Jones M., Fleming S.A. Organic Chemistry

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13.15 Additional Problems 619

13.15 Additional Problems

PROBLEM 13.28 Give names for the following compounds:

PROBLEM 13.29 Write structures for the following compounds:

(a) 2,4,6-trinitrotoluene (TNT)

(b) 4-bromotoluene

(d) m-diiodobenzene

(e) 3-ethylphenol

(f) m-ethylphenol

(g) 1,2,4,5-tetramethylbenzene (durene)

(h) o-deuteriobenzenesulfonic acid (Note the “i” in “deuterio.” It

is not “deutero.”)

(i) p-aminoaniline

(j) m-chlorobenzoic acid

PROBLEM 13.30 Write arrow formalisms for the conversions

of Dewar benzene, Ladenburg benzene (prismane), benzvalene,

and 3,3′-bicyclopropenyl (Fig. 13.4) into benzene. Some of

these tasks may be quite challenging.

PROBLEM 13.31 Which of the following heterocyclic

compounds might be aromatic? Explain. It will help if

PROBLEM 13.32 Pyridine could be called azabenzene. There

are three diazabenzenes. Draw these compounds. Are they

aromatic?

PROBLEM 13.33 Which of the following hydrocarbons might

be aromatic? Explain.

PROBLEM 13.34 Which of the following ions might be

aromatic? Explain.

(e)

(f)

(c)

COOH

(d)

(a)

(b)

(a) (b) (c)

(d) (e)

(a)

(b)

(c)

(d)

(e)

(a) (b)

–

(g)(f )

you start by drawing good Lewis structures for all the

compounds.

Diazocyclopentadiene

620 CHAPTER 13 Conjugation and Aromaticity

PROBLEM 13.35 Design an aromatic heterocyclic compound that

(a) contains one boron atom.

(b) contains one boron and one oxygen atom in a six-membered

ring.

membered ring.

(d) contains two nitrogen atoms in a five-membered ring.

(e) contains three boron and three nitrogen atoms in a six-

membered ring.

PROBLEM 13.36 Try to construct an aromatic heterocycle

in which there is an oxygen and a nitrogen in a neutral six-

membered ring.

PROBLEM 13.37 The molecule 5-bromo-1,3-cyclohexadiene

is very difficult to isolate, but 1-bromo-1,3-cyclohexadiene is

much easier to obtain. Explain why there is so much difference.

PROBLEM 13.38 Carefully explain why azirines of structure 1

have never been isolated, whereas the isomeric azirines of

structure 2 are well known.

PROBLEM 13.42 Most diazo compounds are exceedingly, and

sometimes disastrously unstable. They are notorious explosives.

By contrast, the brilliant, iridescent orange compound diazocy-

clopentadiene is quite stable. Explain.

PROBLEM 13.43 Why is it that (E)-1-phenyl-2-butene reacts

only once with OsO

? Draw the product that is formed.

HBr + C



Ditropyl ether

OsO

PROBLEM 13.39 The dipole moment of THF is 1.7 D, but

that of furan is only 0.7 D. Explain.

PROBLEM 13.40 The heat of hydrogenation of 1,3,5,7-

cyclooctatetraene (COT) is about 101 kcal/mol. The heat of

hydrogenation of cyclooctene is about 23 kcal/mol. Is COT

aromatic? Explain.

PROBLEM 13.41 As early as 1891 the German chemist G.

Merling showed that the bromination of 1,3,5-cycloheptatriene

leads to a liquid dibromide. When the dibromide is heated,

HBr is evolved and a yellow solid of the formula C

(mp 203 °C) can be isolated. This compound is soluble in water,

but cannot be reisolated from water solution. Instead, ditropyl

ether is obtained. Merling could not explain what was happen-

ing, but perhaps you can.

PROBLEM 13.44 Predict the major product(s), if any, for the

bromination (Br

,CH

) of the following compounds:

(a) styrene

(b) 1-ethyl-3-nitrobenzene

(d) 3-phenyl-1-propene

(e) 3-phenylheptane

PROBLEM 13.45 The major product in the bromination

(1 equivalent of Br

in CH

) of (E,E)-1-phenyl-1,3-pentadiene

is (E)-3,4-dibromo-1-phenyl-1-pentene regardless of the reac-

tion conditions (thermodynamic or kinetic control). Show the

mechanism of the reaction and account for this selectivity.

