Jones M., Fleming S.A. Organic Chemistry

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12.16 Additional Problems 569

PROBLEM 12.64 Allylic halides undergo the S

2 reaction

much faster than saturated halides (Section 12.11b). There is

another substitution reaction pathway open to allylic halides

called the S

2′ reaction.

–

2ⴕ

–

As the figure shows, in this simple example it is not possible

to distinguish the two reactions. Devise an experiment that

would allow you to tell if the product were formed through the

2 or S

2′ reaction.

PROBLEM 12.65 Devise an experiment using isotopic labeling

that would determine if the S

2 or S

2′ reaction were occurring

in the reaction of Problem 12.64.

PROBLEM 12.66 Now here’s a much harder problem. The

stereochemistry of the S

2 reaction has been well worked out

and has been described in detail in Section 7.4b. The S

2′

reaction has also been investigated, and this problem asks you

to put yourself in the place of R. M. Magid (b. 1938) and his

co-workers. They used the following experiment to determine

the stereochemistry of the S

2′ reaction. Use their data to

determine whether the nucleophile approaches the double

bond from the same side as the leaving group or from the

opposite side.

(CH

)

(CH

)

(CH

)

Major

Minor

(after deprotonation of the initially

formed ammonium ions)

PROBLEM 12.67 Write an arrow formalism mechanism for the

following bimolecular process. Be sure to account for the stereo-

chemistry of the product.

CH(CH

)

CH(CH

)

OCOPh

PROBLEM 12.68 Prostaglandins are a class of very important

natural products found in all animals. They are responsible for a

wide range of physiological events such as dilation of smooth

cells, aggregation of platelets, pain sensation, cell growth,

hormone regulation, and inflammatory mediation. The

central compound in the formation of prostaglandins is called

prostaglandin H

(PGH

). PGH

is an endoperoxide.

Endoperoxides are formed through a Diels–Alder reaction in

which oxygen (O

) is the dienophile. Based on the structure of

PGH

predict the structure of the organic compound that leads

to PGH

. Hint: Find the six-membered ring, then think of the

reverse Diels–Alder-like reaction that would give O

Prostaglandin H

(PGH

)

Use Organic Reaction Animations (ORA) to answer the

following questions:

PROBLEM 12.69 Select the reaction “1,2-Hydrohalogenation

of diene” that is on the Semester 1B panel. Observe the reac-

tion several times. When the electrophilic HBr reacts with the

nucleophilic diene, an intermediate carbocation is formed.

The animation shows the end carbon being protonated. Draw

the intermediates that could result from protonation of each of

the other carbons. Convince yourself that protonation at the

end is preferred and that protonation at either end is equivalent

in this case. Now go through this exercise with 2-methyl-

1,3-butadiene. Draw each possible intermediate and decide

which is most stable. Don’t forget resonance structures!

PROBLEM 12.70 Chose the LUMO track of the

“1,2-Hydrohalogenation of diene” and stop the animation at the

intermediate (there is only one intermediate in the energy dia-

gram). Notice the nature of this molecular orbital. It is the same

as the LUMO of an allyl cation. Does there appear to be any

more LUMO density on one carbon than the other?

570 CHAPTER 12 Dienes and the Allyl System: 2p Orbitals in Conjugation

PROBLEM 12.71 Look closely at the energy diagrams for the

“1,2-Hydrohalogenation of diene” and “1,4-Halogenation of

diene” reactions. Notice there is a faint pathway in the second

half of the reaction in each case. Why is the product of the

1,4 addition lower in energy than the product of 1,2 addition?

Will the 1,4-addition always be lower? Consider the HBr

addition to 1,3-pentadiene. Is the product of 1,4-addition

lower in this example?

PROBLEM 12.72 Select the “Diels–Alder reaction.” Observe the

reaction several times. Pause the animation just as the bottom

molecule has completely entered the screen. This is the point on

the “Reaction Progress” slide of the energy diagram (the x-axis)

where the reaction is about one-fourth of the way along.The red

ball has not moved. With the reaction paused in this position,

click on LUMO and then HOMO. Based on the information

that you gain from this visualization, which of the two reagents

is the nucleophile? Which is the electrophile? Why?

