18 Aromatic Substitutions

Richard F. Daley and Sally J. Daley www.ochem4free.com

Organic Chemistry

Chapter 18

Aromatic Substitution Reactions 18.1 Mechanism of Aromatic Electrophilic Substitution 914 18.2 The Nitration of Benzene 917 18.3 Halogenation and Sulfonation of Benzene 920 18.4 Friedel-Crafts Alkylation and Acylation 924 18.5 Effects of Monosubstituted Arenes on Substitution 928 18.6 Rate Effects with Monosubstituted Arenes 932 18.7 Classification of Substituents 935 18.8 Friedel-Crafts Acylation 943 Synthesis of o-Benzoylbenzoic Acid 947 18.9 Multiple Substituent Effects 948 18.10 Substitution on Polycyclic Arenes 951 18.11 Diazotization 954 Synthesis of Methyl Orange 957 Sidebar - Sulfa Drugs 958 18.12 Other Diazonium Salt Reactions 961 18.13 Nucleophilic Aromatic Substitution 963 18.14 Benzyne 965 Synthesis of Trypticene 968 18.15 Synthesis Examples 969 Key Ideas from Chapter 18 975

Organic Chemistry - Ch 18 912 Daley & Daley

Copyright 1996-2005 by Richard F. Daley & Sally J. Daley All Rights Reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the prior written permission of the copyright holder.

www.ochem4free.com 5 July 2005


Chapter 18

Aromatic Substitution Reactions

Chapter Outline 18.1 Mechanism of Electrophilic Aromatic Substitution The mechanism of electrophilic substitution of benzene 18.2 The Nitration of Benzene A case study of aromatic electrophilic substitution 18.3 Halogenation and Sulfonation of Benzene The mechanism of chlorination, bromination, and sulfonation of

benzene 18.4 Friedel-Crafts Alkylation Formation of alkyl benzenes 18.5 Effects of Monosubstituted Arenes on Substitution The effects of one substituent on the position of reaction by a second

substituent 18.6 Rate Effects with Monosubstituted Arenes The effect of one substituent on the rate of reaction by a second

substituent 18.7 Classification of Substituents A listing of common substituents showing their directive and rate

controlling effects 18.8 Friedel-Crafts Acylation Formation of acyl benzenes 18.9 Multiple Substituent Effects Predicting the position of substitution when two or more substituents

are on the ring 18.10 Substitution on Polycyclic Arenes Aromatic electrophilic substitution on polycyclic aromatic compounds 18.11 Diazotization Diazotization and the use of the diazonium ion as an electrophile 18.12 Other Diazonium Salt Reactions Replacement of the diazonium ion with a variety of groups 18.13 Nucleophilic Aromatic Substitution Nucleophilic substitution on an aromatic ring 18.14 Benzyne The formation and reaction of the reactive benzyne intermediate 18.15 Synthesis Examples Organic synthesis using aromatic electrophilic substitution reactions



Objectives

✔ Understand the mechanism for aromatic electrophilic substitution reactions

✔ Recognize appropriate electrophiles that will substitute on an aromatic ring

✔ Be able to predict the position of a new substitution on an aromatic ring with one or more existing substituents

✔ Know how the structure of one substituent affects the rate of reaction for adding a second substituent on the ring

✔ Know the diazotization reaction and how the diazonium salts react

✔ Understand the nucleophilic substitution reaction and its mechanism

✔ Be able to use the reactions in this chapter in synthesis

‘Tis true; there’s magic in the web of it. . . —Shakespeare

C ha

foundatio

apter 17 presents the characteristics that make a compound romatic. Understanding those characteristics is the n for this chapter, as it examines the various types of

reactions that occur with aromatic hydrocarbons. The chapter first discusses electrophilic aromatic substitution—a major aromatic hydrocarbon mechanistic type. Electrophilic substitution allows you to directly introduce a variety of functional groups onto the aromatic ring. The chapter then looks at several examples of electrophilic substitutions on benzene and its derivatives. Much of the rest of the chapter discusses how the substituents already on the ring affect the placement of additional substituents.

18.1 Mechanism of Electrophilic Aromatic Substitution Chapter 14 discusses electrophilic addition reactions to the π bond of an alkene. The result of an electrophilic addition reaction is



that a new atom or group of atoms adds to both carbons involved in the double bond. The mechanism proceeds in two steps: The electron cloud of the π bond reacts with the electrophile to form an intermediate carbocation. A nucleophile adds to the intermediate carbocation, thus replacing the π bond with two new σ bonds.

Nu:E E ENu

The AdE2 mechanism.

Benzenoid aromatic compounds also have an electron-rich π bond cloud that is susceptible to attack by an electrophile. As with an alkene, the π electron cloud of the benzenoid aromatic compound reacts with the electrophile and adds the electrophile to one of the carbons in the ring. This reaction produces a carbocation intermediate.

Benzenoid aromatic compounds contain benzene rings or fused benzene rings.

Step 1

The complex formed by the attack of the electrophile The carbocation intermediate, called a σ complex, is not aromatic. The σ complex is written as follows to show the delocalization of the positive charge:

A σ complex is a resonance-stabilized carbocation intermediate.

E

H

+



The carbon receiving the electrophile becomes sp3 hybridized. Having an sp3 hybrid carbon in the ring interrupts the continuous overlap of the p orbitals, which prevents continuous electron delocalization all the way around the ring and causes the benzene ring to lose its aromaticity. At this point, the electrophilic addition reaction with benzene ends its similarity to an electrophilic addition with an alkene. As you may recall from Chapter 17, compounds that can become aromatic will. So it is with the σ complex. The closed shell of benzene is so stable that the σ complex rapidly loses a proton to a base, enabling the reaction to regenerate the aromatic benzene ring. If the nucleophile attacked the aromatic ring the reaction would have formed a substituted cyclohexadiene, and the benzene ring would have permanently lost its aromaticity. However, because the aromatic ring is more stable than the cyclohexadiene ring, an attack of a nucleophile on the ring would require an endothermic process rather than an exothermic process. Thus, the lower energy pathway is the loss of a proton to regenerate the now substituted benzene ring.

Step 2

Base:

Because an electrophile substitutes itself for a hydrogen on the ring, the reaction’s overall type is an electrophilic substitution instead of an electrophilic addition. This reaction type is called an electrophilic aromatic substitution. Electrophilic aromatic

substitution is a reaction in which an electrophile displaces another atom.

Exercise 18.1 The previous illustration of the second step of an electrophilic aromatic substitution shows the reaction of only one of the three resonance structures of the σ complex formed in step one. Show the second step from each of the other two resonance structures of the σ complex. In an electrophilic aromatic substitution reaction, the σ complex is an intermediate, not a transition state. The reaction is not concerted; the bond to the electrophile forms before the bond to the proton breaks. Chemists have gathered much experimental evidence to verify this process. Thus, when drawing a reaction progress



diagram, such as Figure 18.1, place the σ complex ion in a valley between the two transition states.

Reaction Progress

E

H

+

+

E

H

+

+ E

H

+ E

H

Go

G1

G2

‡

‡

Figure 18.1. Reaction progress diagram for electrophilic aromatic substitution. The reaction progress diagram in Figure 18.1 shows that the first step in the mechanism of an electrophilic substitution reaction is the rate-determining step. Note that ∆G1

‡, the free energy of activation for the reaction of benzene and the electrophile to form the σ complex, is greater than ∆G2

‡, the free energy of activation for the reaction of the σ complex with the nucleophile to form the final product. Because the resonance energy in the benzene ring is lost when forming the σ complex, the reaction of the electrophile with benzene to form the σ complex is endothermic, and it proceeds slowly. As the σ complex loses the proton and the benzene ring regains its resonance energy, the reaction is exothermic and proceeds rapidly. Electrophilic aromatic substitution is the most common method used to synthesize substituted aromatic compounds. The reaction directly introduces functional groups onto the benzene ring and works with a variety of electrophilic reagents. Sections 18.2, 18.3, and 18.4 discuss the major electrophilic aromatic substitution reactions. All these reactions follow the mechanism presented in this section.

18.2 The Nitration of Benzene Benzene reacts slowly with hot concentrated nitric acid in an electrophilic aromatic substitution reaction to form nitrobenzene. This reaction is potentially dangerous, however, because nitric acid is a strong oxidizing agent that often explodes in the presence of any material that readily oxidizes. A safer, faster, and more convenient synthesis employs a mixture of concentrated nitric acid and



concentrated sulfuric acid. The concentrated sulfuric acid acts as a catalyst allowing nitration to take place more readily at more moderate temperatures.

The nitration of benzene is the reaction of a benzenoid compound with the ⊕NO2 electrophile.

(85%)

50 - 55oCH2SO4

HNO3NO2

Nitrobenzene

The nitronium ion (⊕NO2) is the electrophile in the nitration of benzene to form nitrobenzene. Although concentrated nitric acid produces the nitronium ion by itself, the equilibrium is so far to the left that the process is slow. Adding concentrated sulfuric acid to the reaction mixture increases the concentration of the nitronium ion, thereby increasing the rate of the nitration reaction. The nitronium ion forms via a pathway similar to the first step in the dehydration of an alcohol.

Dehydration of alcohols is discussed in Section 13.9, page 000.

••

•• ••OSO3HHHO NO2 HO NO2

H

NO2

After the nitronium ion forms, it reacts with benzene to form the σ complex, the first step of the electrophilic aromatic substitution reaction. This step is slow because the σ complex is not aromatic. Additionally, the σ complex is higher in energy than the benzene and the nitronium ion.

NO2 NO2

H

+

σ complex In the next step of the mechanism, the σ complex loses a proton to form nitrobenzene. This step is rapid because the loss of a proton allows the molecule to become aromatic again.



NO2NO2

H

Chemists tested whether the loss of a proton is the fast step or the slow step of an electrophilic aromatic substitution by replacing the hydrogens in benzene with deuterium and then running the reaction. Deuterium (2H abbreviated as D) is an isotope of hydrogen (1H) that contains not only one proton in its nucleus but also one neutron. Thus, deuterium has twice the mass of hydrogen. Because the bond energy between a pair of atoms changes in proportion to the masses of the isotopes involved in that bond, the C—D bond is higher in energy than the C—H bond. This isotope effect is observable in the IR spectrum. The IR absorption of the C—H bond in benzene is approximately 3050 cm–1; whereas the C—D bond of deuteriobenzene is about 2150 cm–1.

As you may recall, wavenumbers are inversely proportional to wavelength. Thus, higher energy = lower wavenumbers. See Section 9.2, page 000 for more details.

