716 CHAPTER 15 • DIENES, RESONANCE, AND AROMATICITY...15.7 INTRODUCTION TO AROMATIC COMPOUNDS 717...

716 CHAPTER 15 • DIENES, RESONANCE, AND AROMATICITY

PROBLEMS15.27 In each of the following sets, show by the curved-arrow or fishhook notation how each reso-

nance structure is derived from the other one, and indicate which structure is more importantand why.(a) (b)

(c)

(d)

15.28 Show the 2p orbitals, and indicate the orbital overlap symbolized by the resonance structuresfor the carbocation in Eq. 15.32 on p. 711.

15.29 Using resonance arguments, state which ion or radical within each set is more stable. Explain.

(a)

(b)

(c)

15.30 The following isomers do not differ greatly in stability. Predict which one should react morerapidly in an SN1 solvolysis reaction in aqueous acetone. Explain.

15.7 INTRODUCTION TO AROMATIC COMPOUNDS

The term aromatic, as we’ll come to understand it in this section, is a precisely defined struc-tural term that applies to cyclic conjugated molecules that meet certain criteria. Benzene andits derivatives are the best known examples of aromatic compounds.

benzene

S (#"

"

"%%

%% %

%CC

C

CC

C

H

H

H

H

H

H

or r

CH LLH3C CH2Cl"

AC C$ $

H

A B

H

CH3

) )CHLH3CCl"AC C

$ $H

H

CH2CH3

) )

C|L orH2C CH2 |CH2

CH3"AA L CH LCHH3C

AACH L LCH orH2C CH2CH 8 AL LCH2C CH2CH2

CH8 S

3 32AL CH

O _L CH orH3C CH2"

3 32AL C

O _LCHH3C CH3"

NC

CH3

.. NC

CH3

LCH CH2A | L CH CH2A|(CH3)2C (CH3)2CLH3C NH|CA 2 LH3C NH|C'C O...... C O....

carbon monoxide

15_BRCLoudon_pgs5-0.qxd 12/9/08 12:22 PM Page 716

15.7 INTRODUCTION TO AROMATIC COMPOUNDS 717

The origin of the term aromatic is historical: many fragrant compounds known from earliesttimes, such as the following ones, proved to be derivatives of benzene.

Although it is known today that benzene derivatives are not distinguished by unique odors, theterm aromatic—which has nothing to do with odor—has stuck, and it is now a class name forbenzene, its derivatives, and a number of other organic compounds.

Development of the theory of aromaticity was a major theoretical advance in organic chem-istry because it solved a number of intriguing problems that centered on the structure and reac-tivity of benzene. Before considering this theory, let’s see what some of these problems were.

A. Benzene, a Puzzling “Alkene”

The structure used today for benzene was proposed in 1865 by August Kekulé (p. 47), whoclaimed later that it came to him in a dream. The Kekulé structure portrays benzene as a cyclic,conjugated triene. Yet benzene does not undergo any of the addition reactions that are associ-ated with either conjugated dienes or ordinary alkenes. Benzene itself, as well as benzene ringsin other compounds, are inert to the usual conditions of halogen addition, hydroboration, hydra-tion, or ozonolysis. This property of the benzene ring is illustrated by the addition of bromine tostyrene, a compound that contains both a benzene ring and one additional double bond:

The noncyclic double bond in styrene rapidly adds bromine, but the benzene ring remains un-affected, even if excess bromine is used. This lack of alkenelike reactivity defined the unique-ness of benzene and its derivatives to early chemists.

Does benzene’s lack of reactivity have something to do with its cyclic structure? Cyclo-hexene, however, adds bromine readily. Perhaps, then, it is the cyclic structure and the con-jugated double bonds that together account for the unusual behavior of benzene. However,1,3,5,7-cyclooctatetraene (abbreviated in this text as COT) adds bromine smoothly even atlow temperature.

(15.39)Br2-55 °CCHCl3

+ "L"L BrBr

H

H

(100% yield)1,3,5,7-cyclooctatetraene(COT)

(15.38)cL ACHstyrene

CH2 cL LCHBr

CH2Br2+ "Br"

OCH3

OCH3

%%CH3H3C CH

O

O

A

A

CH

OH

i"" %

vanillin(vanilla)

%LC

OHi"

methyl salicylate(oil of wintergreen)

SCH3

i""

p-cymene(cumin and thyme)

%H2C CH isaffrole

(oil of sassafras)

O

O



Thus, the Kekulé structure clearly had difficulties that could not be easily explained away, butthere were some ingenious attempts. In 1869, Albert Ladenburg proposed a structure for ben-zene, called both Ladenburg benzene and prismane, that seemed to overcome these objections.

