COO

–

COO

–

Dehydroshikimate

(B)

COO

–

–2

Shikimate-3P

(C)

–2

COO

–

Chorismate

(E)

COO

–

COO

–

Enolpyruvylshikimate-3P

(D)

FIGURE 14.116 Dehydroshikimate

(B) is transformed into chorismate (E).

So far the strategy we guessed at is working—Nature has effectively done two

elimination reactions and seems poised to do a third. But now the side chain that

was added to make D in Figure 14.116 comes into play, and chorismate mutase

684 CHAPTER 14 Substitution Reactions of Aromatic Compounds

–

Prephenate

(F)

Phenylpyruvate

(G)

Phenylalanine

COO

–

COO

–

COO

–

FIGURE 14.118 Finally, prephenate

(F) is transformed into

phenylalanine.

The final elimination occurs in prephenate, with the carboxylate (COO



) directly

attached to the ring acting as trigger. Elimination of this group and the hydroxide

leads to phenylpyruvate (G) (Fig. 14.118). Now the three-carbon chain is elaborated

into the side chain of the amino acid phenylalanine.

COO

–

Tyrosine

This sequence is far from simple,but even though it is laden with bells and whis-

tles,some of which we surely could not anticipate, the general adherence to our early

strategy—build the six-membered ring and then use the hydroxyls as leaving

groups—is apparent.

PROBLEM 14.44 There is another amino acid, tyrosine, that is formed from

prephenate (Fig. 14.117). At first this transformation seems surprising, perhaps

even impossible. Analyze what must be done and explain why it will not be easy.

induces a transformation of E to prephenate,F (Fig. 14.117).You can see this process

in some detail in Chapter 20.

COO

–

COO

–

COO

–

OOC

enzyme

Chorismate

(E)

Prephenate

(F)

FIGURE 14.117 Chorismate (E) is

made into prephenate (F).

PROBLEM 14.43 Write an arrow formalism for the transformation of compound

E into F. Don’t worry about the role of the enzyme.

14.16 Summary 685

(+) (+)

–

FIGURE 14.119 The general mechanism for electrophilic aromatic substitution.

Key Terms

acid chloride (p. 643)

acylium ion (p. 643)

benzyne (p. 681)

Chichibabin reaction (p. 676)

diazonium ion (p. 647)

electrophilic aromatic substitution

(p. 635)

Friedel–Crafts acylation (p. 643)

Friedel–Crafts alkylation (p. 639)

ipso attack (p. 677)

Meisenheimer complex (p. 675)

nucleophilic aromatic substitution (p. 675)

[n]paracyclophane (p. 628)

Sandmeyer reaction (p. 649)

superacid (p. 680)

Reactions, Mechanisms, and Tools

By far the most important reaction in this chapter is elec-

trophilic aromatic substitution by a variety of electrophiles (E



This reaction involves the initial formation of a resonance-

stabilized, but not aromatic, cyclohexadienyl cation.

Deprotonation regenerates an aromatic system (Fig. 14.119).

Nucleophilic aromatic substitution occurs only when

the ring carries substituents that can stabilize the negative

charge introduced through addition of a nucleophile.

Departure of a leaving group regenerates the aromatic system

(Fig. 14.100).

14.16 Summary

New Concepts

This chapter involves the chemistry of aromatic compounds

and is dominated by two fundamental interrelated notions.

First, it is simply difficult to disturb the aromatic sextet, and

second, once aromaticity is disrupted, it is easily regained. In

chemical terms, this means that reactions of benzene and

other simple aromatic compounds generally involve a highly

endothermic first step with formation of a high-energy inter-

mediate. In electrophilic aromatic substitution, the most com-

mon reaction of aromatic compounds, this intermediate is the

cyclohexadienyl cation. Because this intermediate lies far

above benzene in energy, the transition state leading to it is

also high in energy, and the reaction is difficult. Reactions of

the cyclohexadienyl cation that regenerate aromatic systems

are highly exothermic, and therefore have low activation ener-

gy barriers.

Further substitution reactions of an already substituted

benzene are controlled by the properties of the first sub-

stituent. Both resonance and inductive effects are involved, but

the resonance effect usually dominates. Many substituents can

stabilize an intermediate cyclohexadienyl cation through reso-

nance if substitution is in the ortho or para position. The extra

stabilization afforded by the substituent leads to relatively high

rates of reaction. Positively charged substituents generally sub-

stitute meta, because ortho or para substitution places two pos-

itive charges on adjacent positions. The introduction of a

second positive charge means that these substitutions are rela-

tively slow.

