Aniline is successively treated with i) excess bromine ii) $\mathrm{NaNO}_2, \mathrm{HCl}$ iii) $\ma

Question

Aniline is successively treated with i) excess bromine ii) $\mathrm{NaNO}_2, \mathrm{HCl}$ iii) $\mathrm{H}_3 \mathrm{PO}_2$. Major product formed is

Professor PC Thomas Classes · Answer

Bromination gives 2,4,6-tribromoaniline. Diazotization then reduction by $\mathrm{H}_3\mathrm{PO}_2$ removes the $\mathrm{NH}_2$ group, leaving 1,3,5-tribromobenzene.