Question

What happens when aniline reacts with the following:
(A) Bromine water
(B) $(CH_3CO)_2O$ / Pyridine
(C) $HNO_2 + HCl (0^\circ - 5^\circ C)$

Hint:

To get a monobromo derivative of aniline, you must first protect the $-NH_2$ group by acetylation (reaction ii), then brominate, and finally hydrolyze. Direct bromination always yields the tribromo product.

Answer 1

(A) Reaction with Bromine water:
Aniline reacts with bromine water at room temperature to give a white precipitate of 2,4,6-Tribromoaniline. The amino group strongly activates the benzene ring, leading to poly-substitution. \[ C_6H_5NH_2 + 3Br_2(aq) \longrightarrow 2,4,6\text{-Tribromoaniline} + 3HBr \] (B) Reaction with Acetic Anhydride/Pyridine (Acetylation):
Aniline undergoes acetylation to form Acetanilide (N-phenylethanamide). This reaction protects the amino group and decreases the activating power of the $-NH_2$ group. \[ C_6H_5NH_2 + (CH_3CO)_2O \xrightarrow{\text{Pyridine}} C_6H_5NHCOCH_3 + CH_3COOH \] (C) Reaction with $HNO_2 + HCl$ (Diazotization):
Aniline reacts with nitrous acid ($HNO_2$, generated in situ from $NaNO_2 + HCl$) at low temperature ($0-5^\circ C$ or 273-278 K) to form Benzene diazonium chloride. \[ C_6H_5NH_2 + NaNO_2 + 2HCl \xrightarrow{0-5^\circ C} C_6H_5N_2^+Cl^- + NaCl + 2H_2O \]