Which structure is obtained when the curved arrows in the enolate ion below are pushed (the lone pair on O goes down to form C=O and the C=C pi pair moves onto carbon) ?

Question

TestHub · Accepted Answer

The lone pair on O⁻ forms a C=O pi bond. To keep carbon at four bonds the C=C pi pair moves onto the terminal carbon as a lone pair.

The result is the carbanion CH₃-CO-CH₂⁻, written CH₂⁻C(CH₃)=O.

Chemistry - GOC Question with Solution | TestHub

Options:

Doubts & Discussion