The aldol condensation product of acetone on | vedclass

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(D) Acetone $(CH_3COCH_3)$ undergoes aldol condensation in the presence of a base like $Ba(OH)_2$ to form $4-$hydroxy$-4-$methylpentan$-2-$one (diacet

Vedclass · Accepted Answer

$4-$methylpent$-3-$en$-2-$one

Vedclass · Answer

(D) Acetone $(CH_3COCH_3)$ undergoes aldol condensation in the presence of a base like $Ba(OH)_2$ to form $4-$hydroxy$-4-$methylpentan$-2-$one (diacetone alcohol).$2CH_3COCH_3 \xrightarrow{Ba(OH)_2} (CH_3)_2C(OH)CH_2COCH_3$Upon dehydration (heating with acid),the $4-$hydroxy$-4-$methylpentan$-2-$one loses a water molecule to form $4-$methylpent$-3-$en$-2-$one (mesityl oxide).$(CH_3)_2C(OH)CH_2COCH_3 \xrightarrow{\Delta, H^+} (CH_3)_2C=CHCOCH_3 + H_2O$Thus,the final product is $4-$methylpent$-3-$en$-2-$one.

The aldol condensation product of acetone on dehydration gives:

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