See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the starting material: The compound is 2,3-butadien-1-ol derivative — specifically an allene: (CH3)(H)C=C=C(H)[C(CH3)2CH2OH]. The left carbon of the allene bears CH3 and H; the right carbon bears H and the group C(CH3)2CH2OH (a neopentyl-type alcohol substituent: a carbon bearing two methyls and a CH2OH group). Step 2 - Understand allene geometry and reactivity with Br2: Allenes react with Br2 via electrophilic addition. However, in this substrate, the pendant –CH2OH group on the C(CH3)2 unit can participate intramolecularly. The allene has the structure: CH3(H)C=C=C(H)–C(CH3)2–CH2OH. Step 3 - Intramolecular cyclization during bromination: When Br2 reacts with the allene, the bromonium ion forms at one of the double bonds. The hydroxyl group (–CH2OH) can act as an internal nucleophile and attack the bromonium intermediate, forming a cyclic ether (tetrahydropyran or dihydropyran ring). The –OH oxygen attacks to close a six-membered ring, displacing bromide or opening the bromonium ion. Step 4 - Determining which double bond reacts and ring formation: The allene C=C=C: Br2 adds across one of the double bonds generating a vinyl bromide at the remaining double bond position. The oxygen of –CH2OH attacks the activated carbon (bromonium on the terminal allene carbon bearing CH3 and H), forming a six-membered ring: O–CH2–C(CH3)2–C(H)=C(Br)–CH(CH3)–O closure. This produces a dihydropyran ring with a double bond remaining, with Br on the vinylic/allylic position. Step 5 - Identify the product structure as option B: Option (b) shows a six-membered oxygen-containing ring (dihydropyran) with: a double bond in the ring, Br on one of the sp2 carbons (the carbon that also bears H), H on the carbon adjacent to O bearing gem-dimethyl (the C(CH3)2 carbon), and gem-dimethyl (CH3, CH3) on the quaternary carbon next to O. This matches the product of intramolecular trapping of a bromonium ion on the allene by the pendant alcohol oxygen, giving a 2H-dihydropyran with Br at C3 (vinylic position) and the gem-dimethyl group at C2 adjacent to oxygen. Step 6 - Why other options fail: (a) is fully saturated — inconsistent with an allene losing only one degree of unsaturation to give a ring while retaining a double bond. (c) has Br on an sp3 carbon rather than vinylic position, inconsistent with the regiochemistry. (d) lacks the key substituents (no gem-dimethyl clearly placed, no H shown correctly). Therefore, the correct answer is B.