The enolic form of acetone contains

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Question

(B) The enolic form of acetone is prop$-1-$en$-2-$ol,which has the structure $CH_2=C(OH)-CH_3$.In this structure,the bonds are as follows:- $C=C$ bond

Vedclass · Accepted Answer

$9 \sigma$-bonds,$1 \pi$-bond and $2$ lone pairs

Vedclass · Answer

(B) The enolic form of acetone is prop$-1-$en$-2-$ol,which has the structure $CH_2=C(OH)-CH_3$.In this structure,the bonds are as follows:- $C=C$ bond: $1 \sigma$ and $1 \pi$ bond.- $C-C$ bond: $1 \sigma$ bond.- $C-O$ bond: $1 \sigma$ bond.- $O-H$ bond: $1 \sigma$ bond.- $C-H$ bonds: $5 \sigma$ bonds (two on terminal $CH_2$ and three on $CH_3$).Total $\sigma$ bonds = $1+1+1+1+5 = 9$.Total $\pi$ bonds = $1$.Oxygen atom has $2$ lone pairs.Therefore,the correct option is $B$.

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