Because breaking a C—D bond requires more energy than breaking a C—H bond, a reaction whose rate-determining step involves breaking a C—H bond proceeds more slowly when deuterium is present. Thus, replacing C6H6 with C6D6 results in a reduction of the nitration rate if the breaking of a C—H bond is the rate-determining step. With the electrophilic aromatic substitution reaction, chemists measured no difference in the rate of reaction between C6D6 and C6H6. This shows that the rate-determining step is the formation of the σ complex, not the step that breaks the C—H bond. Solved Exercise 18.1 Show a mechanism for the following reaction.

D2SO4D

Solution In this reaction, a deuterium replaces one of the hydrogens on the ring. The D2SO4 is the source of D⊕ electrophile. The formation of the σ complex involves reaction of the ring with the electrophile.

H

DD OSO3D



The DSO4

- anion removes the proton to form the final product.

OSO3DH

D D

Exercise 18.2 Write a mechanism for the formation of a nitronium ion from nitric acid alone.

18.3 Halogenation and Sulfonation of Benzene The conversion of benzene to bromobenzene or chlorobenzene by electrophilic aromatic substitution requires the presence of a Lewis acid in a mixture of benzene and the halogen (usually Cl2 or Br2). The most common laboratory procedure involves adding the halogen to benzene in the presence of some metallic iron. Chemists frequently add this iron by tossing a few carpet tacks into the reaction mixture. To provide the iron in an industrial setting, the reaction is run in iron reaction vessels. The iron itself is not actually the reaction catalyst. The iron reacts with the halogen to form a small amount of iron(III) chloride or iron(III) bromide. These iron halides are the catalysts: 2Fe + 3Cl2 2FeCl3 2Fe + 3Br2 2FeBr3 By themselves, bromine and chlorine are weakly electrophilic. However, neither halogen is electrophilic enough to react with benzene. The iron halides are Lewis acids and form complexes with the halogen atoms.

•• ••••

••

•• ••

•• ••

•• ••

••

Cl ClFeCl3 FeCl3ClCl

•• ••••

••

•• ••

•• ••

•• ••

••

Br BrFeBr3 FeBr3BrBr

The formation of the bromine–iron(III) bromide complex increases the electrophilicity of the bromine to the point that it can attack the benzene ring and form a σ complex. The next step, in which



the - FeBr4 ion removes the proton from the σ complex producing bromobenzene, HBr, and FeBr3, is quick.

••

•• ••

••

••

••••

••FeBr3BrBrH

BrFeBr3Br

Br

Figure 18.2 shows the reaction progress diagram for the bromination of benzene. Note that ∆G2

‡ < ∆G1‡, so formation of the σ complex is

the rate-determining step.

‡

‡

G2

G1

Go

+ Br

H Br

+ Br

H

Br Br

H

+

+

+

HBr

Br2

Br

Reaction Progress Figure 18.2. The reaction progress diagram for the bromination of benzene. Chlorination of benzene follows a similar route to the bromination of benzene. The catalyst, FeCl3, forms from the reaction of iron and chlorine. On some occasions, chemists also use AlCl3 as a catalyst. Section 18.4 discusses the uses of AlCl3 as a strong Lewis acid and its importance in other electrophilic substitutions.

(83%)

FeCl3Cl2

Cl

Chlorobenzene



Fluorination of benzene is a very exothermic reaction. It reacts so rapidly that it requires special conditions and reaction vessels. Even then, the reaction rarely stops with a monofluorination product.

Section 18.12, page 000, describes an indirect method of monofluorinating a benzene by using an intermediate diazonium compound.

Iodine is so unreactive that it requires special techniques to iodinate directly. These techniques involve adding iodine to the reaction mixture in the presence of a strong oxidizing reagent such as nitric acid.

Iodine may also be introduced via the diazonium compound (Section 18.12, page 000).

(86%)

HNO3

I2I

Iodobenzene

Exercise 18.3 An important technique for the introduction of fluorine onto an aromatic ring is via a two-step thallation procedure. Benzene reacts with thallium trifluoracetate (Tl(OOCCF3)3) to form an organothallium intermediate. This intermediate reacts with KF and BF3 to form the fluorobenzene.

KF, BF3Tl(OCCF3)3

OTl(OCCF3)2

O

F

Propose a mechanism for the first step of this reaction. In a sulfonation reaction, benzene reacts with sulfuric acid to produce benzenesulfonic acid.

Benzenesulfonic acid(95%)

H2SO4

SO3H



A sulfonation reaction is readily reversible and yields only low amounts of benzenesulfonic acid because the equilibrium constant favors the substrate. However, good yields of benzenesulfonic acids are obtained by removing the water from the reaction mixture. For example, an azeotropic distillation removes the water along with the unreacted benzene, leaving the benzenesulfonic acid.

Azeotropic distillation is discussed in Section 8.2, page 000. Benzene and water form an azeotrope boiling at 69.4oC having a composition of 91% benzene and 9% water.

Another way to remove the water and increase the yield of benzenesulfonic acid is to use fuming sulfuric acid as the source of the electrophile. The actual electrophile, or sulfonating agent, is sulfur trioxide regardless of whether you use sulfuric acid or fuming sulfuric acid. When you use fuming sulfuric acid, you increase the amount of sulfur trioxide available to act as the electrophile. Because the sulfur trioxide reacts with the water to form sulfuric acid, the sulfur trioxide also removes the water as soon as it forms.

Chemists form fuming sulfuric acid by adding sulfur trioxide to concentrated sulfuric acid.

SO3 + H2O H2SO4

An advantage of the sulfonation reaction is its reversibility. Simply heating benzenesulfonic acid with an aqueous acid removes the sulfonic acid group. Reversibility is a very useful synthetic tool as you can synthesize a benzenesulfonic acid, use it as a synthetic intermediate, and then remove the sulfonic acid group when it is no longer needed. The mechanism for the formation of benzenesulfonic acid follows the same general mechanism as do all electrophilic aromatic substitution reactions. The first step is the attack of benzene on the sulfur of sulfur trioxide, followed by the loss of a proton, which allows the ring to regain its aromaticity.

OSO3H

S

O

O

O

HS

OOH

OH

SO3H

Exercise 18.4 Write a mechanism to explain the presence of SO3 in sulfuric acid.



18.4 Friedel-Crafts Alkylation In 1877, Charles Friedel at the Sorbonne in Paris and James M. Crafts of the Massachusetts Institute of Technology collaborated on the development of a method to produce alkylbenzenes and acylbenzenes. Their method, the Friedel-Crafts reaction, is one of the most useful synthetic methods in organic chemistry because it allows the introduction of a carbon side chain to benzene.

The Friedel-Crafts reaction uses a carbocation or acylium ion as the electrophile with benzenoid aromatic compounds.

The formation of an alkyl benzene in a Friedel-Crafts alkylation involves benzene, an alkyl halide, and a Lewis acid. Frequently, the Lewis acid used is aluminum chloride.

CH2CH3CH3CH2Cl

AlCl3

(84%)Ethylbenzene

The mechanism for an alkylation reaction begins as the Lewis acid, in this case AlCl3, reacts with the alkyl halide to form a complex. This complex is the reacting species for a primary alkyl halide. For a tertiary alkyl halide, the complex tends to dissociate forming a free carbocation. In addition, if a carbocation-like rearrangement of the alkyl group can occur, it will.

Carbocation rearrangements are introduced in Section 12.4, page 000.

•• ••

••••

••CH3CH2 Cl

AlCl3 CH3CH2 Cl AlCl3

The benzene ring then undergoes electrophilic attack by the complex to form a σ complex—completing the first step of the electrophilic aromatic substitution reaction. Immediately following the first step, the σ complex undergoes the second step and loses a proton to form the alkyl benzene.



CH2CH3

H

CH2CH3

AlCl4CH3CH2 Cl AlCl3••

••

Rearrangement of the alkyl group can occur during a Friedel-Crafts alkylation, so the product of the alkylation is almost never exclusively a primary alkyl benzene, unless the primary carbocation cannot rearrange. For example, the reaction of 1-chloropropane with benzene in the presence of aluminum chloride produces isopropylbenzene and propyl benzene in a 2:1 ratio.

33%

67%

AlCl3

CH3CH2CH2Cl

CH(CH3)2

CH2CH2CH3

Isopropylbenzene

Propylbenzene The electrophile for the isopropylbenzene product is an isopropyl complex that formed when the alkyl halide lost the chlorine ion and a hydride shift occurred.

CH3CHCH2 Cl AlCl3

H

AlCl3CH3CH2CH2 Cl

••

•• ••

••••

CH3CHCH3

Cl AlCl3••••

••



The driving force for the rearrangement is the stability of the carbocations (3o > 2o > 1o). The rearrangement of the carbocation is a major limitation of the synthetic utility of the Friedel-Crafts reaction. The reaction generally works well if the desired product comes from the more stable carbocation.

As you may recall from Section 12.5 (page 000), more highly substituted carbocations are more stable.

Solved Exercise 18.2 What product forms in the following reaction?

CH2Cl

AlCl3

Solution In the presence of a Lewis acid catalyst like AlCl3, benzyl chloride forms the benzyl cation. The benzyl cation is the electrophile that reacts with the benzene to form diphenylmethane.

CH2Cl

AlCl3

CH2

Diphenylmethane Exercise 18.5 Propose a step-by-step mechanism for the synthesis of cyclopentylbenzene from benzene, chlorocyclopentane, and aluminum chloride. Lewis acids other than AlCl3 catalyze Friedel-Crafts reactions. These catalysts include most any Lewis acid that can form a carbocation from an alkyl halide. Some examples are FeCl3, ZnCl2, and BF3. Chapters 12 through 14 cover a number of reactions involving carbocations. Electrophilic attack on benzene by a carbocation is simply another of those reactions. Any reagent that forms a carbocation, not just the ones specifically mentioned in this chapter, can catalyze a Friedel-Crafts alkylation. For example, a mixture of propene and liquid hydrogen fluoride, being used as both solvent and proton donor, reacts with benzene to produce isopropylbenzene.



(75%)

CH(CH3)2

HF

CH3CH CH2

Isopropylbenzene

Exercise 18.6 Write a step-by-step mechanism for the reaction of cyclopentene with benzene in liquid HF. Alcohols in acidic media undergo reactions that are comparable to the reactions of alkyl halides with Lewis acids. Just like the alkyl halides, alcohol substrates also readily rearrange as they lose the —OH group to form the carbocation. For example, isobutyl alcohol with boron trifluoride (BF3) as a catalyst, reacts with benzene to produce tert-butylbenzene. The reaction produces tert-butylbenzene instead of isobutylbenzene because a hydride shift in the isobutyl alcohol occurs at the same time that it loses the —OH group, thus producing the more stable carbocation intermediate.

