The correct sequence of reactions to convert benzene to 3-bromobenzenamine is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → nitration, bromination, reduction
In 3-bromobenzenamine (m-bromoaniline) the Br and \( \ce{NH2} \) groups are meta to each other. The –NO2 group is meta-directing, so the nitro group must be in the ring when bromine enters, and it is reduced last. First, nitration:
\( \displaystyle \ce{C6H6 + HNO3} \xrightarrow{\text{conc. }\ce{H2SO4},\ \text{323-333 K}} \ce{C6H5NO2 + H2O} \)
Then bromination puts Br meta to \( \ce{NO2} \):
\( \displaystyle \ce{C6H5NO2 + Br2} \xrightarrow{\ce{FeBr3}} m\text{-}\ce{BrC6H4NO2 + HBr} \)
Last, reduction turns –\( \ce{NO2} \) into –\( \ce{NH2} \) without moving the bromine:
\( \displaystyle m\text{-}\ce{BrC6H4NO2 + 6[H]} \xrightarrow{\ce{Sn/HCl}} m\text{-}\ce{BrC6H4NH2 + 2H2O} \)
Option (1) nitration, reduction, bromination — Incorrect. Reducing first gives aniline; the strongly activating, ortho/para-directing –\( \ce{NH2} \) group then gives 2,4,6-tribromoaniline with bromine.
Option (2) reduction, nitration, bromination — Incorrect. Benzene has no group that can be reduced, so the sequence cannot start with reduction.
Option (4) bromination, nitration, reduction — Incorrect. Br is ortho/para-directing, so nitration of bromobenzene gives o- and p-bromonitrobenzene, which reduce to o- and p-bromoaniline.
NCERT (2026-27 edition):
Reduction of Nitro Compounds – “Nitro compounds are reduced to amines by passing hydrogen gas in the presence of finely divided nickel, palladium or platinum and also by reduction with metals in acidic medium.”