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Question

List- I List- II
A) C6H5NO2 Sn + HCl−−−−−−→ i) Azobenzene
B) C6H5NO2 Zn + NH4Cl−−−−−−−−→ ii) Hydrazobenzene
C) C6H5NO2 LiAlH4−−−−→ iii) Phenyl hydroxyl amine
D) C6H5NO2 Zn + KOH(alc.)−−−−−−−−−→ iv) Amino benzene

A
A - ii, B - iv, C - i, D - iii
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B
A - iv, B - iii, C - ii, D - i
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C
A - iv, B - ii, C - iii, D - i
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D
A - iv, B - iii, C - i, D - ii
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Solution

The correct option is D A - iv, B - iii, C - i, D - ii
Nitrobenzene can be reduced by using various products.
The products of reduction depend on the reducing agent used.
Reduction of nitrobenzene with tin in presence of HCl gives aniline (aminobenzene).
C6H5NO2 Sn + HCl−−−−−C6H5NH2

Reduction of nitrobenzene with zinc in presence of ammonium chloride gives phenyl hydroxyl amine.
C6H5NO2 Zn + NH4Cl−−−−−−−C6H5NHOH

Reduction of nitrobenzene with lithium aluminum hydride gives azobenzene.
C6H5NO2 Zn + NH4Cl−−−−−−−C6H5N=NC6H5

Reduction of nitrobenzene with zinc in the presence of alcoholic potassium hydroxide gives hydrazobenzene.
C6H5NO2 Zn + KOH−−−−−−C6H5HNNHC6H5

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