Question

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

Answer: The synthetic pathway for obtaining m-Nitrobenzenesulfonic acid involves the following two steps: 1. Nitration of benzene: React benzene with a mixture of concentrated HNO\(_3\) and H\(_2\)SO\(_4\) to introduce a nitro group, giving nitrobenzene. 2. Sulfonation of nitrobenzene: React nitrobenzene with concentrated H\(_2\)SO\(_4\) to introduce a sulfonic acid group at the meta-position relative to the nitro group, giving m-Nitrobenzenesulfonic acid.

Step-by-step solution

(a) Synthesis of 1-Bromo-3-nitrobenzene

To synthesize 1-Bromo-3-nitrobenzene, we can start with benzene as the source of the aromatic ring. The plan is to first introduce the nitro group and then add the bromo group. 1. Nitration of benzene: React benzene with a mixture of concentrated HNO\(_3\) and H\(_2\)SO\(_4\) to introduce a nitro group, giving nitrobenzene. 2. Bromination of nitrobenzene: React nitrobenzene with Br\(_2\) and FeBr\(_3\) to introduce a bromo group at the meta-position relative to the nitro group, giving 1-Bromo-3-nitrobenzene.

(b) Synthesis of 1-Bromo-4-nitrobenzene

To synthesize 1-Bromo-4-nitrobenzene, we can start with benzene as the source of the aromatic ring. The plan is to first introduce the bromo group and then add the nitro group. 1. Bromination of benzene: React benzene with Br\(_2\) and FeBr\(_3\) to introduce a bromo group, giving bromobenzene. 2. Nitration of bromobenzene: React bromobenzene with a mixture of concentrated HNO\(_3\) and H\(_2\)SO\(_4\) to introduce a nitro group at the para-position relative to the bromo group, giving 1-Bromo-4-nitrobenzene.

(c) Synthesis of 2,4,6-Trinitrotoluene (TNT)

For the synthesis of 2,4,6-Trinitrotoluene (TNT), we can start with toluene. The plan is to introduce three nitro groups onto the aromatic ring. 1. Nitration of toluene: React toluene with a mixture of concentrated HNO\(_3\) and H\(_2\)SO\(_4\) at low temperature to introduce a nitro group in the ortho- and para- positions relative to the methyl group, giving a mixture of 2-nitrotoluene and 4-nitrotoluene. 2. Separation of isomers: Separate the 2-nitrotoluene and 4-nitrotoluene by chromatography or fractional crystallization. 3. Second nitration of 2-nitrotoluene: React 2-nitrotoluene with a mixture of concentrated HNO\(_3\) and H\(_2\)SO\(_4\) to introduce a second nitro group at the 4-position, giving 2,4-dinitrotoluene. 4. Third nitration of 2,4-dinitrotoluene: React 2,4-dinitrotoluene with a mixture of concentrated HNO\(_3\) and H\(_2\)SO\(_4\) to introduce a third nitro group at the 6-position, giving 2,4,6-Trinitrotoluene (TNT).

(d) Synthesis of m-Chlorobenzoic acid

To synthesize m-Chlorobenzoic acid, we can start with benzene as the source of the aromatic ring. The plan is to first introduce a nitro group, reduce it to an amino group, and then perform a diazonium coupling reaction to introduce the chloro group, followed by oxidation of the side chain. 1. Nitration of benzene: React benzene with a mixture of concentrated HNO\(_3\) and H\(_2\)SO\(_4\) to introduce a nitro group, giving nitrobenzene. 2. Reduction of nitrobenzene: React nitrobenzene with Sn and HCl, followed by neutralization with excess NaOH, to reduce the nitro group to an amino group, giving aniline. 3. Diazotization of aniline: React aniline with NaNO\(_2\) and HCl to form a diazonium salt. 4. Sandmeyer reaction: React the diazonium salt with CuCl to introduce a chloro group at the position of the amino group, giving chlorobenzene. 5. Side-chain oxidation: React chlorobenzene with KMnO\(_4\) and H\(_2\)O in the presence of acidic conditions to oxidize the side-chain to introduce a carboxylic acid group, giving m-Chlorobenzoic acid.

(e) Synthesis of p-Chlorobenzoic acid

To synthesize p-Chlorobenzoic acid, we can start with benzene as the source of the aromatic ring. The plan is to first introduce a carboxylic acid group, and then introduce the chloro group. 1. Side-chain oxidation: React benzene with KMnO\(_4\) and H\(_2\)O in the presence of acidic conditions to introduce a carboxylic acid group, giving benzoic acid. 2. Chlorination of benzoic acid: React benzoic acid with SOCl\(_2\) to form an acid chloride, and then react the acid chloride with AlCl\(_3\) to introduce a chloro group at the para-position relative to the carboxylic acid group, giving p-Chlorobenzoic acid.

(f) Synthesis of p-Dichlorobenzene

To synthesize p-Dichlorobenzene, we can start with benzene and perform two electrophilic aromatic substitutions. 1. Chlorination of benzene: React benzene with Cl\(_2\) in the presence of FeCl\(_3\) to introduce a chloro group, giving chlorobenzene. 2. Second chlorination: React chlorobenzene with Cl\(_2\) in the presence of FeCl\(_3\) to introduce a second chloro group at the para-position relative to the first chloro group, giving p-Dichlorobenzene.

(g) Synthesis of m-Nitrobenzenesulfonic acid

To synthesize m-Nitrobenzenesulfonic acid, we can start with benzene. The plan is to first introduce a nitro group, and then introduce the sulfonyl group. 1. Nitration of benzene: React benzene with a mixture of concentrated HNO\(_3\) and H\(_2\)SO\(_4\) to introduce a nitro group, giving nitrobenzene. 2. Sulfonation of nitrobenzene: React nitrobenzene with concentrated H\(_2\)SO\(_4\) to introduce a sulfonic acid group at the meta-position relative to the nitro group, giving m-Nitrobenzenesulfonic acid.