See image — Isomerism and Stereochemistry Chemistry Question

💡 Solution & Explanation

# Solution: Deuterium Exchange in α,β-Unsaturated Ketone **Step 1: Identify the reactive sites** The starting material is an $\alpha,\beta$-unsaturated ketone with: - An $\alpha$-CH$_2$ group (between Ph and C=O) - An $\alpha$-hydrogen on the double bond - A methyl group on the $\beta$-position **Step 2: Understand OD⁻/D₂O conditions** Basic deuterium conditions promote: - **Enolization**: The $\alpha$-CH₂$ is acidic and readily forms an enolate anion - **Deuterium exchange**: The enolate abstracts deuterium from D₂O, replacing all hydrogens at the $\alpha$-carbon **Step 3: Determine which hydrogens exchange** - The $\alpha$-CH₂$ group (adjacent to C=O) has **2 exchangeable hydrogens** → becomes **CD₂** after prolonged treatment - The $\beta$-position has only $\alpha$-hydrogens on the double bond (one H) → becomes **D** (exchanges) - The methyl group (**CH₃**) is not $\alpha$ to the carbonyl and remains unchanged in prolonged basic conditions **Step 4: Confirm structure (B)** $$\text{Ph-CO-CD=CD-CH}_3$$ This matches option **(B)**, which shows: - CD₂ at the $\alpha$-position (two deuteria exchanged) - D on the $\beta$-carbon of the double bond (one deuterium exchanged) - CH₃ at the terminal position (unchanged) **Answer: (B)** — The $\alpha$-CH₂$ undergoes complete deuteration (→ CD₂), and the $\alpha$-hydrogen of the $\beta$-carbon also exchanges (→ D).