See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: This reaction involves intramolecular electrophilic addition (hydrochlorination) of HCl across the terminal alkene of a molecule that already contains a cyclohexane ring tethered via an ethylene chain to a vinyl group (but-3-en-1-yl chain on cyclohexane). The starting material is 3-butenylcyclohexane or more precisely a compound where a cyclohexane ring bears a -CH2-CH2-CH=CH2 side chain (the structure shown is HO-cyclohexane... wait, re-reading: the starting material is simply CH2-CH2-CH=CH2 attached to a cyclohexane, i.e., 4-(but-3-en-1-yl)cyclohex-1-ene or similar). Step 1: Identify the starting material. The structure shown is a six-membered ring (cyclohexene) bearing a -CH2CH2CH=CH2 side chain, making it a vinyl-terminated chain that can cyclize. Step 2: HCl adds to the terminal alkene. The proton (H+) adds to the terminal CH2 (less substituted carbon, Markovnikov), generating a secondary carbocation at the internal carbon (the carbon adjacent to the ring chain). Step 3: The carbocation intermediate can be captured intramolecularly. The carbocation forms at the carbon that is 3 carbons away from the ring, allowing ring closure to form a new six-membered ring fused to the existing cyclohexene ring, producing a decalin (bicyclo[4.4.0]decane) framework. Step 4: The chloride ion then attacks the carbocation. In the major product, Markovnikov addition and the most stable carbocation placement give Cl at a secondary position on the newly formed ring (not at the ring junction). The double bond from the original cyclohexene ring is retained in the product. Step 5: Evaluating options: - Option (a): Has OH and methyl group - inconsistent with HCl addition product (no source of OH or extra methyl). - Option (b): Cl at ring junction with retained double bond - ring junction carbocation would be tertiary but product geometry and regiochemistry disfavor this. - Option (c): Cl at ring junction, fully saturated - requires double bond consumption not consistent with mechanism. - Option (d): Decalin framework with Cl on the second ring at a non-junction (secondary) carbon and a double bond retained in the first ring - this is consistent with Markovnikov addition followed by intramolecular cyclization where Cl ends up at the secondary carbon of the newly formed ring, with the original double bond intact. The major product (d) results from Markovnikov protonation of the terminal alkene, formation of a secondary carbocation, intramolecular ring closure to give a six-membered ring, and chloride capture at the secondary carbon, retaining the double bond from the original ring. Therefore, the correct answer is D.