The enolic form of acetone contains:

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Question

(A) The keto form of acetone is $CH_3-CO-CH_3$. The enolic form is $CH_2=C(OH)-CH_3$.
Counting the bonds in the enolic form $(CH_2=C(OH)-CH_3)$:
- S

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The keto form of acetone is $CH_3-CO-CH_3$. The enolic form is $CH_2=C(OH)-CH_3$.
Counting the bonds in the enolic form $(CH_2=C(OH)-CH_3)$:
- Sigma $(\sigma)$ bonds: $3$ (in $CH_3$) + $1$ ($C$-$C$) + $1$ ($C$=$C$) + $1$ ($C$-$O$) + $1$ ($O$-$H$) + $2$ (in $CH_2$) = $9$ $\sigma$ bonds.
- Pi $(\pi)$ bonds: $1$ (in $C$=$C$).
- Lone pairs: Oxygen atom in the hydroxyl group $(-OH)$ has $2$ lone pairs.
Thus,the enolic form contains $9$ $\sigma$ bonds,$1$ $\pi$ bond,and $2$ lone pairs.

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