See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Step 1 - Concept: In the electrophilic addition of Cl2 to an alkene (halohydrin formation in water), the first step is attack of the alkene pi electrons on Cl2. The intermediate formed is NOT an open carbocation but rather a cyclic halonium ion — specifically a bridged chloronium ion, where the chlorine bridges both carbons of the former double bond, bearing a positive charge on the chlorine. Step 2 - Why a cyclic chloronium ion forms: Chlorine, like bromine, has lone pairs that can donate back to the adjacent carbocation, forming a three-membered ring (episulfonium-like bridged halonium). This bridged chloronium ion is more stable than an open carbocation because the positive charge is delocalized onto the electronegative chlorine which stabilizes it through bridging, and it prevents rotation, explaining the anti addition stereochemistry observed in halohydrin formation. Step 3 - Identifying option D: Option (d) shows a cyclohexane ring with one carbon bearing a wedge H, a +Cl (positively charged chlorine bridging), and a dash H — this represents the cyclic chloronium ion intermediate where Cl bridges across the two carbons of the original double bond with a formal positive charge on Cl. Step 4 - Why other options fail: - Option (a) shows a protonated alcohol (oxocarbenium) on cyclohexane — this is a product-side species, not the initial intermediate from Cl2 attack. - Option (b) shows an open carbocation adjacent to Cl — this would be less stable than the bridged chloronium ion because the positive charge on carbon is not stabilized by bridging, and chlorine at a distance cannot donate electrons effectively. - Option (c) similarly shows an open carbocation with Cl, same reasoning as (b); the open carbenium ion is less stable than the bridged chloronium. Step 5 - Conclusion: The bridged cyclic chloronium ion (option d) is the most stable intermediate because the three-membered ring with positively charged Cl distributes charge and prevents formation of a fully exposed carbocation. Therefore, the correct answer is D.