Q.The product obtained when nitrobenzene is reduced using Zn/NaOH is :

Complete reduction of nitrobenzene to aniline needs a strong acidic reducing system such as $Sn/HCl$ (or equivalently $Fe/HCl$ or catalytic $H_2/Ni$); Zn/NH$_4$Cl only partially reduces the nitro group.

Nitrobenzene reduction proceeds through several stages: $C_6H_5NO_2 \rightarrow C_6H_5NO \rightarrow C_6H_5NHOH \rightarrow C_6H_5NH_2$. Tin with concentrated hydrochloric acid ($Sn/HCl$) supplies enough reducing power to carry the reduction all the way to aniline ($C_6H_5NH_2$) — the standard laboratory method for this conversion. Zn dust with neutral $NH_4Cl$ solution (option b) is a much milder reducing system that arrests the sequence at the N-phenylhydroxylamine stage rather than going to completion. Zn(Hg)/NaOH (option a, Clemmensen …

Zn/NaOH is a strongly reducing alkaline system that reduces nitrobenzene completely through nitroso- and azoxy-/azo- intermediates to hydrazobenzene, $C_6H_5-NH-NH-C_6H_5$.

The reduction of nitrobenzene proceeds through a well-defined sequence of intermediates, $C_6H_5NO_2 \rightarrow C_6H_5NO \rightarrow C_6H_5NHOH \rightarrow$ (coupling) $\rightarrow C_6H_5-N=N(\to O)-C_6H_5$ (azoxybenzene) $\rightarrow C_6H_5-N=N-C_6H_5$ (azobenzene) $\rightarrow C_6H_5-NH-NH-C_6H_5$ (hydrazobenzene), and which product is isolated depends entirely on the strength and pH of the reducing system used: Zn/$NH_4Cl$ (neutral, mild) stops at phenylhydroxylamine; Sn/HCl or Fe/HCl (acidic, strong) goes all the way to aniline; Zn dust in strongly alkaline NaOH solution is the classical reagent that carries the reduction through to the fully reduced coupled produc …

The reducing agent used on nitrobenzene changes the final product: acidic reduction (Sn/HCl or Fe/HCl) gives aniline directly; alkaline reduction (Zn dust + $NaOH$) proceeds through a coupling pathway to give hydrazobenzene, via successive intermediates.

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