Question

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 1

The reaction involved is the Claisen-Schmidt condensation:
$2\text{C}_6\text{H}_5\text{CHO} + \text{CH}_3\text{COCH}_3 \xrightarrow{\text{NaOH}} \text{C}_6\text{H}_5\text{CH} = \text{CHCOCH} = \text{CHC}_6\text{H}_5 + \text{H}_2\text{O}.$
Step 1: Reaction Stoichiometry
2 moles of benzaldehyde ($\text{C}_6\text{H}_5\text{CHO}$) react with 1 mole of acetone ($\text{CH}_3\text{COCH}_3$) to form 1 mole of dibenzalacetone ($\text{C}_6\text{H}_5\text{CH} = \text{CHCOCH} = \text{CHC}_6\text{H}_5$).
Given that 87 g of acetone is used, the molar mass of acetone is:
$\text{Molar mass of } \text{CH}_3\text{COCH}_3 = 12 + (1 \times 3) + 12 + 16 + 1 = 58 \, \text{g/mol}.$
$\text{Moles of acetone} = \frac{\text{Mass of acetone}}{\text{Molar mass of acetone}} = \frac{87}{58} \approx 1.5 \, \text{moles}.$
Step 2: Calculate moles of benzaldehyde required
From the reaction stoichiometry, 2 moles of benzaldehyde are required for every 1 mole of acetone. Thus, the moles of benzaldehyde needed are:
$\text{Moles of benzaldehyde} = 2 \times \text{Moles of acetone} = 2 \times 1.5 = 3 \, \text{moles}.$
Step 3: Calculate the mass of benzaldehyde required
The molar mass of benzaldehyde ($\text{C}_6\text{H}_5\text{CHO}$) is:
$\text{Molar mass of benzaldehyde} = (6 \times 12) + (5 \times 1) + 12 + 16 = 106 \, \text{g/mol}.$
$\text{Mass of benzaldehyde} = \text{Moles of benzaldehyde} \times \text{Molar mass of benzaldehyde} = 3 \times 106 = 318 \, \text{g}.$
Step 4: Verify with product formation
The reaction produces 1 mole of dibenzalacetone for every 2 moles of benzaldehyde. For 1.5 moles of acetone, 1.5 moles of dibenzalacetone are formed, which corresponds to 351 g (given).
Final Answer: 318 g.