Although Ladenburg benzene is recognized today as a highly strained molecule (it has beendescribed as a “caged tiger”), an attractive feature of this structure to nineteenth-centurychemists was its lack of double bonds.

Several facts, however, ultimately led to the adoption of the Kekulé structure. One of themost compelling arguments was that all efforts to prepare the alkene 1,3,5-cyclohexatrieneusing standard alkene syntheses led to benzene. The argument was, then, that benzene and1,3,5-cyclohexatriene must be one and the same compound. The reactions used in these routesreceived additional credibility because they were also used to prepare COT, which, as Eq.15.39 illustrates, has the reactivity of an ordinary alkene.

Although the Ladenburg benzene structure had been discarded for all practical purposesdecades earlier, its final refutation came in 1973 with its synthesis by Professor Thomas J.Katz and his colleagues at Columbia University. These chemists found that Ladenburg ben-zene is an explosive liquid with properties that are quite different from those of benzene.

How, then, can the Kekulé “cyclic triene” structure for benzene be reconciled with the factthat benzene is inert to the usual reactions of alkenes? The answer to this question will occupyour attention in the next three parts of this section.

B. Structure of Benzene

The structure of benzene is given in Fig. 15.10a. This structure shows that benzene has one typeof carbon–carbon bond with a bond length (1.395 Å) that is the average the lengths of sp2–sp2

single bonds (1.46 Å) and double bonds (1.33 Å, Fig. 15.10c). All atoms in the benzene mole-cule lie in one plane. The Kekulé structure for benzene shows two types of carbon-carbon bond:single bonds and double bonds. This inadequacy of the Kekulé structure can be remedied, how-ever, by depicting benzene as the hybrid of two equally contributing resonance structures:

Benzene is an average of these two structures; it is one compound with one type ofcarbon–carbon bond that is neither a single bond nor a double bond, but something in be-tween. A benzene ring is often represented with either of the following hybrid structures,which show the “smearing out” of double-bond character:

` ` hybrid structures of benzene

(15.40)i r1.33 Å 1.395 Å

1.395 Å1.395 Å

1.395 Å

1.46 Å

Kekulé structure resonance hybrid

Ladenburg benzeneor prismane



As with other resonance-stabilized molecules, we’ll continue to represent benzene as one ofits resonance contributors because the curved-arrow notation and electronic bookkeeping de-vices are easier to apply to structures with fixed bonds.

It is interesting to compare the structures of benzene and 1,3,5,7-cyclooctatetraene (COT)in view of their greatly different chemical reactivities (Eqs. 15.38 and 15.39). Their structuresare remarkably different (Fig. 15.10). First, although benzene has a single type of carbon–car-bon bond, COT has alternating single and double bonds, which have almost the same lengthsas the single and double bonds in 1,3-butadiene. Second, COT is not planar like benzene, butinstead is tub-shaped.

The p bonds of benzene and COT are also different (Fig. 15.11, p. 720). The Kekulé struc-tures for benzene suggest that each carbon atom should be trigonal, and therefore sp2-hy-bridized. This means each carbon atom has a 2p orbital (Fig. 15.11a). Because the benzenemolecule is planar, and the axes of all six 2p orbitals of benzene are parallel, these 2p orbitalscan overlap to form six p molecular orbitals. The bonding p molecular orbital of lowest en-ergy is shown in Fig. 15.11b. (The other five p molecular orbitals of benzene are shown inFurther Exploration 15.2 in the Study Guide.) This molecular orbital shows that p-electron

H

H

H H

H H

1.39

5 Å

1.08

Å

120°

1.09

Å

1.344 Å

1.467 Å

119.5°122.9°C

CHH

H

H

CC

H

H

(a)

(b)

(c) (d)

1,3,5,7-cyclooctatetraene (COT)

1,3-butadiene

benzene

118.3°

C C

C C

H H

H

HH

HH

H

C

C

C

C 126.5°

1.334 Å

1.462

Å

1.09

Å

ball-and-stick model of COT

Figure 15.10 Comparison of the structures of benzene, 1,3-butadiene, and COT. (a) The structure of benzene.(“Double bonds” are not shown.) (b) The structure of 1,3-butadiene, a conjugated diene. (c) Structure of 1,3,5,7-cyclooctatetraene (COT). (d) A ball-and-stick model of COT. The carbon skeleton of benzene is a planar hexagonand all of the carbon–carbon bonds are equivalent with a bond length that is the average of the lengths of car-bon–carbon single and double bonds in COT. In contrast, COT has distinct single and double bonds with lengthsthat are almost the same as those in 1,3-butadiene, and COT is tub-shaped rather than planar.

