See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Step 1 (Concept): Terminal alkynes are weakly acidic (pKa ~25). A strong base like LiNH2 (pKa of conjugate acid ~38) deprotonates the terminal alkyne to form a lithium acetylide (carbanion/organolithium salt). Step 2 (Formation of A): CH3 — CH2 — CH2 — C≡CH + LiNH2 → CH3 — CH2 — CH2 — C≡C⁻ Li⁺ + NH3. So compound (A) is the lithium acetylide: CH3 — (CH2)2 — C≡C — Li. Step 3 (Reaction with (CH3)2SO4): Dimethyl sulfate, (CH3)2SO4, is a methylating agent. It donates a methyl group (CH3) to the nucleophilic acetylide carbon. The acetylide anion attacks the electrophilic methyl group of dimethyl sulfate via an SN2 reaction. Step 4 (Formation of B): CH3 — (CH2)2 — C≡C⁻ Li⁺ + (CH3)2SO4 → CH3 — (CH2)2 — C≡C — CH3 + LiOSO3CH3. Compound (B) is therefore CH3 — (CH2)2 — C≡C — CH3, which is 2-heptyne (pent-2-yne extended: hex-2-yne). Step 5 (Why other options fail): - Option (a): SO3H group cannot come from (CH3)2SO4 methylation; this would require sulfonation chemistry. - Option (c): CH2 — O — S(=O)(=O) — H is a sulfonate ester/acid fragment, not the product of simple C-alkylation by dimethyl sulfate on an acetylide. - Option (d): CH3 — CH2 — C≡C — CH2 is an incomplete structure (carbene/carbanion) and does not represent a stable neutral molecule; also the carbon count and connectivity are wrong. Option (b) correctly represents the product of methylation of the terminal acetylide carbon, giving an internal alkyne with a methyl group added at the terminal carbon. Therefore, the correct answer is B.