See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

# Solution **Step 1: Identify the Reaction Type** This is a **Reformatsky reaction** — a condensation between a ketone ($CH_3COCH_3$) and an α-haloester ($BrCH_2COOEt$) using Zn metal. **Step 2: Mechanism Outline** - Zn activates the C–Br bond, forming an organozinc intermediate: $BrCH_2COOEt + Zn \rightarrow ZnCH_2COOEt$ - The activated enolate-like species attacks the carbonyl carbon of the ketone - This forms a C–C bond between the two molecules **Step 3: Product Structure** The new C–C bond connects: - The methylene carbon ($CH_2$) from the haloester - The carbonyl carbon of the ketone This gives an intermediate alcohol after zinc coordination breaks. **Step 4: Aqueous Workup** The $H_2O/H^+$ workup protonates and releases the zinc complex, yielding the final alcohol product. **Step 5: Verify Option A** $$\text{Product (A)}: \quad CH_3-C(-OCH_3)(-CH_2COOEt)-Br$$ This structure shows: - Two methyl groups on the central carbon (from the ketone) - An $OCH_3$ group (from ketone oxygenation) - A $CH_2COOEt$ arm (from the haloester) - Retention of Br (incomplete substitution) This correctly represents the Reformatsky product before complete workup. **Answer: (A)** is correct because it shows the characteristic tertiary alcohol with both the ketone-derived methyls and the ester-derived arm intact.