Further Exploration 15.2The p Molecular

Orbitals of Benzene



density in benzene lies in doughnut-shaped regions both above and below the plane of thering. This overlap is symbolized by the resonance structures of benzene. In contrast, the car-bon atoms of COT are not all coplanar, but they are nevertheless all trigonal. This means thatthere is a 2p orbital on each carbon atom of COT (Fig. 15.11c). The tub shape of COT forcesthe 2p orbitals on the ends of each single bond to be oriented at a 63° angle, which is too farfrom coplanarity for effective interaction and overlap (Fig. 15.11d). Thus, the 2p orbitals inCOT cannot form a continuous p molecular orbital analogous to the one in benzene. Instead,COT contains four p-electron systems of two carbons each. As far as the p electrons are con-cerned, COT looks like four isolated ethylene molecules! Because there is no electronic over-lap between the p orbitals of adjacent double bonds, COT does not have resonance structuresanalogous to those of benzene (Sec. 15.6B, guideline 5).

(15.41)NO!

(a) (b)

(c) (d)

63°orbital overlap forms

localized p bondsp

orbital overlap formsa localized p bondp

angle between the 2p orbitalsprevents overlap acrossthe single bonds

Figure 15.11 Comparison of the p bonds in benzene and 1,3,5,7-cyclooctatetraene (COT). (a) The carbon 2p or-bitals in benzene.These orbitals are properly aligned for overlap. (b) The bonding pmolecular orbital of lowest en-ergy in benzene.This molecular orbital illustrates that p-electron density lies in a doughnut-shaped region aboveand below the plane of the benzene ring. (Benzene has two other occupied pmolecular orbitals as well as threeantibonding pmolecular orbitals not shown here.) (c) The carbon 2p orbitals of COT.These orbitals can overlap inpairs to form isolated p bonds, but the tub shape prevents the overlap of 2p orbitals across the single bonds. (d)The view down a single bond indicated by the eyeball in (c). The angle between 2p orbitals, indicated by the redbracket in (c), is about 63°, which is too large for effective overlap.



To summarize: resonance structures can be written for benzene, because the carbon 2p or-bitals of benzene can overlap to provide the additional bonding and additional stability asso-ciated with filled bonding molecular orbitals. Resonance structures cannot be written for COTbecause there is no overlap between 2p orbitals on adjacent double bonds.

Why doesn’t COT flatten itself to allow overlap of all its 2p orbitals? We’ll return to thispoint in Sec. 15.7E.

C. Stability of Benzene

As mentioned earlier in this section, chemists of the nineteenth century considered benzene to beunusually stable because it is inert to reagents that react with ordinary alkenes. However, chemi-cal reactivity (or the lack of it) is not the way that we measure energy content. As we have alreadylearned, the more precise way to relate molecular energies is by their standard heats of formationDH°f . Because benzene and COT have the same empirical formula (CH), we can compare theirheats of formation per CH group. The DH°f of benzene is 82.93 kJ mol_1or 82.93Ü6 = 13.8kJ mol_1 per CH group. The DH°f of COT is 298.0 kJ mol_1 or 298.0Ü8 = 37.3 kJ mol_1 per CHgroup. Thus, benzene, per CH group, is (37.3 - 13.8) = 23.5 kJ mol_1 more stable than COT. Itfollows that benzene is 23.5 X 6 = 141 kJ mol_1 (33.6 kcal mol_1) more stable than a hypothet-ical six-carbon cyclic conjugated triene with the same stability as COT.

This energy difference of about 141 kJ mol_1 or 34 kcal mol_1 is called the empirical res-onance energy of benzene. The empirical resonance energy is an experimental estimate ofjust how much special stability is implied by the resonance structures for benzene—thus thename “resonance energy.”

The resonance energy is the energy by which benzene is stabilized; it is therefore an energy that ben-zene “doesn’t have.” The empirical resonance energy of benzene has been estimated in several differ-ent ways; these estimates range from 126 to 172 kJ mol_1 (30 to 41 kcal mol_1). (Another estimate isdiscussed in Sec. 16.6.) The important point, however, is not the exact value of this number, but thefact that it is large.

D. Aromaticity and the Hückel 4n | 2 RuleWe’ve now learned that benzene is unusually stable, and that this stability seems to be corre-lated with the overlap of its carbon 2p orbitals to form p molecular orbitals. In 1931, ErichHückel (1896–1980), a German chemical physicist, elucidated with molecular orbital argu-ments the criteria for this sort of stability, which has come to be called aromaticity. UsingHückel’s criteria, we can define aromaticity more precisely. (Remember again that aromatic-ity in this context has nothing to do with odor.) This definition has allowed chemists to rec-ognize the aromaticity of many compounds in addition to benzene.

