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Aldehydes Ketones and Carboxylic Acids question

2024 · 5 Apr · Shift 2 · Q27
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  5. /2024 · 5 Apr · Shift 2 · Q27

Aldehydes Ketones and Carboxylic Acids question

2024 · 5 Apr · Shift 2 · Q27

JEE MainChemistryAldehydes Ketones and Carboxylic AcidsNumerical+4 / −1
In the Claisen-Schmidt reaction to prepare 351 g351 \mathrm{~g}351 g of dibenzalacetone using 87 g87 \mathrm{~g}87 g of acetone, the amount of benzaldehyde required is ‾\underline{\hspace{2cm}}​ g. (Nearest integer)
Numerical answer
View written solutionFree

Correct answer: 318

  1. Write the Claisen–Schmidt reaction

Dibenzalacetone is formed by condensation of 1 mole of acetone with 2 moles of benzaldehyde:

Acetone+2 Benzaldehyde→Dibenzalacetone+2H2O\text{Acetone} + 2\,\text{Benzaldehyde} \rightarrow \text{Dibenzalacetone} + 2H_2OAcetone+2Benzaldehyde→Dibenzalacetone+2H2​O

So, the mole ratio is:

Acetone:Benzaldehyde=1:2\text{Acetone} : \text{Benzaldehyde} = 1:2Acetone:Benzaldehyde=1:2
  1. Calculate moles of acetone given

Molar mass of acetone, (CH3)2CO(CH_3)_2CO(CH3​)2​CO:

3×12+6×1+16=58 g mol−13\times 12 + 6\times 1 + 16 = 58\ \text{g mol}^{-1}3×12+6×1+16=58 g mol−1

Given mass of acetone = 87 g87\ \text{g}87 g

Moles of acetone=8758=1.5 mol\text{Moles of acetone} = \frac{87}{58} = 1.5\ \text{mol}Moles of acetone=5887​=1.5 mol
  1. Find moles of benzaldehyde required

From stoichiometry:

1 mol acetone requires 2 mol benzaldehyde1\ \text{mol acetone requires } 2\ \text{mol benzaldehyde}1 mol acetone requires 2 mol benzaldehyde

Therefore, 1.51.51.5 mol acetone requires:

1.5×2=3 mol benzaldehyde1.5 \times 2 = 3\ \text{mol benzaldehyde}1.5×2=3 mol benzaldehyde
  1. Convert moles of benzaldehyde to mass

Molar mass of benzaldehyde, C6H5CHO=C7H6OC_6H_5CHO = C_7H_6OC6​H5​CHO=C7​H6​O:

7×12+6×1+16=106 g mol−17\times 12 + 6\times 1 + 16 = 106\ \text{g mol}^{-1}7×12+6×1+16=106 g mol−1

Required mass:

3×106=318 g3 \times 106 = 318\ \text{g}3×106=318 g
  1. Check with dibenzalacetone mass given

Molar mass of dibenzalacetone, C17H14OC_{17}H_{14}OC17​H14​O:

17×12+14×1+16=234 g mol−117\times 12 + 14\times 1 + 16 = 234\ \text{g mol}^{-1}17×12+14×1+16=234 g mol−1

Given product mass = 351 g351\ \text{g}351 g

Moles of dibenzalacetone=351234=1.5 mol\text{Moles of dibenzalacetone} = \frac{351}{234} = 1.5\ \text{mol}Moles of dibenzalacetone=234351​=1.5 mol

This matches the 1.51.51.5 mol acetone used, confirming consistency.

  1. Final answer
318 g\boxed{318\ \text{g}}318 g​
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