See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: Terminal alkynes can be deprotonated by strong bases like NaNH2 to form acetylide anions, which are excellent nucleophiles that undergo SN2 reactions with primary alkyl halides. Step 1 - First deprotonation and alkylation: Ethyne (HC≡CH) reacts with NaNH2 (excess) to form the acetylide anion HC≡C⁻ (sodium acetylide). This acetylide anion then reacts with I-CH2-(CH2)2-CH3 (1-iodobutane, a primary alkyl halide) via SN2, attaching the butyl group to one end of the triple bond. This gives HC≡C-CH2-(CH2)2-CH3, which is 1-hexyne (a terminal alkyne with 6 carbons total). Step 2 - Second deprotonation and alkylation: Since NaNH2 is in excess and 1-hexyne is a terminal alkyne, the remaining terminal C-H (pKa ~25) is again deprotonated by excess NaNH2 to give the acetylide anion ⁻C≡C-CH2-(CH2)2-CH3. This anion reacts again with another equivalent of I-CH2-(CH2)2-CH3 (1-iodobutane) via SN2, attaching a second butyl group to the other end of the triple bond. This gives CH3-(CH2)2-CH2-C≡C-CH2-(CH2)2-CH3. Step 3 - Protonation: H⁺ simply protonates any remaining anion (workup), leaving the product unchanged as CH3(CH2)3-C≡C-(CH2)3CH3. Product identification: The product is butyl-C≡C-butyl, i.e., a symmetrical internal alkyne with 4 carbons on each side of the triple bond. Counting: C4-C≡C-C4 = 10 carbons total. Numbering from one end: C1-C2-C3-C4-C5≡C6-C7-C8-C9-C10, so the triple bond is at C5-C6, giving 5-decyne. Why other options fail: - (a) 6-iodo-1-hexyne: No iodide would remain in the product; the iodide is displaced in the SN2 reaction. - (b) 1-hexyne: This is only the intermediate after the first alkylation; since NaNH2 and the alkyl iodide are both in excess, the reaction proceeds to double alkylation. - (d) 1-iodo-1-hexene: This is not consistent with any step in the mechanism; no reduction or addition to give a vinyl iodide occurs here. Therefore, the correct answer is C.