(CH3)2CHCH2OH

(64%)

C(CH3)3

BF3

tert-Butylbenzene

18.5 Effects of Monosubstituted Arenes on Substitution So far, you have investigated only those reactions that involve various electrophilic reagents with benzene. This section begins the discussion of electrophilic aromatic substitution reactions that occur with benzene rings already bearing a substituent. An electrophilic substitution with a monosubstituted benzene greatly affects the outcome of the reaction compared to an electrophilic substitution on a nonsubstituted benzene. The substituent already on the benzene ring affects the reaction in two important ways: it affects the regiochemistry of the incoming electrophile and the rate of the reaction. For example, the nitration of toluene produces three products in differing amounts. Only 3% of the product forms with the nitro group in the position meta to the methyl group, whereas the



remaining 97% forms in the ortho and para positions (63% ortho and 34% para.

34%3%63%p-Nitrotoluenem-Nitrotolueneo-Nitrotoluene

H2SO4

HNO3 ++

CH3

NO2

CH3

NO2

CH3

NO2

CH3

Because toluene produces predominantly ortho and para substitution products, the methyl substituent is called an ortho, para directing group.

An ortho, para directing group guides an incoming electrophile to either the ortho or para position on the ring.

Exercise 18.7 Statistically what percentage of each nitrotoluene product would you expect to get? In the nitration of (trifluoromethyl)benzene, on the other hand, 91% of the product forms with the nitro group in the position meta to the trifluoromethyl group. The other 9% of the product is divided with 6% in the ortho position and 3% in the para position.

CF3 CF3

NO2

CF3

NO2

CF3

NO2

+ +HNO3

H2SO4

o-Nitro(trifluoro-methyl)benzene

6% 91% 3%

m-Nitro(trifluoro-methyl)benzene

p-Nitro(trifluoro-methyl)benzene

(Trifluoromethyl) benzene

Because substitution in (trifluoromethyl)benzene occurs primarily at the meta position, the trifluoromethyl group is called a meta director.

A meta directing group guides an incoming electrophile to the meta position on the ring.

The nitration of toluene and (trifluoromethyl)benzene illustrates how the substituent already present on the benzene ring affects the regioselectivity of an electrophile in a substitution reaction.



The existing substituent gains its influence over the location of an incoming group by either donating electron density to or withdrawing electron density from the π electron cloud. Electron donating substituents (e.g. CH3—) direct incoming electrophiles primarily to the ortho and para positions. Electron-withdrawing substituents (e.g. CF3—) direct incoming electrophiles to the meta position. The directing effect of the existing substituent group on the arene is based on the interaction of that group with the positive charge of the σ complex. When the substituent group stabilizes the σ complex by its electron-donating qualities, it directs the incoming group primarily to the ortho and para positions. When the substituent group destabilizes the σ complex by withdrawing electron density from the benzene ring, the electrophile reacts mostly to the meta position. The following discussion of toluene gives a practical explanation as to why the ortho- and para-substituted σ complexes are more stable than the meta-substituted σ complex.

Ortho Substitution CH3

NO2

H

CH3NO2

H

CH3NO2

H

Para Substitution CH3

H NO2

CH3

H NO2

CH3

H NO2 As you look at the three resonance contributors of the σ complex for the ortho and para substitutions, notice that both complexes contain one 3o carbocation (the one with positive charge on the carbon bearing the methyl group) and two 2o carbocations. The tertiary carbocation is much more stable than either of the two secondary carbocations, so the tertiary carbocation adds overall stability to the ortho- and para-substituted σ complexes.



Para complexOrtho complex

CH3

H NO2

CH3NO2

H

Site of the more stabletertiary carbocation.

On the other hand, the three resonance structures with the meta-substituted σ complex are all secondary carbocations.

Meta Substitution CH3

NO2

H

CH3

NO2

H

CH3

NO2

H

The meta-substituted σ complex is much less stable than the ortho- and para-substituted σ complexes.

A methyl group is an electron-donating group, and although it activates all three positions relative to benzene, it activates the ortho and para positions more than the meta positions. This increased reactivity at the ortho and para sites directs the incoming electrophiles primarily to those positions. All alkyl groups are electron donating; thus, they are all ortho, para directing groups. In contrast to the methyl group, the trifluoromethyl group is strongly electron withdrawing. Because of the high electronegativity of the fluorines, the C—F bond is quite polar with the positive end of the dipole at the carbon.

The donating character of alkyl groups is introduced in Section 7.5, page 000.

CF

FF

When a resonance contributor from a trifluoromethyl σ complex has the positive charge on the carbon bearing the trifluoromethyl group, the positive charge from the trifluoromethyl dipole and the positive charge from the resonance contributor repel each other. This repelling behavior destabilizes that carbocation, and thus destabilizes the whole



σ complex. Both the ortho and the para σ complexes have one resonance contributor with a positive charge on the carbon bearing the CF3 group; whereas, none of the resonance contributors in the meta σ complex has a positive charge on the carbon bearing the CF3 group.

Ortho Substitution CF3

NO2

H

CF3NO2

H

CF3NO2

H

Para Substitution CF3

H NO2

CF3

H NO2

CF3

H NO2

Meta Substitution CF3

NO2

H

CF3

NO2

H

CF3

NO2

H

The positive charge on the carbon of the ring and the partial positive charge on the trifluoromethyl group strongly destabilize the ortho and para σ complexes.

+

These positive sitesrepel one another.

CF3NO2

H

CF3

H NO2

+

Ortho complex Para complex

An electrophilic attack at the meta position leads to a more stable intermediate σ complex than does an electrophilic attack at



either the ortho or para positions. Although the trifluoromethyl group deactivates all three positions in comparison to benzene, it deactivates the meta position less when compared to either the ortho or para positions.

18.6 Rate Effects with Monosubstituted Arenes Formation of the σ complex is the rate-determining step in electrophilic substitution reactions. Factors that affect the stability of the σ complex also affect the rate at which it forms. Thus, substituents on a benzene ring not only affect the position of an incoming electrophile takes on that ring, but also affect the rate of reaction. Some substituents speed up the rate of reaction in comparison to benzene’s rate, and some slow it down. As with regioselectivity, the more stable the σ complex, the easier the σ complex forms, and the faster it forms. The data in Table 18.1 shows the relative rates for the nitration of benzene, toluene, and (trifluoromethyl)benzene. Note that the rate of reaction for the ortho and para positions of toluene is about one million times faster than for the ortho and para positions of (trifluoromethyl)benzene. Note also that the meta position of toluene is over 30,000 times as reactive as the meta position of (trifluoromethyl)benzene. Observe that the rates for reaction at the ortho position are lower than for the para position. However, experimentally there is twice as much ortho product as para product. This is due to the fact that there are two ortho hydrogens and only one para hydrogen.

Relative Rates Benzene Toluene (Trifluoromethyl)benzene

Ortho 1 44 4.6 x 10–6

Meta 1 2.4 69. x 10–6

Para 1 59 4.5 x 10–6 Table 18.1. Relative rates for the nitration of benzene, toluene, and (trifluoromethyl)benzene. Recall from Section 18.5 that, when one resonance contributor in a σ complex possesses more stability than the rest of the resonance contributors, the more stable contributor increases the overall stability of the entire σ complex. In fact, it increases the stability of the whole chemical species. Looking again at the example of toluene, both the ortho σ complex and the para σ complex have one resonance contributor that is more stable than the other contributors. None of

An activating ortho, para directing group reacts faster than benzene and directs an incoming electrophile to the ortho and para positions.



the meta σ complex contributors are more stable than the others. Nevertheless the meta σ complex benefits from the electron-donating ability of the methyl group. Thus, the σ complex from toluene is more stable than the σ complex formed from benzene, and it reacts faster than benzene in electrophilic aromatic substitution reactions. The presence of the methyl group increases the rate of reaction of toluene at all three sites, especially the ortho and para positions. Because of this increased rate of reaction in comparison to benzene, the methyl group is called an activating ortho, para director. The greater the stability of the product of a reaction in a family of related reactions, in this case the intermediate σ complex, the less energy of activation required to form it. Figure 18.3 shows the relationship of the energies of activation for the formation of benzene’s one σ complex and toluene’s three. All three σ complexes from the reaction of toluene are more stable than the σ complex from benzene; therefore, they need less energy to form. Figure 18.3 also shows the energy relationship among the various σ complexes of toluene. Recall from Section 18.5 that 63% of the product forms in the ortho position and 34% in the para position. Thus, the ortho position has nearly twice as much substitution as does the para position. Because there are two ortho sites and only one para site, you would expect exactly a 2:1 ratio of product if both had identical reactivity. However, the methyl group sterically hinders the ortho position slightly, which causes the reaction to require more energy to place the electrophile in an ortho position.

Reaction Progress

Para

Meta

Ortho

+ NO2

CH3

NO2+

G1

NO2

H

+

‡

Figure 18.3. The relative activation energies for the formation of the σ complexes of benzene and ortho, meta, and para substitutions in toluene.



As you saw in Section 18.5, the trifluoromethyl group is electron-withdrawing in electrophilic aromatic substitution reactions and destabilizes the σ complexes in comparison to the σ complex of benzene. The trifluoromethyl group destabilizes the ortho σ complex and para σ complex the most because both have a resonance contributor with the positive charge on the carbon bearing the CF3 group. Although the meta σ complex is the most stable, it is less stable than benzene. The CF3 electron-withdrawing group destabilizes the meta σ complex because the electron-withdrawing group is only one carbon away from the destabilized ortho and para resonance contributors. All three σ complexes are less stable than benzene and therefore take longer to react. Thus, the CF3 group is called a deactivating meta director.

A deactivating meta directing group reacts slower than benzene and directs an incoming electrophile to the meta position.

Figure 18.4 shows the relationship of the activation energies and the σ complexes of benzene and the ortho, meta, and para positions of (trifluoromethyl)benzene.

NO2

H

+

G1

+ NO2

CF3

NO2+

Ortho

Meta

Para

Reaction Progress

‡

Figure 18.4. The relative activation energies for the formation of the σ complexes of benzene and ortho, meta, and para substitutions in (trifluoromethyl)benzene. Exercise 18.8 The rate of nitration of 1,3-dimethylbenzene (m-xylene) is 100 times as fast as 1,4-dimethylbenzene (p-xylene). Predict the product(s) of nitration for both xylenes. If there are more than one isomer, which would you expect to be the major product? Explain the difference in the relative rates.

