Q26 / 90 · JEE Main 2024 (05 Apr Shift 2) · Chemistry
In the Claisen-Schmidt reaction to prepare \(351 \mathrm{~g}\) of dibenzalacetone using \(87 \mathrm{~g}\) of acetone, the amount of benzaldehyde required is ______ g. (Nearest integer)
Correct answer: 318
Solution
Claisen Schmidt reaction
mw of benzaldehyde $=106$ $106 \times 3=318 \mathrm{gm}$. Benzaldehyde is required to give 1.5 mole (or $351 \mathrm{gm}$ ) product
mw of benzaldehyde $=106$ $106 \times 3=318 \mathrm{gm}$. Benzaldehyde is required to give 1.5 mole (or $351 \mathrm{gm}$ ) product
Hence, the correct answer is 318.