Conjugation and

Aromaticity

571

13.1 Preview

13.2 The Structure of Benzene

13.3 A Resonance Picture of

Benzene

13.4 The Molecular Orbital Picture

of Benzene

13.5 Quantitative Evaluations of

Resonance Stabilization in

Benzene

13.6 A Generalization of

Aromaticity: Hückel’s 4n ⴙ 2

Rule

13.7 Substituted Benzenes

13.8 Physical Properties of

Substituted Benzenes

13.9 Heterobenzenes and Other

Heterocyclic Aromatic

Compounds

13.10 Polynuclear Aromatic

Compounds

13.11 Introduction to the Chemistry

of Benzene

13.12 The Benzyl Group and Its

Reactivity

13.13 Special Topic:

The Bio-Downside, the

Mechanism of Carcinogenesis

by Polycyclic Aromatic

Compounds

13.14 Summary

13.15 Additional Problems

BAD AIR Forest fires are a significant source of polycyclic aromatic hydrocarbons,

which make up the solid, cancer-causing particulates that come from combustion.

572 CHAPTER 13 Conjugation and Aromaticity

1,2- and

1,4-Additions

excess

1,2- and

1,4-Additions

Diels–Alder

/Pt

Diels–Alder

HBr

FIGURE 13.1 Acyclic and cyclic

dienes have similar chemical

properties.

/Pt

Very slowly formed

Benzene

“1,3,5-Cyclohexatriene”

No reaction

Diels–Alder

No reaction

WEB 3D

FIGURE 13.2 Benzene

(“1,3,5-cyclohexatriene”) does not

undergo most of the reactions of

normal alkenes.

Vikram Seth was born in Calcutta in 1952 and educated in India, Britain, and the United States. This

brilliant novel gets the MJ 10-star recommendation.

Curious, though, isn’t it, um, Patwardhan, that the number, er, six

should be, um, embodied in one of the most, er, er, beautiful, er, shapes

in all nature: I refer, um, needless to say, to the, er, benzene ring with

its single, and, er, double carbon bonds. But is it, er, truly symmetrical,

Patwardhan, or, um, asymmetrical? Or asymmetrically symmetrical,

perhaps. . . .

—VIKRAM SETH,

A SUITABLE BOY

13.1 Preview

Chapter 12 was devoted to the consequences of conjugation—the overlap of 2p

orbitals in acyclic systems. Now we will see what happens when conjugated systems

are cyclized, turned into rings. Nothing much changes if the atoms of the rings do

not maintain orbital overlap throughout the ring—if they are not fully conjugated.

1,3-Cyclohexadiene is not especially different from 1,3-hexadiene, for example

(Fig. 13.1).There are small differences in chemical and physical properties, but the

two molecules clearly belong to the same chemical family. Hydrogenation is easy,

addition reactions abound, the Diels–Alder reaction is common, and so on.

When the conjugation is continued around the 1,3-cyclohexadiene ring to give

what we might call “1,3,5-cyclohexatriene,” matters change (Fig. 13.2). None of

the reactions shown in Figure 13.1 is easy, and this molecule appears to be remark-

ably unreactive. It is clearly in a different class from 1,3-cyclohexadiene and the

acyclic dienes.

13.2 The Structure of Benzene 573

1,3,5-Cycloheptatriene

1,3,5,7-Cyclooctatetraene

Cyclobutadiene

These are normal polyenes (for example, they undergo

addition reactions and hydrogenate easily)

This molecule is extraordinarily unstable—it can

only be detected at very low temperature

WEB 3D

FIGURE 13.3 Many cyclic polyenes are either normal in their chemical properties or exceptionally unstable.

