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

In a reaction of aniline,a coloured product $C$ was obtained as shown below:
$A$ $\xrightarrow{NaNO_2/HCl} B$ $\xrightarrow{N,N-dimethylaniline, \text{cold}} C$
The structure of $C$ would be:

Which of the following amines will not give $N_2$ gas on treatment with nitrous acid $(NaNO_2 + HCl)$?

Match the following:

List-$I$ (Amine)	List-$II$ ($pK_b$ value)
$A$. $N,N$-Dimethylaniline	$I$. $9.30$
$B$. Aniline	$II$. $8.92$
$C$. $N$-Ethylethanamine	$III$. $9.38$
$D$. $N$-Methylaniline	$IV$. $3.00$

The correct answer 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

Arrange the following:
$(i)$ In decreasing order of the $pK_{b}$ values:
$C_{2}H_{5}NH_{2}, C_{6}H_{5}NHCH_{3}, (C_{2}H_{5})_{2}NH$ and $C_{6}H_{5}NH_{2}$
$(ii)$ In increasing order of basic strength:
$C_{6}H_{5}NH_{2}, C_{6}H_{5}N(CH_{3})_{2}, (C_{2}H_{5})_{2}NH$ and $CH_{3}NH_{2}$
$(iii)$ In increasing order of basic strength:
$(a)$ Aniline,$p-$nitroaniline and $p-$toluidine
$(b)$ $C_{6}H_{5}NH_{2}, C_{6}H_{5}NHCH_{3}, C_{6}H_{5}CH_{2}NH_{2}$
$(iv)$ In decreasing order of basic strength in gas phase:
$C_{2}H_{5}NH_{2}, (C_{2}H_{5})_{2}NH, (C_{2}H_{5})_{3}N$ and $NH_{3}$
$(v)$ In increasing order of boiling point:
$C_{2}H_{5}OH, (CH_{3})_{2}NH, C_{2}H_{5}NH_{2}$
$(vi)$ In increasing order of solubility in water:
$C_{6}H_{5}NH_{2}, (C_{2}H_{5})_{2}NH, C_{2}H_{5}NH_{2}$