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(C) $1$. The reaction of aniline with $NaNO_2/HCl$ at $0-5^{\circ}C$ is a diazotization reaction,which produces benzene diazonium chloride $(C_6H_5N_2

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$(A)$: Benzonitrile,$(B)$: Benzaldehyde

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(C) $1$. The reaction of aniline with $NaNO_2/HCl$ at $0-5^{\circ}C$ is a diazotization reaction,which produces benzene diazonium chloride $(C_6H_5N_2^+Cl^-)$.$2$. The reaction of benzene diazonium chloride with $KCN$ (in the presence of $CuCN$) yields benzonitrile $(C_6H_5CN)$. Thus,$(A)$ is benzonitrile.$3$. The reduction of benzonitrile with $SnCl_2/HCl$ followed by hydrolysis $(H_3O^+)$ is the Stephen reduction,which converts the nitrile group $(-CN)$ into an aldehyde group $(-CHO)$. Thus,$(B)$ is benzaldehyde.$4$. Therefore,$(A)$ is benzonitrile and $(B)$ is benzaldehyde.

List-$I$ (Compounds)	List-$II$ (Catalyst/Reagents)
$A$. $R-CONH_2$	$I$. $NaOH$ (aqueous)
$B$. $C_6H_5NO_2$	$II$. $H_2 / Ni$
$C$. $R-C \equiv N$	$III$. $LiAlH_4, H_2O$
$D$. $N$-alkylphthalimide	$IV$. $Sn,HCl$

Identify $A$ and $B$ in the following reaction sequence:
Aniline $\xrightarrow[KCN]{NaNO_2/HCl} (A)$ $\xrightarrow{SnCl_2/HCl/H_3O^+} (B)$

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Identify $A$ and $B$ in the following reaction sequence:Aniline $\xrightarrow[KCN]{NaNO_2/HCl} (A)$ $\xrightarrow{SnCl_2/HCl/H_3O^+} (B)$