An organic compound $A$ reacts with $NH_3$ on heating to give $B$. Compound $B$ reacts with $Br_2$ in the presence of $KOH$ to give $CH_3CH_2NH_2$. What is $A$?

Match List-$I$ with List-$II$:

List-$I$	List-$II$
$A$. Benzenesulphonyl chloride	$I$. Test for primary amines
$B$. Hoffmann bromamide reaction	$II$. Anti Saytzeff
$C$. Carbylamine reaction	$III$. Hinsberg reagent
$D$. Hoffmann orientation	$IV$. Known reaction of isocyanates

Choose the correct answer from the options given below.

$STATEMENT-1$: 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 ammonolysis of alcohol is represented by the reaction $xCH_3OH + yNH_3 \xrightarrow[Al_2O_3/ThO_2]{300-400^{\circ}C} \text{Products}$. The products formed are:

Which one of the following reaction sequences will give an azo dye?

$A$ is a neutral organic compound $(M.F.: C_8H_9ON)$. On treatment with aqueous $Br_2 / HO^{(-)}$, $A$ forms a compound $B$ which is soluble in dilute acid. $B$ on treatment with aqueous $NaNO_2 / HCl$ $(0-5^{\circ}C)$ produces a compound $C$ which on treatment with $CuCN / NaCN$ produces $D$. Hydrolysis of $D$ produces $E$ which is also obtainable from the hydrolysis of $A$. $E$ on treatment with acidified $KMnO_4$ produces $F$. $F$ contains two different types of hydrogen atoms. The structure of $A$ is