In this chapter, we will uncover the sources of these differences in stability

between cyclic polyenes and begin to see the chemical and physical consequences

of this special kind of conjugative interaction of 2p orbitals. We will encounter the

molecule benzene (Fig.13.2), in which the overlap of six carbon 2p orbitals in a ring

has great consequences for both structure and reactivity. We will also see a general-

ization of the properties that make benzene so stable, and will learn how to predict

which cyclic polyenes should share benzene’s stability and which should not.

But not all ring compounds containing double bonds behave in this way. Some

cycloalkenes react as expected; cycloheptatriene and cyclooctatetraene are examples

of this kind of cyclic molecule. Others are exceptionally unstable; cyclobutadiene is

the archetype of this kind of cyclic polyene (Fig. 13.3).

ESSENTIAL SKILLS AND DETAILS

1. The properties that determine whether or not a molecule is properly termed “aromatic”

are simple to write down but are a bit more difficult to apply. It is very important that

you be able to think about these qualities, to see why they apply to certain molecules

and why they do not apply to others. Understanding the hard-to-define quality called

aromaticity is critical.

Chemists have argued for years about the special stability called aromaticity that

attends cyclic, planar polyenes with the proper numbers of π electrons.The wonderful

quote that begins this chapter is not so far from an accurate representation of remarks

heard at many a chemical conference! However, even though it is not easy to define this

slippery concept precisely, especially in marginal cases, it is not difficult to understand

the simple examples, to see why the number of π electrons is important, and to learn

which molecules are likely to exhibit this property.

2. This chapter introduces the generic aromatic substitution reaction. We will see this

reaction again many, many times in Chapter 14, but it would be a very good idea if you

arrived at that chapter knowing the basics.

3. The Frost circle allows you to determine quickly the relative energies of the π

molecular orbitals for any planar, cyclic, fully conjugated polyene. Remember: The

polygon must be inscribed vertex down.

4. Aromaticity is not magic! It is just another stabilizing effect that can be augmented or

opposed by other effects.

13.2 The Structure of Benzene

The molecule in Figure 13.2 that might be called “1,3,5-cyclohexatriene” was iso-

lated from the thermal rendering of whale blubber in the nineteenth century.

Although the formula was established as C

[or better, (CH)

], there remained a

serious problem: What was its structure? There are many possibilities, among them

the four shown in Figure 13.4. Two of these, Dewar benzene and Ladenburg

Bicyclo[2.2.0]hexa-2,5-diene

(Dewar benzene)

Tetracyclo[2.2.0.0

2,6

3,5

]hexane

(Ladenburg benzene or prismane)

3,3'-Bicyclopropenyl

Tricyclo[3.1.0.0

2,6

]hex-3-ene

(benzvalene)

WEB 3D

FIGURE 13.4 Four possible structures

for benzene.

574 CHAPTER 13 Conjugation and Aromaticity

C C

FIGURE 13.5 The structure

of benzene, drawn as

1,3,5-cyclohexatriene.

H H

OCH

H H

Traditional, or Kekule

´,

structures

The same structures written by J. J. Loschmidt in 1861

FIGURE 13.6 Loschmidt’s

formulations of benzene structures

(see footnote).

benzene, are named for the people who suggested them, Sir James Dewar

(1842–1923) and Albert Ladenburg (1842–1911). The third, benzvalene, is more

cryptically named.The last compound in Figure 13.4, 3,3′-bicyclopropenyl, was first

made only in 1989 by a group headed by W. E. Billups (b. 1939) at Rice University.

The German chemist Friedrich August Kekulé von Stradonitz (1829–1896) sug-

gested the correct ring structure.Legend has it that Kekulé was provoked by a dream

of a snake biting its tail. Kekulé describes dozing and seeing long chains of carbon

atoms twisting and turning until one “gripped its own tail and the picture whirled

scornfully before my eyes.”Kekulé’s suggestion for the structure of benzene is shown

in cyclohexatriene form in Figure 13.5.

Kekulé gets the lion’s share of the credit, but in the mid-1800s there were others hot on the trail of structur-

al theory in general and a decent representation of benzene in particular. One such chemist was Alexandr M.

