JEE Main · 2021 · 24 Feb 2021 · Shift 2 · Compounds Containing Nitrogen

The nitro group deactivates the ring and directs an incoming electrophile to the meta position, so the bromine must be introduced while that group is still in place: $\mathrm{B}{\mathrm{r}}_2/Fe$ converts nitrobenzene into $m$-bromonitrobenzene. Reducing first would give aniline, whose amino group is strongly activating and ortho-para directing, and the bromines would then end up in the wrong positions. $Sn/HCl$ now reduces the nitro group to give $m$-bromoaniline, $\mathrm{NaN}{\mathrm{O}}_2/HCl$ at $273$-$278$ K converts the amino group into a diazonium group, and the Sandmeyer reaction with $CuBr/HBr$ replaces that diazonium group by bromine. The sequence required is $\mathrm{B}{\mathrm{r}}_2/Fe$, then $Sn/HCl$, then $\mathrm{NaN}{\mathrm{O}}_2/HCl$, then $CuBr/HBr$.