See image — Isomerism and Stereochemistry Chemistry Question

💡 Solution & Explanation

**Analysis: Enolization of Acetone in $D_2O$** **Step 1: Understand enolization mechanism** Acetone undergoes keto-enol tautomerism. In $D_2O$, deuterium replaces hydrogen at exchangeable positions. **Step 2: Identify the enol form structure** The enol form of acetone is: $CH_2=C(OH)-CH_3$ **Step 3: Determine deuterium exchange sites** In $D_2O$ (acidic or basic conditions): - The **OH group** exchanges: $OH \rightarrow OD$ - The **α-hydrogens** (on carbons adjacent to C=O) exchange with $D_2O$ **Step 4: Apply to the enol form** Starting from enol: $CH_2=C(OH)-CH_3$ Both methyl groups ($-CH_3$) attached to the C=C double bond are α-carbons. These hydrogens are acidic and exchange: - Left $CH_2$ becomes $CD_2$ - Right $CH_3$ becomes $CD_3$ - $OH$ becomes $OD$ **Step 5: Write the final product** $$CD_2=C(OD)-CD_3$$ **Answer: (D)** $CD_2=C(OD)-CD_3$ The key is recognizing that all exchangeable hydrogens (both α-hydrogens and the enolic OH) undergo deuterium exchange in $D_2O$.