Butlerov (1828–1886) from Kazan, Russia, who may well have independently conceived of the ring structure

for benzene, and who surely contributed greatly, if quite invisibly in the West, to the development of much of

modern chemistry. Another, even more obscure contributor, was the Austrian chemist–physicist Johann Josef

Loschmidt (1821–1895) who, four years before Kekulé’s proposal in 1865, clearly wrote a ring structure for

benzene. Figure 13.6 shows how forward-looking were the structures proposed by this too-little-known

Austrian schoolteacher.

PROBLEM 13.1 Suppose Dewar, Ladenburg, and Kekulé had had a

C NMR spec-

trometer. How many signals would the various candidates shown in Figure 13.4

show in their spectra? How many would cyclohexatriene show?

But there were problems with the Kekulé formulation

as well as with the four

shown in Figure 13.4. Benzene didn’t react with halogens or hydrogen halides like

any self-respecting polyene, for example. Hydrogenation was much slower than for

13.3 A Resonance Picture of Benzene 575

other alkenes and required severe conditions. Benzene was clearly a remarkably sta-

ble compound, and the cyclohexatriene structure couldn’t account for this.

Moreover, it eventually became known that the structure of benzene was that of

a regular hexagon (i.e., equal sided). Our 1,3,5-cyclohexatriene is immediately ruled

out as a structural possibility, because it must contain alternating long single bonds

and short double bonds. We might expect normal double bonds of about 1.33 Å and

single bonds of about the length of the bond in 1,3-butadiene: 1.47 Å.

Instead, benzene has a uniform carbon–carbon bond distance of 1.39 Å. Note that

the measured bond distance, 1.39 Å, is just about the average of the double-bond

distance, 1.33 Å, and the single-bond distance, 1.47 Å (Fig. 13.7). At this point, you

can probably see the beginnings of the answer to these structural problems. Kekulé’s

static cyclohexatriene formulation doesn’t take account of resonance stabilization.

C(2)

C(3)

1.33 A

⬚

Identical bond lengths (1.39

⬚

How can this be?

1.47 A

⬚

The structure expected for

1,3,5-cyclohexatriene

(bond length differences

exaggerated)

Real structure

WEB 3D

FIGURE 13.7

Any 1,3,5-cyclohexatriene structure

must contain alternating long single

bonds and short double bonds

(shown here with exaggerated short

and long bonds). Such is not the

case for benzene, which is a

regular hexagon with a uniform

carbon–carbon bond distance

of 1.39 Å.

(a)

(b)

FIGURE 13.8 These two Kekulé

forms together provide a real

structure for benzene. The dashed

lines represent the overlap implied in

the line structures.

PROBLEM 13.2 One early problem with the Kekulé structure was that only one

1,2-disubstituted isomer of benzene could be found for any given substituent.

That is, there is only one 1,2-dimethylbenzene (o-xylene). Explain why this

observation would have been hard for Kekulé to explain.

13.3 A Resonance Picture of Benzene

Perhaps the remarkable stability of benzene can be explained just by realizing that

delocalization of electrons is strongly stabilizing.There are really two Kekulé forms

for benzene. They are resonance forms, differing from each other only in the distri-

bution of electrons but not in the positions of atoms. An orbital picture for cyclo-

hexatriene should have influenced us in this direction. Each carbon is hybridized sp

and there is, therefore, a 2p orbital on every carbon. The structure in Figure 13.8a

576 CHAPTER 13 Conjugation and Aromaticity

stresses overlap between the 2p orbitals on C(1) and C(2), C(3) and C(4),

and C(5) and C(6), but not between C(2) and C(3), C(4) and C(5), or C(6)

and C(1). This specificity is suspicious, because there is a 2p orbital on every

carbon atom and the drawing of Figure 13.8a takes no account of the symmetry

of the situation. If there is overlap between the orbitals on C(1) and C(2),

then there must be equivalent overlap between the 2p orbitals on C(2) and C(3)

as well, as shown in Figure 13.8b. Drawing both Kekulé forms emphasizes

this, and produces a regular hexagon as the combination of the two Kekulé reso-

nance forms.

