Amines — NCERT Solutions

Aniline reacts with benzoyl chloride by benzoylation:

$$\mathrm{C_6H_5NH_2 + C_6H_5COCl \rightarrow C_6H_5NHCOC_6H_5 + HCl}$$

The product is benzanilide (also called N-phenylbenzamide).

The isomers of $\mathrm{C_3H_9N}$ are:

1. $\mathrm{CH_3CH_2CH_2NH_2}$ — propan-1-amine
2. $\mathrm{CH_3CH(NH_2)CH_3}$ — propan-2-amine
3. $\mathrm{CH_3CH_2NHCH_3}$ — N-methylethanamine
4. $\mathrm{(CH_3)_3N}$ — N,N-dimethylmethanamine

On treatment with nitrous acid, primary aliphatic amines liberate nitrogen gas. Therefore, the isomers that liberate $\mathrm{N_2}$ are:

• propan-1-amine
• propan-2-amine

(i) 3-Methylaniline into 3-nitrotoluene

First convert the amino group to a diazonium salt, then replace the diazo group by nitro group.

$$3\text{-methylaniline} \xrightarrow{NaNO_2/HCl,\ 273-278\ K} 3\text{-methylbenzenediazonium chloride}$$
$$\xrightarrow{CuNO_2} 3\text{-nitrotoluene}$$

(ii) Aniline into 1,3,5-tribromobenzene

Aniline is strongly activating and bromine water gives 2,4,6-tribromoaniline. To obtain 1,3,5-tribromobenzene, first convert aniline into the diazonium salt and then replace the diazonium group by hydrogen, followed by bromination under suitable conditions.

A direct textbook route is:

1. Diazotisation of aniline:

$$C_6H_5NH_2 \xrightarrow{NaNO_2/HCl,\ 273-278\ K} C_6H_5N_2^+Cl^-$$

2. Replacement by H:

$$C_6H_5N_2^+Cl^- \xrightarrow{H_3PO_2/H_2O} C_6H_6$$

3. Bromination of benzene under appropriate conditions gives bromobenzene, and further bromination gives the tribromo product.

Since the question asks for conversion, the key idea is to use diazotisation and substitution of the diazonium group to control the ring substitution. The final product is 1,3,5-tribromobenzene.

If written in step form:
$$C_6H_5NH_2 \rightarrow C_6H_5N_2^+Cl^- \rightarrow C_6H_6 \rightarrow C_6H_5Br \rightarrow 1,3,5\text{-tribromobenzene}$$

An aromatic compound that on treatment with aqueous ammonia and heating forms a compound which on heating with $\mathrm{Br_2}$ and KOH gives a product of formula $\mathrm{C_6H_7N}$ is identified as follows:

• A must be benzene only if the chapter’s standard sequence is interpreted through the preparation of aniline.
• B is aniline.
• C also has molecular formula $\mathrm{C_6H_7N}$, so it is aniline.

Thus the structures/names are:

• A: benzene
• B: benzenamine (aniline)
• C: benzenamine (aniline)

This is the compound discussed in the chapter where an amide on Hofmann bromamide degradation gives aniline.

(i)

Carbylamine reaction:
$$C_6H_5NH_2 + CHCl_3 + 3KOH\xrightarrow{alc.,\ heat} C_6H_5NC + 3KCl + 3H_2O$$
Product: phenyl isocyanide.

(ii)

Reduction of benzenediazonium chloride:
$$C_6H_5N_2Cl + H_3PO_2 + H_2O \rightarrow C_6H_6 + N_2 + H_3PO_3 + HCl$$
Product: benzene.

(iii)

Sulphonation of aniline:
$$C_6H_5NH_2 + H_2SO_4\, (conc.) \rightarrow C_6H_5NH_3^+HSO_4^-$$
On heating, it gives sulphanilic acid:
$$C_6H_5NH_2 \xrightarrow{conc.\ H_2SO_4,\ 453-473K} p\text{-aminobenzenesulphonic acid}$$

(iv)

Reduction by ethanol:
$$C_6H_5N_2Cl + C_2H_5OH \rightarrow C_6H_6 + N_2 + CH_3CHO + HCl$$
Product: benzene.

(v)

Bromination of aniline:
$$C_6H_5NH_2 + 3Br_2(aq) \rightarrow 2,4,6\text{-tribromoaniline} + 3HBr$$
Product: 2,4,6-tribromoaniline.

(vi)

Acetylation of aniline:
$$C_6H_5NH_2 + (CH_3CO)_2O \rightarrow C_6H_5NHCOCH_3 + CH_3COOH$$
Product: acetanilide or N-phenylethanamide.

(vii)

Diazonium fluoroborate route and replacement by nitro group:
$$C_6H_5N_2Cl \xrightarrow{HBF_4} C_6H_5N_2BF_4 \xrightarrow{NaNO_2/Cu,\ \Delta} C_6H_5NO_2$$
Product: nitrobenzene.

Gabriel phthalimide synthesis works only for alkyl halides because the phthalimide anion must attack by nucleophilic substitution.

Aromatic primary amines cannot be prepared by this method because aryl halides do not undergo nucleophilic substitution with the anion formed by phthalimide. Hence the required replacement of halogen by the phthalimide group does not take place.

(i) Aromatic primary amines with nitrous acid

Aromatic primary amines react with nitrous acid at 273–278 K to form diazonium salts:

$$\mathrm{C_6H_5NH_2 + NaNO_2 + 2HCl \xrightarrow{273-278\,K} C_6H_5N_2Cl + NaCl + 2H_2O}$$

(ii) Aliphatic primary amines with nitrous acid

Aliphatic primary amines form unstable diazonium salts which decompose to alcohols with evolution of nitrogen:

$$\mathrm{RNH_2 + HNO_2 \rightarrow ROH + N_2 + H_2O}$$

So, aromatic amines give stable diazonium salts in the cold, while aliphatic amines give alcohols and nitrogen gas.

(i) Why are amines less acidic than alcohols of comparable molecular masses?

Amines are less acidic because nitrogen is less electronegative than oxygen. Therefore the $N-H$ bond is less polar than the $O-H$ bond, so amines do not release $H^+$ as easily as alcohols.

(ii) Why do primary amines have higher boiling point than tertiary amines?

Primary amines have two N-H bonds, so they can form more intermolecular hydrogen bonds than tertiary amines, which have no N-H bond. Stronger intermolecular association raises the boiling point.

(iii) Why are aliphatic amines stronger bases than aromatic amines?

In aliphatic amines, alkyl groups show the +I effect, increasing electron density on nitrogen and making the lone pair more available for protonation. In aromatic amines, the lone pair on nitrogen is delocalised into the benzene ring by resonance, so it is less available to accept a proton. Hence aliphatic amines are stronger bases than aromatic amines.