See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the starting material: The substrate is 1-(3-hydroxy-3-methylbutyl)cyclohex-1-ene. The molecule has a tertiary alcohol [C(OH)(CH3)2] at the end of a two-carbon tether attached to C1 of cyclohex-1-ene. Step 2 - Protonation and carbocation formation: Conc. H2SO4 protonates the tertiary OH, and water leaves to generate a tertiary carbocation at the C bearing two methyl groups: cyclohexenyl-CH2CH2-C+(CH3)2. Step 3 - Carbocation rearrangement via intramolecular cyclization: The tertiary carbocation is positioned such that the double bond of the cyclohexene ring (C1=C2) can attack it intramolecularly. The pi electrons of the endocyclic double bond attack the tertiary carbocation in a 5-exo or 6-endo fashion. Given the two-carbon tether (CH2CH2), the double bond carbon attacks the cationic center to form a new C-C bond, generating a bicyclic system. This is a transannular or ring-closure cationic cyclization. The attack of C2 of the cyclohexene double bond onto the tertiary carbocation creates a new six-membered ring fused to the original cyclohexane (now cyclohexyl) ring, producing a bicyclic carbocation at C1 of the original cyclohexene (now a tertiary bridgehead carbocation or secondary carbocation). Step 4 - The resulting bicyclic carbocation loses a proton (elimination, Zaitsev) to give the most stable alkene. The new bicyclic product is a fused bicyclic alkene. With the gem-dimethyl group at the junction and the double bond within the ring, this matches option (a): a fused bicyclic structure (decalin-type) with a quaternary carbon bearing two methyl groups at one position and a double bond in the ring. Step 5 - Why other options fail: - Option (b): Also a bicyclic product but the gem-dimethyl and double bond placement differs; this would require a different carbocation pathway. - Option (c): Simple intermolecular elimination to give an open-chain diene/alkene; not favored when intramolecular cyclization can generate a stable bicyclic tertiary carbocation. - Option (d): Another open-chain elimination product; same reasoning as (c), intramolecular cyclization is kinetically and thermodynamically preferred (Markovnikov, ring formation). The major product is the bicyclic alkene formed by intramolecular cationic olefin cyclization followed by deprotonation, giving the fused bicyclic structure in option (a). Therefore, the correct answer is A.