JEE Main · 2024 · 29 Jan 2024 · Shift 2 · Alcohols Phenols and Ethers

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Match List I with List II:
List-I (Compound): A. Ethanol B. Phenol C. m-Nitrophenol D. p-Nitrophenol
List-II (pKₐ value): I. 10.0 II. 15.9 III. 7.1 IV. 8.3
Choose the correct answer from the options given below:
  1. (A)A-I, B-II, C-III, D-IV
  2. (B)A-IV, B-I, C-II, D-III
  3. (C)A-III, B-IV, C-I, D-II
  4. (D)A-II, B-I, C-IV, D-III✓ Correct

Answer: (D)

Solution

Acidity increases as the conjugate base becomes more stable. Ethanol's alkoxide has no resonance stabilisation, making ethanol the weakest acid here, with the highest (15.9). Phenol's conjugate base, phenoxide, is stabilised by delocalisation of the negative charge into the aromatic ring, giving .

A nitro group withdraws electron density and stabilises the phenoxide further, lowering relative to phenol. At the para position, is in direct conjugation with the ring oxygen, so the negative charge delocalises strongly onto the nitro oxygens, making p-nitrophenol the most acidic of the set (). At the meta position this direct resonance path is unavailable, and the nitro group acts mainly through the weaker inductive effect, so m-nitrophenol is less acidic than the para isomer but still more acidic than phenol ().

The correct matching is Ethanol–15.9, Phenol–10.0, m-Nitrophenol–8.3, p-Nitrophenol–7.1.

JEE Main · 2024 · 29 Jan 2024 · Shift 2 · Single correct option

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