See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: The reaction of an organic dihalide with sodium metal in dry ether is a Wurtz-type reaction that forms a new C-C bond by reductive coupling, eliminating the two halide leaving groups. Step 1 - Identify the starting material: The substrate is 1-chloro-3-bromocyclobutane. The Cl and Br substituents are on C1 and C3 of the cyclobutane ring (1,3-positions, i.e., on opposite carbons of the four-membered ring). Step 2 - Reaction mechanism: Na metal donates electrons to each carbon bearing a halogen, generating carbanion/radical intermediates at C1 and C3. These two carbon centers are on the same cyclobutane ring, so intramolecular coupling between C1 and C3 forms a new C-C bond bridging those two carbons. Step 3 - Determine the product: When C1 and C3 of cyclobutane are connected by a new bond, a bicyclic compound results. C1-C2-C3 forms one three-membered ring, and C1-C4-C3 forms another three-membered ring, both sharing the new C1-C3 bond. This is bicyclo[1.1.0]butane, which consists of two fused cyclopropane rings sharing a common (central) C-C bond - depicted as option (b). Step 4 - Why other options fail: - (a) Cyclobutadiene would require loss of four hydrogens and formation of two double bonds, which does not occur under Wurtz conditions. - (c) Cyclobutene would require elimination of HX to form a double bond; the Wurtz reaction forms C-C bonds, not double bonds via elimination. - (d) Methylenecyclopropane would require ring contraction and formation of an exocyclic double bond, which is not consistent with intramolecular Wurtz coupling of 1,3-dihalocyclobutane. The high yield (97%) is consistent with the favorable intramolecular Wurtz coupling forming the strained but well-known bicyclo[1.1.0]butane. Therefore, the correct answer is B.