Question

Complete the following equations:
(i) $\mathrm{C_6H_5NH_2 + CHCl_3 + KOH(alc)} \;\longrightarrow$
(ii) $\mathrm{C_6H_5NH_2 + Br_2(aq)} \;\longrightarrow$
(iii) $\mathrm{C_6H_5NH_2 + (CH_3CO)_2O} \;\longrightarrow$
(iv) $\mathrm{C_6H_5NH_2 + H_2SO_4(conc.)} \;\longrightarrow$
(v) $\mathrm{C_6H_5NH_2 + CH_3COCl \xrightarrow{\,base\,}} \;\longrightarrow$

Hint:

Aniline is strongly activating and ortho/para–directing; in aqueous bromine it gives 2,4,6-tribromoaniline instantly. Protecting the \(-NH_2\) group by acetylation (acetanilide) moderates its reactivity for controlled EAS. Carbylamine forms foul-smelling isocyanides only from primary amines.

Answer 1

(i) Carbylamine reaction (isocyanide test):
\[ \mathrm{C_6H_5NH_2 + CHCl_3 + 3KOH \;\longrightarrow\; C_6H_5NC + 3KCl + 3H_2O}\,. \]
(ii) Bromination in water (tribromination):
\[ \mathrm{C_6H_5NH_2 + 3Br_2(aq) \;\longrightarrow\; 2,4,6\!-\!C_6H_2Br_3NH_2\,(s) + 3HBr}\,. \]
(iii) Acetylation with acetic anhydride:
\[ \mathrm{C_6H_5NH_2 + (CH_3CO)_2O \;\longrightarrow\; C_6H_5NHCOCH_3\ (acetanilide) + CH_3COOH}\,. \]
(iv) Protonation by conc.\ sulfuric acid:
\[ \mathrm{C_6H_5NH_2 + H_2SO_4 \;\longrightarrow\; C_6H_5NH_3^+\ HSO_4^- \ (anilinium\ hydrogensulfate)}\,. \]
(v) Acetylation with acetyl chloride (base present):
\[ \mathrm{C_6H_5NH_2 + CH_3COCl \xrightarrow{\,base\,} C_6H_5NHCOCH_3\ (acetanilide) + HCl}\,, \] (the base neutralises HCl as \(\mathrm{base\cdot HCl}\)).