A given nitrogen-containing aromatic compound | vedclass

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(B) The reaction sequence is as follows:$1$. Compound $A$ reacts with $Sn/HCl$ (a reducing agent) to form an amine. This indicates that $A$ is a nitro

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Nitrobenzene

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(B) The reaction sequence is as follows:$1$. Compound $A$ reacts with $Sn/HCl$ (a reducing agent) to form an amine. This indicates that $A$ is a nitro compound,specifically nitrobenzene $(C_6H_5NO_2)$.$2$. Nitrobenzene $(C_6H_5NO_2)$ is reduced to aniline $(C_6H_5NH_2)$.$3$. Aniline reacts with $HNO_2$ (nitrous acid) at $0-5^{\circ}C$ to form benzenediazonium chloride $(C_6H_5N_2^+Cl^-)$,which is the unstable compound $B$.$4$. Benzenediazonium chloride $(B)$ undergoes a coupling reaction with phenol $(C_6H_5OH)$ in a basic medium to form $p$-hydroxyazobenzene $(C_6H_5-N=N-C_6H_4OH)$,which is the coloured compound $C$ with molecular formula $C_{12}H_{10}N_2O$.Therefore,compound $A$ is nitrobenzene.

Column-$I$	Column-$II$
$A$. Ammonolysis	$1$. Amine with lesser number of carbon atoms
$B$. Gabriel phthalimide synthesis	$2$. Detection test for primary amine
$C$. Hoffmann Bromamide reaction	$3$. Reaction of phthalimide with $KOH$ and $R-X$
$D$. Carbylamine reaction	$4$. Reaction of alkyl-halides with $NH_3$

$A$ given nitrogen-containing aromatic compound $A$ reacts with $Sn/HCl,$ followed by $HNO_2$ to give an unstable compound $B$. $B,$ on treatment with phenol,forms a beautiful coloured compound $C$ with the molecular formula $C_{12}H_{10}N_2O.$ The structure of compound $A$ is

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