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(C) The given reaction is an aldol condensation of acetophenone $(C_6H_5COCH_3)$.Step $1$: Two molecules of acetophenone react in the presence of a ba

Vedclass · Accepted Answer

$1,3-$diphenyl$-2-$methylprop$-2-$en$-1-$one

Vedclass · Answer

(C) The given reaction is an aldol condensation of acetophenone $(C_6H_5COCH_3)$.Step $1$: Two molecules of acetophenone react in the presence of a base $(OH^-)$ to form a $\beta$-hydroxy ketone (aldol product,$A$).$2 C_6H_5COCH_3 \xrightarrow{OH^-} C_6H_5C(OH)(CH_3)CH_2COC_6H_5$ (Product $A$)Step $2$: Upon heating $(\Delta)$,the $\beta$-hydroxy ketone undergoes dehydration to form an $\alpha,\beta$-unsaturated ketone (Product $B$).$C_6H_5C(OH)(CH_3)CH_2COC_6H_5 \xrightarrow{\Delta} C_6H_5C(CH_3)=CHCOC_6H_5$ (Product $B$)Thus,the final product $(B)$ is $1,3$-diphenyl-$2$-methylprop-$2$-en-$1$-one.

Column-$I$ (Reagent)	Column-$II$ (Product) and Column-$III$ (Name)
$A$. Ammonia $(NH_{3})$	$i$. $CH_{3}CH=NCH_{3}$; $m$. Acetaldehyde hydrazone
$B$. Amine $(CH_{3}NH_{2})$	$ii$. $CH_{3}CH=NNH_{2}$; $n$. Acetaldehyde oxime
$C$. Hydroxyl amine $(NH_{2}OH)$	$iii$. $CH_{3}CH=NH$; $o$. Acetaldehyde imine
$D$. Hydrazine $(NH_{2}NH_{2})$	$iv$. $CH_{3}CH=NOH$; $q$. Acetaldehyde methyl imine

The structure of the main product $(B)$ of the following reaction will be:

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