PROBLEM 13.46 Predict the major product(s), if any, for the

reaction of HBr in THF (solvent) with each of the following:

(a) styrene

(b) (Z)-2-phenyl-2-hexene

(d) 3-phenyl-1-propene

(e) naphthalene

PROBLEM 13.47 Predict the product of the reaction with NBS

(CCl

, hν) for each of the following compounds:

(a) 3-phenyloctane

(b) 3-methyl-1-phenylbutane

(d) isopropylbenzene

(e) phenylcyclohexane

PROBLEM 13.48 Predict the major product in the reaction

between 1 equivalent of NaI and 1,5-dichloro-1-phenylhexane.

Explain the regioselectivity.

PROBLEM 13.49 Construct the π molecular orbitals for ben-

zene (p. 578) from the π molecular orbitals of allyl. Remember:

Only combinations of orbitals of comparable energy need be

considered.

13.15 Additional Problems 621

–

Anthracene

1.37 A

⬚

1.42 A

⬚

PROBLEM 13.50 Vogel and Roth’s synthesis of the bridged

cyclodecapentaene (1) is outlined below. Supply the missing

reagents (a, b, c, and d) and an arrow formalism for the

last step.

PROBLEM 13.52 Draw cyclopropene. How many types of

hydrogens are there in this molecule? Which hydrogen is more

acidic? Explain why. Use the table inside the back cover of this

book to estimate the pK

for each type of hydrogen.

PROBLEM 13.53 Explain why the 1,2-bond of anthracene is

shorter than the 2,3-bond.

PROBLEM 13.54 The nitronium ion (



) is known and

can react with simple benzenes to give nitrobenzenes. Write an

arrow formalism mechanism for this reaction.

PROBLEM 13.56 The compound shown below was synthesized

by Professor I. M. Dimm, who was astonished to find that it

was exceedingly stable. Prof. Dimm had expected world renown

for making what he thought would be a most unstable com-

pound. Dimm reasoned that with six double bonds and a triple

bond in the molecule, there would be 16 π electrons and thus,

no aromatic character (4n, n  4; not 4n  2). Where was his

reasoning wrong?

BrBr

H Cl

C(2)

Pyridine Pyrrole Imidazole

Cu(NO

)

acetic anhydride

0 ⬚C

PROBLEM 13.51 Pyridine and imidazole are modest Brønsted

bases at nitrogen, whereas pyrrole is not.

PROBLEM 13.55 The molecule shown below undergoes reac-

tion with cupric nitrate in acetic anhydride (a source of



)

to give a high yield of the substituted product shown. Write a

mechanism for this reaction.

In fact, pyrrole is protonated only in strong acid, and protonation

occurs at C(2), not on nitrogen. First, explain why pyridine and

imidazole are basic and pyrrole is not. Then, explain why eventu-

al protonation of pyrrole is at carbon, not nitrogen. Caution! Yo u

will have to write full Lewis structures to answer this question.

622 CHAPTER 13 Conjugation and Aromaticity

COOH

1. Na/NH

ether, –78 ⬚C

2. CH

3. H

O/H

PROBLEM 13.57 Provide a mechanism for the following

reaction sequence:

PROBLEM 13.58 There are many aromatic rings in nature.

You have learned about the nucleotide bases. Draw adenine,

guanine, cytosine, and thymine (p. 615). Circle any aromatic

rings in these molecules. You might need to draw resonance

structures to see the aromaticity.

PROBLEM 13.59 Even though naphthalene and anthracene

are polynuclear aromatic compounds, they undergo the

Diels–Alder reaction. For example, anthracene readily reacts

with maleic anhydride. However, the reaction of naphthalene

with maleic anhydride proceeds efficiently only under high

pressure.

100 C

9.6 kbar, 17 h

xylene

140 C, 10 min

(a)

(b)

(c)

C(CH

)

CH(CH

)

(a) Draw the structures of the Diels–Alder adducts formed in

these reactions.

(b) Explain why anthracene is so much more reactive than

naphthalene in the Diels–Alder reaction.

Hint: The calculat-

ed resonance energies of naphthalene and anthracene are

61 kcal/mol and 84 kcal/mol, respectively.

Use Organic Reaction Animations (ORA) to answer the

following questions:

PROBLEM 13.61 Select the reaction “Benzylic oxidation” on the

Semester 1B panel. Observe the reaction several times. The ani-

mation uses oxygen (O

) as the initiator of the reaction, which

also occurs in Nature. One of the reasons for the common prac-

tice of packing organic foods over nitrogen (N

) is to avoid the

reaction of O

with benzylic (and allylic) positions in natural

products present in food. Draw each step of the reaction

between O

and 2-phenylpropane. Include a termination reac-

tion. Label your steps as initiation, propagation, or termination.

