JEE Main 2024 — 5 April, Shift 2 — Question 84

The Claisen–Schmidt reaction for preparing dibenzalacetone is

$\mathrm{CH_3COCH_3 + 2\,C_6H_5CHO \rightarrow C_6H_5CH{=}CHCOCH{=}CHC_6H_5 + 2H_2O}$

Given molar masses: $M_{\mathrm{acetone}} = 58.08\,\mathrm{g\,mol^{-1}},\; M_{\mathrm{benzaldehyde}} = 106.12\,\mathrm{g\,mol^{-1}},\; M_{\mathrm{dibenzalacetone}} = 234.28\,\mathrm{g\,mol^{-1}}$

Moles of dibenzalacetone required, $n = \dfrac{351}{234.28} = 1.498\,\mathrm{mol}$

Stoichiometry: $1\,\text{mol \;acetone} \rightarrow 1\,\text{mol\; dibenzalacetone},\; 2\,\text{mol \;benzaldehyde} \rightarrow 1\,\text{mol \;dibenzalacetone}$

Moles of benzaldehyde required $= 2 \times 1.498 = 2.996\,\mathrm{mol}$

Mass of benzaldehyde required, $m = 2.996 \times 106.12 = 317.97\,\mathrm{g}$

Thus, amount of benzaldehyde required ( in nearest integer) is ${318\,\mathrm{g}}$.