Question

Nitrobenzene gives N-phenylhydroxylamine by
A. Sn/HCl
B. \[{H_2}/Pd - C\]
C. Zn/NaOH
D. Zn/\[N{H_4}Cl\]

Answer 1

Hint: The conversion of nitrobenzene to N-phenylhydroxyamine is done with the help of a reducing agent. Here, the double-bonded oxygen group present in a nitro group is converted to oxime or a hydroxyl group.

Complete Step by Step Solution:
Nitrobenzene is converted to N-phenylhydroxylamine in presence of zinc dust along with ammonium chloride. It acts as a mild reducing agent which reduces the double bond oxygen present in the nitro group into an oxime. The zinc donates the electron pair to convert into a divalent cation.

The reaction between nitrobenzene and zinc dust in the presence of ammonium chloride is shown below.

Image: Conversion of nitrobenzene to N-phenylhydroxyl amine

The reaction mechanism is shown below:
\[Ph - N{O_2} \to Ph - N = O \overset{2e^{-}/Zn^{2+}}{\rightarrow} PhNH(OH)\]

Here, Ph is the phenyl group
The nitrogen atom present in the nitro group is partially positive in nature. The zinc atom transfers its one electron to the nitrogen as a result one electron moves to oxygen and another electron to the nitrogen atom in \[N = O\]. The reaction of a proton with both \[ - {O^ - }\] forms N, N-hydroxybenzene which is very unstable due to the presence of two hydroxyl groups. By losing water molecules it forms the intermediate product nitrosobenzene which on reacting with zinc ion forms N-phenyl hydroxylamine.
Thus, Nitrobenzene gives N-phenylhydroxylamine by Zn/\[N{H_4}Cl\].

Therefore, the correct option is D.

Note: Reduction is defined as a reaction where there is a decrease in the oxidation state of the central atom. In the given reaction ammonium chloride is used as a promoter for zinc dust.