JEE Main · 2022 · 28 Jun 2022 · Shift 1 · Compounds Containing Nitrogen

Aniline is the starting material, and its $-\mathrm{NH}{}_2$ group is a strongly activating, ortho- and para-directing substituent, so stirring it with bromine water brominates the ring exhaustively at both ortho positions and the para position to give 2,4,6-tribromoaniline; the ring is so activated that no Lewis-acid catalyst such as $\mathrm{FeBr}{}_3$ is required. The three bromine atoms now sit at the 2-, 4- and 6-positions relative to nitrogen, so the amino group that directed them must itself be removed to expose a symmetrical ring. This is achieved by diazotisation: cold treatment with $\mathrm{NaNO}{}_2$ and $\mathrm{HCl}$ converts $-\mathrm{NH}{}_2$ into the 2,4,6-tribromobenzenediazonium salt. Warming that salt with hypophosphorous acid, $\mathrm{H}{}_3\mathrm{PO}{}_2$, in water reductively replaces the diazonium group by hydrogen, a deamination, so the carbon that once bore nitrogen now carries only $-\mathrm{H}$. What is left is a benzene ring carrying three bromine atoms at positions that are mutually meta to one another, which is precisely 1,3,5-tribromobenzene. Note that no Sandmeyer step is involved and bromine is not a meta-director; the symmetrical 1,3,5 pattern arises only because the amino group first steered bromination to its three ortho and para sites and was then stripped away.