Jones M., Fleming S.A. Organic Chemistry

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14.3 Diels–Alder Reactions 629

PROBLEM 14.2 Why are boat-shaped benzenes less stable than flat ones?

(CH

)

[7]Paracyclophane

160 ⬚C

CCCF

(>52%)

WEB 3D

FIGURE 14.11 [7]Paracyclophane

undergoes the Diels–Alder reaction

at a lower temperature than benzene.

PROBLEM 14.3 Use an arrow formalism to show how [7]paracyclophane forms

the product shown in Figure 14.11 through a Diels–Alder reaction.

We really needn’t go to the trouble of making paracyclophanes (unless one is

interested in what happens to benzene’s π system as it deforms), because there are

naturally occurring molecules that are less aromatic than benzene and thus should

be more reactive. Recall how we reached a quantitative determination of the reso-

nance energy for benzene through measurements of the heats of hydrogenation (or

combustion) for the compounds cyclohexene and cyclohexadiene.We then used these

data to estimate the heat of hydrogenation of the hypothetical molecule 1,3,5-cyclo-

hexatriene. Finally, we compared our estimate with the experimental value deter-

mined by hydrogenating the real molecule benzene (p. 581).Table 14.1 gives similarly

determined values for a variety of heteroaromatic compounds.

There is less delocalization energy in furan than benzene (Table 14.1) and preser-

vation of aromaticity is less important. Diels–Alder reactions are common for furans,

which make excellent trapping agents for reactive π systems (Fig. 14.12).

TABLE 14.1

Resonance Energies of Some

Heteroaromatic Compounds

and Benzene

Compound Resonance

Energy (kcal/mol)

Benzene 33

Thiophene 29

Pyridine 23

Pyrrole 21

Furan 16

NC CNCC

OOC

(94%)

(61%)

COOCH

OAc

110 ⬚C

12 h

25 ⬚C

12 h

OAc

FIGURE 14.12 Two Diels–Alder

reactions of furans.

has been detected), the ring becomes warped, and must assume a boat shape

(Fig. 14.10), which severely destabilizes the molecule. In this case, the models tell all.

Make a model of a paracyclophane with n  10 and then reduce the number of bridg-

ing methylenes one by one.The benzene ring will increasingly deform as this happens.

The source of the destabilization should also become apparent.Try Problem 14.2.

As anticipated, these destabilized benzenes are much more reactive in the

Diels–Alder reaction than their undistorted relatives (Fig. 14.11).

630 CHAPTER 14 Substitution Reactions of Aromatic Compounds

COOCH

.. ..

140 ⬚C

OOC C C COOCH

COOCH

(43%)

OOC

COOCH

OOC

COOCH

OOC

COOCH

(new bonds shown in boldface)

OOC

COOCH

Pyrroles, like furans, are also 1,3-dienes and one might suspect that the

Diels–Alder reaction would occur as with furan. This suspicion is correct, but the

aromaticity of pyrrole introduces complications. Such reactions are often slow, and

aromaticity is sometimes regained through a retro Diels–Alder reaction.With those

hints, you should be able to tackle Problem 14.4.

WORKED PROBLEM 14.4 Write a mechanism for the following reaction:

ANSWER All those hints in the text surely point directly at a Diels–Alder reaction.

In this case, that leads to intermediate A.

Now, with luck, you can see the symmetry in A. Because Diels–Alder reactions

are reversible, there are two possible reverse Diels–Alder routes for A to take. In

one, the reaction simply reverses to give the starting materials. In the other, the

reverse Diels–Alder reaction proceeds in the other direction, and acetylene is lost

to give the observed product. One imagines that it is the irreversible loss of the

acetylene gas that drives the reaction toward the product shown.The arrows show

the product-forming pathway.

14.4 Substitution Reactions of Aromatic Compounds 631

PROBLEM SOLVING

Problem 14.4 is an example of the “hide the cyclohexene” game we warned you

about (p. 554). In this example, the cyclohexene is disguised not only by all the

groups (chemical “spinach”) hanging onto it, but by the presence of a second

double bond. Problems in which a forward Diels–Alder reaction (usually rather

easy to find) is followed by a reverse Diels–Alder in another direction (much

harder to find) may be encountered.

Summary

It is not easy to disrupt the aromatic sextet,and few reactions of benzene do so per-

manently. It takes a very reactive reagent to overcome the high activation energy

barriers surrounding the benzene ring and usually there is a path available that

regains the lost aromaticity. In this section, you saw two reactions that do “dearo-

matize” benzene; in the next sections you will see many that don’t.

