Question

The enolic for of acetone contains:

• A. $9$-sigma, $1$ pi-bond and $2$ lone pairs
• B. $8$-sigma, $2$ pi-bond and $2$ lone pairs
• C. $10$-sigma, $1$ pi-bond and $1$ lone pair
• D. $9$-sigma, $2$ pi-bond and $1$ lone pair

Answer 1

The correct option is A $9$-sigma, $1$ pi-bond and $2$ lone pairs

The enolic form of acetone contains $9$-sigma, $1$ pi-bond and $2$ lone pairs. Acetone shows tautomerism with keto and enol form.Acetone is a carbon double bonded to an oxygen. Then that carbon is further bonded to $2$ more carbons which each have $3$ hydrogen. So when it goes to the enolic form, the oxygen becomes protonated and the double bond moves to between the central carbon and one of the other carbons. Now count bonds. Central carbon has $1$ pi and $3$ sigmas. The double bonded carbon has $2$ sigma. The other carbon has $3$ sigma. The oxygen has $1$ sigma and $2$ lone pair. $3+2+3+1=9$ sigma and $1$ pi and $2$ lone pair