18.7 Classification of Substituents



Methyl and trifluoromethyl groups are only two of the many substituents that affect the way electrophiles react with benzene in substitution reactions. In regard to regioselectivity, all substituent groups are either ortho, para directors or meta directors. Each group varies as to how it affects the rate of electrophilic substitution. In general, ortho, para directors activate the aromatic ring with respect to benzene, and meta directors deactivate the ring. The ortho, para directors all share the ability to stabilize the σ complex. For example, the following four resonance contributors can be written for para substitution on anisole.

OCH3

H E

OCH3

H E

OCH3

H E

••OCH3

EH

••••••

•• •• ••

The first three resonance structures are identical to the ones drawn for para-substituted toluene in Section 18.5. The fourth resonance structure, however, is particularly stable because the nonbonding pair of electrons on the oxygen atom helps stabilize the positive charge. In fact, this is the major resonance contributor because all the atoms have filled orbitals in their valence shells. All meta directors inductively destabilize the σ complex because they have a partial positive or a full positive charge on the atom attached to the ring thereby discouraging substitution at sites ortho or para to the substituent. By default, the reactive site is the meta position because it is not directly destabilized by the substituent. For example, the least stable resonance contributor for nitrobenzene has a positive charge on the carbon bearing the nitrogen. The nitro group nitrogen also has a formal positive charge.

N

EH

O O•• •••• ••

••



Table 18.2 summarizes the rate and orientation effects of the most common substituents. The list places the substituents in order of decreasing rate of substitution. Amines are the most activating group, and nitro groups are the strongest deactivating group.

Effect on Rate Substituent Product Orientation

Very strongly activating ••

NH2

Ortho, Para

••

NHR

Ortho, Para

••

NR2

Ortho, Para

••

••

OH

Ortho, Para

Strongly activating

••

NHCR

O••

••

Ortho, Para

••

••

OR

Ortho, Para

OCR

O••

••

••

••

Ortho, Para

Activating R

Ortho, Para

Ar

Ortho, Para

CH=CR2

Ortho, Para

Reference H

Deactivating ••

••

••

X

(X=F,Cl,Br,I)

Ortho, Para

••

••

••

CH2X

Ortho, Para

Strongly deactivating ••

••

CR

O

Meta



Effect on Rate Substituent Product Orientation

COH

O••

••

••

••

Meta

COR

O••

••

••

••

Meta

CH

O••

••

Meta

CCl

O••

••

••

••

••

Meta

••C N Meta

SO3H

Meta

Very strongly deactivating CF3

Meta

NH3 Meta

NO2

Meta

NR3 Meta

Table 18.2. Classification of substituents in electrophilic aromatic substitution reactions. The following five rules summarize Table 18.2:

1) Activating substituents are ortho, para directors. 2) Ortho, para directors, except for alkyl, aryl, and vinyl

groups, have nonbonding electrons on the atom attached to the aromatic ring.

3) Deactivating substituents are meta directors. 4) Meta directing groups have at least a partial positive charge

on the atom that bonds to the ring carbons. 5) Halogens are an exception to the above rules. They are

deactivating, but are ortho, para directing groups, and they have nonbonding electrons.



Phenol is an example of an ortho, para director with nonbonding electrons. In a nitration substitution reaction of phenol, both the ortho σ complex and the para σ complex have four resonance contributors. The first three are identical to the three shown in Section 18.5 for either toluene or (trifluoromethyl)benzene. In the fourth one, the oxygen contributes a pair of nonbonding electrons to the electron deficient carbon to form a carbon—oxygen double bond. All the atoms in this resonance contributor have an octet of electrons, so this contributor is more stable than the other three.

Ortho Substitution •• •• ••

••••••

•• OHNO2

H

OHNO2

H

OHNO2

H

OHNO2

H

Major contributor

Para Substitution

•• •• •••• •• •• ••

OH

H NO2

OH

H NO2

OH

H NO2

OH

H NO2Major contributor

The stabilization of the positive charge of the σ complex by the oxygen makes it the major resonance contributor. This stability provided by the major contributor permits the σ complex to form rapidly. The rate of formation of the σ complex from phenol in an electrophilic substitution is much faster than the reaction of the σ complex of benzene. In fact, the nitration of phenol is approximately 50,000 times faster than the nitration of benzene. The nonbonding pairs of electrons on the oxygen cannot stabilize the meta σ complex of phenol because none of the resonance contributors have a positive charge on the carbon bearing the OH group. Thus, the meta σ complex has only three resonance contributors. The rate of reaction for the meta position, although slower than the ortho or para positions, is still 10.4 times faster than the rate of reaction with benzene.

Meta Substitution



NO2

H

OH

NO2

H

OH

NO2

H

OH ••••••

••••••

Beyond what Section 18.6 presents about substitution reactions on a ring containing an electron-withdrawing substituent, you need to know that the greater the partial positive charge on the substituent, the more deactivating the group. The halogens are an exception to the general rules of how substituents on benzene behave. The halogen substituents are electronegative; thus, they are electron withdrawing. They induce a partial positive charge on the carbon bearing the halogen and thereby deactivate the ring. However, unlike other deactivating groups, halogens direct the electrophile to the ortho and para positions because they have nonbonding pairs of electrons available to stabilize the σ complex. The stabilization is less significant than with a phenol or aniline, even though oxygen and nitrogen have electronegativities equal to or greater than the halogens. The lower stabilization by the halogens is because the nonbonding electrons are further from the ring and thus less likely to be donated to stabilize the σ complex. In the nitration of bromobenzene for example, the bromine deactivates the ring causing the reaction to be 20 times slower than the nitration of benzene. The primary products are o-bromonitrobenzene and p-bromonitrobenzene.

60%1%39%

++H2SO4

HNO3

Br

NO2

Br

NO2

Br

NO2

Br

The halogens direct the incoming electrophile to the ortho and para positions because their nonbonding electrons stabilize the ortho and para σ complexes but not the meta σ complex. The major resonance contributor for the halogen-bearing σ complex has the positive charge on the halogen atom. All atoms in this structure have an octet of electrons; thus, it is the most stable resonance contributor.



••

••

••

••••

••

••

••

••

••Br

NO2

H

Br

NO2

H

Br

NO2

H

Br

NO2

H

Both resonancecontributors havefull octets on all atoms.

Exercise 18.9 Aniline (C6H5NH2) is more reactive towards electrophilic substitution than is acetanilide (C6H5NHCOCH3). Explain this difference in reactivity. When using highly activating substituent groups, limiting the number of incoming substituents to only one is difficult. For example, phenol reacts rapidly with bromine in water to form a quantitative yield of 2,4,6-tribromophenol.

(100%)

Br2H2O

OH

BrBr

Br

OH

2,4,6-Tribromophenol

Solved Exercise 18.3 Complete the following reactions showing the major monosubstitution product(s) from each reaction. a)

BrBr2

Fe

Solution



Bromine is a weakly deactivating ortho, para directing group. Thus, you will get a mixture of ortho and para substitution products. Because bromine is large, there will be a larger fraction of para than ortho product.

BrBr2

Fe

Br

Br

Br

Br

+

b)

OCH3CH3CH2CH2Cl

AlCl3

Solution The methoxy group is an activating ortho, para directing group. Thus, you will get a mixture of ortho and para substitution. Because the oxygen is small, the ortho product will likely predominate. In this case, the electrophile will rearrange to form a secondary carbocation. Thus, the product is an isopropyl-substituted anisole.

OCH3

+CH3CH2CH2Cl

AlCl3

OCH3

CHCH3

CH3

OCH3

CH3CH

CH3 c)

COCH3

O

HNO3

H2SO4

Solution The ester functional group is a deactivating meta directing group. Thus, the product will have a nitro group substituted meta to the ester group.



COCH3

O

HNO3

H2SO4

COCH3

O

NO2

18.8 Friedel-Crafts Acylation Section 18.4 describes the Friedel-Crafts alkylation reaction. As you may recall, the reaction has a limitation in that, if possible, an alkyl group will rearrange to form the more stable carbocation. A second limitation of the Friedel-Crafts alkylation is that it does not reliably produce a monoalkyl benzene. The reaction produces a slightly activated aromatic ring, which then undergoes a second alkylation at a higher rate than does benzene.

The product is more reactivethan the substrate, so it readilyundergoes further reaction.

CH3

AlCl3

CH3Cl

CH3

CH3

CH3

CH3

+AlCl3

CH3Cl

A third limitation of the Friedel-Crafts reaction is that it does not proceed well with deactivated benzene rings. Any group more deactivating than the halogens normally gives a low yield or no reaction at all in the Friedel-Crafts reaction. A variation of the Friedel-Crafts reaction is the acylation reaction. A Friedel-Crafts acylation reaction involves the reaction of an acyl halide or an acid anhydride and a Lewis acid with benzene to yield an acylbenzene. Both the acyl halide and the anhydride work well in this reaction because each possesses a good leaving group. The acyl halide has a halide ion leaving group and the acid anhydride has a carboxylate ion leaving group. The Friedel-Crafts acylation produces a deactivated ring and, because deactivated rings cannot undergo a second acylation, the reaction stops at that point.



CH3CCl

O

AlCl3

CCH3

O

The product is less reactivethan the substrate, so it doesnot undergo further reaction.

Unlike an alkylation reaction, which needs only catalytic amounts of AlCl3 to react, an acylation reaction requires one mole of AlCl3 per mole of acyl halide. Acylation reactions need this stoichiometric amount of aluminum chloride because the aluminum chloride first forms an acid/base complex with the carbonyl group of the acyl halide. When running an acylation reaction, chemists usually allow the acid/base complex to form before they add the benzene to the reaction mixture.

O

CH3CH2CCl

O AlCl3

CH3CH2CClAlCl3

••••••

Acid-base complex The acid/base complex is relatively stable in low polarity solvents. In higher polarity solvents, however, it ionizes and forms a resonance-stabilized acyl (or acylium) cation.

•• O AlCl3

CH3CH2CCl

CH3CH2C O

Acyl cation

••••

••

O

CH3CH2C

An acyl cation reacts with benzene in much the same way as any other electrophile.



••••

••••

CCH2CH3

OCCH2CH3

H

O

CH3CH2C

O••

••

AlCl4

Chemists use anhydrides to generate the acylium ion less frequently than they use acyl halides because only a few anhydrides are commonly available. In addition, acid anhydrides are expensive and only half the material is available for reaction. The other half is thrown away. The following mechanism shows the use of an anhydride in a Friedel-Crafts acylation reaction.

