Answer
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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 .
Note:
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.
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 .
Note:
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.
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