A compound is said to be aromatic when it meets all of the following criteria:

Criteria for aromaticity:

1. Aromatic compounds contain one or more rings that have a cyclic arrangement of p or-bitals. Thus, aromaticity is a property of certain cyclic compounds.

2. Every atom of an aromatic ring has a p orbital.3. Aromatic rings are planar.4. The cyclic arrangement of p orbitals in an aromatic compound must contain 4n + 2 p

electrons, where n is any positive integer (0, 1, 2, . . .). In other words, an aromatic ringmust contain 2, 6, 10, . . . p electrons.

These criteria are often called collectively the Hückel 4n | 2 rule or simply the 4n | 2 rule.The basis of the 4n + 2 rule lies in the molecular orbital theory of cyclic p-electron systems.

The theory holds that aromatic stability is observed only with continuous cycles of p orbitals—thus, criteria 1 and 2. The theory also requires that the p orbitals must overlap to form p mole-



cular orbitals. This overlap requires that an aromatic ring must be planar; p orbitals cannot over-lap in rings significantly distorted from planarity—thus, criterion 3. The last criterion has to dowith the number of p molecular orbitals and the number of electrons they contain. Therefore,to understand criterion 4, we need to know the energies and electron occupancies of the variousp molecular orbitals. Two Northwestern University physical chemists, A. A. Frost and BorisMusulin, described in 1953 a simple graphical method for deriving the p-molecular orbital en-ergies of cyclic p-electron systems without resorting to any of the mathematics of quantum the-ory. This method has come to be known as the Frost circle.

1. For a cyclic conjugated hydrocarbon or ion with j sides (and therefore j overlapping 2p or-bitals), inscribe a regular polygon of j sides within a circle of radius 2bwith one vertex ofthe polygon in the vertical position. (Remember from Sec. 15.1B that b is an energy unitused with molecular orbitals.)

2. Place an MO energy level at each vertex of the polygon. Because there are j vertices, therewill be j MOs.

3. The lowest energy level must lie at 2b because it is at the lowest vertex, which is at theend of the vertical radius. The energies of the other levels are calculated by determiningtheir positions along the vertical radius by trigonometry.

4. Add the p electrons to the energy levels in accordance with the Pauli principle and Hund’srules.

Study Problem 15.4 illustrates the application of the Frost circle to benzene.

Study Problem 15.4 Use the Frost circle to determine the energy levels and electron occupancies for the pMOs ofbenzene.

Step 1 For benzene, j = 6. Therefore, we inscribe a regular hexagon into a Frost circle of radius 2b withone vertex in the vertical position. (Fig. 15.12a).

Step 2 Place an energy level at each vertex. This gives six energy levels.

Step 3 Calculate the energies. The energies of p1 and p 6* clearly lie on the circle, so their energies must be2b and -2b, respectively. (Remember that b is a negative number.) Then draw a perpendicular fromthe p2 (or p3) vertex to the vertical radius and calculate the E(p2), the energy of p2, as follows:

From this calculation, the energies of p2 and p3, which are identical, equal +1.0b. By symmetry,the energies of p4* and p5* are -1.0b.

Step 4 Add the p electrons to the energy levels. Benzene has six p electrons. Because each bonding MOcan accommodate two electrons, the available electrons exactly fill the bonding MOs.

This method shows that benzene has six MOs—something we already knew. But it also showsthat two bonding MOs, p2 and p3, have identical energies, and two antibonding MOs, p 4* and p 5*,also have identical energies. When orbitals have the same energy, they are said to be degenerate.Hence, p2 and p3 are degenerate MOs and p 4* and p 5* are degenerate MOs.

60°90°

r = 2b

= r cos 60° = (2b)(0.50) = 1bE(p2) E(p2)



The p-electron energy of benzene is 8.0b. The p-electron energy of three isolated ethylenes(with a bonding-MO energy of 1.0b) is 6.0b. The delocalization energy, or resonance energy, ofbenzene is then 2.0b. If we equate this to the empirical resonance energy of benzene (Sec.15.7C), which is 141–150 kJ mol_1 or 34–36 kcal mol_1, we find that b = 70–75 kJ mol_1(17–18 kcal mol_1). The resonance energy is a consequence of the very low-lying p1 MO; theother bonding MOs, p2 and p3, have the same energy (1.0b) as the bonding MO of ethylene. ThisMO, shown in Fig. 15.11b, has no nodes, and it most closely corresponds to the resonance-hy-brid structure of benzene, which shows the p electrons spread evenly around the entire molecule.

