JEE Advanced 2025 — Paper 1 — CHEM-IV solution

JEE Advanced2025Paper 1CHEM-IV

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 options.

List-I	List-II
(P) Stephen reaction	(1) Toluene $(i) CrO_{2} Cl_{2} / CS_{2}; (ii) H_{3} O^{+}$
(Q) Sandmeyer reaction	(2) Benzoic acid $(i) PCl_{5}; (ii) NH_{3}; (iii) P_{4} O_{10}, Δ$
(R) Hoffmann bromamide degradation reaction	(3) Nitrobenzene $(i) Fe, HCl; (ii) HCl, NaNO_{2}, 273-278 K, H_{2} O$
(S) Cannizzaro reaction	(4) Toluene $(i) Cl_{2} / h ν, H_{2} O; (ii) Tollen’s reagent; (iii) SO_{2} Cl_{2}; (i v) NH_{3}$
	(5) Aniline $(i) (CH_{3} CO)_{2} O, Pyridine; (ii) HNO_{3}, H_{2} SO_{4}, 288 K; (iii) aq. NaOH$

A
P $\to$ 2; Q $\to$ 4; R $\to$ 1; S $\to$ 3
B
P $\to$ 2; Q $\to$ 3; R $\to$ 4; S $\to$ 1
C
P $\to$ 5; Q $\to$ 3; R $\to$ 4; S $\to$ 2
D
P $\to$ 5; Q $\to$ 4; R $\to$ 2; S $\to$ 1

Solution

Work out the major product of each List-II sequence, then match to the named reaction whose starting material it is.

List-II (1): Toluene + CrO $_{2}$ Cl $_{2}$ /CS $_{2}$ then aqueous workup is the Etard reaction, giving benzaldehyde (PhCHO). Benzaldehyde without an $α$ -H undergoes Cannizzaro reaction. So (1) feeds (S).

List-II (2): Benzoic acid → (PCl $_{5}$ ) PhCOCl → (NH $_{3}$ ) PhCONH $_{2}$ → (P $_{4}$ O $_{10}$ , $Δ$ ) dehydration to PhCN. The nitrile undergoes the Stephen reaction (SnCl $_{2}$ /HCl) to yield PhCHO. So (2) feeds (P).

List-II (3): Nitrobenzene → (Fe/HCl) aniline → (HCl, NaNO $_{2}$ , 273–278 K) benzenediazonium chloride. This is the substrate for the Sandmeyer reaction. So (3) feeds (Q).

List-II (4): Toluene → (Cl $_{2}$ /h $ν$ , H $_{2}$ O) benzaldehyde → (Tollen’s) benzoic acid → (SO $_{2}$ Cl $_{2}$ ) PhCOCl → (NH $_{3}$ ) PhCONH $_{2}$ . Benzamide undergoes Hoffmann bromamide degradation. So (4) feeds (R).

$∴$ P $\to$ 2, Q $\to$ 3, R $\to$ 4, S $\to$ 1. Option (B).

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