See image — Isomerism and Stereochemistry Chemistry Question

💡 Solution & Explanation

## Tautomerization of $TCH_2\text{-}C(=O)\text{-}CH_2H$ under $D^+$ Catalyst **Step 1: Identify the tautomerization type** This is keto-enol tautomerism. The starting material has a ketone group ($C=O$) with an α-hydrogen on the adjacent carbon. **Step 2: Mechanism under acid catalyst ($D^+$)** - The carbonyl oxygen accepts a proton: $C=O \rightarrow C^+\text{-}OH$ - An α-hydrogen (from $CH_2H$ group) is abstracted by a base, forming a double bond between the α-carbon and carbonyl carbon - This generates the enol form: $CH_2=C(OH)\text{-}CH_3$ **Step 3: Tautomerization to enol product** Under deuterated acid ($D^+$) conditions, the equilibrium shifts toward the enol tautomer $CH_2=C(OH)\text{-}CH_3$ rather than reverting to the keto form. The deuterium substitutes the acidic hydroxyl proton. **Step 4: Why option (C) is correct** The product $CH_2=C\text{-}CH_3$ represents the enol form (vinyl alcohol equivalent). This is the stable tautomer under these conditions. **Why others are wrong:** - **(A)**: Keto form with $OH$ group—reversion to original form, not tautomerization - **(B)**: Deuterium on $OH$—deuterium exchanges but doesn't show the actual tautomeric shift - **(D)**: Not all products form; the reaction favors the enol under these conditions **Answer: (C)** $CH_2=C\text{-}CH_3$ (enol tautomer)