The enolic form of acetone contains

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(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 $B$.

Similar Questions

Hybridization of $1$ and $2$ carbon atoms in $\mathop {CH_2}\limits^1 = \mathop C\limits^2 = CH_2$

How is the hybridization of carbon in an organic compound determined based on its structure? Explain.

Which of the following compounds contains carbon atoms with both $sp$ and $sp^2$ hybridization?

Number of $sp$ hybridised carbon atoms in the given compound $CH_2=CH-C\equiv C-CH_2-CN$ is:

Which of the following compounds do not have $sp^3$ carbon atom$(s)$?
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$III$. Buta-$1,3$-diene
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$V$. Naphthalene

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