See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Concept: Sodium acetylide (HC≡CNa) is a strong nucleophile. When it reacts with a dihaloalkane or a molecule bearing two different leaving groups, it will selectively attack the more reactive (better leaving group) site via SN2. Step 1: Identify the reagents. HC≡CNa is sodium acetylide, a carbanion nucleophile. The substrate is Cl–CH2–CH2–CH2–I, which has both a chloro group and an iodo group. Step 2: Determine selectivity. In SN2 reactions, iodide (I⁻) is a much better leaving group than chloride (Cl⁻) because C–I bond is weaker and iodide is more stable as a leaving group. Therefore, the acetylide nucleophile attacks the carbon bearing iodine (the primary carbon at the end of the chain). Step 3: Write the reaction. HC≡C⁻ attacks the –CH2–I end, displacing I⁻ to give: H–C≡C–CH2–CH2–CH2–Cl The chloride end remains intact because it is a poorer leaving group and is not attacked under these conditions. Step 4: Evaluate options. (a) H–C≡C–CH2–CH2–CH2–I: This would require displacement of Cl, which is less reactive — not the major product. (b) CH2=CH–CH2–I: This is allyl iodide, which is unrelated to this reaction mechanism. (c) H–C≡C–CH2–CH2–CH2–Cl: This results from displacement of I⁻ (better leaving group) by the acetylide — correct major product. (d) CH2=CH–CH2–Cl: Allyl chloride, unrelated to this reaction. Therefore, the correct answer is C.