Question

The reduction of nitrobenzene in the presence of $\mathrm{Zn}$/Ammonium chloride gives:

• A. Aniline
• B. Azobenzene
• C. Hydrazobenzene
• D. N-phenyl hydroxylamine

Answer 1

The correct option is D

N-phenyl hydroxylamine

Explanation for correct option:

(D) N-phenyl hydroxylamine

The reduction of nitrobenzene in the presence of Zn/Ammonium chloride

• Zinc reduction can also be aided by the use of ammonium salts such as ammonium formate and acetate.
• In order for the Zinc reduction of nitrobenzene to continue at temperatures close to ambient, promoters such as ammonium chloride act as electrolytes and complex-forming agents.
• The reagent, $\mathrm{Zn}/{\mathrm{NH}}_4\mathrm{Cl}$, reduces nitrobenzene and produces N-phenyl hydroxylamine in a neutral media, which means $\mathrm{Zn}$ dust and aqueous ${\mathrm{NH}}_4\mathrm{Cl}$ are present in the solution.
• The chemical reaction can be depicted as:

Explanation for incorrect options:

(A) Aniline

• An amino functional group is substituted for one of the benzene hydrogens in aniline, a primary arylamine. $\mathrm{Sn}$ and concentrated $\mathrm{HCl}$ reduce nitrobenzene to aniline. $\mathrm{Zn}$ or $\mathrm{Fe}$ can alternatively be used in place of $\mathrm{Sn}$.
• The aniline salt is then mixed with aqueous $\mathrm{NaOH}$ to liberate the aniline.
• Hence, the reduction of nitrobenzene in the presence of Zn/Ammonium chloride does not give aniline.

(B) Azobenzene

1. Two phenyl rings connected by an $\mathrm{N}=\mathrm{N}$ double bond make up the photoswitchable chemical compound known as azobenzene.
2. In order to convert nitrobenzene into azobenzene, a reduction is done in the presence of $\mathrm{Zn}/\mathrm{NaOH}$ in ${\mathrm{CH}}_3\mathrm{OH}$.
3. Hence, the reduction of nitrobenzene in the presence of Zn/Ammonium chloride does not give azobenzene.

(C) Hydrazobenzene

1. Two aniline groups connected by their nitrogen atoms form the aromatic organic molecule hydrazobenzene.
2. A method for converting nitrobenzene to hydrazobenzene involves contacting nitrobenzene in a solution of $50–80$ percent aqueous alcohol with a sodium amalgam that contains $0.1–0.2\%$ sodium in the presence of carbon.
3. Hence, the reduction of nitrobenzene in the presence of Zn/Ammonium chloride does not give hydrazobenzene.

Therefore, the correct option is D i.e. N-phenyl hydroxylamine.