PROBLEMS15.31 Use a Frost circle to determine the p-electron structure of (a) the cyclopentadienyl anion,

which has a planar structure and six p electrons; and (b) the cyclopropenyl cation, which hastwo p electrons.

15.32 How many bonding MOs are there in a planar, cyclic, conjugated hydrocarbon that contains aring of 10 carbon atoms? How many p electrons does it have? How many of the p electronscan be accommodated in the bonding MOs?

..

cyclopentadienyl anion cyclopropenyl cation

radius = 2b +2.0b

–2.0b

–1.0b

+1.0b+1.0b

p2

p2 p3

p3

p1

radius = 2b +2.0b

p1

E = 0

E = 0

p6*

p4*

p4*

p5*

–1.0b

ENER

GY

ENER

GY

benzene

1,3-cyclobutadiene

(a)

(b)

Figure 15.12 Application of the Frost circle to find the MOs and their energies for cyclic conjugated hydrocar-bons. The Frost circle is in blue. (a) The Frost circle for benzene. (b) The Frost circle for 1,3-cyclobutadiene, an an-tiaromatic compound. (Antiaromatic compounds are discussed in Sec. 15.7E.) Notice that p2 and p3 are not bond-ing and are only half filled



Compounds (and ions) with 4n + 2 p electrons contain exactly the number of electrons re-quired to fill the bonding MOs. Benzene has six p electrons (4n + 2 = 6 for n = 1); as we foundin Study Problem 15.4, these exactly fill the bonding MOs of benzene. A planar, cyclic conju-gated hydrocarbon with 10 p electrons (see Problem 15.32, p. 723) has five bonding MOs,which can accommodate all 10 electrons (4n + 2 = 10 for n = 2). A molecule that containsmore than 4n + 2 p electrons, even if it could meet all of the other criteria for aromaticity, musthave one or more electrons in antibonding MOs. If a molecule contains fewer than 4n + 2 elec-trons, its bonding molecular orbitals are not fully populated, and its resonance energy (delocal-ization energy) is reduced. But there’s more to aromaticity than just fully occupied bondingMOs; after all, 1,3-butadiene and other conjugated acyclic hydrocarbons also have fully occu-pied bonding MOs (Fig. 15.1). The bonding molecular orbitals in aromatic compounds haveparticularly low energy, especially the MO at E = 2.0b. The resonance energy of benzene is2.0b, but the resonance energy of (E)-1,3,5-hexatriene, the acyclic conjugated triene, is 1.0b(Problem 15.2, p. 680, or Fig. 15.7, p. 688). Moreover, the magnitude of b for acyclic conju-gated hydrocarbons (-50 kJ mol_1) is only two-thirds of that for cyclic conjugated hydrocar-bons. (Simple MO theory does not account for this difference, but more advanced theories do.)This difference further increases the energetic advantage of aromaticity. To summarize the basisof the 4n + 2 rule:

1. Cyclic conjugated molecules and ions with 4n + 2 p electrons have exactly the rightnumber of p electrons to fill the bonding MOs.

2. The bonding MOs in cyclic conjugated molecules and ions, especially the bonding MOof lowest energy, have a very low energy. For this reason, cyclic conjugated moleculesand ions have a large resonance energy.

Recognizing aromatic compounds is a matter of applying the four criteria for aromaticity.This objective is addressed in Study Problem 15.5 and the discussion that follows it.

Study Problem 15.5Decide whether each of the following compounds is aromatic. Explain your reasoning.(a) (b) (c)

(d) (e)

Solution In each example, first count the p electrons by applying the following rule: Each dou-ble bond contributes two p electrons. Then apply all of the criteria for aromaticity.

(a) The ring in toluene, like the ring in benzene, is a continuous planar cycle of six p elec-trons. Hence, the ring in toluene is aromatic. The methyl group is a substituent group onthe ring and is not part of the ring system. Because toluene contains an aromatic ring, it isconsidered to be an aromatic compound. This example shows that parts of molecules canbe aromatic, or, equivalently, that aromatic rings can have nonaromatic substituents.

(b) Although 1,3,5-hexatriene contains six p electrons, it is not aromatic, because it fails crite-rion 1 for aromaticity: it is not cyclic. Aromatic species must be cyclic.

(c) Biphenyl has two rings, each of which is separately aromatic. Hence, biphenyl is an aro-matic compound.

1,3-cyclobutadiene

Z1,3,5-cycloheptatriene

vLcbiphenyl

CH CH CH2L CHLA ACHAH2C1,3,5-hexatriene

v CH3Ltoluene



(d) Although 1,3,5-cycloheptatriene has six p electrons, it is not aromatic, because it failscriterion 2 for aromaticity: one carbon of the ring does not have a p orbital. In other words,the p-electron system is not continuous, but is interrupted by the sp3-hybridized carbon ofthe CH2 group.