PROBLEM 13.62 Watch the “Benzylic oxidation” reaction using

the SOMO track. What does SOMO stand for? Stop the reac-

tion as O

is approaching the benzylic hydrogen. Notice the

aromatic ring. What positions are showing SOMO character?

Why these carbons? Does this help you understand the reso-

nance structures? Explain.

PROBLEM 13.63 One of the final products from the benzylic

oxidation of 2-phenylpropane is phenol. See if you can show a

mechanism for the formation of phenol from the peroxide

product shown in the animation. Remember that the

oxygen–oxygen bond is very weak. The mechanism will require

a rearrangement and a second radical oxidation process.

PROBLEM 13.64 We answered questions about the “Stabilized

alkene halogenation” reaction in Chapter 10. This reaction is at

the top of the Semester 1B page. Watch it again to remind your-

self about the pathway. Bromination in this animation is shown to

occur via a syn addition. Bromination usually proceeds through an

anti addition. Why would syn addition be favored over anti addi-

tion for this reaction? Notice the benzylic-like intermediate in the

reaction. Do you suppose the napthylmethyl cation (like the one

in this animation) is more stable than a benzyl cation? Why?

PROBLEM 13.60 A certain alkylbenzene was known to have

one of the isomeric structures shown below.

Explain how the structure of the unknown compound could be

determined by oxidation with KMnO

Substitution Reactions

of Aromatic Compounds

623

14.1 Preview

14.2 Hydrogenation of Aromatic

Compounds

14.3 Diels–Alder Reactions

14.4 Substitution Reactions of

Aromatic Compounds

14.5 Carbon–Carbon Bond

Formation: Friedel–Crafts

Alkylation

14.6 Friedel–Crafts Acylation

14.7 Synthetic Reactions We Can

Do So Far

14.8 Electrophilic Aromatic

Substitution of

Heteroaromatic Compounds

14.9 Disubstituted Benzenes:

Ortho, Meta, and Para

Substitution

14.10 Inductive Effects in Aromatic

Substitution

14.11 Synthesis of Polysubstituted

Benzenes

14.12 Nucleophilic Aromatic

Substitution

14.13 Special Topic: Stable

Carbocations in “Superacid”

14.14 Special Topic: Benzyne

14.15 Special Topic: Biological

Synthesis of Aromatic Rings;

Phenylalanine

14.16 Summary

14.17 Additional Problems

AROMATIC SUBSTITUTION The most common reaction for aromatic compounds

is substitution.This process is much like a soccer substitution; one group (in this

case, a person) replaces another.

624 CHAPTER 14 Substitution Reactions of Aromatic Compounds

Energy

Reaction progress

Aromatic

compound

Transition

state

Nonaromatic

compound

Activation

energy

FIGURE 14.1 A typical Energy versus

Reaction progress diagram for the

conversion of an aromatic compound

into a nonaromatic compound.

Aromaticity is lost in this reaction,

and there is a high activation barrier.

Energy

Reaction progress

Aromatic

compound

Transition

state

Nonaromatic

compound

Activation

energy

FIGURE 14.2 A typical Energy versus

Reaction progress diagram for the

conversion of a nonaromatic compound into

an aromatic compound. Aromaticity is

regained in this reaction, and there is a

correspondingly low activation barrier.

Even electrons, supposedly the paragons of unpredictability, are tame and

obsequious little creatures that rush around at the speed of light, going

precisely where they are supposed to go. They make faint whistling

sounds that when apprehended in varying combinations are as pleasant as

the wind flying through a forest, and they do exactly as they are told. Of

this, one can be certain.

—MARK HELPRIN,

WINTER’S TALE

14.1 Preview

In Chapters 12 and 13, we examined the structural consequences of conjugation,

the overlap of 2p orbitals. This story culminates in the enormously stable aromatic

molecules, with the parent compound, benzene, being the prototype. In this chap-

ter, we will focus mostly on substitution reactions of aromatic compounds, an idea

introduced in Section 13.11a (p. 606).The chemistry of aromatic compounds is dom-

inated by a tendency to retain the stable aromatic sextet of electrons. It takes a great

deal of energy to destroy this arrangement, and it is very easily regained.

