See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: Addition of excess HBr to a terminal alkyne follows Markovnikov's rule for each addition step. Step 1 - Identify the starting material: 3-methylbut-1-yne, CH3-CH(CH3)-C≡CH. It is a terminal alkyne. Step 2 - First equivalent of HBr adds across the triple bond following Markovnikov's rule. The H adds to the terminal carbon (C1, which has more H's) and Br adds to the internal carbon (C2 of the triple bond, i.e., the carbon bearing the isobutyl group). This gives the vinyl bromide: CH3-CH(CH3)-C(Br)=CH2. Step 3 - Second equivalent of HBr adds to the vinyl bromide intermediate again following Markovnikov's rule. The bromine already on the carbon makes that carbon electron-deficient/stabilized carbocation on the adjacent carbon. By Markovnikov addition, H adds to CH2 (the terminal =CH2) and the second Br adds to the carbon already bearing one Br (the more substituted carbon, which can better stabilize positive charge due to the adjacent alkyl group). This gives the gem-dibromide: CH3-CH(CH3)-CBr2-CH3. Step 4 - The product is 2,2-dibromo-3-methylbutane: CH3-CH(CH3)-CBr2-CH3, which corresponds to option (c). Why other options fail: - Option (a) shows a vicinal dibromide (Br on adjacent carbons), which would result from anti-Markovnikov addition or bromine addition, not HBr addition. - Option (b) is only the monoaddition product (intermediate), not the final product with excess HBr. - Option (d) shows gem-dibromide at the terminal carbon, which would require anti-Markovnikov addition in both steps. Therefore, the correct answer is C.