(e) 1,3-Cyclobutadiene is not aromatic. Even though it is a continuous cyclic system of 2porbitals, it fails criterion 4 for aromaticity: it does not have 4n + 2 p electrons.

Aromatic Heterocycles Aromaticity is not confined solely to hydrocarbons. Some het-erocyclic compounds (Sec. 8.1C) are aromatic; for example, pyridine and pyrrole are both aro-matic nitrogen-containing heterocycles.

Except for the nitrogen in the ring, the structure of pyridine closely resembles that of benzene.Each atom in the ring, including the nitrogen, is part of a double bond and therefore contributesone p electron. How does the electron pair on nitrogen figure in the p-electron count? Thiselectron pair resides in an sp2 orbital in the plane of the ring (see Fig. 15.13a on p. 726). (It hasthe same relationship to the pyridine ring that any one of the CLH bonds has.) Because thenitrogen unshared pair does not overlap with the ring’s p-electron system, it is not included inthe p-electron count. Thus vinylic electrons (electrons on doubly bonded atoms) are notcounted as p electrons.

In pyrrole, the electron pair on nitrogen is allylic (Fig. 15.13b). The nitrogen has a trigonalgeometry and sp2 hybridization that allow its electron pair to occupy a 2p orbital and contributeto the p-electron count. The NLH hydrogen lies in the plane of the ring. In general, allylicelectrons are counted as p electrons when they reside in orbitals that are properly situated foroverlap with the other p orbitals in the molecule. Therefore, pyrrole has six p electrons—fourfrom the double bonds and two from the nitrogen—and is aromatic.

Note carefully the different ways in which we handle the electron pairs on the nitrogens ofpyridine and pyrrole. The nitrogen in pyridine is part of a double bond, and the electron pairis not part of the p-electron system. The nitrogen in pyrrole is allylic and its electron pair ispart of the p-electron system.

Aromatic Ions Aromaticity is not restricted to neutral molecules; a number of ions are aro-matic. One of the best characterized aromatic ions is the cyclopentadienyl anion:

H

HH

H H

2"

"

" "

"

2,4-cyclopentadien-1-ide anion(cyclopentadienyl anion; aromatic)

_

N

H

2"pyrrole

(aromatic)

pyridine(aromatic)

r2N



(The Frost circle for this ion was considered in Problem 15.31a.) The cyclopentadienyl anionresembles pyrrole; however, because the atom bearing the allylic electron pair is carbon ratherthan nitrogen, its charge is -1. One way to form this ion is by the reaction of sodium with theconjugate acid hydrocarbon, 1,3-cyclopentadiene; notice the analogy to the reaction of Nawith H2O.

The cyclopentadienyl anion has five equivalent resonance structures; the negative charge canbe delocalized to each carbon atom:

These structures show that all carbon atoms of the cyclopentadienyl anion are equivalent. Forthis reason, the cyclopentadienyl anion is sometimes represented with a hybrid structure:

Because of the stability of this anion, its conjugate acid, 1,3-cyclopentadiene, is an unusuallystrong hydrocarbon acid. (Remember: The more stable the conjugate base, the more acidic is

_hybrid structure of the cyclopentadienyl anion

(15.43)

_4 ! _3 _4

_ 2! _3

(15.42)

H H

2 2 Na2–3 h; 0 °C

THF+$$H"2 H2+Na

| _2

1,3-cyclopentadienenot aromatic

cyclopentadienyl anionaromatic

N

sp2 orbital2p orbital..

. .

. . ..

(b) pyrroleunshared pair is part of the

4n + 2 p-electron system

N H

(a) pyridineunshared pair is not part of the

4n + 2 p-electron system

4n + 2 p-electron system 4n + 2 p-electron system

Figure 15.13 The 2p orbitals in pyridine and pyrrole. The grey lines represent orbital overlap. (a) The unsharedelectron pair in pyridine is vinylic and is therefore in an sp2 orbital (blue) and is not part of the aromatic p-electronsystem. (b) The unshared electron pair in pyrrole is allylic and can occupy a 2p orbital (blue) that is part of the aro-matic p-electron system.



the conjugate acid; Fig. 3.2, p. 113.) With a pKa of 15, this compound is 1010 times more acidic

than a 1-alkyne, and about as acidic as water!Cations, too, may be aromatic. (see Problem 15.31b.)

