When $methyl$ $iodide$ is heated with $ammonia$,the product obtained is

Consider the following reaction sequence starting from $p$-nitrotoluene:
$(i)$ $\text{Sn/HCl; OH}^-$
(ii) $(\text{CH}_3\text{CO})_2\text{O}$
(iii) $\text{Br}_2/\text{AlBr}_3$
(iv) $\text{H}_3\text{O}^+$
When the final product $(P)$ is subjected to Carius analysis using $\text{AgNO}_3$, $1.0 \text{ g}$ of the product $(P)$ will produce . . . . . . $\text{g}$ of the precipitate of $\text{AgBr}$. (Nearest Integer)
(Given: molar mass in $\text{g mol}^{-1}$: $\text{C} = 12, \text{H} = 1, \text{O} = 16, \text{N} = 14, \text{Br} = 80, \text{Ag} = 108$)

Predict the stereochemistry of the product when a $d$ and $l$-amine mixture reacts with an $l$-acid.

$9.3 \ g$ of pure aniline upon diazotisation followed by coupling with phenol gives an orange dye. The mass of orange dye produced (assume $100\%$ yield/conversion) is . . . . . $g$. (nearest integer)

In the reaction sequence $CH_3CH_2Br$ $\xrightarrow{AgCN} A$ $\xrightarrow{H_3O^+} HCOOH + B; B$ $\xrightarrow{CHCl_3/KOH} A$ $\xrightarrow{Reduction} C$,what are $A, B,$ and $C$ respectively?

$STATEMENT-1$: 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.