+RC

OAlCl3

RC OCR

O O

Cl3Al OCR

O

Exercise 18.10 In some low polarity solvents, the acid/base complex does not readily form the acylium ion. Thus, the acid/base complex is the reacting species. Write a mechanism for this reaction. (Hint: Refer to Chapter 8 for a starter.) In contrast to Friedel-Crafts alkylation reactions, the cations in Friedel-Crafts acylation reactions do not rearrange. Thus, the acyl group in the acyl benzene product has the same structure as the acyl group in the acyl halide substrate. Rearrangement does not occur because the acylium ion is resonance stabilized. Thus, the acylium ion is much more stable than a carbocation. The lack of rearrangement of the carbon skeleton of the acyl group makes the Friedel-Crafts acylation a useful synthesis of alkyl benzenes. In addition, the acylation reaction does not bring about a second substitution reaction, whereas most alkylation reactions give significant di- and trisubstitution products. When chemists want to form a compound with a —CH2— adjacent to the benzene ring, they first do an acylation reaction to form a ketone, then they reduce the carbonyl of the ketone to a —CH2— group. This synthetic strategy is called an acylation-reduction sequence.

The acylation-reduction sequence involves the synthesis of a 1-phenyl ketone followed by the conversion of the carbonyl to a —CH2— group.



AlCl3

RCCl

O

reduceCR

O

CH2R

The most common reaction that chemists use to reduce the carbonyl group to an alkyl group is the Clemmensen reduction. A Clemmensen reduction reaction involves heating the ketone with a zinc-mercury amalgam in concentrated HCl. Many ketones that are stable in hot acids are reduced by the Clemmensen reduction.

The Clemmensen reduction is a method used to convert the carbonyl group to a —CH2— group.

(71% overall)Propylbenzene

AlCl3 HClZn(Hg)CH3CH2CCl

OCCH2CH3

O

CH2CH2CH3

An alternative procedure to a Clemmensen reduction is the Wolff-Kishner reduction. The procedure for a Wolff-Kishner reduction involves heating the ketone with hydrazine (NH2NH2) and potassium hydroxide in a high boiling alcohol solvent. The Wolff-Kishner reduction works with a variety of ketones and aldehydes that are stable in hot, concentrated base.

The Wolff-Kishner reduction is another method used to convert the carbonyl group to a —CH2— group.

CH2C(CH3)3CC(CH3)3

O

AlCl3

(CH3)3CCCl

O

NH2NH2, KOH

Diethylene glycolreflux

(76% overall)(2,2-Dimethyl-1-propyl)benzene

Both the Clemmensen reduction and the Wolff-Kishner reduction are very specific for reducing an aldehyde or a ketone to a methylene group. Neither reaction reduces the carbonyl group of carboxylic acid derivatives C—O π bonds. The choice between the two reduction processes depends on other groups present in the molecule. If there are acid sensitive functional groups, then use the Wolff-Kishner reaction. For base sensitive functional groups, use the Clemmensen reaction.



Exercise 18.11 Propose a synthesis for each of the following substituted aromatics. a) tert-Butylbenzene b) 2-Methyl-1-phenylpropane c) Butylbenzene d) Toluene e) Neopentylbenzene [PhCH2C(CH3)3] Sample solution c)

AlCl3

CH3CH2CH2CCl

O

CH2CH2CH2CH3

CCH2CH2CH3

O

NH2NH2, KOHDiethylene glycolreflux

Synthesis of o-Benzoylbenzoic Acid

(55%)

O

O

OHAlCl3

O

O

O

Phthalic anhydride o-Benzoylbenzoic acid

To a dry 50 mL round-bottom flask, add 1.65 g (0.011 mol) of phthalic anhydride, 7 mL of dry benzene, and 3.4 g (0.025 mol) of anhydrous aluminum chloride. The apparatus must be dry because aluminum chloride reacts rapidly with water to release hydrogen chloride, thus, destroying the aluminum chloride. Fit the flask with a trap to collect the hydrogen chloride produced by the reaction. Stir at room temperature for 15 minutes. Warm to 50-60oC for 5-10 minutes then reflux for an hour. Cool the flask in an ice bath then add 15 g of crushed ice. Next, add concentrated hydrochloric acid until the solution clears (about 3 mL). Place the entire mixture in a 200 mL flask and distill until the distillate becomes clear. Cool the residue in the flask in ice, filter the product, and wash it with 5 mL of cold water. Dissolve the solid in 10 mL of 10% sodium carbonate. Filter any insoluble residue.



Cautiously acidify with concentrated hydrochloric acid (until a pH of 1-2 is reached), adding a little acid at a time with stirring. Continue stirring until any oil that separates becomes crystalline. The yield of the monohydrate of the acid is 2.7 g (55%), mp 94oC (loses water). Discussion Questions 1. The Friedel-Crafts acylation does not work well with a carboxylic acid because an

acylium ion cannot readily form from the acid. Explain why the acylium ion cannot readily form.

2. What is the purpose of dissolving the crude product in sodium carbonate solution?

18.9 Multiple Substituent Effects When a benzene ring has two or more substituents, all the substituents exert their combined effects on the reactivity of the ring and in the placement of any incoming electrophiles. In most cases, multiple substituents affect an electrophilic aromatic substitution reaction in one of the following four ways. 1) All available sites are equivalent. This means that a substitution at any one of these sites gives the same product.

CH3

CH3

CH3

CH3

BrBr2Fe

All sites areequivalent

2) All sites have comparable reactivity, but one site is more sterically hindered than the other. The reaction then takes place at the less sterically hindered site.

Less sterically hindered

More sterically hindered

Fe

Br2

C(CH3)3

CH3

Br

C(CH3)3

CH3



3) The directing effects of the groups reinforce one another. For example, in p-nitrotoluene the nitro group is a meta directing group, the methyl group is an ortho, para directing group, and the two groups are para to each other. Thus, both groups direct to the same pair of carbon atoms as the preferred site of reaction.

NO2

CH3

NO2

CH3

BrBr2Fe

Ortho to the CH3and meta to the NO2

4) The substituents have directing influences that oppose one another. When this occurs, the substituent with the greatest influence controls the outcome of the reaction. If the difference between the two groups is small, then the reaction does not select one site in preference to the other. When the difference is large, the more activating substituent provides the greater influence. An activating group always directs the reaction in preference to a deactivating group. This is true even if the activating group is weak and the deactivating group is strong because there is a higher rate of reaction for the activating group.

Deactivated by Cl

Activated by NH2

Br2Fe

Cl

NH2

Br

Cl

NH2

Solved Exercise 18.4 Predict the major product(s) from the bromination of the following compounds. a)



Cl

NHCCH3

O

Solution Because of the nonbonding electrons on the nitrogen attached to the ring, the amide group is a strongly activating ortho, para director. The chlorine is a weakly deactivating ortho, para director. Thus, the sites ortho to the amide group are more reactive than the sites ortho to the chlorine.

Br2

FeCl

NHCCH3

O

Cl

NHCCH3

O

Br

b)

COCH3

O

O2N

Both the ester and the nitro groups are deactivating meta directors. Because they are meta to each other, they both direct towards the same site on the ring.

COCH3

O

O2N COCH3

O

O2N

Br

FeBr2

Exercise 18.12 Predict the major mononitration product(s) for each of the following aromatic compounds.

a) m-Dichlorobenzene b) p-Methylphenol c) m-Nitroanisole d) 4-Ethylbenzoic acid e) 1,3-Dimethylbenzene f) 2,6-Dichloroanisole



Sample solution b) The —OH group is a much stronger activating group than the —CH3 group, so the —OH group directs the position of the nitro group.

H2SO4

HNO3

CH3

OH

NO2

CH3

OH

18.10 Substitution on Polycyclic Arenes Polycyclic aromatic hydrocarbons also undergo electrophilic aromatic substitution reactions with the same reagents that react with benzene and the benzene derivatives. In terms of relative rate of reaction, most polycyclic aromatics react more rapidly than benzene. Because they lack the symmetry of benzene, the product mixture from reactions with polycyclic aromatic hydrocarbons is usually more complex than the product mixture from a similar reaction with benzene. Naphthalene, a ten carbon bicyclic aromatic hydrocarbon, has two positions available for attack by the electrophile in a nitration reaction. These positions are on C1 and C2. For both positions, naphthalene has four equivalent sites around the two rings. Naphthalene needs much milder conditions for the reaction to proceed than does benzene. A weakly acidic mixture of acetic acid/acetic anhydride acts well as a solvent for the reaction. The reaction forms two products in a 91:9 ratio.

Naphthalene

1

2

3

410

5

6

7

89

HNO3

CH3COOH(CH3CO)2O

warm

NO2

NO2

+

1-Nitronaphthalene 2-Nitronaphthalene91% 9%



To understand why C1 is so much more reactive than C2, examine the resonance contributors for the σ complexes that lead to each of the two products. Attack at C1 produces five resonance contributors.

Reaction at C1 NO2H

NO2HNO2H

NO2HNO2H

The first two contributors possess two factors that give them more stability than the other three resonance contributors and the resonance contributors in the C2 σ complex. Both resonance contributors have one ring in which the aromaticity is undisturbed, and each of the two aromatic contributors in the other ring also has an allylic resonance. Thus, these two contributors have a special stability due to the delocalization of the positive charge between C2 and C4. On the other hand, only one of the resonance contributors for the reaction at C2 has aromatic character in either ring. The remaining resonance contributors are not aromatic. Formation of the C2 σ complex is a higher energy pathway than electrophilic attack at C1.

Reaction at C2

H

NO2

H

NO2

H

NO2

H

NO2

H

NO2



Due to steric hindrance at the C1 position, larger substituents tend to be more stable at the C2 position. For example, formation of naphthalenesulfonic acid under mild conditions gives naphthalene-1-sulfonic acid, but at high temperatures, the product formed is naphthalene-2-sulfonic acid. This reaction is another example of kinetic versus thermodynamic control of a reaction. Because of its size, a sulfonic acid group on C1 comes within the van der Waals radius of the hydrogen at C8. But, when the sulfonic acid group is on C2, it has no such steric interference because the molecule has sufficient room for the sulfonic acid group between the hydrogens at C1 and C3. Thus, the product of the higher energy pathway is actually more stable than the product of the lower energy pathway.

Thermodynamic versus kinetic control of a reaction is discussed in Section 16.4, page 000.

