JEE Advanced 2023 — Paper 1 — CHEM-IV solution

JEE Advanced2023Paper 1CHEM-IV

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

List-I	List-II
(P) Etard reaction	(1) Acetophenone $Zn-Hg, HCl$
(Q) Gattermann reaction	(2) Toluene $(i) KMnO_{4}, KOH, Δ (ii) SOCl_{2}$
(R) Gattermann-Koch reaction	(3) Benzene $CH_{3} Cl anhyd. AlCl_{3}$
(S) Rosenmund reduction	(4) Aniline $NaNO_{2} / HCl 273-278 K$
	(5) Phenol $Zn, Δ$

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

Solution

Each item in List-II produces a reactant required for a named reaction in List-I. Identify the product of each transformation and pair it with the reaction that uses it.

(1) Acetophenone + Zn(Hg)/HCl. Clemmensen reduction converts the ketone to ethylbenzene. This is not a starting material for any of the four named reactions listed.

(2) Toluene then (i) KMnO $_{4}$ /KOH/ $Δ$ , (ii) SOCl $_{2}$ . KMnO $_{4}$ oxidises the methyl group to benzoic acid; SOCl $_{2}$ converts it to benzoyl chloride. Benzoyl chloride is the substrate for the Rosenmund reduction ( $H_{2}$ /Pd-BaSO $_{4}$ gives benzaldehyde). So S $\to$ 2.

(3) Benzene + CH $_{3}$ Cl / anhyd. AlCl $_{3}$ . Friedel-Crafts alkylation gives toluene. Toluene is the substrate for the Etard reaction (CrO $_{2}$ Cl $_{2}$ converts toluene to benzaldehyde). So P $\to$ 3.

(4) Aniline + NaNO $_{2}$ /HCl, 273-278 K. Diazotisation gives benzenediazonium chloride. This is the substrate for the Gattermann reaction (Cu/HCl or Cu/HBr gives the aryl halide). So Q $\to$ 4.

(5) Phenol + Zn / $Δ$ . Phenol is reduced to benzene. Benzene undergoes the Gattermann-Koch reaction with CO/HCl/AlCl $_{3}$ to give benzaldehyde. So R $\to$ 5.

Mapping: P $\to$ 3, Q $\to$ 4, R $\to$ 5, S $\to$ 2. Answer: (D).

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