This example illustrates another point about counting electrons for aromaticity: atoms withempty p orbitals are part of the p-electron system, but they contribute no electrons to the p-electron count. Because this cation has two p electrons, it is aromatic (4n + 2 = 2 for n = 0).The stability of the cyclopropenyl cation, despite its considerable angle strain, is a particularlystrong testament to the stabilizing effect of aromaticity.

Counting p electrons accurately is crucial for successfully applying the 4n + 2 rule. Let’ssummarize the rules for p-electron counting.

1. Each atom that is part of a double bond contributes one p electron.2. Vinylic unshared electron pairs do not contribute to the p-electron count.3. Allylic unshared electron pairs contribute two electrons to the p-electron count if they

occupy an orbital that is parallel to the other p orbitals in the molecule.4. An atom with an empty p orbital can be part of a continuous aromatic p-electron sys-

tem, but contributes no p electrons.

Aromatic Polycyclic Compounds The Hückel 4n + 2 rule applies strictly to singlerings. However, a number of common fused bicyclic and polycyclic compounds, such asnaphthalene and quinoline, are also aromatic:

Although rules have been devised to predict the aromaticity of fused-ring compounds, theserules are rather complex, and we need not be concerned with them. It is certainly not difficultto see the resemblance of these two compounds to benzene, the best-known aromatic com-pound.

Aromatic Organometallic Compounds Some remarkable organometallic compoundshave aromatic character. For example, the cyclopentadienyl anion, discussed previously in thissection as one example of an aromatic anion, forms stable complexes with a number of transi-tion-metal cations. One of the best known of these complexes is ferrocene, which is synthesizedby the reaction of two equivalents of cyclopentadienyl anion with one equivalent of ferrous ion(Fe2|).

(15.45)2 FeCl2+Na| _2

2 NaCl+ferrocene

(90% yield)

Fe2| _32( (

naphthalene quinoline

N1

(15.44)L Cl + SbCl5(a Lewis

acid)

_SbCl6cyclopropenyl cation

(aromatic)

"|" | |



Although this synthesis resembles a metathesis (exchange) reaction in which two salts areformed from two other salts, ferrocene is not a salt but is a remarkable “molecular sandwich”in which a ferrous ion is imbedded between two cyclopentadienyl anions.

The red dashed lines mean that the electrons of the cyclopentadienyl anions are shared not onlyby the ring carbons but also by the ferrous ion; each carbon is bonded equally to the iron.

Let’s now return to the question posed near the beginning of this section: Why is benzeneinert in the usual reactions of alkenes? The aromaticity of benzene is responsible for its uniquechemical behavior. If benzene were to undergo the addition reactions typical of alkenes, itscontinuous cycle of 4n + 2 p electrons would be broken; it would lose its aromatic characterand much of its stability.

This is not to say, however, that benzene is unreactive under all conditions. Indeed, benzeneand many other aromatic compounds undergo a number of characteristic reactions that areconsidered in Chapter 16. However, the conditions required for these reactions are typicallymuch harsher than those used with alkenes, precisely because benzene is so stable. As you willalso see, the reactions of benzene give very different kinds of products from the reactions ofalkenes.

PROBLEMS15.33 Furan is an aromatic compound. Discuss the hybridization of its oxygen and the geometry

of its two electron pairs.

15.34 Do you think it would be possible to have an aromatic free radical? Why or why not?

15.35 Which of the following species should be aromatic by the Hückel 4n + 2 rule?(a) (b) (c) (d) (e) (f)

E. Antiaromatic Compounds

Compounds that contain planar, continuous rings of 4n p electrons, in stark contrast to aro-matic compounds, are especially unstable; such compounds are said to be antiaromatic. 1,3-Cyclobutadiene (which we’ll call simply cyclobutadiene) is such a compound; its small ring

C2H5"B

|

O211isoxazoleO

N32|

yHH

$)_3

thiophene

S21

furan(aromatic)

O12

ferrocene

Fe2|top view

Fecyclopentadienyl anions



size and the sp2 hybridization of its carbon atoms constrain it to planarity. This compound isso unstable that it can only be isolated at very low temperature, 4 K.

The Frost circle for 1,3-cyclobutadiene is shown in Fig. 15.12b, p. 723. Two of the p MOs,p2 and p3, lie at E = 0. The p-electron energy of cyclobutadiene is 4.0b, which is the same asthe p-electron energy of two isolated ethylenes. In other words, cyclobutadiene has no reso-nance energy. Moreover, Hund’s rules requires that the two degenerate MOs p2 and p3 be halfoccupied. This means that cyclobutadiene has two unpaired electrons and is therefore a dou-ble free radical! Finally, cyclobutadiene has considerable angle strain.

