See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: Oxymercuration-demercuration (Hg(OAc)2/H2O followed by NaBH4) is an anti-Markovnikov-free, Markovnikov-selective hydration of alkenes that proceeds via a mercurinium ion intermediate with anti addition of water, followed by NaBH4 reduction that replaces Hg with H with retention of configuration at carbon. Step 1 - Identify the substrate: The starting material is 1-methylcyclohex-1-ene, a cyclohexene ring with a methyl group on C1 (one of the double bond carbons). Step 2 - Regioselectivity: Oxymercuration follows Markovnikov's rule. Water (the nucleophile) adds to the more substituted carbon of the double bond. C1 bears the methyl group, making it the more substituted carbon. Therefore, OH adds to C1 (the carbon bearing CH3), giving a tertiary alcohol at C1. Step 3 - Stereochemistry of oxymercuration: The mercurinium ion forms on one face of the double bond, and water attacks the more substituted carbon from the opposite face (anti addition). This means the OH and the Hg(OAc) add in an anti fashion across the double bond. After NaBH4 reduction, Hg is replaced by H with retention of configuration. The net result is anti addition of OH and H across the double bond. Step 4 - Product structure: The product is 1-methyl-1-hydroxycyclohexane... wait, re-examining: C1 has CH3 and becomes the carbon bearing OH (Markovnikov), and C2 gets H. This gives 1-methylcyclohexan-1-ol as the regiochemical product. However, looking at option (c), it shows 1,1-dimethylcyclohexanol which is incorrect. Option (d) shows a cyclohexane with CH3 and OH both on C1 (wedge bonds) and H on C1 and H on C2 on dash bonds, representing 1-methylcyclohexan-1-ol with defined stereochemistry from anti addition. Step 5 - Why option (d) is correct: In 1-methylcyclohex-1-ene, C1 (bearing CH3) and C2 are the double bond carbons. Markovnikov addition places OH on C1. The anti addition means H (from reduction of C-Hg bond) and OH are trans to each other as added across the double bond. Option (d) correctly depicts 1-methylcyclohexan-1-ol with the CH3 and OH on C1 shown on wedge, and H on C2 on dash, consistent with the stereochemical outcome of anti addition. Why other options fail: - (a) shows OH at C4 (far from double bond), incorrect regiochemistry. - (b) retains the double bond and shows OH at C2, incorrect - the double bond should be consumed. - (c) shows 1,1-dimethylcyclohexanol, which would require two methyl groups; impossible from this substrate. Therefore, the correct answer is D.