See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: This problem involves alkylation of a terminal alkyne using a strong base to form an acetylide anion, followed by an SN2 reaction with an alkyl halide. Step 1 - Reagent A: The terminal alkyne (THP-O-CH2-C≡CH) has a terminal alkynyl proton (pKa ~25). Treatment with NaNH2 (a strong base, pKa of NH3 ~38) deprotonates the terminal alkyne to form the acetylide anion: THP-O-CH2-C≡C⁻ Na⁺. This is common to all four options. Step 2 - Reagent B: The acetylide anion must alkylate with an alkyl halide X-CH2-CH2-CH2-Br to introduce the -CH2-CH2-CH2-Br chain via SN2 attack at the carbon bearing X. The product contains the -C≡C-CH2-CH2-CH2-Br chain, meaning the acetylide attacks the carbon bearing the leaving group X, displacing it. For SN2 reactivity, we need a good leaving group at the site of attack. The leaving group order for SN2 is I > Br > Cl >> F. Option (a) uses Cl (poor leaving group, SN2 slower than I), option (b) uses F (very poor leaving group, essentially no SN2), option (c) uses I (excellent leaving group, best SN2 reactivity), and option (d) uses I on both ends (I-CH2-CH2-CH2-I), which risks double alkylation or other side reactions. Option (c): I-CH2-CH2-CH2-Br — The acetylide anion selectively attacks the carbon bearing iodine (better leaving group, primary carbon) via SN2, displacing I⁻ and forming THP-O-CH2-C≡C-CH2-CH2-CH2-Br. The terminal Br remains intact, giving the desired product. Why other options fail: - (a): Cl is a poorer leaving group than I; reaction is slower and less efficient. - (b): F is a very poor leaving group; SN2 essentially does not occur. - (d): Using I-CH2-CH2-CH2-I risks the product acetylide (after first alkylation) further reacting with another equivalent of the dihalide or the product itself undergoing further reactions, leading to mixtures. Therefore, the correct answer is C.