Because aromatic compounds are so stable,so low in energy, they are often sur-

rounded by especially high energy barriers, and the activation energies for their

reactions are usually formidably large. The transformation of an aromatic com-

pound into a nonaromatic compound is most often highly endothermic (Fig. 14.1),

so the transition state lies close to the relatively high energy nonaromatic prod-

uct (recall the Hammond postulate,p.351). By contrast, the conversion of a nonaro-

matic compound into an aromatic molecule is likely to be very exothermic (Fig. 14.2).

Mark Helprin (b. 1947) is an American author. Among his wonderfully imaginative novels are Winter’s Tale,

A Soldier of the Great War, and Memoir from Antproof Case.

14.1 Preview 625

Energy

Reaction progress

Aromatic

compound 1

Aromatic

compound 2

Transition

state 2

Nonaromatic

intermediate

Transition

state 1

FIGURE 14.3 A typical Energy versus

Reaction progress diagram for the

conversion of one aromatic compound

into another. A combination of

Figures 14.1 and 14.2.

In such a case, the transition state will resemble the starting material. So, if we

again recall the Hammond postulate (and why it works), we will expect that only

small activation barriers will need to be surmounted in such reactions. A com-

plete energy diagram for a typical conversion of one aromatic compound into

another is shown in Figure 14.3. Notice that it is a combination of Figures 14.1

and 14.2.

An electrophile = Lewis acid

BHB

–

FIGURE 14.4 A generic aromatic

substitution reaction.

ESSENTIAL SKILLS AND DETAILS

1. The great thermodynamic and kinetic stability of benzenes results in reactions that

ultimately preserve the aromatic sextet of π electrons.The classic aromatic substitution

reaction, introduced in Chapter 13, is greatly elaborated here. Knowing the mechanism

of electrophilic aromatic substitution and manipulating the details in order to

synthesize polysubstituted compounds are the central tasks of this chapter.

In this chapter, we will see many examples of a general reaction of aromatic com-

pounds in which E



, an electrophile, a “lover of electrons” (a Lewis acid, which is

an electron-pair acceptor), substitutes for one of the hydrogens on a benzene ring.

Note that the aromatic sextet, here represented by the circle in the hexagon, is

retained in the overall reaction (Fig. 14.4). Later in this chapter,we take up the ques-

tion of how an initial substitution on the ring affects further reaction.Some groups

make a second substitution reaction easier, some make it harder.We begin with reac-

tions that are unusual for benzene rings, simple additions in which aromaticity

is lost.

626 CHAPTER 14 Substitution Reactions of Aromatic Compounds

Energy

Measured at

–28.6 kcal/mol

Delocalization or

resonance energy

–32.9 kcal/mol

Measured at

–49.3 kcal/mol

Estimated at

–82.2 kcal/mol

3 H

Energy of

cyclohexane

FIGURE 14.5 The determination of

the delocalization, or resonance

energy, of benzene. Note the

exaggerated short double bonds and

long single bonds in cyclohexatriene.

2. Some substituents increase the rate of a subsequent substitution and generally direct it

to the ortho and para positions. Other substituents slow the rate and direct further

substitution to the meta position. Understanding the dependence of substitution

position and rate on the structure of the initial substituent is essential.

3. The resonance-stabilized cyclohexadienyl cation is an intermediate in the central

reaction of this chapter. Be sure you understand why three—and only three—of the

carbons in the ring share the positive charge.

4. The nitrobenzene–aminobenzene (aniline) pair is one important key to doing aromatic

syntheses. Be sure you exit this chapter being able to interconvert these two compounds

and being able to use both nitrobenzene and aniline in synthesis.

5. Making bonds between carbon and the aromatic ring generally requires either

Friedel–Crafts alkylation or Friedel–Crafts acylation. There are subtle differences

between the two reactions, and you’ll need to know these differences to take advantage

of Friedel–Crafts reactions in synthesis.

14.2 Hydrogenation of Aromatic Compounds

If one were to try to predict the reactions of benzene from first principles, one would

probably start by examining the reactions of other π systems and applying them to

aromatic compounds. For example, we know that hydrogenation of cyclohexene is

exothermic by 28.6 kcal/mol (p. 580). It is no great surprise to find that absorption

of 3 mol of hydrogen by benzene to give cyclohexane is exothermic as well; indeed

we examined this reaction in some detail in Chapter 13.Although the reaction is very

exothermic,it is far less exothermic than a guess based on the value of 28.6 kcal/mol

for a simple alkene would suggest. The difference between the estimated value and

the measured value was taken as a measure of the delocalization energy of benzene,

approximately 33 kcal/mol (Fig. 14.5).

