The product $(A)$ of the given reaction is:

Amongst the compounds given,the one that would form a brilliant colored dye on treatment with $NaNO_2$ in dil. $HCl$ followed by addition to an alkaline solution of $\beta$-naphthol is

The product of this Hoffmann bromamide reaction is:

$STATEMENT-1$: Aniline on reaction with $NaNO_2 / HCl$ at $0^{\circ} C$ followed by coupling with $\beta$-naphthol gives a dark blue coloured precipitate.
$STATEMENT-2$: The colour of the compound formed in the reaction of aniline with $NaNO_2 / HCl$ at $0^{\circ} C$ followed by coupling with $\beta$-naphthol is due to the extended conjugation.

The final product $R$ for the reaction is:

Match the following:

List-$I$	List-$II$
$A$. Amide	$I$. Carbylamine reaction
$B$. Nitrile	$II$. Hinsberg's reagent
$C$. $C_6H_5SO_2Cl$	$III$. Hofmann's bromamide reaction
$D$. $1^{\circ}$-Amine	$IV$. $LiAlH_4$