Identify substrate $A$ in the following conversion.
$A + \underset{\text{(excess)}}{CH_3 I} \xrightarrow{\Delta} C_2 H_5(CH_3)_3 N^+ I^-$

The relative basic strength of the compounds is correctly shown in the option.

Identify the product $Z$ in the following reaction:
$C_6H_5NH_2$ $\xrightarrow{(Ac)_2O} X$ $\xrightarrow{Br_2/CCl_4} Y$ $\xrightarrow{HOH} Z$

Given below are two statements: one is labelled as Assertion $(A)$ and the other is labelled as Reason $(R)$.
Assertion $(A)$: Experimental reaction of $CH_{3}Cl$ with aniline and anhydrous $AlCl_{3}$ does not give $o-$ and $p-$methylaniline.
Reason $(R)$: The $-NH_{2}$ group of aniline becomes deactivating because of salt formation with anhydrous $AlCl_{3}$ and hence yields $m-$methylaniline as the product.
In the light of the above statements,choose the most appropriate answer from the options given below.

$CH_3-NH_2 \xrightarrow{HNO_2} ?$

$4-$Nitrotoluene ($para-$nitrotoluene) on reduction with $Fe / HCl$ and then electrophilic bromination with an excess amount of $Br_2$ gives