14.4 Substitution Reactions of Aromatic Compounds

14.4a Why Benzene Doesn’t Undergo Simple Electrophilic Addition

Reactions

We are first going to do some thought experiments, beginning with

a “What if ”question.This process will lead us to an understanding of why benzene

reacts so differently from simple alkenes.In particular, we are going to see why ben-

zene does not undergo simple addition of HX (hydrohalogenation) as alkenes do.

Then we will go on to see why benzene reacts as it does.

What if benzene reacted as the simple alkenes do to give addition products with

HX reagents? The answer to this hypothetical question is more complicated than it

appears at first.Let’s work through a prediction for the addition of HBr to benzene,

a reaction that in fact does not occur. By analogy to alkene reactions, the first step

should be the addition of a proton to the π system of benzene to give an interme-

diate cyclohexadienyl carbocation. Note that aromaticity is lost (Fig. 14.13).

–

A nonaromatic

cyclohexadienyl cation

Benzene

FIGURE 14.13 Protonation of

benzene to give a cyclohexadienyl

cation.

–

H H

FIGURE 14.14 Addition of bromide

would give a cyclohexadienyl

bromide.

Simple addition of the bromide ion to the cation, again by analogy to the addi-

tion of HBr to an alkene, would generate a cyclohexadienyl bromide (Fig. 14.14).

632 CHAPTER 14 Substitution Reactions of Aromatic Compounds

Summary structures

(+)

FIGURE 14.15 The cyclohexadienyl cation is resonance stabilized.

The positive charge is shared by the three carbons marked with dots.

But we have forgotten to look closely at the structure of the carbocation formed by

the addition of a proton. Just as is the case with simple dienes, the positive charge is

not localized on one atom, but instead is shared by two other carbons.The resonance

picture of Figure 14.15 shows a better representation of the cyclohexadienyl cation.

Look carefully again at the shorthand used to indicate the delocalized species.

The dashed loop of Figure 14.15 indicates delocalization quite clearly,but it does not

do a good job of showing which atoms share the positive charge. To get this infor-

mation, you must either draw out the separate resonance forms, or use a single res-

onance form with the other positive positions labeled as (). Keep in mind that the

dashed loop does not mean that all the atoms surrounding the loop are partially pos-

itive. In Figure 14.15, the positions sharing the positive charge are shown with dots.

CONVENTION ALERT

So, addition of bromide could take place at any one of the three carbons shar-

ing the positive charge to produce two different products (Fig. 14.16). Yet, neither

of these compounds is the product of the reaction! In fact, there is no apparent

reaction at all, and benzene is recovered unchanged from the attempted addition

reaction. One searches in vain for the predicted cyclohexadienyl bromides.

–

5-Bromo-1,3-

cyclohexadiene

3-Bromo-1,4-

cyclohexadiene

5-Bromo-1,3-

cyclohexadiene

FIGURE 14.16 Capture of the

cyclohexadienyl cation at the three

possible positions bearing positive

charge would give two

cyclohexadienyl bromides.

14.4b Electrophilic Deuteration of Benzene In this section, we see how ben-

zene does react. What we will see is the result of an energetic imperative to preserve the

aromatic sextet.Benzene reacts not by addition,which would destroy aromatic stabiliza-

tion, but by substitution reactions that preserve the aromatic sextet and stabilization.

The prototypal example is substitution of each hydrogen by deuterium. If sufficiently

acidic conditions are used, we can see that the apparent nonreaction of HX with ben-

zene is an illusion. As we saw in Chapter 13, when benzene is treated with D

, the

14.4 Substitution Reactions of Aromatic Compounds 633

recovered benzene has incorporated deuterium, with the amount of incorporation

depending on the strength of acid used and the time of the reaction. Ultimately, excess

deuteriosulfuric acid will produce fully exchanged benzene (Fig. 14.17).

WEB 3D

FIGURE 14.17 Hydrogen–deuterium

exchange in benzene.

OSO

OHHOSO

HHOS

–

OSO H

Bisulfate ion

–

S O

FIGURE 14.18 Protonation of

benzene by sulfuric acid, followed by

addition of the bisulfate ion, would

produce a pair of addition products.

Because deuterium must become incorporated onto the ring, perhaps the pro-

posed reaction shown in Figure 14.13 is not wrong after all. If benzene is protonat-

ed to produce the cyclohexadienyl cation shown in Figure 14.18, there are two

possible further reactions.