Less interferenceSteric interference

SO3H

H

H

SO3HH

Naphthalene-1-sulfonic acid Naphthalene-2-sulfonic acid Using 100% sulfuric acid at a reaction temperature of less than 75oC, the yield of the 1-sulfonic acid is 98%. When the reaction is heated above 165oC, the yield of the 2-sulfonic acid is 88%. Exercise 18.13 Following an analysis similar to that above, find the most reactive site for electrophilic substitution on anthracene.

Anthracene



18.11 Diazotization Chemists synthesize aryl diazonium salts from aryl amines. An aryl diazonium salt

has the formula ArN2

⊕. An aryl diazonium salt is made from an aryl amine.

NH2 N NNaNO2, HCl

H2O, 0-5oCCl

••

Benzene diazonium chloride The electrophile used to react with the amine to form the aryl diazonium salt is the mild ⊕NO electrophile. The following mechanism shows how the aryl diazonium salt forms from HCl and NaNO2. The ⊕NO ion is resonance-stabilized, so it forms relatively easily.

++ NaClHO N OHClNaNO2••

••

••

••

••••

••+ H HO N O

H

••

••

••••HO N O

••N O

HO N O

H

•••• ••

••••

••

••N O

+ H2O

After the ⊕NO electrophile forms, it reacts in acid/base fashion with the amine nitrogen. The intermediate then loses water and forms the diazonium salt; thus, completing the diazotization process. Compare the following diazotization process with the substitution and elimination reactions of Chapters 12 and 13.

Diazotization is the name of the process that forms a diazonium salt.



••••N N O

H

H

N N O

H

NH2

N N N N OH2 N N OHH

H

ON ••••

••

••

••

••

•• ••

•• •• ••

••

••••••

•• ••

Once formed, diazonium salts can be used as synthetic intermediates for a number of substituted aromatic compounds. For example, nucleophiles displace the —N2

⊕ group, so aryl diazonium salts can be used in reactions to prepare a number of substituted aromatic compounds. Section 18.12 (page 000) discusses the reaction of diazonium salts with nucleophiles. Diazonium salts can also be used as electrophiles to substitute onto another aromatic system. This type of electrophilic aromatic substitution reaction is called azo coupling. Although diazonium salts are relatively weak electrophiles, they do react well with strongly activated aromatic rings. Chemists most frequently use the —OH and the —NR2 groups to activate the ring. The products formed in azo coupling reactions contain the azo functional group (–N=N–) and have the general structure of Ar–N=N–Ar. The general mechanism for the reaction follows a similar mechanism to the other electrophilic substitution reactions studied in this chapter.

Azo coupling is the electrophilic substitution reaction of a diazonium salt with an activated aromatic ring to form a product with the structure Ar–N=N–Ar.



••••

H2O

4-PhenylazophenolAn azo compound

N N

OH

N NH

OH

OHN N

••

••••

••

••

••••

••

••

••

Most azo compounds are highly colored and make excellent dyes. These dyes, called azo dyes, are usually more stable and retain their bright colors better than most other dyes. Following are a few examples of azo dyes.

••N N

O2N

+

Alizarin yellow

N N

O2N

OH

COOH

OH

COOH

(82%)

••N N

HO3S

+

Methyl orange

N N

HO3S

N(CH3)2

N(CH3)2

(70%)



••N N

O2N

HO

N N

O2N

HO+

Para red(76%)

Synthesis of Methyl Orange

Methyl orange

Sulfanilic acid

(70%)

N

HO3S

N

N(CH3)2

2) NaNO2, HCl < 5oC

1) Na2CO3

N(CH3)2

NH2

HO3S

Dissolve 0.6 g of anhydrous sodium carbonate in 50 mL of water. Add 1.75 g (10 mmol) of sulfanilic acid to the solution and heat until it dissolves. Cool the solution to room temperature and add 0.8 g of sodium nitrite. Cool in an ice bath until the temperature is below 10oC then add 2.5 mL of concentrated hydrochloric acid. A precipitate of diazonium salt will form. Keep this reaction mixture cold. In a separate flask, dissolve 1.3 mL (10 mmol) of N,N-dimethyl aniline in 1.0 mL of glacial acetic acid. Add this solution to the reaction mixture. Keep cold for about 15 minutes. During this time a red precipitate will form. Add 15 mL of 10% sodium hydroxide. Check to see if the solution is basic. If not, add more 10% sodium hydroxide until the solution is basic. Heat this solution to boiling and boil until most of the solid is dissolved. Carefully add 5 g of sodium chloride to the boiling solution. Allow the mixture to cool, then cool it further in ice. Filter the solution. Wash the flask with two ice-cold portions of saturated sodium chloride solution and pour these through the methyl orange in the filter. Dissolve the solid in 150 mL of boiling water. All of the solid will not dissolve, but the sodium chloride contaminant will dissolve. Boil the solution for 5-10 minutes. Cool the solution to room temperature then place in an ice bath. When cold, filter the solid and allow it to dry. Yield of methyl orange is 2.15 g (70%), m.p. decomposes. Discussion Questions



1. The pH of the reaction mixture is important to the course of the coupling reaction with N,N-dimethylaniline. The rate of the coupling reaction increases with increasing pH. Explain this behavior.

2. Methyl orange, and all other azo compounds, is brightly colored. What molecular features of azo compounds explain the colors of these molecules?

Exercise 18.14 Propose a synthesis for each of the following compounds.

a)

CH3

NN

CH3

OCH3

b)

N N

SO2NH2(CH3)2N

c)

NN

NO2

NH2

d)

Cl

N N

N(CH3)2

Sample Solution c)

NO2

NH2

NO2

NH2

NNNH2NaNO2, H2SO4

H2O, 0 - 5oC

[SIDEBAR] Sulfa Drugs The introduction of sulfa drugs in the 1930s hailed the beginning of modern drug therapy. Before their introduction, even a minor bacterial infection could become potentially life-threatening. Because at first no one understood how sulfa drugs worked, even physicians considered them as almost magical.



Researchers knew that many bacteria absorbed a number of dyes. With some dyes, the entire bacteria absorbed the dye; with other dyes, only specific portions of the bacteria absorbed it. Because of this behavior, researchers used dyes as a standard method to study the structure of bacteria. They began to wonder if they might find a dye that was toxic to the bacteria. To solve this puzzle, a group of researchers at the German dyestuff manufacturer I. G. Farbenindustrie investigated the various dyes in their collection. They screened thousands of dyes and found that a number of them were active both in vitro and in vivo, respectively meaning “in glass” and “in life.” They did the test tube or petri dish (in vitro) screening first. Any promising substances found there, they tested using living organisms—laboratory animals or human volunteers (in vivo). The researchers next decided to determine why some dyes were active both in vitro and in vivo, some were active in vitro only, and some were inactive in both cases. As they conducted this work, they made an unexpected discovery. They discovered that p-[(2,4-diaminophenyl)azo]benzenesulfonamide, which was inactive in vitro, was very active in vivo!

H2N

N N

SO2NH2

NH2

p-[(2,4-Diaminophenyl)azo]benzenesulfonamide

In 1932, Gerhard Domagk tested p-[(2,4-diaminophenyl)azo]benzenesulfonamide, later marketed under the trade name Prontosil, on a ten-month-old boy sick with a dangerous staphylococcal infection. The boy rapidly recovered. Domagk was awarded the 1939 Nobel Prize in medicine or physiology for this work, although he was forced to decline the Prize by the Nazi German Government. So, Prontosil worked. It had saved a child’s life. But how? Research had already shown that it was inactive in vitro. Why then was it active in vivo? Identifying the mechanism by which Prontosil combats bacterial infections was a major step in the development of pharmacology. The goal of the I. G. Farben research group was to find a dye toxic to bacteria. Prontosil had healed a child with a bacterial infection, yet, obviously it was not toxic to the bacteria. In vitro studies had proven that fact. In conclusion, the researchers decided that Prontosil’s toxicity observed in vivo must be from something



else—something that had nothing to do with it being a dye. As they tracked what happened to Prontosil in the body, they found that Prontosil undergoes a cleavage that produces sulfanilamide. Sulfanilamide was the substance responsible for Prontosil’s biological activity. Animals have the necessary enzymes to do this sort of cleavage, but bacteria do not, thus the explanation for Prontosil’s inactivity in vitro and activity in vivo.

Sulfanilamide

in vivoH2N

SO2NH2H2N

N N

SO2NH2

NH2

The explanation found by researchers for how sulfanilamide works was in its structure. Sulfanilamide and p-aminobenzoic acid (PABA) are so similar that the bacteria mistake sulfanilamide for PABA. Bacteria require PABA to biosynthesize folic acid, a coenzyme involved in the transfer of single carbon units in biosynthesis. In the body, sulfanilamide slows the biosynthesis of folic acid, which, in turn, slows the bacterial growth, thus allowing the body’s natural defenses to affect a cure.

p-Aminobenzoic acidSulfanilamide

H2N

COOH

H2N

SO2NH2

Once researchers recognized that sulfanilamide was the active agent against bacteria, they began synthesizing analogs. Between 1935 and 1946, they synthesized over 5,000 compounds. Of these thousands of compounds, they found two that were more effective than the others. These were sulfathiazole and sulfadiazine.

SulfadiazineSulfathiazole

H2NN

NSO2NH

H2NS

NSO2NH



As researchers found more effective medications, such as penicillins and tetracyclines, physicians moved away from sulfa drugs, although veterinarians still widely use them. Sulfa drugs made a large impact on the lives of people. In the United States alone, the estimate is that sulfa drugs reduced the death rate from pneumonia by at least 25,000-30,000 deaths per year in the 1940s and 1950s.

18.12 Other Diazonium Salt Reactions In addition to the azo coupling reaction, diazonium salts undergo a substitution reaction at the carbon bearing the diazo group. This reaction follows a mechanism outside the scope of this book. The most common example of the azo substitution reaction is the Sandmeyer reaction. In a Sandmeyer reaction, the diazonium salt reacts with copper(I) chloride, bromide, or cyanide to replace the —N2

⊕ group on the ring. Chemists often choose the Sandmeyer reaction as the best method to prepare aryl chlorides, bromides, or nitriles in part because it permits precise control of where the substituent will be on the product.

The Sandmeyer reaction converts a diazonium salt to an aryl chloride, bromide, or nitrile.