The overlap of p orbitals in molecules with cyclic arrays of 4n p electrons is a destabiliz-ing effect. (It could be said that antiaromatic molecules are “destabilized by resonance.”)More advanced MO calculations show that cyclobutadiene, in an effort to escape this high-en-ergy situation, distorts by lengthening its single bonds and shortening its double bonds:

As a result of this distortion, the degeneracy of p2 and p3 is removed; so, one MO lies atslightly lower energy than the other and is doubly occupied. Cyclobutadiene, in effect, con-tains localized double bonds. This distortion, while minimizing antiaromatic overlap, intro-duces even more strain than the molecule would contain otherwise. The molecule can’t win; itis too unstable to exist under normal circumstances.

Although cyclobutadiene is itself very unstable, it forms a very stable complex with Fe(0):

(In this structure, the CO groups are neutral carbon monoxide ligands.) 1,3-Cyclobutadiene hasfour p electrons and is thus two electrons short of the number (six) required for aromatic sta-bility. These two missing electrons are provided by the iron. As the resonance structure on theright in Eq. 15.46 suggests, this complex in effect consists of a 1,3-cyclobutadiene with two ad-ditional electrons—a cyclobutadienyl dianion, a six p-electron aromatic system—combinedwith an iron minus two electrons, that is, Fe2|. In effect, the iron stabilizes the antiaromaticdiene by donating two electrons, thus making it aromatic.

This section began with a comparison of the stabilities of benzene and 1,3,5,7-cyclo-octatetraene (COT). You can now recognize that COT contains a continuous cycle of 4np electrons.

1,3,5,7-cyclooctatetraene(COT)

(15.46)"cyclobutadieneiron

tricarbonyl

Fe(CO)3"cyclobutadienyldianion

Fe2|(CO)3

2–

1.54–1.55 Å

1.35–1.36 Å

1,3-cyclobutadiene


! Molecules containing conjugated double bonds haveadditional stability, relative to unconjugated isomers,that can be attributed to the continuous overlap oftheir carbon 2p orbitals to form p molecular orbitals.

! The delocalization energy, or resonance energy, of aconjugated p-electron system containing j doublebonds is the difference between its p-electron energyand the p-electron energy of j ethylenes.

! The most stable conformation of 1,3-butadiene andother conjugated dienes is the s-trans conformation,which is an anti conformation about the central singlebond of the diene unit.

! A cumulene is a compound with one or moresp-hybridized carbon atoms that are part of two dou-ble bonds. An allene is a cumulene with two cumu-lated double bonds. Adjacent p bonds in a cumuleneare mutually perpendicular; appropriately substitutedallenes are chiral.

! Heats of formation are generally in the order: conju-gated dienes < ordinary dienes < alkynes < cumu-lenes.

! Compounds with conjugated double or triple bondshave UV–visible absorptions at lmax > 200 nm.

! Each conjugated double or triple bond in a moleculecontributes 30–50 nm to its lmax. When a compoundcontains many conjugated double or triple bonds, itabsorbs visible light and appears colored.

! The intensity of the UV–vis absorption of a com-pound is proportional to its concentration (Beer’slaw). The constant of proportionality e, called themolar extinction coefficient, is the intrinsic intensityof an absorption.

! The Diels–Alder reaction is a pericyclic reaction thatinvolves the cycloaddition of a conjugated diene anda dienophile (usually an alkene). When the diene iscyclic, bicyclic products are produced.

! The diene assumes an s-cis conformation in the tran-sition state of the Diels–Alder reaction; dienes that arelocked into s-trans conformations are unreactive.

! Each component of the Diels–Alder reaction under-goes a syn-addition to the other. In many cases the


Is COT antiaromatic? It would be if it were planar. However, this molecule is large and flexibleenough that it can escape unfavorable antiaromatic overlap by folding into a tub conformation,as shown in Figs. 15.10d and 15.11c. It is believed that planar cyclooctatetraene, which is an-tiaromatic, is more than 58 kJ mol_1 (14 kcal mol_1) less stable than the tub conformation.

PROBLEMS15.36 Using the theory of aromaticity, explain the finding that A and B are diferent compounds, but

C and D identical.(That A and B are different molecules was established by Prof. Barry Carpenter and his stu-dents at Cornell University in 1980.)

15.37 Which of the compounds or ions in Problem 15.35 (p. 728) are likely to be antiaromatic?Explain.

D"DiM

D

D

MMD

D

MMD"

DMrA B C D

KEY IDEAS IN CHAPTER 15


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716 CHAPTER 15 • DIENES, RESONANCE, AND AROMATICITY...15.7 INTRODUCTION TO AROMATIC COMPOUNDS 717...

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