JEE Main · 2025 · 29 Jan 2025 · Shift 2 · Aldehydes Ketones and Carboxylic Acids

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In the Claisen-Schmidt reaction to prepare, dibenzalacetone from 5.3 g benzaldehyde, a total of 3.51 g of product was obtained. The percentage yield in this reaction was ______ %.

Answer: 60

Solution

The Claisen–Schmidt condensation combines 2 mol of benzaldehyde with 1 mol of acetone to give dibenzalacetone:

2 C_{6} H_{5} C H O + C H_{3} C O C H_{3} \to C_{6} H_{5} C H = C H - C O - C H = C H C_{6} H_{5} + 2 H_{2} O

. Moles of benzaldehyde taken

= 5.3/106 = 0.05

mol, theoretically yielding

0.05/2 = 0.025

mol of dibenzalacetone (molar mass 234 g/mol), i.e. a theoretical mass of

0.025 \times 234 = 5.85

g. The actual yield obtained was 3.51 g, so

% yield = \frac{3.51}{5.85} \times 100 = 60%.

JEE Main · 2025 · 29 Jan 2025 · Shift 2 · Integer answer

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