NH2 X

NaNO2, H2SO4

H2O, 0 - 5oC

CuX

(X = Cl, Br or CN)

The following are some examples of the Sandmeyer reaction:

1-Bromo-3-nitrobenzene(87%)

CuBrH2O, 0 - 5oC

NaNO2, H2SO4

NO2

Br

NO2

NH2

(75%)2-Chlorobenzonitrile

CuCNH2O, 0 - 5oC

NaNO2, H2SO4

Cl

NH2

Cl

CN



4-Chlorobenzenesulfonamide(71%)

Cl

NH2O2S

NH2

NH2O2S

NaNO2, H2SO4

H2O, 0 - 5oCCuCl

Another example of an azo substitution reaction is the Schiemann reaction. To run a Schiemann reaction, fluoboric acid (HBF4) is added to the diazotization reaction medium. Heating the reaction mixture then forms the aryl fluoride.

Chemists use Schiemann reactions to produce aryl fluorides that are otherwise difficult to produce.

NH2 FNaNO2, H2SO4

H2O, 0 - 5oC

1) HBF4

2) heat

Fluorobenzene(83%)

To prepare an aryl iodide, simply stir the diazonium salt with KI at room temperature until the aryl iodide forms.

NH2CH3KI

ICH3

H2O, 0 - 5oC

NaNO2, H2SO4

3-Iodotoluene(82%)

Another important reaction involving diazonium salts is the formation of phenols. In this reaction, you need only to warm the diazonium salt reaction mixture.

3-Methylphenol(74%)

NH2

CH3

H2O, 0 - 5oC

NaNO2, H2SO4

CH3

OHwarm

The final substitution reaction discussed here that uses a diazonium salt is the replacement of the —N2

⊕ group with a



hydrogen. In this reaction, the diazonium salt is mixed with hypophosphorus acid (H3PO2).

(68%)

NH2

Br

Br BrH2O, 0 - 5oC

NaNO2, H2SO4H

Br

Br

Br

H3PO2

2,4,6-Tribromobenzene

Using deuteriohypophosphorus acid (D3PO2) instead of H3PO2, allows the selective placement of a deuterium on the ring.

NH2

H2O, 0 - 5oNaNO2, H2SO4

DD3PO2

(81%)Deuterobenzene

Exercise 18.15 Propose a synthesis of 3-chloroethylbenzene from benzene.

18.13 Nucleophilic Aromatic Substitution Aryl halides that bear one or more strongly electron-withdrawing groups in the ortho or para positions to the halide readily undergo nucleophilic substitution reactions. These electron-withdrawing groups must be groups that withdraw electron density due to resonance rather than inductively. Inductive withdrawing groups include the halogens and the —CF3 group. Resonance electron-withdrawing groups include those groups with a positive charge on the atom attached to the ring; for example, carbonyl groups and the —NO2 group. The most commonly used electron-withdrawing group is the nitro group. The rate of reaction depends on how strongly electron withdrawing the ortho and para groups are.



80oCNaOH H

NO2

OH

NO2

NO2

Cl

NO2

2,4-Dinitrophenol(95%)

The mechanism for nucleophilic substitution reactions is analogous to the mechanism for electrophilic substitution reactions. The hydroxide ion attacks the carbon bearing the halogen, which causes the ring to lose its aromaticity. The electron-withdrawing substituents then stabilize the negative charge by their partial or full positive charge adjacent to the ring. Finally, the ring loses the halogen, restoring aromaticity to the ring and forming a phenol.

••N

NO2

OHCl O

ON

NO2

OHCl O

ON

NO2

OHCl O

O

OH

Cl

N

NO2

O

O OH

N

NO2

O

ON

NO2

OHCl O

O

•••• ••

•• ••

•••• ••

••••

••

••

•• ••

•• ••••

•• •• •• •••• ••

••

•• ••

•• •• •• ••••

••

••••

••

••

••

•• •• ••

•• •• ••

•• •• •• •• •• ••

••

•••••• •• ••

••

•• ••

••

••

••

The rate-determining step in nucleophilic aromatic substitution reactions is the step that forms the carbon—nucleophile bond. The rate of reaction depends on which halogen is present on the ring and the number and strength of the electron withdrawing groups that are ortho or para to that halogen. The relative rate of reaction for the halides in nucleophilic aromatic substitution reactions is the opposite of their rates in aliphatic nucleophilic substitutions. Thus, aryl fluorides are the most reactive halide leaving groups and aryl iodides the least reactive. Aryl fluorides are the most reactive because fluorine is the most electronegative, so it is the best leaving group.



Fluorine is also the smallest of the halides, thus giving less steric hindrance to the reaction of the nucleophile than the other halides. Table 18.3 shows the relative rates of reaction for the halogens.

Aryl Halide Rate Ar—F 310 Ar—Cl 1.0Ar—Br 0.8Ar—I 0.4

Table 18.3. Relative rates of reactions of 4-halonitrobenzenes with sodium methoxide in methanol at 50oC. By increasing the number of electron-withdrawing groups on the benzene ring, you dramatically increase the rate of reaction. The increase in reaction rate occurs because each electron-withdrawing group helps stabilize the negative charge. Table 18.4 shows how increasing the number of nitro groups affects the rate.

Substitution Rate None 1.4 x 10–8 4-Nitro 1 2,4-Dinitro 2.9 x 105 2,4,6-Trinitro Too fast

to measure Table 18.4. Relative rate of reaction for various chloronitrobenzenes with sodium methoxide in methanol at 50oC. Exercise 18.16 Reaction of 1,2-dichloro-3,5-dinitrobenzene with sodium methoxide in methanol produces a single product, C7H5ClN2O5. What is the structure of the product?

18.14 Benzyne Although aryl halides without electron-withdrawing substituents also undergo nucleophilic substitution reactions, they require extreme conditions or very strong bases. For example the commercial “Dow process” used for making phenol involves heating chlorobenzene with sodium hydroxide at 350oC.



350oCNaOH

OHCl

H

(98%)Phenol

An example of a nucleophilic substitution reaction that proceeds with a very strong base is the reaction of chlorobenzene with sodium amide (NaNH2) in liquid ammonia to produce aniline. Because the amide ion is such a strong base, the reaction does not require high temperatures.

Cl NH2

NaNH2

NH3, -33oC

Aniline(100%)

The mechanism for a nucleophilic substitution reaction on an aryl halide without an electron-withdrawing group is different from the mechanism described in Section 18.13 for a nucleophilic substitution reaction on an aryl halide with an electron-withdrawing group. Chemists had no reason to suspect a different mechanism until they discovered that the substitution takes place not only at the carbon bearing the halide but also at the carbon adjacent to the halide bearing carbon. For example, the reaction of 4-bromotoluene with sodium amide yields a 50:50 mixture of 3-methylaniline and 4-methylaniline.

(p-Toluidine) (m-Toluidine)3-Methylaniline4-Methylaniline

50%50%

NH3, –33oC

NaNH2

NH2

CH3

NH2

CH3

Br

CH3

+



This nucleophilic substitution reaction follows a similar pathway to an elimination-addition reaction. The amide ion, acting as a base, first removes a proton from one of the carbons adjacent to the halide. Then the halide departs. Both the halide and a proton leave from adjacent carbons causing the compound to become a neutral intermediate called benzyne. Chemists use the name benzyne because the structure is written with a triple bond.

A benzene ring that has undergone an elimination reaction that results in an additional weak π bond is called a benzyne.

CH3

a Benzyne

The two carbons involved in the “triple” bond of benzyne are sp2 hybridized. The overlap of their bonding orbitals is ineffective because the shape of the benzene ring directs them 60o apart rather than parallel to each other, which is best for an sp2 hybridized bond. Thus, benzyne is unstable and reactive. In fact, benzyne is so reactive, it cannot be isolated.

Poor overlap

CH3

After benzyne forms, the amide ion can attack it on either end of its weak, reactive triple bond. This part of the reaction produces a benzene anion. The benzene anion then removes a proton from an ammonia molecule giving the final products, 3-methylaniline and 4-methylaniline.

NH2

CH3

••

••

•• ••

••••NH2

H NH2

H

CH3

NH2

CH3

3-Methylaniline



NH2

CH3

NH2

CH3

H

CH3

H NH2NH2

••

••

••••

•• ••

4-Methylaniline Benzyne is also a very reactive dienophile in a Diels-Alder reaction. The benzyne is prepared via a Grignard reagent, although an organolithium works as well. o-Bromofluorobenzene, when treated with magnesium, first forms the Grignard from the bromide. The bromine is more reactive than the fluorine in a Grignard reaction. The Grignard then loses FMgBr to form the benzyne. Because benzyne is so reactive, the benzyne must be generated in the presence of a diene, so it can undergo a Diels-Alder reaction. See Section 16.6, page

000 for the Diels-Alder reaction.

(78%)

Mg, THF

F

Br

F

MgBr

o-Bromofluorobenzene

Synthesis of Trypticene

(70%)

(CH3)2CHCH2CH2NO2NH2

COOH

Anthranilic acidTrypticene



Dissolve 355 mg (2 mmol) of anthracene in 2.5 mL of 1,2-dimethoxyethane. Add 0.35 mL of isoamyl nitrite. Bring this mixture to a boil. Add a solution of 400 mg (2.9 mmol) of anthranilic acid dissolved in 2 mL of 1,2-dimethoxyethane dropwise over a period of about 30 minutes. Then add another 0.35 mL of isoamyl nitrite to the reaction mixture. Add a second 400 mg (2.9 mmol) of anthranilic acid dissolved in 2 mL of 1,2-dimethoxyethane dropwise. Reflux for an additional 15 minutes. Add 1.5 mL of ethanol and a solution of 0.5g of sodium hydroxide in water. Filter the resulting mixture and wash the solid with an ice-cold mixture of 4:1 methanol and water. Collect the solid and dry in an oven to a constant weight. Dissolve the solid product in 4 mL of triglyme. Add 200 mg of maleic anhydride and reflux the mixture for 5 minutes. Cool the solution to about 100oC and add 2 mL of ethanol and 500 mg of sodium hydroxide in 5 mL of water. Cool the solution in ice. Filter the crystals that form and wash with 4 mL of ice cold 4:1 methanol and water mixture. Dry the product in an oven. The yield of trypticene is 356 mg (70%), mp 255oC. Discussion Questions 1. The reaction of isoamyl nitrite with anthranilic acid is a diazotization reaction.

Propose a mechanism for formation of benzyne from diazotized anthranilic acid. 2. The purpose for reaction of the crude solid material with maleic anhydride is to

remove unwanted starting material. What is the structure of the product of this reaction?

Exercise 18.17 What structure would you expect from the formation of a Grignard reagent from 1-bromo-2-fluoro-4,5-dimethylbenzene in the presence of furan?

