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I apologize for the inconvenience. It seems there might be a technical issue preventing the reactions from displaying properly. Let's simplify it further:

**Case 1: NaOH(aq)**
1. Acetone (\( \text{CH}_3\text{COCH}_3 \)) reacts with hydroxide ion (\( \text{OH}^- \)) to form enolate ion (\( \text{CH}_3\text{COCH}_2^- \)) and water (\( \text{H}_2\text{O} \)).
2. Enolate ion (\( \text{CH}_3\text{COCH}_2^- \)) rearranges to form enol form (\( \text{CH}_2=\text{C(OH)}\text{CH}_3 \)).
3. Enol form (\( \text{CH}_2=\text{C(OH)}\text{CH}_3 \)) accepts a proton from another hydroxide ion (\( \text{OH}^- \)) to form keto-enol equilibrium (\( \text{CH}_2=\text{C}(\text{OH}_2)\text{CH}_3 \)).

**Case 2: NaOD/D2O**
1. Acetone (\( \text{CH}_3\text{COCD}_3 \)) reacts with deuteride ion (\( \text{OD}^- \)) to form deuterated enolate (\( \text{CH}_3\text{COCD}_2^- \)) and deuterium oxide (\( \text{D}_2\text{O} \)).
2. Deuterated enolate (\( \text{CH}_3\text{COCD}_2^- \)) rearranges to form deuterated enol (\( \text{CH}_3\text{COC(D)D}^- \)).
3. Deuterated enol (\( \text{CH}_3\text{COC(D)D}^- \)) rearranges to form deuterated enol form (\( \text{CH}_2=\text{C(OD)CD}_3 \)).
4. Deuterated enol form (\( \text{CH}_2=\text{C(OD)CD}_3 \)) accepts a deuterium ion (\( \text{OD}^- \)) to form keto-enol equilibrium (\( \text{CH}_2=\text{C}(\text{OD}_2)\text{CD}_3 \)).

I hope this format is visible to you. Let me know if you need any further assistance or clarification!