As shown in Figure 14.18, the bisulfate anion produced in the first step of the reac-

tion could add to give two cyclohexadienyl compounds. We imagined a related reaction

in Section 14.4a, where the Lewis base was bromide, Br



. In this reaction, aromaticity

is lost, and we would not expect this process to be favorable energetically. Alternatively,

a proton might be lost to reverse the initial addition reaction as shown in Figure 14.19.

In this scenario, the aromatic sextet is regenerated. In the absence of an isotope such as

deuterium, this reaction is invisible because it simply regenerates starting material.

HOS

(+) (+)

–

FIGURE 14.19 The protonation

reaction is reversible. Bisulfate could

remove a proton to regenerate benzene.

634 CHAPTER 14 Substitution Reactions of Aromatic Compounds

loss of D

O DD

–

(+) (+)

loss of H

Deuteriobenzene

FIGURE 14.20 If deuteriosulfuric

acid is used, the initial addition

must generate a deuterated

cyclohexadienyl cation. Reversal of

the reaction will regenerate benzene,

but loss of a proton will give

deuteriobenzene. In an excess of

, all the protons of benzene

can eventually be exchanged for

deuterons.

If it is a deuteron that is added to produce an isotopically labeled cyclohexadienyl

cation, either a deuteron or a proton could be lost to regenerate benzene (Fig. 14.20).

Energy

Reaction progress

H D

–

OSO

DHSO

and

exothermic

FIGURE 14.21 Both proton removal

and addition are exothermic reactions

for the cyclohexadienyl cation, but

the regaining of aromaticity by

proton removal is much more so.

If the proton is lost, deuteriobenzene will appear as product, because one hydrogen

will have been exchanged for a deuterium. Repetition of the reaction can serve to

exchange all the hydrogens for deuterons.If the cyclohexadienyl cation is formed,why

aren’t normal addition products, such as the dienes in Figure 14.18, observed? The

answer is that loss of a proton (or deuteron) regenerates an aromatic system in a high-

ly exothermic reaction.Addition to the relatively high-energy carbocation would still

be exothermic, but not nearly as much as the substitution reaction (green); no aro-

matic molecule would be generated.The addition reaction (red) is much less exother-

mic than the rearomatization pathway and cannot compete with it (Fig. 14.21).

Summary

We have just seen one example of substitution of a benzene ring in which aro-

maticity is initially lost only to be regained later. Next we generalize, discovering

many variations on this theme.The main points to remember are, first, that aro-

maticity is not easily disrupted, and second, that once the aromatic stabilization

is lost, it is easily regained.

14.4 Substitution Reactions of Aromatic Compounds 635

The reaction is always referred to as electrophilic aromatic substitution,because

the emphasis is on the agent doing the substituting, the electrophile. Fair enough,

but in every electrophilic reaction there must be a nucleophile as well. In this case,

the π system of the benzene ring acts as the nucleophile. In orbital terms, it is one

of the degenerate filled molecular orbitals of benzene that overlaps with the empty

orbital on the electrophile (Fig. 14.22).

14.4c Electrophilic Aromatic Substitution: The General Reaction

The exchange reaction of benzene in deuterated acid serves as a prototype for a

host of similar reactions in which an electrophile, E



, is substituted for a hydrogen

on the ring:

–

As in the substitution reaction of D for H on the aromatic ring,the general reac-

tion is completed by the removal of a proton from the intermediate cyclohexadienyl

cation by a base, regenerating the aromatic benzene ring (Fig. 14.23).

Nucleophile Electrophile

Energy

Nucleophile

(filled orbital)

Electrophile

(empty orbital)

FIGURE 14.22 One of benzene’s degenerate HOMOs overlaps with the empty orbital on



in a Lewis base–Lewis acid reaction.

–

(+) (+)

FIGURE 14.23 The general aromatic substitution mechanism involves formation

of a cyclohexadienyl cation, followed by proton removal, regenerating the

aromatic ring.

(100%)

Benzene-

sulfonic acid

99% H

WEB 3D

FIGURE 14.24 Sulfonation of

benzene.

(+) (+)

The resonance-

stabilized

cyclohexadienyl

cation intermediate

–

FIGURE 14.25 The SO

molecule

acts as an electrophile and adds to

benzene.

636 CHAPTER 14 Substitution Reactions of Aromatic Compounds

PROBLEM 14.5 Write a mechanism for the formation of benzenesulfonic acid

when benzene is treated with SO

. Hint: Sulfur trioxide (SO

) can be

written as a sulfur atom doubly bonded to each of the three oxygens.

