See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Concept: When a 1,ω-dihalide (a dihalide with halogens on both ends of a carbon chain) is treated with zinc dust, an intramolecular dehalogenation (Reformatsky-type ring closure) occurs, forming a cyclic hydrocarbon. This reaction is known as the Freund reaction. Step 1: Identify the reagent and substrate. 1,3-Dibromopropane is BrCH2CH2CH2Br — bromine atoms are on C1 and C3, separated by one carbon (C2). Step 2: Reaction with zinc dust. Zinc acts as a reducing metal that abstracts both bromine atoms simultaneously (or in a concerted-like fashion), forming a zinc dibromide (ZnBr2) as a byproduct and creating a new C–C bond between C1 and C3. Step 3: The new C–C bond between C1 and C3 closes the three-carbon chain into a three-membered ring, yielding cyclopropane. Overall: BrCH2CH2CH2Br + Zn → cyclopropane + ZnBr2 Why other options fail: (a) Propene would require elimination of HBr (one bromine and one hydrogen), not both bromines; this would need a base, not Zn dust. (b) Propane would require full reduction of both C–Br bonds to C–H, which needs a stronger reducing agent (e.g., H2/catalyst or LiAlH4), not simply Zn dust. (d) 3-Bromopropane would mean only one bromine is removed, which is not the typical outcome of Zn dust with a 1,3-dihalide under these conditions. Therefore, the correct answer is C.