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Reduction of nitrobenzene in the presence of $Zn/N{H_4}Cl$ gives:
A. hydrazobenzene
B. aniline
C. azobenzene
D. N- phenyl hydroxylamine

Last updated date: 13th Jun 2024
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Hint: The zinc dust along with ammonium chloride act as a mild reducing agent and they help to reduce a double bonded oxygen group (as in the case of a nitro group) to oxime (or simply to a hydroxyl group). The zinc acts as an electron pair donor to get converted into its bivalent cation.

Complete step by step answer:
The reaction of nitrobenzene with zinc dust in the presence of ammonium chloride is as follows:
$Ph - N{O_2}\xrightarrow{{2{e^ - }/Z{n^{2 + }}}}Ph - N = O\xrightarrow{{2{e^ - }/Z{n^{ + 2}}}}Ph - NH(OH)$
Here, $Ph = $ Phenyl group i.e. benzene with a free valency.
In the above reaction mechanism, the nitrogen atom of the nitro group is partially positive charged and there is a transfer of one electron from the zinc atom to the nitrogen centre and due to this, one electron of the $N = O$ bond jumps over the oxygen atom and the other to the nitrogen atom. Thus, on reaction of protons with both the $ - {O^ - }$ centers give a N,N-dihydroxy benzene which is unstable because of the presence of two hydroxyl groups over it. Die to this, it suffers de hydration and there is a loss of one water molecule. This forms the nitrosobenzene (the middle compound) which again undergoes reaction with two electrons from the zinc dust to produce N- phenyl hydroxylamine (the third product).

So, the correct answer is OptionD .

The ammonium salts such as ammonium chloride act as a promoter of the zinc reduction of nitrobenzene. $Zn/N{H_4}Cl$acts as a mild reducing agent and helps in the reduction of nitrobenzene. The reduction of any functional group means to decrease the oxidation state of the central atom of the substituent.