When used in an aprotic medium,SO

is a strong enough electrophile to react with

benzene without protonation (Fig. 14.25). Under these circumstances the sulfonation

reaction looks a little different.Addition leads to a cyclohexadienyl cation,but now there

is also a negative charge in the molecule.As in the general mechanism for electrophilic

aromatic substitution, a base now removes a proton to give the aromatic product.The

in this particular reaction is the base that will remove the very acidic hydrogen to

restore aromaticity. When the reaction is run in H

we find that even sulfuric acid

is basic enough to deprotonate the cyclohexadienyl cation.The structure of the prod-

uct of this reaction, benzenesulfonic acid, gives students problems because the formu-

las, SO

H or SO

OH hide the detail.The full structure is drawn out in Figure 14.26.

14.4d Sulfonation When benzene is treated either with very concentrated

sulfuric acid or with a solution of sulfur trioxide in sulfuric acid, called oleum,

benzenesulfonic acid (Fig. 14.24) is the product. When oleum is used, the sulfur

trioxide is protonated to produce the electrophile, .HOS

Three representations of benzenesulfonic acid

HSO

–

(+) (+)

FIGURE 14.26 Removal of a proton

from the cyclohexadienyl cation

intermediate regenerates the

aromatic ring.

Sulfonation of benzene is reversible, especially at high temperature. Treatment

of benzenesulfonic acid in steam regenerates the parent benzene (Fig. 14.27).

H H

⌬

FIGURE 14.27 Sulfonation is

reversible.

WORKED PROBLEM 14.6 Write a mechanism for the reaction in Figure 14.27.

ANSWER The mechanism for the transformation of benzenesulfonic acid into

benzene is the reverse of the mechanism for formation of benzenesulfonic acid

from benzene (Figs. 14.25 and 14.26). Protonation gives a cyclohexadienyl cation

that can either lose a proton to revert to benzenesulfonic acid, or lose the sulfonic

acid group to give benzene and protonated sulfuric acid, which will lose its pro-

ton to water.

14.4 Substitution Reactions of Aromatic Compounds 637

OHHO

14.4e Nitration In concentrated nitric acid, often in the presence of sulfuric

acid, benzene is converted into nitrobenzene (Fig. 14.28). In strong acid, nitric acid

HNO

HO HONO

HNO

–

WEB 3D

FIGURE 14.28 Nitration of benzene.

is protonated to give (Fig. 14.29). Loss of water can generate a powerful

Lewis acid, the nitronium ion,



, that can act as the electrophile in an aromatic

HO NO

Nitronium

ion

+++

FIGURE 14.29 In strong acid, nitric

acid becomes protonated, then loses

water to form the nitronium ion.

638 CHAPTER 14 Substitution Reactions of Aromatic Compounds

substitution reaction (Fig. 14.30). Indeed, stable nitronium salts are known and can

also effect nitration. As usual, the reaction is completed by deprotonation of the

cyclohexadienyl cation by one of the weak Lewis bases present in the mixture

(water or even nitric acid).

(80–90%)

HNO

(87%)

–

–20 ⬚C

FIGURE 14.30 Two nitration

reactions. In each case the electrophile

is the nitronium ion,



14.4f Halogenation Unlike alkenes, aromatic rings fail to react with Br

. It takes a catalyst, usually FeBr

or FeCl

, to initiate the reaction. Chlorine, by

itself, is not a strong enough electrophile (Lewis acid) to react with the weakly

nucleophilic benzene. The function of the catalyst is to convert chlorine into a

stronger Lewis acid that can react with the aromatic compound. Ferric chloride

(FeCl

) does this by forming a complex with chlorine (Fig. 14.31).

FeCl

–

FIGURE 14.31 Complex formation

from Cl

and FeCl

Benzene is nucleophilic enough to displace the good leaving group FeCl



from

the complex (Fig. 14.32). Displacement generates a cyclohexadienyl

cation that can be deprotonated to give chlorobenzene. The first step of this reac-

tion may look strange, but it has a most familiar analogy in the protonation of an

alcohol to convert OH into the good leaving group (Fig. 14.32). The FeCl

plays the role of the proton in the alcohol reaction by converting the Cl into the

good leaving group . The mechanism of bromination using Br

and

FeBr

is similar.

FeCl

–

FeCl

–

Cl Cl

Analogy

ROH

–

HCl

WEB 3D

FIGURE 14.32 Chlorination of benzene.

PROBLEM 14.7 Write a mechanism for the bromination of benzene using Br

/FeBr