So, our picture of benzene has now been elaborated somewhat to show the

orbitals. The 2p orbitals on the six carbons, extending above and below the plane

containing the carbons and their attached hydrogens, overlap to form a circular

cloud of electron density above and below the plane of the ring.

A convention has been adopted that attempts to show the cyclic overlap of

the six 2p orbitals. Instead of drawing out each double bond as in the Kekulé

forms, a circle is inscribed inside the ring (Fig. 13.9). The cyclic overlap of 2p

orbitals is nicely evoked by this formulation, but a price is paid. The Kekulé

forms are easier for bookkeeping purposes: for keeping track of bonds making

and breaking in chemical reactions. We will use the Kekulé form most often in

this book.

FIGURE 13.9 There is a ring of electron density above and below the six-membered

benzene ring. The circle in the center of the benzene ring is evocative of the cyclic overlap

of the six 2p orbitals.

CONVENTION ALERT

WORKED PROBLEM 13.3 There are three other resonance forms for benzene,

which can be included to get a slightly better electronic description of the mol-

ecule than is provided by the pair of Kekulé forms alone. In these other resonance

forms, overlap between two p orbitals on “across the ring” carbons is taken into

account. Draw these three resonance forms, called Dewar forms, showing the

orbitals involved in the “cross-ring” bond. Be careful! This problem uses bonding

that you probably haven’t seen before.

ANSWER Any resonance form represents one electronic description for a mol-

ecule and ignores all others. For example, in one resonance form for the allyl

cation, overlap between the 2p orbitals on C(1) and C(2) is emphasized, and the

2p orbital on C(3) is written as if it did not overlap. This error is taken into

(continued)

13.3 A Resonance Picture of Benzene 577

For the central bond to be a σ bond, atoms would have to move, and that cannot

be done in resonance forms. Only electrons can move.

PROBLEM 13.4 Do you think the Dewar forms will contribute strongly to the

benzene structure? Why or why not?

Summary structure

The two Kekulé forms of benzene shown in Figure 13.8 make similar approx-

imations and do a good job of giving the structure of benzene only when taken in

combination.

In the Dewar forms, 1,4-overlap is explicitly emphasized. There are three Dewar

forms that differ in which 1,4-overlap (1,4; 3,6; or 2,5) is counted.

Notice that this business is tricky. By now, your eye is trained to see this central

bond as a

σ bond, and it isn’t! As in the Kekulé forms, it is 2p/2p, π overlap:

One Dewar

resonance form

One Kekule´

resonance form

account in the other resonance form that shows overlap between C(3) and C(2),

and ignores any overlap with the 2p orbital on C(1).

578 CHAPTER 13 Conjugation and Aromaticity

PROBLEM 13.5 There is a real compound (CH)

called Dewar benzene containing

a pair of fused (they share an edge) cyclobutene rings (Dewar benzene is correctly

named bicyclo[2.2.0]hexa-2,5-diene). This molecule is not flat, and therefore not

a resonance form of benzene, but a separate molecule potentially in equilibrium

with benzene (see figure below). Make a good three-dimensional drawing of

Dewar benzene.

Energy

Three antibonding

molecular orbitals

Three bonding

molecular orbitals

Nonbonding

FIGURE 13.10 The π molecular

orbitals of benzene.

WEB 3D

13.4 The Molecular Orbital Picture of Benzene

The raw material for a calculation of the molecular orbitals of benzene is six 2p

orbitals in a ring. Linear combination of these atomic orbitals leads to a set of six

molecular orbitals: three bonding and three antibonding.Their symmetries and rel-

ative energies are shown in Figure 13.10.

There are six electrons to go into the system, one from each carbon, and these

electrons naturally occupy the three lowest-energy, bonding molecular orbitals. The

antibonding molecular orbitals are empty. If we compare benzene to a system of three

double bonds not tied together into a ring (1,3,5-hexatriene is a good model) we

can see that there are substantial energetic savings involved in the ring structure.