An organic compound $A$ on reduction gives compound $B$,which on reaction with trichloromethane and caustic potash forms $C$. The compound $C$ on catalytic reduction gives $N$-methylbenzenamine. The compound $A$ is,

Match List-$I$ with List-$II$.

List-$I$ (Mixture of Compounds)	List-$II$ (Reagent used to distinguish)
$A$. Diethyl amine + Ethyl amine	$I$. Bromine water
$B$. Acetaldehyde + Acetone	$II$. $CHCl_3 + KOH, \Delta$
$C$. Ethanol + Phenol	$III$. Neutral $FeCl_3$
$D$. Benzoic acid + Cinnamic acid	$IV$. Ammonical silver nitrate

Which of the following reactions does $NOT$ yield an amine?

Which of the following species would not be involved in the Hoffmann rearrangement shown below?

Which among the following represents reagent $A$ in the given reaction?
$C_6H_5N_2^+Cl^- + A \xrightarrow{NaOH} \text{Product (Orange-red dye)}$

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 $\text{g mol}^{-1}$: $\text{C} = 12, \text{H} = 1, \text{O} = 16, \text{N} = 14, \text{Br} = 80, \text{Ag} = 108$)