Question

The product obtained when nitrobenzene is reduced using Zn/NaOH is:
A. azobenzene
B. Aniline
C. hydrazobenzene
D. azoxybenzene

Answer 1

Hint: The reduction of nitro compounds yields amines. These amines can be primary, secondary, or tertiary. Also the reduction of bulky benzene with nitro, can yield a compound with two benzene rings.

Complete answer: When nitro compounds are reduced, then amines are formed. The bulky nitrobenzene is reduced using zinc, and sodium hydroxide.
The reaction is such that 2 moles of nitrobenzene are involved along with 10 hydrogen atoms, for reduction, with Zn in alkaline medium (NaOH). The reaction results in forming a compound with 2 benzene rings and secondary amines. This compound is called as hydrazobenzene, the reaction is as follows:
$\begin{array}{rl}& 2C_6{H}_5-N{O}_2\underset{10[H}{\overset{]}{\to }}Zn/NaOH{C}_6{H}_5-NH-NH-C_6{H}_5\\ & nitrobenzenehydrazobenzene\end{array}$
The reduction of $N{O}_2$is in such a way that it gets converted to amines, the amines are attached with themselves and then benzene rings are attached.
Hence, option C is correct as hydrazobenzene is formed by reduction of nitrobenzene with Zn and NaOH.

Additional information: the IUPAC name for hydrazobenzene is 1,2-Diphenylhydrazine. It is a chemical used to make benzidine dye of orange color. The decomposition leads to toxic fumes of nitrogen oxides, also it is considered as a carcinogen.

Note: when nitrobenzene is treated with methanolic solution of zinc and sodium hydroxide then, azobenzene is formed, this happens because alcohol, removes the hydrogen present in amine and convert to simple nitrogen resulting in the azobenzene. The reaction is, $\begin{array}{rl}& 2C_6{H}_5-N{O}_2\underset{C{H}_{3}OH}{\overset{Zn/NaOH}{\to }}{C}_6{H}_5-N=N-C_6{H}_5\\ & nitrobenzeneazobenzene\end{array}$