See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the starting material: The zigzag structure shown is but-2-ene (CH3-CH=CH-CH3). Step 2 - Reaction with Br2/CCl4: Addition of Br2 across the double bond of but-2-ene gives 2,3-dibromobutane (CH3-CHBr-CHBr-CH3). This is a vicinal dibromide. Step 3 - Reaction with alc. KOH (step i): Alcoholic KOH causes E2 elimination of one molecule of HBr from the vicinal dibromide. This gives a vinyl bromide, specifically 2-bromobutene (CH3-CBr=CH-CH3 or CH3-CH=CBr-CH3). The major product of elimination from 2,3-dibromobutane is 2-bromobutene (an internal vinyl bromide). Step 4 - Reaction with NaNH2 (step ii): NaNH2 is a very strong base that causes dehydrohalogenation (elimination of HBr) from the vinyl bromide. Elimination from an internal vinyl bromide (CH3-CBr=CH-CH3) with NaNH2 follows the rule that the base removes a proton from the less substituted carbon adjacent to the C=C, giving an internal alkyne. The product is but-2-yne: CH3-C≡C-CH3. Step 5 - Why the major product is but-2-yne (option B): The two-step elimination sequence (vicinal dibromide → vinyl bromide → alkyne) starting from but-2-ene (an internal alkene) produces an internal alkyne. Both eliminations proceed to give the thermodynamically and stereoelectronically preferred internal triple bond, yielding CH3-C≡C-CH3 as the major product. Why other options fail: - (a) 1,3-butadiene: would require a very different reaction pathway involving diene formation, not consistent with these reagents. - (c) 1-butyne (CH3-CH2-C≡CH): would require a terminal alkyne, but starting from an internal alkene (but-2-ene), both eliminations favor the internal alkyne. - (d) CH3-CH=C=CH2 (1,2-butadiene / allene): allenes are not the expected product of this two-step elimination sequence. Therefore, the correct answer is B.