See image — Alcohols Phenols and Ethers Chemistry Question

💡 Solution & Explanation

Concept: We need to prepare both stereoisomers (or a racemic mixture) of 1-methylcyclohexanol — specifically, the compound where OH and CH3 are both at C1 of cyclohexane, meaning 1-methylcyclohexan-1-ol in both its stereochemical forms (the two structures shown have OH on wedge and OH on dash, with CH3 group, representing the two enantiomers or diastereomers of 1-methylcyclohexanol). Step 1 - Analyze option (d): Reaction of cyclohexene with MCPBA gives cyclohexene oxide (1,2-epoxycyclohexane) via electrophilic epoxidation. The epoxide is then opened by CH3MgBr (a Grignard reagent). The Grignard attacks the epoxide in an SN2 fashion at one of the carbons. Since cyclohexene oxide is symmetric, attack by CH3MgBr at either carbon gives trans-2-methylcyclohexanol... However, reconsidering: if the epoxide formed is 1,2-epoxycyclohexane and CH3MgBr attacks, it opens at C1 or C2 giving trans-2-methylcyclohexan-1-ol. But the target shows 1-methylcyclohexanol structures. Actually, re-examining: the two structures shown both have OH and CH3 on the same carbon (C1), making them 1-methylcyclohexan-1-ol (a tertiary alcohol). Wait — looking again at the structures: OH is at top of ring and CH3 is at bottom of ring on the same carbon, indicating 1-methylcyclohexan-1-ol. For option (d): MCPBA on cyclohexene gives 1,2-epoxycyclohexane; CH3MgBr opens the epoxide — Grignard reagents open epoxides at the less hindered carbon via SN2, giving after protonation trans-2-methylcyclohexan-1-ol. The two wedge/dash structures shown could represent the two enantiomers of trans-2-methylcyclohexan-1-ol (racemic mixture), since the epoxide is achiral/symmetric and the Grignard attack gives a racemic trans product. This matches option (d) perfectly — the two structures are the (1R,2S) and (1S,2R) enantiomers (racemic trans-2-methylcyclohexan-1-ol). Step 2 - Why other options fail: - Option (a): Cyclohexanone + CH3Li gives 1-methylcyclohexanol (tertiary alcohol), but only one compound, not two stereoisomers shown as separate structures, and the product is the same tertiary alcohol regardless. - Option (b): 1-methylcyclohexene + Hg(OAc)2/NaBH4 (oxymercuration) gives Markovnikov addition, producing 1-methylcyclohexanol (tertiary), not two distinguishable stereoisomers as shown. - Option (c): Cyclohexene + BH3 then NaOH/H2O2 gives cyclohexanol; CH3Br would not introduce methyl at the alcohol carbon to give the desired product cleanly. - Option (d): MCPBA epoxidizes cyclohexene to give 1,2-epoxycyclohexane; CH3MgBr opens this symmetrical epoxide via anti (SN2) attack to give racemic trans-2-methylcyclohexan-1-ol — the two enantiomers depicted in the question. Therefore, the correct answer is D.