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Chapter 22: Problem 40

Starting with benzene, toluene, or phenol as the only sources of aromatic rings, show how to synthesize the following. Assume in all syntheses that mixtures of ortho-para products can be separated into the desired isomer. (a) 1-Bromo-3-nitrobenzene (b) 1-Bromo-4-nitrobenzene (c) \(2,4,6\)-Trinitrotoluene (TNT) (d) \(m\)-Chlorobenzoic acid (e) \(p\)-Chlorobenzoic acid (f) \(p\)-Dichlorobenzene (g) \(m\)-Nitrobenzenesulfonic acid

Short Answer

Expert verified
Answer: The correct order of reactions is as follows: 1. Nitration of benzene to form nitrobenzene 2. Reduction of nitrobenzene to aniline 3. Diazotization and hydrolysis to form p-chlorobenzoic acid

Step by step solution

01

Nitration of benzene

We start with the nitration of benzene by using a mixture of concentrated nitric acid (HNO3) and concentrated sulfuric acid (H2SO4) to form nitrobenzene.
02

Bromination of nitrobenzene

Nitro group is a meta-directing group. So, performing bromination using Br2 and a Lewis acid catalyst like AlBr3 (the catalyst can be generated in-situ) will lead to the formation of 1-Bromo-3-nitrobenzene. (b) 1-Bromo-4-nitrobenzene
03

Bromination of benzene

We start with the bromination of benzene using Br2 and a Lewis acid catalyst like AlBr3 to form bromobenzene.
04

Nitration of bromobenzene

Bromine is an ortho-para directing group, so we can perform nitration using a mixture of concentrated nitric acid (HNO3) and concentrated sulfuric acid (H2SO4). The para product is preferred due to steric hindrance at the ortho position, leading to 1-Bromo-4-nitrobenzene. (c) 2,4,6-Trinitrotoluene (TNT)
05

Nitration of toluene

Toluene is nitrated using a mixture of concentrated nitric acid (HNO3) and concentrated sulfuric acid (H2SO4) to form 2-nitrotoluene, with the nitro group ortho to the methyl group.
06

Second nitration

Nitrate 2-nitrotoluene again with concentrated HNO3 and H2SO4. As nitro groups are meta-directing, the second nitro group will be 1,3 to the first one, forming 2,4-dinitrotoluene.
07

Final nitration

Perform another nitration with concentrated HNO3 and H2SO4 on 2,4-dinitrotoluene. The third nitro group will be para to the second, creating 2,4,6-trinitrotoluene (TNT). (d) m-Chlorobenzoic acid
08

Amination of benzene

We start with the nitration of benzene as in the previous step, followed by a reduction using \ce{Sn/HCl} to form aniline.
09

Acylation of aniline

Acylate aniline with acetyl chloride (\ce{CH3COCl}) and \ce{AlCl3}. This forms an amide, which directs electrophilic substitution to the meta position and protects the amine.
10

Chlorination via Friedel-Crafts

Perform a Friedel-Crafts chlorination using Cl2 and AlCl3 to form m-chloro-N-acetylaniline.
11

Hydrolysis of amide

Add a base like NaOH to hydrolyze the amide to form m-chloroaniline and a carboxylate ion; then protonate the carboxylate ion with an acid like HCl, which will provide m-chlorobenzoic acid. (e) p-Chlorobenzoic acid
12

Nitration of benzene

Start with nitrating benzene using a mixture of concentrated HNO3 and H2SO4 to form nitrobenzene.
13

Reduction of nitrobenzene to aniline

Nitro group in nitrobenzene can be reduced to an amine group with reducing agents like \ce{Sn/HCl} or \ce{Fe/HCl} to form aniline.
14

Diazotization and hydrolysis

Diazotize aniline using NaNO2 and HCl, followed by hydrolysis of the diazonium ion with cuprous chloride in HCl to form p-chlorobenzoic acid. (f) p-Dichlorobenzene
15

Chlorination of benzene

We start with the chlorination of benzene using Cl2 and AlCl3 to form chlorobenzene.
16

Second chlorination

Perform another chlorination reaction on chlorobenzene using Cl2 and AlCl3. The second chlorine atom will be para to the first one, leading to p-dichlorobenzene. (g) m-Nitrobenzenesulfonic acid
17

Nitration of benzene

We start with the nitration of benzene with a mixture of concentrated nitric acid (HNO3) and concentrated sulfuric acid (H2SO4) to form nitrobenzene.
18

Sulfonation of nitrobenzene

Sulfonate nitrobenzene by heating it with concentrated sulfuric acid. Because nitro groups are meta-directing, the sulfonic acid group will attach meta to the nitro group, forming m-nitrobenzenesulfonic acid.

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Most popular questions from this chapter

The first widely used herbicide for the control of weeds was 2,4 -dichlorophenoxyacetic acid (2,4-D). Show how this compound might be synthesized from phenol and chloroacetic acid by way of the given chlorinated phenol intermediate.

Addition of tert-butylbenzene to the strongly acidic solvent \(\mathrm{HF} / \mathrm{SbF}_{5}\) followed by aqueous work-up gives benzene. Propose a mechanism for this dealkylation reaction. What is the other product of the reaction?

Other groups besides \(\mathrm{H}^{+}\)can act as leaving groups in electrophilic aromatic substitution. One of the best is the trimethylsilyl group, \(\mathrm{Me}_{3} \mathrm{Si}-\). For example, treatment of \(\mathrm{Me}_{3} \mathrm{SiC}_{6} \mathrm{H}_{5}\) with \(\mathrm{CF}_{3} \mathrm{COOD}\) rapidly forms \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{D}\). What are the properties of a siliconcarbon bond that allows you to predict this kind of reactivity?

Propose a synthesis for 3,5 -dichloro-2-methoxybenzoic acid starting from phenol.

Phenol is the starting material for the synthesis of \(2,3,4,5,6\)-pentachlorophenol, known alternatively as pentachlorophenol or more simply as penta. At one time, penta was widely used as a wood preservative for decks, siding, and outdoor wood furniture. Draw the structural formula for pentachlorophenol, and describe its synthesis from phenol.
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