Write the structures of compounds A, B and C in the following reactions:
$C_{6} H_{5} {N O}_{2} F e / H C l - ---- \to A H {N O}_{2}; 273 K - ------ \to B C_{6} H_{5} O H - ---- \to C$

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Solution

Solution:-
$C_{6} H_{5} - N O_{2} F e / H C l - ---- \to C_{6} H_{5} - N H_{2} (A) H N O_{2} - --- \to 273 K {C_{6} H_{5} N_{2}}^{+} {C l}^{-} (B) C_{6} H_{5} O H - ---- \to O H - C_{6} H_{4} - N = N - C_{6} H_{5} (C)$
Therefore,
$A$ is Aniline.
$B$ is Benzenediazonium chloride.
$C$ is $4$ -hydroxyazobenzene.

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