A Level Chemistry 9701: Phenylamine Preparation, Diazotisation and Azo Dye Formation

Phenylamine, $\text{C}_6\text{H}_5\text{NH}_2$ , is prepared from nitrobenzene and then converted into an azo dye.

(a) Nitrobenzene, $\text{C}_6\text{H}_5\text{NO}_2$ , is heated under reflux with tin and an excess of concentrated hydrochloric acid, and the mixture is then treated with excess sodium hydroxide solution. Give an overall equation for the reduction (using $[\text{H}]$ to represent hydrogen from the reducing agent), and state the purpose of adding the sodium hydroxide solution. [3]

(b) The phenylamine obtained in (a) is treated with nitrous acid, $\text{HNO}_2$ (generated in situ from sodium nitrite and dilute hydrochloric acid), at a temperature kept below $10\,^\circ\text{C}$ . Give an equation for this reaction, name the type of reaction and the organic product formed, and explain why the temperature must be kept below $10\,^\circ\text{C}$ . [3]

(c) The cold solution from (b) is added to an alkaline solution of phenol. State what is observed, give the structural formula of the azo compound formed (coupling occurs at the position on the phenol ring para to the -OH group), and explain why this compound is coloured. [3]

Part (a): Reducing nitrobenzene to phenylamine

Tin and excess concentrated hydrochloric acid, heated under reflux, act as a reducing agent for the nitro group. Overall, six hydrogen atoms (represented as $[\text{H}]$ ) are added, converting $-\text{NO}_2$ into $-\text{NH}_2$ and releasing two molecules of water:

$\text{C}_6\text{H}_5\text{NO}_2 + 6[\text{H}] \rightarrow \text{C}_6\text{H}_5\text{NH}_2 + 2\text{H}_2\text{O}$

Because the reaction mixture is strongly acidic, the phenylamine formed is immediately protonated to its ammonium salt (phenylammonium chloride). Adding excess sodium hydroxide solution afterwards neutralises the excess hydrochloric acid and deprotonates this salt, releasing free phenylamine, $\text{C}_6\text{H}_5\text{NH}_2$ , which can then be separated (e.g. by steam distillation).

Part (b): Diazotisation

Phenylamine reacts with nitrous acid, generated in situ from sodium nitrite and dilute hydrochloric acid, in a reaction called diazotisation:

$\text{C}_6\text{H}_5\text{NH}_2 + \text{HNO}_2 + \text{HCl} \rightarrow \text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^- + 2\text{H}_2\text{O}$

The organic product is benzenediazonium chloride, $\text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^-$ . The temperature must be kept below $10\,^\circ\text{C}$ because the diazonium ion is thermally unstable: above this temperature it decomposes rapidly (releasing nitrogen gas and forming phenol), so the diazonium salt would break down before it could be used in a further reaction.

Part (c): Coupling with phenol to form an azo dye

The cold benzenediazonium chloride solution is added to phenol dissolved in sodium hydroxide solution (as the phenoxide ion, which is strongly activated towards electrophilic attack). The diazonium ion acts as an electrophile and couples onto the phenol ring at the position para to the -OH group, giving a bright yellow/orange azo compound:

$\text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^- + \text{C}_6\text{H}_5\text{OH} \rightarrow \text{C}_6\text{H}_5\text{-N=N-C}_6\text{H}_4\text{-OH} + \text{HCl}$

This azo compound is coloured because the $-\text{N=N}-$ azo linkage joins the pi systems of the two benzene rings into one long, continuously delocalised pi system. Extending the delocalisation over more atoms lowers the energy gap between the highest occupied and lowest unoccupied molecular orbitals, so the compound can absorb photons of visible light (rather than only ultraviolet light, as an isolated benzene ring would); the light that is not absorbed is reflected or transmitted, giving the compound its visible colour.

Final answers

(a) $\text{C}_6\text{H}_5\text{NO}_2 + 6[\text{H}] \rightarrow \text{C}_6\text{H}_5\text{NH}_2 + 2\text{H}_2\text{O}$ ; NaOH neutralises excess HCl and liberates free phenylamine from its ammonium salt.
(b) $\text{C}_6\text{H}_5\text{NH}_2 + \text{HNO}_2 + \text{HCl} \rightarrow \text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^- + 2\text{H}_2\text{O}$ ; diazotisation, forming benzenediazonium chloride; kept below $10\,^\circ\text{C}$ because the diazonium salt decomposes above this temperature.
(c) Yellow/orange colour observed; product is $\text{C}_6\text{H}_5\text{-N=N-C}_6\text{H}_4\text{-OH}$ ; coloured because the extended delocalised pi system across both rings and the azo link narrows the HOMO-LUMO gap into the visible region.