At low temperature, in sufficiently polar, but nonnucleo-

philic solvents, some carbocations are stable. Even under

these superacid conditions, rearrangements through hydride

or alkyl shifts to the most stable possible carbocation are

common.

686 CHAPTER 14 Substitution Reactions of Aromatic Compounds

Wolff–Kishner reduction

/base

ethylene glycol

Clemmensen reduction

Zn/Hg/HCl

R Cl

Friedel–Crafts alkylation; carbocationic

rearrangements occur to give the most

stable ion; the possible structures of

R are therefore limited

AlCl

Wolff–Kishner

Clemmensen

reduction

1. Alkanes

Sn/HCl

Reduction of nitro groups

catalytic

hydrogenation

2. Alkylbenzenes

3. Anilines

4. Benzenes

Reversal of sulfonation reaction

O/H

The diazonium ion can be made from nitrobenzene

5. Benzenesulfonic Acids

HOSO

Simple sulfonation

6. Cyanobenzenes

Sandmeyer reaction

CuCN

7. Cyclohexanes

High-pressure catalytic hydrogenation

/Pt

polar solvent, pressure

Syntheses

This chapter provides many new syntheses of aromatic com-

pounds. Although we can now make all sorts of substituted

benzenes, there is still a unifying theme: Each product ultimately

derives from a reaction of benzene with an electrophile.

14.16 Summary 687

8. Deuterated Benzenes

See also methods for making benzenes in item 4

DOSO

9. Diazonium Ions

HONO/HCl

–

10. Halobenzenes

Aromatic halogenation

/AlBr

/AlCl

Sandmeyer reaction

CuCl

Sandmeyer reaction

CuBr

Schiemann reaction

HBF

11. Nitrobenzenes

COOOH

Oxidation of aniline

Simple nitration

HONO

HOSO

12. Phenols

13. Phenyl Ketones

AlCl

Friedel–Crafts acylation; no rearrangements occu

688 CHAPTER 14 Substitution Reactions of Aromatic Compounds

14.17 Additional Problems

PROBLEM 14.45 Explain why the trifluoromethyl group is

meta-directing in electrophilic aromatic substitution. Would you

expect CF

to be activating or deactivating? Why?

PROBLEM 14.46 Show where the following compounds would

undergo further substitution in electrophilic aromatic substitution.

There may be more than one position for further substitution in

some cases. Explain your reasoning.

(a) (b) (c)

(d)

(e)

(f)

(g)

(h)

PROBLEM 14.47 Predict the position of further substitution in

biphenyl. Explain your reasoning.

Biphenyl

COOH

Diclofenac

PROBLEM 14.48 Show specific reagent(s) and product(s) for

the following reactions with anisole (methoxybenzene):

(a) nitration

(b) alkylation

(d) monobromination

(e) acylation

PROBLEM 14.49 Show the product(s) for chlorination (Cl

FeCl

) of the following compounds:

(a) toluene

(b) chlorobenzene

(d) 2,4-dinitrophenol

(e) diphenylether

PROBLEM 14.50 Diclofenac is a nonsteroidal antiinflammato-

ry drug (NSAID) that is effective for treating pain. In an effort

to better understand its enzyme selectivity, it has been nitrated.

A single product was obtained. Predict the product and explain

your choice.

PROBLEM 14.51 Predict the product for the first reaction shown

on the next page. What mechanism do you suppose is involved

in making the product you have predicted? Based on your

answer to the first reaction, what product do you predict for the

second reaction shown?

Common Errors

The most common conceptual error is to attempt to analyze the

course of aromatic substitution reactions by looking at the start-

ing materials and products. The various resonance and inductive

effects discussed in this chapter are far more important in the

charged intermediates and in the transition states leading to

those intermediates. When analyzing one of these reactions,

always look first at the cyclohexadienyl intermediate (a cation

in electrophilic aromatic substitution and an anion in the less

common nucleophilic reactions), and then consider the transi-

tion state leading to it.

Another potential error is failing to keep up with the prolif-

erating synthetic detail. Be sure to look at each new reaction in

this and subsequent chapters with an eye to using it in synthesis.

Keep up your file cards!