Which of the following best describes the movement of electrons (curved arrows) during the nucleophilic addition that converts an ${\mathrm{enolate}}_{\mathrm{ion}}$ to its $\mathrm{keto}$ form?

• A. The lone pair on the $\alpha$ carbon attacks the carbonyl carbon, forming a new carbon-carbon bond and generating a cyclic intermediate.
• B. A lone pair on the enolate oxygen forms a double bond with the adjacent carbon, and the $\pi$ electrons from the carbon-carbon double bond move to capture a proton, resulting in the $\mathrm{keto}$ form.
• C. The $\pi$ electrons of the carbon-carbon double bond move to form a new bond with a proton, while the lone pair on the oxygen remains unchanged.
• D. A lone pair on the enolate oxygen forms a double bond with the adjacent carbon, while the carbon-carbon double bond electrons move to the $\alpha$ carbon to form a carbanion.

Answer: (B)

Verified step by step guidance

1. Understand the structure of the enolate ion: it has a negatively charged oxygen atom and a carbon-carbon double bond between the α-carbon and the carbonyl carbon.
2. Recall that the keto form is the neutral, carbonyl-containing form, where the double bond is between carbon and oxygen, and the α-carbon is fully saturated (no double bond).
3. Identify the electron movement during tautomerization: the lone pair on the enolate oxygen forms a double bond with the adjacent carbon, restoring the carbonyl group (C=O).
4. Simultaneously, the π electrons from the carbon-carbon double bond move to the α-carbon, which then captures a proton (H⁺), converting the double bond into a single bond and neutralizing the charge.
5. Summarize the curved arrow movement: from the lone pair on oxygen to form the C=O double bond, and from the C=C π bond to the α-carbon to accept a proton, resulting in the keto form.