18.15 Synthesis Examples Planning a synthesis to prepare a benzene compound with several substituents from benzene requires that you take into account the directive influences of each group you want to place on the ring. These directive influences determine the reaction sequence. One sequence may produce a low yield or even stop the synthesis before it is complete, whereas another sequence produces the desired product in good yield. This section works through three syntheses of substituted benzenes. To propose a synthesis for p-bromoacetophenone from benzene, use the principles of retrosynthetic analysis. First ask yourself, “What is the immediate precursor of this target molecule?”



p-Bromoacetophenone

?

Br

CCH3

O

Your answer, “A benzene compound that contains one of the desired substituents. But which one?” To determine which one, consider each group as the only substituent on the benzene ring. Then think how that substituent would direct an incoming group. If the substituent were the ketone group, it would direct any incoming groups to the meta position. Thus, you cannot brominate acetophenone because you would get the meta isomer of bromoacetophenone. If the only substituent were the bromo group, it would direct an incoming group to the ortho and para positions. Thus, the immediate precursor is bromobenzene.

CH3CCl

O

AlCl3Br Br

CCH3

O

4-Bromoacetophenone Your first step in the reaction process is to brominate the benzene. Your second step is to add the ketone group. The following reaction sequence illustrates the complete synthesis of p-bromoacetophenone:

FeBr2 CH3CCl

O

AlCl3Br Br

CCH3

O

The second synthesis is of 1,3,5-tribromobenzene from benzene. Bromine is an ortho, para director, but in 1,3,5-tribromobenzene the three bromine groups are meta to each other. Your problem in this synthesis is to find a functional group that will both force the bromines to substitute onto the benzene ring meta to each other and then allow you to remove it from the molecule. Two possible groups are the amine group and the sulfonic acid group, as a hydrogen ion electrophile easily replaces them on a benzene ring. Recall from Table



18.2 (page 000), however, that the sulfonic acid group brominates meta to itself, which would be in the wrong position. An amine group, on the other hand, is an ortho, para director, so it would place the bromine in the correct position. In fact, an amine group is such a strong activating group that it usually causes trisubstitution on a benzene ring.

H2O, NaHCO3

Br2

NH2

Br

Br

BrNH2

2,4,6-Tribromobenzene

To remove the amine, form the diazonium salt then treat it with H3PO2. To form the aniline, reduce the nitrobenzene. The reduction of nitrobenzene to form aniline is a reaction that you have not previously studied. This reaction is quite simple in that zinc or tin metal with acid reduces the —NO2 group.

NO2 NH2Zn (or Sn)

HCl

The complete synthesis of 1,3,5-tribromobenzene is as follows:

H2O, 0 - 10oC

NaNO2, H2SO4H3PO2Br

Br

Br

H2SO4

HNO3

NO2

HClZn

H2O, NaHCO3

Br2

NH2

Br

Br

Br

NH2



The third target molecule for you to synthesize is 4-nitro-3-propylphenol from benzene.

OH

O2N

?

4-Nitro-3-propylphenol

4-Nitro-3-propylphenol has three possible immediate

precursors: 4-nitrophenol, 2-propylnitrobenzene, and 3-propylphenol. However, 4-nitrophenol will not work because it would give 4-nitro-2-propylphenol in a Friedel-Crafts reaction. A Friedel-Crafts reaction is the reaction you use to add the propyl group. 2-Propylnitrobenzene will not work either because the propyl and nitro group would direct the phenol to C5 instead of para to the nitro group. Nitration of 3-propylphenol, on the other hand, places the nitro group in the correct position—the para position. This is the precursor that you must use, even though the reaction also gives some nitration ortho to the –OH group and para to the propyl group.

OH

O2N

OH

O2N O2N

OH

Next, you must determine the precursor to 3-propylphenol. The only method covered so far that synthesizes a phenol is via a diazonium salt from an aniline, in this case, 3-propylaniline. To



convert 3-propylaniline to 3-propylphenol, warm the diazotization reaction mixture.

OHNH2warmNaNO2, H2SO4

H2O, 0 - 10oC

3-Propylphenol 3-Propylaniline has two substituents that are meta to each other, but both substituents are ortho, para directing groups. Thus, you cannot start with the desired substituent. Instead, the first substituent on the ring must be a meta directing group to direct the second substitution at its proper position. It must also be a group that will then react to form the desired group. The two possibilities are a nitro group, which readily converts to an aniline, and an acyl group, which readily converts to an alkyl group. Since Friedel-Crafts acylations do not take place on deactivated aromatic rings, the sequence you must use is an acylation followed by a nitration. To do the Friedel-Crafts acylation, start with benzene and make propiophenone. Then nitrate the propiophenone to produce 3-nitropropiophenone.

NO2

O

HNO3

H2SO4

O

CH3CH2CCl

O

AlCl3

1-(3-Nitrophenyl)-1-propanone At this point, you must convert the nitro group to an aniline and the carbonyl to a —CH2— group. The conversion from the nitro compound to an aniline is similar to the Clemmensen reduction of a ketone. Thus, you can combine the two reductions into one step.



HClZn(Hg)

O

NO2 NH2

3-Propylaniline The following reaction sequence shows the complete synthesis of 4-nitro-3-propylphenol:

CH3COOH

HNO3

OH

O2N

warm

H2O, 0 - 10oC

NaNO2, H2SO4OH NH2

HClZn(Hg)

AlCl3

CH3CH2CCl

O

O

H2SO4

HNO3NO2

O

Exercise 18.18 When planning a synthesis, knowing what not to do is as important as knowing what to do. The following syntheses have a flaw. What product would you expect to obtain from each of the following schemes? Additionally, propose an alternative synthesis that would prepare the products shown. You may use any stable monosubstituted benzene molecule as a starting material. a)

Br Br

Cl

CH3CH2CH2

Cl2

FeAlCl3

CH3CH2CH2Cl

b)



OCH3 OCH3

Br

NO2HNO3

H2SO4

Br2

Fe

c)

COOH COOH

CCH2CH3

O

O2NAlCl3

CH3CH2CCl

O

HNO3

H2SO4

Key Ideas from Chapter 18 ❑ Aromatic molecules are electron-rich and susceptible to

electrophilic attack by positively charged reagents. Unlike alkenes, which add electrophiles, aromatic molecules undergo electrophilic substitution.

❑ The intermediate in the mechanism for electrophilic aromatic

substitution is called a σ complex. The σ complex is a resonance-stabilized carbocation.

❑ The mechanism for an electrophilic aromatic substitution

involves an attack by the electrophile followed by the loss of a proton to restore the aromaticity of the ring.

❑ Substitution of the aromatic ring of benzene and its derivatives

is possible by treatment with HNO3 and another strong acid (usually H2SO4). This reaction generates the nitronium ion (⊕NO2) electrophile and produces an aromatic nitrate (ArNO2.)

❑ To substitute a bromine or chlorine atom, heat benzene with

the halogen in the presence of a Lewis acid. In most cases this reaction must be run in situ with elemental Fe, which produces small amounts of FeX3, the Lewis Acid catalyst.

❑ The methods used to place bromine and chlorine on a benzene

ring do not usually work for fluorine and iodine.



❑ Benzenesulfonic acids are synthesized by reaction of an arene

with fuming sulfuric acid. Fuming sulfuric acid is a solution of SO3 in concentrated H2SO4.

❑ The Friedel-Crafts reaction either alkylates or acylates an

aromatic ring. Alkylation occurs when an alkyl halide reacts with an arene in the presence of a Lewis acid. Acylation requires an acyl halide instead of an alkyl halide.

❑ Any method for generating a carbocation works well for the

Friedel-Crafts reaction. ❑ Because the Friedel-Crafts reaction generates a carbocation,

the carbon skeleton tends to rearrange whenever possible.

❑ The Friedel-Crafts reaction does not occur if there is an electron-withdrawing group on the ring. Halogens are the exceptions because they have nonbonding electrons to stabilize the σ complex.

❑ The Clemmensen reduction uses zinc amalgam and HCl to

convert many ketone or aldehyde carbonyl groups to CH2 groups.

❑ The Wolff-Kishner reduction also converts many ketone or

aldehyde groups to CH2 groups using NH2NH2 and a strong base such as KOH.

❑ The presence of a substituent on the ring determines the

regiochemistry of the second substituent being placed on the ring.

❑ Electron-donating substituents direct incoming electrophiles to

the ortho and/or para positions on the ring. ❑ Electron-withdrawing substituents direct incoming

electrophiles to the meta position on the ring. ❑ Electron-donating substituents increase the rate of reaction for

the incoming electrophile. ❑ Electron-withdrawing substituents reduce the rate of reaction

for the incoming electrophile.



❑ Halide substituents are an exception to the rules for how current substituents affect reactions. Halide substituents are electron withdrawing and slow the rate of reaction. But they direct new substituents to the ortho and para positions, however, because the nonbonding electrons on the halogen stabilize the resonance structures.

❑ When a benzene ring contains two or more substituents, you

must consider their combined influences to determine where an incoming electrophile will go. The two groups either direct to the same location or they compete to direct the incoming electrophile to different locations. Generally, an electron-donating group exerts a greater influence than does an electron-withdrawing group.

❑ Naphthalene has two possible positions for substitution: C1

and C2. Of the two positions, C1 is the more reactive because two of the resonance contributors do not remove the aromaticity of the second ring. Substitution at C2 has only one such resonance contributor.

❑ Diazotization produces a diazonium salt. A diazonium salt is an

arene bearing an —N2⊕ substituent.

❑ Azo coupling is the electrophilic substitution of a diazonium

salt on an aromatic ring bearing an electron-donating group. Many azo compounds, such as methyl orange and para red, are used as dyes.

❑ The Sandmeyer reaction reacts copper(I) bromide, chloride, or

cyanide with a diazonium salt to form an aryl bromide, chloride, or nitrile.

❑ When a diazonium salt reacts with HBF4, it produces an aryl

fluoride. This reaction is known as the Schiemann reaction. ❑ A reaction of a diazonium salt with potassium iodide produces

an aryl iodide. ❑ Warming an aqueous reaction solution containing a diazonium

salt to room temperature or slightly warmer produces a phenol. ❑ To replace a diazo group with hydrogen, react the salt with

hypophosphorus acid (H3PO2).




❑ Aryl halides bearing strongly electron-withdrawing groups undergo nucleophilic aromatic substitution with strong nucleophiles or with weaker nucleophiles only under extreme reaction conditions.

❑ Benzyne is a benzene ring with an additional π bond between

two adjacent carbons. This additional π bond is very reactive and can undergo an addition reaction, or it can act as a dienophile in a Diels-Alder reaction.

❑ Aromatic synthesis requires that you take into account the

directive influences of any substituents already on the ring.

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18 Aromatic Substitutions