Animated Solution for Chemistry - Organic Chemistry: Correct option(s) for the following sequence of reactions is(are)

The problem presents a fascinating web of organic transformations, starting from a simple aromatic hydrocarbon and branching into various nitrogen-containing compounds. Let's break down this reaction sequence step by step to uncover the identities of the unknown reagents and products.

The Benzylic Bromination Our journey begins with toluene (${\text{PhCH}}_3$). The first reaction subjects toluene to bromine (${\text{Br}}_2$) in the presence of light. This is a classic condition for free radical substitution. Because the benzylic radical is highly stabilized by resonance with the aromatic ring, the substitution occurs exclusively at the side chain, not on the ring.

This gives us our first intermediate, Product P, which is benzyl bromide (${\text{Ph-CH}}_2\text{Br}$).

The Oxidation and Amidation Pathway From benzyl bromide, the reaction splits. Let's follow the downward path first. Benzyl bromide is treated with alkaline potassium permanganate (${\text{KMnO}}_4$) followed by acidic hydrolysis (${\text{H}}_3{\text{O}}^{+}$). Alkaline ${\text{KMnO}}_4$ is a vigorous oxidizing agent that oxidizes any benzylic carbon (having at least one hydrogen) all the way to a carboxylic acid. Thus, Product T is benzoic acid ($\text{Ph-COOH}$).

Next, benzoic acid is heated with ammonia (${\text{NH}}_3$). This is a standard method for preparing amides. The reaction eliminates a molecule of water, converting the carboxylic acid into benzamide (${\text{Ph-CONH}}_2$), which is Product U.

The Reduction to Amine Now, we need to convert benzamide ($U$) into Product R. Looking at the other pathway, $R$ is also formed from benzyl bromide ($P$). If we reduce benzamide using a strong reducing agent like lithium aluminum hydride (${\text{LiAlH}}_4$), it converts the amide group ($-{\text{CONH}}_2$) into a primary amine group ($-{\text{CH}}_2{\text{NH}}_2$).

Therefore, Reagent W is ${\text{LiAlH}}_4$, and Product R is benzylamine (${\text{Ph-CH}}_2{\text{NH}}_2$). The IUPAC name for benzylamine is phenylmethanamine.

Let's verify this by looking at the horizontal path from $P$ to $R$. Benzyl bromide ($P$) reacts with Reagent Q, followed by catalytic hydrogenation (${\text{H}}_2,\text{Pd/C}$). To get an amine after reduction, the intermediate must be a nitro compound. Reacting benzyl bromide with silver nitrite (${\text{AgNO}}_2$) yields phenylnitromethane (${\text{Ph-CH}}_2{\text{NO}}_2$), because ${\text{AgNO}}_2$ is covalent and attacks via the nitrogen atom. Thus, Reagent Q is ${\text{AgNO}}_2$.

The Direct Paths and the Foul Smell Finally, benzylamine ($R$) is treated with chloroform (${\text{CHCl}}_3$) and potassium hydroxide ($\text{KOH}$). This is the infamous carbylamine reaction, a definitive test for primary amines. It produces an isocyanide, which is characterized by an intensely foul smell.

Thus, Product S is benzyl isocyanide (${\text{Ph-CH}}_2\text{NC}$).

The reaction scheme also shows a direct path from benzyl bromide ($P$) to benzyl isocyanide ($S$) using Reagent V. To convert an alkyl halide to an isocyanide, we must use silver cyanide ($\text{AgCN}$). Like ${\text{AgNO}}_2$, $\text{AgCN}$ is predominantly covalent, leaving the nitrogen lone pair available to act as the nucleophile. Therefore, Reagent V is $\text{AgCN}$.

By meticulously tracing each pathway, we have identified all the unknowns: Q = ${\text{AgNO}}_2$ R = phenylmethanamine W = ${\text{LiAlH}}_4$ V = $\text{AgCN}$

Matching these with the given options, we find that options (C) and (D) are the correct choices.