Often, conditions under which alkenes are swiftly transformed into alkanes are quite

without effect on benzene and other simple aromatic compounds (Fig. 14.6a). Indeed,

benzene can be used as solvent in many hydrogenation reactions. We can eventually

14.2 Hydrogenation of Aromatic Compounds 627

5% Pd/C

25 ⬚C, H

Rh/ Pt oxides,

OHOH

(100%)

(b)

(a)

FIGURE 14.6 (a) Benzene does not hydrogenate under the same conditions as alkenes.

(b) More active catalysts can be used to hydrogenate the aromatic ring.

make the reaction occur by using high temperature and pressure, or by using especially

active catalysts, such as rhodium, ruthenium or platinum oxide (Fig. 14.6b).

The hydrogenation reaction is highly exothermic, by 49.3 kcal/mol in the case

of the parent benzene. Even though the combination of benzene and three hydro-

gen molecules is higher in energy than cyclohexane by a substantial amount, ben-

zene is protected by high activation barriers. Heat, pressure, and/or special catalysts

are needed to help overcome these barriers. Benzene rests in a valley, separated from

other valleys—even lower energy valleys—by especially high mountains.

In the language of chemistry, we say that benzene is both thermodynamically

and kinetically very stable. The valley containing benzene by itself is low (thermo-

dynamics), and the mountains surrounding it (the activation barriers, kinetics) are

very high. Benzene is even protected by these barriers from undergoing reactions

that are, ultimately, exothermic.The example of benzene plus hydrogen nicely illus-

trates this point, as the combination of these molecules lies higher in energy than

the product of hydrogenation, cyclohexane (Fig. 14.7).

Energy

Reaction progress

Transition

state

Activation energy; this high

barrier prevents formation

of cyclohexane under usual

conditions

Cyclohexane is

more stable than

benzene 3 H

3 H

FIGURE 14.7 The activation energy

barrier for cyclohexane formation is

high and makes this reaction difficult

under normal conditions.

PROBLEM 14.1 Explain why the only product of hydrogenation of benzene is

cyclohexane. Cyclohexadienes and cyclohexene must be formed first. Why can

they never be isolated, even if the reaction is interrupted before completion?

628 CHAPTER 14 Substitution Reactions of Aromatic Compounds

200 ⬚C

(40%)

WEB 3D

FIGURE 14.8 Benzenes will undergo

the Diels–Alder reaction, but only

with reactive dienophiles under

forcing conditions.

Energy

Reaction progress

Higher activation

energy for benzene

Lower activation energy

for a destabilized benzene

Benzene Dienophile

Destabilized

benzene

dienophile

Diels–Alder

adduct to

benzene

FIGURE 14.9 If benzene is

destabilized, the energy barrier for

reaction should be reduced.

(CH

)

[n ]Paracyclophanes

For large n, the ring

can be flat

A small n forces the

ring into a boat form

(CH

)

FIGURE 14.10 1,4-Bridged benzenes:

[n]paracyclophanes.

To see if all this makes sense, let’s try a thought experiment. Suppose we were

set the task of finding a way to make the Diels–Alder reaction possible for benzenes.

We might first try to use as reactive (high energy) a dienophile as possible, as in

Figure 14.8; but there is another way. We might try instead to destabilize the ben-

zene so that surmounting the activation energy barriers for Diels–Alder reaction

becomes easier. Figure 14.9 shows this idea in graphic form.

There are several nice practical examples of this idea. Imagine a benzene

ring bridged by a chain of methylene groups between the 1- and 4-positions

(Fig.14.10).Such a molecule is called an [n]paracyclophane,where n gives the

number of methylene groups in the bridge. As long as there are enough meth-

ylene groups (more than 7) the molecule behaves like a simple benzene. But as

n diminishes (molecules with n  5, 6, and 7 are known and the n  4 isomer

14.3 Diels–Alder Reactions

Benzene and other aromatic compounds contain a conjugated π system. Such mol-

ecules are known to undergo the Diels–Alder reaction (p. 544). Yet, it is extremely

difficult to induce the Diels–Alder reaction for simple benzenes,and yields are often

poor. Harsh conditions (temperature as high as 200 °C, for example) and very reactive

dienophiles such as hexafluoro-2-butyne are necessary (Fig. 14.8). The reason for

the lack of reactivity is simple: Benzene is too stable (lies in a deep energy well) and

the activation energies for this reaction are significant.