See image — Reaction Mechanism Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the starting material: The compound is 1-(dibromomethyl)cyclohex-2-ene, i.e., a cyclohexene ring where the carbon at position 1 bears a -CHBr2 group and the double bond is between C2 and C3. Step 2 - Reaction conditions: NaOH (50%, aqueous/alcoholic), heat (Delta). These conditions promote elimination reactions. With a gem-dihalide (CHBr2), NaOH first eliminates one HBr to give a vinyl bromide or, more relevantly here, the gem-dibromide can undergo double dehydrohalogenation or intramolecular cyclization. Step 3 - Mechanism: NaOH abstracts a proton from CHBr2 to generate a carbanion/carbene intermediate, or stepwise: first elimination of HBr from CHBr2 gives a -CBr= species (a vinylidene/carbene equivalent). The resulting alpha-elimination of the gem-dibromide under strong base gives a carbene (:CBr- after first HBr loss gives -CH=CBr, but with gem-dibromide on the CH, strong base causes alpha-elimination to form a dihalocarbene or after loss of Br- a carbenoid). More precisely: NaOH abstracts H from CHBr2 to give :CBr2 (dibromocarbene) attached to the ring carbon? No - the CHBr2 is exocyclic. Under strong base, CHBr2 undergoes alpha-elimination: base removes H from CHBr2 to give [CBr2]^- which loses Br^- to form :CBr2 (dibromocarbene in situ) attached... actually the CHBr2 is a substituent, so base acts on it: NaOH + CHBr2-R → :CBr2 (free carbene) is not correct since CHBr2 is attached to the ring. Step 4 - Correct mechanism: The exocyclic CHBr2 group on the cyclohexene undergoes intramolecular reaction. NaOH eliminates HBr from CHBr2 to give an exocyclic =CBr2 or a carbenoid. The vinylic/allylic system then undergoes intramolecular cyclization. With NaOH (50%), the gem-dibromide loses HBr to give an allylic bromide with an exocyclic double bond, then intramolecular SN2 or further elimination, OR: double elimination of both HBr molecules gives an allene/cyclopropane. The key reaction: base-induced alpha-elimination of CHBr2 generates :CBr2 (carbene) which inserts into the nearby C=C double bond of the cyclohexene ring intramolecularly, forming a bicyclic product with a cyclopropane ring fused to the cyclohexane, retaining one Br. Step 5 - Product: The carbene (:CBr2 equivalent, generated in situ from the CHBr2 group on C1 of cyclohexene) adds to the adjacent C2=C3 double bond intramolecularly. This cyclopropanation of the double bond within the ring forms a bicyclo[3.1.0] or bicyclo[4.1.0] system (norcarane-type). Since the CHBr2 is on C1 and the double bond is C2-C3, the carbene carbon bridges C2 and C3, forming a bicyclo[4.1.0] system (bicycloheptane with a cyclopropane ring fused to cyclohexane) with one Br remaining on the bridge carbon. Step 6 - This bicyclo[4.1.0]heptyl bromide (norcarane with Br on the one-carbon bridge) corresponds to option (a), which shows a bicyclic structure with a small bridge bearing Br. Step 7 - Why other options fail: (b) and (c) show Br in different positions (endo or on larger bridge) inconsistent with the one-carbon bridge bearing the Br; (d) shows an adamantane-type skeleton which requires more carbons and rearrangement not supported here. Therefore, the correct answer is A.