See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: Electrophilic addition of bromine (Br2) to an alkene proceeds via an anti addition mechanism through a bromonium ion intermediate. When isotopically labeled 82Br2 is used, both bromine atoms added are 82Br. However, the starting alkene here is a vinylidenecyclohexane-type compound (methylenecyclohexane with substituents), where the double bond is exocyclic to the ring. Step 1 - Identify the substrate: The starting material is 1-methylenecyclohexane (exocyclic alkene on a cyclohexane ring) with a substituent (propyl/ethyl group on the exocyclic carbon). The double bond is between C1 of the ring and the exocyclic carbon (=CH-). Step 2 - Bromonium ion formation: 82Br2 approaches the double bond, and the more electrophilic 82Br forms a bridged bromonium ion across C1 (ring carbon) and the exocyclic carbon. Step 3 - Nucleophilic attack: The bromide ion (82Br-) attacks the bromonium ion from the back (anti). Since C1 of the ring is more substituted (tertiary, part of the ring), the 82Br- preferentially attacks C1 of the ring from the back face. Step 4 - Stereochemical outcome: The bromonium ion forms on one face; the nucleophilic 82Br- attacks from the opposite face at C1, giving anti addition. The exocyclic carbon gets one 82Br (from bromonium), and C1 gets 82Br- (from nucleophile), both labeled. The H on the exocyclic carbon remains. Step 5 - Product analysis: The major product has both bromines as 82Br (since 82Br2 is used exclusively). At the ring carbon C1: one 82Br added anti to the other 82Br on the exocyclic carbon. Option (b) shows Br82 on wedge at C1, Br on the exocyclic position, H on dash, and Br82 on dash - representing the anti addition product where both added bromines are 82Br and the arrangement is anti (trans diaxial-like addition). The non-labeled Br in option (b) on the exocyclic carbon actually represents the 82Br from the bromonium, and the arrangement in (b) correctly shows the anti addition stereochemistry with both bromines being 82Br. Why other options fail: - Option (a): Shows a non-labeled Br at top and Br82 labels at bottom - incorrect isotope labeling pattern, suggesting syn addition or wrong regiochemistry. - Option (c): Has a non-labeled Br at the bottom and Br82 at top - wrong isotope distribution. - Option (d): Missing the wedge Br, showing wrong stereochemical arrangement. Option (b) correctly represents anti addition of 82Br2 with the bromonium mechanism, giving the trans (anti) dibromide with correct isotope labeling. Therefore, the correct answer is B.