See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Concept: Zinc-copper couple (Zn-Cu) with heat is a classic reagent for dehalogenation reactions. When a 1,3-dihalide (or vicinal/geminal dihalides in certain arrangements) is treated with Zn-Cu, it undergoes intramolecular coupling to form a cyclopropane ring. Step 1 – Identify the substrate: The starting material is Ph–CH(Br)–CH2–CH2Br, which is a 1,3-dibromide (bromine on C1 and C3 relative to each other, with phenyl on C1). Step 2 – Reaction with Zn-Cu: Zn-Cu couple removes both bromine atoms in a reductive elimination/intramolecular coupling process. The two terminal carbons bearing the bromines (C1 and C3) form a new C–C bond between them. Step 3 – Product formation: The new C–C bond between C1 (bearing Ph) and C3 closes the three-carbon chain into a three-membered ring, giving phenylcyclopropane. Step 4 – Why other options fail: - Option (a): Ph–CH=CH–CH2–Br would result from only one dehalogenation (elimination of one HBr), not the Zn-Cu intramolecular coupling. - Option (c): Ph–CHBr–CH=CH2 is a partial elimination product, not what Zn-Cu gives with a 1,3-dibromide. - Option (d): Ph–C≡C–CH3 would require an alkyne formation, which is not consistent with a 1,3-dibromide and Zn-Cu conditions. Therefore, the correct answer is B.