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.

List-$I$ (Compounds)	List-$II$ (Reactions)
$A. \ CH_3(CH_2)_3NH_2$	$(i)$ Alkaline hydrolysis
$B. \ CH_3C\equiv CH$	$(ii)$ With $KOH$ (alc.) and $CHCl_3$ produces bad smell
$C. \ CH_3CH_2COOCH_3$	$(iii)$ Gives white ppt. with ammoniacal $AgNO_3$
$D. \ CH_3CH(OH)CH_3$	$(iv)$ With Lucas reagent cloudiness appears after $5 \ minutes$

$A$ 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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