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A $\xrightarrow{reduction}$ B $\xrightarrow{CHCl_3/KOH}$ C $\xrightarrow{reduction}$ N-methyl aniline, then A is :
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A
Aniline, $C_6H_5NH_2$
B
Nitrobenzene, $C_6H_5NO_2$
C
$CH_3NH_2$
D
Phenyl isocyanide, $C_6H_5NC$
Detailed Solution
Reduction of an isocyanide gives a secondary amine with a methyl group on nitrogen: $C_6H_5N\equiv C + 4[H] \rightarrow C_6H_5NHCH_3$. So C is phenyl isocyanide.
An isocyanide is formed from a primary amine by the carbylamine reaction: $C_6H_5NH_2 + CHCl_3 + 3KOH \rightarrow C_6H_5NC + 3KCl + 3H_2O$. So B is aniline.
Aniline is formed by reduction of nitrobenzene: $C_6H_5NO_2 + 6[H] \xrightarrow{Sn/HCl} C_6H_5NH_2 + 2H_2O$. So A is nitrobenzene.
Sequence: $C_6H_5NO_2 \rightarrow C_6H_5NH_2 \rightarrow C_6H_5NC \rightarrow C_6H_5NHCH_3$
So A is nitrobenzene.
An isocyanide is formed from a primary amine by the carbylamine reaction: $C_6H_5NH_2 + CHCl_3 + 3KOH \rightarrow C_6H_5NC + 3KCl + 3H_2O$. So B is aniline.
Aniline is formed by reduction of nitrobenzene: $C_6H_5NO_2 + 6[H] \xrightarrow{Sn/HCl} C_6H_5NH_2 + 2H_2O$. So A is nitrobenzene.
Sequence: $C_6H_5NO_2 \rightarrow C_6H_5NH_2 \rightarrow C_6H_5NC \rightarrow C_6H_5NHCH_3$
So A is nitrobenzene.