See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Step 1 – Identify the starting material: The substrate is a tertiary alcohol containing a methylenecyclohexane (exocyclic alkene) unit tethered by a two-carbon chain to a quaternary carbon bearing the OH and gem-dimethyl groups (plus an additional methyl, making it effectively a tert-amyl or neopentyl-type alcohol). The molecule has the skeleton needed to form a bicyclic (decalin-type) product via an acid-catalyzed cyclization. Step 2 – Acid-catalyzed dehydration/cyclization mechanism: Treatment with H2SO4 and heat first protonates the OH to form a carbocation (or promotes E1 dehydration). Because the molecule contains both an alkene (exocyclic double bond on the cyclohexane ring) and a tertiary carbocation center, an intramolecular Prins-type or carbocation cyclization can occur. The carbocation formed after loss of water is tertiary and positioned to undergo intramolecular electrophilic addition to the nearby exocyclic double bond. Step 3 – Ring closure: The tertiary carbocation attacks the exocyclic double bond intramolecularly, forming a new six-membered ring. This creates a bicyclic (decalin) skeleton. The newly formed carbocation after ring closure is then stabilized by loss of a proton (elimination) to give the thermodynamic alkene product. Step 4 – Regiochemistry and methyl group placement: The gem-dimethyl groups end up on one carbon of the new ring (from the quaternary carbon of the original alcohol), and the remaining methyl group ends up on an adjacent carbon. The double bond forms between the ring-junction region to give the most stable (trisubstituted) alkene, consistent with Zaitsev's rule. This gives a 4,4,5-trimethyl (or equivalent numbering) octahydronaphthalene with the double bond at the ring junction or adjacent position — matching option (b). Step 5 – Why other options fail: - (a) places the gem-dimethyl and extra methyl in a different connectivity that does not match the regiochemistry of the carbocation closure. - (c) and (d) have incorrect double bond positions or methyl group distributions inconsistent with the Markovnikov/thermodynamic product of this specific cyclization. Option (b) correctly represents the bicyclic product with gem-dimethyl on one carbon, an additional methyl on an adjacent carbon of the same ring, and the double bond at the thermodynamically favored position in the decalin framework. Therefore, the correct answer is B.