See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: Oxymercuration-demercuration (Hg(OAc)2 followed by NaBH4) normally adds water across a double bond in a Markovnikov fashion. However, when the substrate contains both a hydroxyl group and an alkene tethered appropriately, intramolecular oxymercuration can occur, forming a cyclic ether (tetrahydropyran or tetrahydrofuran ring) preferentially. Step 1: Identify the substrate. The molecule is 6-methylhept-6-en-1-ol (or equivalently, the alcohol has a terminal OH at one end and a trisubstituted/1,1-disubstituted alkene at the other end: CH2=C(CH3)-(CH2)4-OH). Counting carbons: the OH is on C1, and the alkene is at C6 (with a methyl branch), giving a 6-carbon chain between OH and the double bond carbon. Step 2: Intramolecular oxymercuration. The hydroxyl oxygen acts as the nucleophile attacking the more substituted carbon of the double bond (Markovnikov selectivity) via a mercurinium ion intermediate. The chain length between OH and the alkene allows formation of a 6-membered ring (tetrahydropyran): the oxygen attacks the tertiary carbon of the alkene (C(CH3)=CH2), forming a six-membered cyclic oxymercury intermediate. This gives a tetrahydropyran ring with a -CH2HgOAc group and a methyl substituent on the carbon bearing the oxygen. Step 3: NaBH4 reduction removes the mercury substituent (reductive demercuration), replacing C-Hg with C-H. Step 4: The product A is 2-methyl-tetrahydropyran (specifically, 2-methyltetrahydro-2H-pyran), a six-membered cyclic ether with a methyl group at the 2-position (the carbon bearing the ring oxygen), with a -CH2- group (formerly =CH2) now as a -CH3 or remaining as part of the ring. Re-examining: The substrate CH2=C(CH3)-(CH2)4-OH: O is at one end, alkene at the other. For intramolecular attack, O attacks the electrophilic mercurinium. The more substituted alkene carbon (C(CH3)) is more electrophilic in oxymercuration. Distance: O-C1-C2-C3-C4-C5=C(CH3) with CH2. O attacking C5 (the internal carbon bearing CH3) through a 6-membered transition state (O, C1, C2, C3, C4, C5) gives a tetrahydropyran ring. The exo-CH2HgOAc becomes CH3 after NaBH4. Product: 2-methyl-tetrahydropyran (tetrahydropyran with methyl at C2). Why other options fail: Intermolecular Markovnikov addition would give an acyclic alcohol, not a cyclic ether. Five-membered ring (THF) formation would be less favored than six-membered (THP). Anti-Markovnikov product is not formed under these conditions. Therefore, the correct answer is B.