In the Claisen-Schmidt reaction to prepare di | vedclass

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(C) The chemical reaction is: $2 \ C_6H_5CHO + CH_3COCH_3 \rightarrow C_6H_5CH=CHCOCH=CHC_6H_5 + 2H_2O$.Moles of benzaldehyde used $= \frac{5.3 \ g}{1

Vedclass · Accepted Answer

$60$

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(C) The chemical reaction is: $2 \ C_6H_5CHO + CH_3COCH_3 ightarrow C_6H_5CH=CHCOCH=CHC_6H_5 + 2H_2O$.Moles of benzaldehyde used $= \frac{5.3 \ g}{106 \ g/mol} = 0.05 \ mol$.According to the stoichiometry,$2 \ mol$ of benzaldehyde produces $1 \ mol$ of dibenzalacetone.Therefore,theoretical moles of dibenzalacetone $= \frac{0.05}{2} = 0.025 \ mol$.Molar mass of dibenzalacetone $(C_{17}H_{14}O) = (17 	imes 12) + (14 	imes 1) + 16 = 234 \ g/mol$.Theoretical yield in grams $= 0.025 \ mol 	imes 234 \ g/mol = 5.85 \ g$.Actual yield $= 3.51 \ g$.Percentage yield $= \frac{	ext{Actual yield}}{	ext{Theoretical yield}} 	imes 100 = \frac{3.51}{5.85} 	imes 100 = 60 \%$.

In the Claisen$-$Schmidt reaction to prepare dibenzalacetone from $5.3 \ g$ benzaldehyde,a total of $3.51 \ g$ of product was obtained. The percentage yield in this reaction was $......... \%.$

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