See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: When a molecule contains both an alkene and a hydroxyl group, treatment with Br2 can lead to intramolecular cyclization if the geometry is favorable, forming a cyclic bromonium intermediate that is then trapped intramolecularly by the hydroxyl group rather than by bromide ion. Step 1 - Identify the substrate: The starting material is 2-methylene-7-heptanol (or equivalently 2-methyleneheptan-1-ol depending on numbering). It has a terminal alkene (CH2=C(CH3)-) at one end and a primary alcohol (-CH2OH) at the other end of a six-carbon chain. Step 2 - Bromine reacts with the alkene to form a bromonium ion intermediate: Br2 attacks the electron-rich double bond to give a cyclic bromonium ion at the alkene carbons. Step 3 - Count the atoms for intramolecular cyclization: From the bromonium ion carbon to the oxygen of the -OH group, there are enough atoms to form a favorable ring. The hydroxyl oxygen can act as an internal nucleophile attacking the bromonium ion carbon in an anti fashion. Counting the chain: O-C-C-C-C-C-C(bromonium) gives a 6-membered ring transition state (tetrahydropyran ring), which is highly favorable. Step 4 - Ring closure: The oxygen attacks the more substituted carbon of the bromonium ion (Markovnikov-like intramolecular attack), displacing bromide and forming a six-membered cyclic ether (tetrahydropyran). The methyl group and a -CH2Br group end up on the same carbon (C2 of the ring, which bears the oxygen), giving a 2-methyl-2-(bromomethyl)tetrahydropyran. Step 5 - Product identification: This matches option (c): a tetrahydropyran ring with a methyl group and a -CH2Br group on the carbon bearing the oxygen. Why other options fail: - (a) and (d): Simple addition of Br2 across the double bond without cyclization; this would require the intramolecular OH nucleophile to be ignored, which is not the case given the favorable 6-membered ring geometry. - (b): Gem-dibromide formation is not expected from simple alkene + Br2; this would require a different mechanism. Therefore, the correct answer is C.