Reduction of nitrobenzene by which of the following reagent does not give aniline?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(Sn/NH_4OH\)
Nitrobenzene is converted to aniline by a six-electron reduction of the –\( \ce{NO2} \) group. This goes all the way to the amine only with a metal in acidic medium or by catalytic hydrogenation:
\( \displaystyle \ce{C6H5NO2 + 6[H]} \xrightarrow{\ce{Sn + HCl}\text{ or }\ce{Fe + HCl}} \ce{C6H5NH2 + 2H2O} \)
\( \displaystyle \ce{C6H5NO2 + 3H2} \xrightarrow{\ce{Pd}} \ce{C6H5NH2 + 2H2O} \)
In a neutral or basic medium the reduction stops at intermediate stages or the intermediates combine with one another. In basic medium nitrobenzene gives coupled products such as azoxybenzene, azobenzene and hydrazobenzene, and in neutral medium (Zn/\( \ce{NH4Cl} \)) N-phenylhydroxylamine, \( \ce{C6H5NHOH} \).
Option (1) \(Sn/HCl\) — Gives aniline. Tin and hydrochloric acid reduce nitrobenzene to aniline (as the anilinium salt, freed by alkali).
Option (2) \(Fe/HCl\) — Gives aniline. Iron and HCl reduce the nitro group in the same way; this is the method preferred in practice.
Option (3) \(H_2-Pd\) — Gives aniline. Hydrogen gas over finely divided palladium reduces nitrobenzene to aniline.
Option (4) \(Sn/NH_4OH\) — Does not give aniline. In a basic medium the reduction stops at intermediate or coupled products, not the amine.
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. Nitroalkanes can also be similarly reduced to the corresponding alkanamines.”