JEE Main 2025 Apr 2 Shift 1, Chemistry Q19: Carboxylic Acid Derivatives

For nucleophilic acyl substitution, the rate is governed by how good a leaving group the X attached to the acyl carbon is. Ranking leaving groups:

$C{l}^{-}$ (acid chloride) $>$ $RCO{O}^{-}$ (anhydride) $>$ $R{O}^{-}$ (ester) $>$ $N{H}_2^{-}$ (amide)

Equivalently, resonance donation from X into the carbonyl reduces its electrophilicity in the reverse order:

Amide (strong N $\to$ C=O resonance) $>$ ester $>$ anhydride $>$ acid chloride (weakest donor) — meaning the acid chloride is the most reactive electrophile.

Therefore the hydrolysis order is:

$$\text{acid chloride (p)}>\text{anhydride (q)}>\text{ester (r)}>\text{amide (s)}$$

$\therefore$ Option (4) is correct.