Match List-I with List-II | List-I Conversion | List-II Reagents, Conditions used | |-------------------|-----------------------------------| | (A) (see structure) | (I) Warm, $\mathrm{H}_{2} \mathrm{O}$ | | (B) (see structure) | (II) (a) $\mathrm{NaOH}, 368 \mathrm{~K}$; (b) $\mathrm{H}_{3} \mathrm{O}^{+}$ | | (C) (see structure) | (III) (a) $\mathrm{NaOH}, 443 \mathrm{~K}$; (b) $\mathrm{H}_{3} \mathrm{O}^{+}$ | | (D) (see structure) | (IV) (a) $\mathrm{NaOH}, 623 \mathrm{~K}$, 300 atm ; (b) $\mathrm{H}_{3} \mathrm{O}^{+}$ | Choose the correct answer from the options given below :

JEE Main · 2025 · 07 Apr 2025 · Shift 2 · Alcohols Phenols and Ethers

Match List-I with List-II

List-I Conversion	List-II Reagents, Conditions used
(A) (see structure)	(I) Warm, $H_{2} O$
(B) (see structure)	(II) (a) $NaOH, 368 K$ ; (b) $H_{3} O^{+}$
(C) (see structure)	(III) (a) $NaOH, 443 K$ ; (b) $H_{3} O^{+}$
(D) (see structure)	(IV) (a) $NaOH, 623 K$ , 300 atm ; (b) $H_{3} O^{+}$

Choose the correct answer from the options given below :

Answer: A → IV, B → III, C → II, D → I

Solution

Turning an aryl halide into a phenol with hydroxide grows easier as more electron-withdrawing nitro groups sit ortho and para to the halogen, since they stabilise the intermediate that forms when hydroxide attacks. Plain chlorobenzene has no such help and needs the forcing Dow conditions, fused alkali at

623

K and

300

atm. A single para nitro group lets the reaction run at

443

K, two nitro groups at the ortho and para positions bring it down to

368

K, and three nitro groups activate the ring so strongly that merely warming with water suffices. Thus chlorobenzene pairs with the

623

300

atm conditions, the mononitro compound with

443

K, the dinitro compound with

368

K, and the trinitro compound with warm water.

JEE Main · 2025 · 07 Apr 2025 · Shift 2 · Match the columns

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JEE Main · 2025 · 07 Apr 2025 · Shift 2 · Alcohols Phenols and Ethers

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