Question 15

The major products of List-II reactions serve as reactants for the reactions in List-I:

1. Path (1): Toluene $\xrightarrow{CrO_2Cl_2/CS_2, H_3O^+}$ Benzaldehyde. Benzaldehyde lacks $\alpha$-hydrogens and undergoes the Cannizzaro reaction (S). Thus, S $\to$ 1.

2. Path (2): Benzoic acid $\xrightarrow{PCl_5} C_6H_5COCl \xrightarrow{NH_3} C_6H_5CONH_2 \xrightarrow{P_4O_{10}, \Delta} C_6H_5CN$ (Benzonitrile). Benzonitrile is the starting material for the Stephen reaction (reduction to benzaldehyde). Thus, P $\to$ 2.

3. Path (3): Nitrobenzene $\xrightarrow{Fe/HCl} C_6H_5NH_2 \xrightarrow{NaNO_2/HCl, H_2O} C_6H_5N_2^+Cl^-$ (Benzene diazonium chloride). This salt is the reactant for the Sandmeyer reaction. Thus, Q $\to$ 3.

4. Path (4): Toluene $\xrightarrow{Cl_2/h\nu, H_2O} C_6H_5CHO \xrightarrow{\text{Tollen's}} C_6H_5COOH \xrightarrow{SO_2Cl_2, NH_3} C_6H_5CONH_2$ (Benzamide). Benzamide is the reactant for the Hoffmann bromamide degradation reaction. Thus, R $\to$ 4.

Matching: P-2, Q-3, R-4, S-1. Correct Option: B