See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: Alkoxymercuration-demercuration is the alcohol analogue of oxymercuration-demercuration. Instead of water, an alcohol (here CH3OH) acts as the nucleophile. The reaction follows Markovnikov's rule with anti addition (mercury goes to the less substituted carbon, methoxy group goes to the more substituted carbon), and the demercuration step with NaBH4 proceeds with overall Markovnikov regioselectivity without carbocation rearrangement. Step 1 – Identify the alkene: The starting material is 1-methylcyclohex-1-ene, which has a trisubstituted double bond between C1 (bearing CH3) and C2. Step 2 – Regioselectivity: In alkoxymercuration, the electrophilic mercury attacks to form a mercurinium ion. The nucleophile (CH3OH/methanol) attacks the more substituted (more electrophilic, more stable partial positive) carbon, which is C1 (the carbon bearing the methyl group). This places the OCH3 group on C1 (the tertiary carbon), consistent with Markovnikov addition. Step 3 – Demercuration: NaBH4/HO- replaces the C–Hg bond with C–H at C2. The overall result is addition of OCH3 to the more substituted carbon (C1) and H to the less substituted carbon (C2). Step 4 – Product: The product is 1-methoxy-1-methylcyclohexane, where both the methyl group and the OCH3 group are on the same carbon (C1), giving a tertiary ether. This corresponds to option (b). Why other options fail: - Option (a): OCH3 at C2 (less substituted) would be anti-Markovnikov, not the major product. - Option (c): OH instead of OCH3; the nucleophile is CH3OH, not water, so an ether forms, not an alcohol. - Option (d): OH at C2 is both anti-Markovnikov and uses the wrong nucleophile. Therefore, the correct answer is B.