In the Claisen-Schmidt reaction to prepare 351 g of dibenzalacetone using 87 g of acetone, the amoun — JEE Mains Chemistry Past Papers Chemistry Question

💡 Solution & Explanation

Claisen Schmidt reaction H O C 2 mole + O CH3 CH3 1 mole NaOH ⊕ Θ — CH = CH — C — CH = CH — O Dibanzal acetone 1 mole 3 mole 1.5 mole 351 gm = 1.5 mole mw of benzaldehyde = 106 106 × 3 = 318 gm. Benzaldehyde is required to give 1.5 mole (or 351 gm) product