In the Claisen-Schmidt reaction to prepare 351 g of dibenzalacetone using 87 g of acetone, the amount of benzaldehyde required is g. (Nearest integer)
Answer
318
Solution
2C6H5CHO + CH3COCH3 NaOH
(C6H5CH=CH)2CO + 2H2O
Acetone: 8758 = 1.5 mol
Dibenzalacetone ( C17H14O, M = 234): 351234 = 1.5 mol
∵ 1 mol product needs 2 mol benzaldehyde
⇒ benzaldehyde = 2 × 1.5 = 3 mol
∴ mass = 3 × 106 = 318 g
Answer: 318