JEE Main 2026 Jan 24 Shift 2, Chemistry Q9: Preparation of Carboxylic acids

JEE Main2026Jan 24, Shift 2Chemistry

A student has planned to prepare acetanilide from aniline using acetic anhydride. The student has started from 9.3g of aniline. However, the student has managed to obtain 11 g of dry acetanilide. The % yield of this reaction is :-

A
81.5%
B
97.5%
C
59.5%
D
72.5%

Solution

Aniline ( $C_{6} H_{5} NH_{2}$ , MW = 93) reacts with acetic anhydride to give acetanilide ( $C_{6} H_{5} NHCOCH_{3}$ , MW = 135) in a 1:1 stoichiometry.

Moles of aniline (limiting reagent) $= \frac{9.3}{93} = 0.1$ mol.

Theoretical mass of acetanilide $= 0.1 \times 135 = 13.5$ g.

Moles of acetanilide actually obtained $= \frac{11}{135} \approx 0.0815$ mol.

% yield $= \frac{0.0815}{0.1} \times 100 = 81.5%$ .

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JEE Main 2026 Jan 24 Shift 2, Chemistry Q9: Preparation of Carboxylic acids

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