The given compound A is C₆H₅CONH₂, an amide which undergoes Hoffmann Bromamide synthesis which gives the compound B, primary amine C₆H₅NH₂ aniline. Aniline reacts with NaNO₂ and HCl at 0°C to give compound C Benzene Diazonium Chloride, a diazo salt. Benzene diazonium chloride reacts with ethanol, CH₃CH₂OH to give a compound D hydrocarbon Benzene by elimination of N₂⁺Cl^- and replacement with H. The given compound B is aniline because only aniline reacts with Br₂ water to give a white precipitate of compound E, 2,4,6-tribromophenylamine.

The reactions are as follows.

2,4,6-tribromophenylamine

OR

a(i) Aniline is treated with NaNO₂ and HCl at 0°C to form benzene diazonium chloride. This salt on treatment with HBF₄ or Fluroboric acid gives Fluorobenzene and has toxic gas BF₃ as by-product.

(ii)Benzamide is converted to Benzylamine by Hoffmann bromamide synthesis, where an amide is treated with Br₂ water and aqueous or ethanolic NaOH to give benzylamine, a primary amine.

(iii)Ethanamine, which is a primary amine, can be converted to N,N-Diethylethanamine which is a tertiary amine by ammonolysis of alkyl halides. An alkyl or benzyl halide on reaction with an ethanolic solution of ammonia undergoes nucleophilic substitution reaction in which the halogen atom is replaced by an amino (–NH₂) group. The reaction is carried out in a sealed tube at 100C. The primary amine thus obtained behaves as a nucleophile and can further react with alkyl halide to form secondary and tertiary amines, and finally quaternary ammonium salt. Ethanamine, C₂H₅—NH₂, the primary amine is further treated with ethyl halide, C₂H₅X, leading to the substitution of the –H groups on ethanamine with the ethyl groups on the halide, first forming secondary amine N-ethylethanamine (C₂H₅)₂NH, and then tertiary amine N, N-Diethylethanamine (C₂H₅)₃N.

C₂H₅—NH₂ + C₂H₅X → (C₂H₅)₂NH + C₂H₅X → (C₂H₅)₃N

(b)(i) CH₃CH₂CN, ethyl nitrile, on partial hydrolysis with mild heat with NaOH or KOH gives compound A, ethyl amide, CH₃CH₂CONH₂.

CH₃CH₂CN + NaOH/KOH → CH₃CONH₂

Ethyl amide on treatment with NaOH and Br₂ water undergoes Hoffmann Bromamide degradation to form compound B, a primary amine, ethyl amine, or CH₃CH₂NH₂.

CH₃CONH₂ + Br₂ + NaOH → CH₃CH₂NH₂.

(ii) CH₃CH₂Br, ethyl bromide, when treated with KCN, undergoes nucleophilic substitution reaction to form CH₃CH₂CN. CH₃CH₂CN on treatment with LiAlH₄, a strong reducing agent, forms compound A, CH₃CH₂NH₂, ethyl amine, primary amine. CH₃CH₂NH₂ on treatment with nitrous acid at 0C, forms compound B which is a diazo salt, ethyldiazonium ion or ethyldiazonium chloride, CH₃CH₂N₂⁺Cl^-