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(D) The reaction of $2$ molecules of propanone in the presence of $Ba(OH)_2$ is an aldol condensation reaction.Step $1$: Two molecules of propanone un

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$4-$Methylpent$-3-$en$-2-$one

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(D) The reaction of $2$ molecules of propanone in the presence of $Ba(OH)_2$ is an aldol condensation reaction.Step $1$: Two molecules of propanone undergo aldol condensation to form $4-$hydroxy$-4-$methylpentan$-2-$one (product $A$).$2 CH_3COCH_3 \xrightarrow{Ba(OH)_2} (CH_3)_2C(OH)CH_2COCH_3$ (Product $A$)Step $2$: Upon heating $( \Delta )$ and dehydration $(-H_2O)$,product $A$ loses a water molecule to form an $\alpha, \beta-$unsaturated ketone,which is $4-$methylpent$-3-$en$-2-$one (product $B$).$(CH_3)_2C(OH)CH_2COCH_3 \xrightarrow{\Delta, -H_2O} (CH_3)_2C=CHCOCH_3$ (Product $B$)Thus,the correct product $B$ is $4-$methylpent$-3-$en$-2-$one.

Identify product $B$ in the following sequence of reactions:
$2 \ CH_3COCH_3$ $\xrightarrow{Ba(OH)_2} A$ $\xrightarrow{\Delta, -H_2O} B$

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Identify product $B$ in the following sequence of reactions:$2 \ CH_3COCH_3$ $\xrightarrow{Ba(OH)_2} A$ $\xrightarrow{\Delta, -H_2O} B$