Reaction Sequence Analysis for Toluene Derivatives and Carbylamine Test

To determine the correct options for the given reaction scheme, we analyze each step logically:

1. Formation of Compound $\mathbf{P}$: Reaction of toluene ($\text{PhCH}_3$) with $\text{Br}_2$ in the presence of light undergoes free radical halogenation: $\text{PhCH}_3 + \text{Br}_2 \xrightarrow{h\nu} \text{PhCH}_2\text{Br}\ (\mathbf{P}) + \text{HBr}$ Thus, $\mathbf{P}$ is benzyl bromide ($\text{PhCH}_2\text{Br}$).

2. Formation of Compounds $\mathbf{T}$ and $\mathbf{U}$:

   • Vigorous oxidation of toluene using alkaline $\text{KMnO}_4$ followed by acidification gives benzoic acid: $\text{PhCH}_3 \xrightarrow[\text{ii. }\text{H}_3\text{O}^+]{\text{i. }\text{KMnO}_4/\text{KOH, heat}} \text{PhCOOH}\ (\mathbf{T})$
   • Heating benzoic acid ($\mathbf{T}$) with ammonia yields benzamide: $\text{PhCOOH} + \text{NH}_3 \xrightarrow{\Delta} \text{PhCONH}_2\ (\mathbf{U})$

3. Formation of Compound $\mathbf{R}$ and Reagent $\mathbf{W}$: Benzamide ($\mathbf{U}$) is reduced to a primary amine, phenylmethanamine (benzylamine, $\text{PhCH}_2\text{NH}_2$), using lithium aluminium hydride ($\text{LiAlH}_4$): $\text{PhCONH}_2\ (\mathbf{U}) \xrightarrow{\text{LiAlH}_4\ (\mathbf{W})} \text{PhCH}_2\text{NH}_2\ (\mathbf{R})$ Thus, $\mathbf{W} = \text{LiAlH}_4$ and $\mathbf{R} = \text{phenylmethanamine}$.

4. Reagent $\mathbf{Q}$: Benzyl bromide ($\mathbf{P}$) reacts with silver nitrite ($\text{AgNO}_2$) via nucleophilic substitution through the nitrogen atom (due to the covalent nature of $\text{AgNO}_2$) to form phenylnitromethane, which on catalytic reduction ($\text{H}_2/\text{Pd-C}$) produces phenylmethanamine ($\mathbf{R}$): $\text{PhCH}_2\text{Br}\ (\mathbf{P}) \xrightarrow{\text{AgNO}_2\ (\mathbf{Q})} \text{PhCH}_2\text{NO}_2 \xrightarrow{\text{H}_2/\text{Pd-C}} \text{PhCH}_2\text{NH}_2\ (\mathbf{R})$ (Note: $\text{KNO}_2$ would primarily form benzyl nitrite, $\text{PhCH}_2\text{ONO}$, making $\mathbf{Q} = \text{AgNO}_2$).

5. Carbylamine Test to form Compound $\mathbf{S}$ and Reagent $\mathbf{V}$:

   • Treatment of phenylmethanamine ($\mathbf{R}$) with chloroform ($\text{CHCl}_3$) and base ($\text{KOH}$) gives the foul-smelling benzyl isocyanide ($\mathbf{S}$): $\text{PhCH}_2\text{NH}_2\ (\mathbf{R}) + \text{CHCl}_3 + 3\text{KOH} \xrightarrow{\Delta} \text{PhCH}_2\text{NC}\ (\mathbf{S}) + 3\text{KCl} + 3\text{H}_2\text{O}$
   • Direct conversion of benzyl bromide ($\mathbf{P}$) to benzyl isocyanide ($\mathbf{S}$) requires silver cyanide ($\text{AgCN}$), as $\text{AgCN}$ is covalent and nucleophilic attack occurs through nitrogen: $\text{PhCH}_2\text{Br}\ (\mathbf{P}) \xrightarrow{\text{AgCN}\ (\mathbf{V})} \text{PhCH}_2\text{NC}\ (\mathbf{S})$ Thus, $\mathbf{V} = \text{AgCN}$.

Conclusion:

• $\mathbf{Q} = \text{AgNO}_2$
• $\mathbf{R} = \text{phenylmethanamine}$
• $\mathbf{W} = \text{LiAlH}_4$
• $\mathbf{V} = \text{AgCN}$

Comparing with the given choices:

• Option (A) is incorrect since $\mathbf{Q} = \text{AgNO}_2$.
• Option (B) is incorrect since $\mathbf{R} = \text{phenylmethanamine}$ and $\mathbf{V} = \text{AgCN}$.
• Option (C) is correct.
• Option (D) is correct.

Correct Options: (C) and (D)