Question 15

MATRIX MATCHHARD

The major products obtained from the reactions in List-II are the reactants for the named reactions mentioned in List-I. Match each entry in List-I with the appropriate entry in List-II and choose the correct option.

List - I

Stephen reaction

Sandmeyer reaction

Hoffmann bromamide degradation reaction

Cannizzaro reaction

List-II

Toluene $\xrightarrow{\text{(i) } CrO_2Cl_2/CS_2 \text{ (ii) } H_3O^+}$

Benzoic acid $\xrightarrow{\text{(i) } PCl_5 \text{ (ii) } NH_3 \text{ (iii) } P_4O_{10}, \Delta}$

Nitrobenzene $\xrightarrow{\text{(i) } Fe, HCl \text{ (ii) } HCl, NaNO_2 (273-278 K), H_2O}$

Toluene $\xrightarrow{\text{(i) } Cl_2/h\nu, H_2O \text{ (ii) } \text{Tollen's reagent} \text{ (iii) } SO_2Cl_2 \text{ (iv) } NH_3}$

Aniline $\xrightarrow{\text{(i) } (CH_3CO)_2O, \text{Pyridine} \text{ (ii) } HNO_3, H_2SO_4, 288 K \text{ (iii) } aq. NaOH}$

Correct Match:

P → 2

Q → 3

R → 4

S → 1

Detailed Solution

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

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.
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.
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.
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

Question 15

List - I

List-II

Correct Match:

Detailed Solution

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