Question

Bonds and lone pairs present in enolic form of butanone are explain

Answer 1

Answer: There are $12\sigma,9\pi \text{ and }2\text{ lone pairs}$ present in enolic form of butanone.

Explanation: Butanone is a chemical compound having molecular formula $C_4H_8O$ . This is the ketonic form.

The enolic form of butanone is formed by tautomerization.

Tautomers are the isomers having same molecular formula having different connectivity.

There is a double bond present between carbon and oxygen atom in ketonic form but in enolic form, the double bond is present between carbon and carbon atom.

Number of sigma bonds present in enolic form of butanone = 12

Number of pi- bonds present in enolic form of butanone = 1

Number of lone pairs present in enolic form of butanone = 2

The two forms of butanone are shown in the image attached.