See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: When a terminal alkyne reacts with a methoxide nucleophile (MeO⁻) in methanol (MeOH), the methoxide acts as a nucleophile and adds across the triple bond via a conjugate/nucleophilic addition mechanism. This is a nucleophilic addition to an activated alkyne (the phenyl group activates the beta carbon toward nucleophilic addition). Step 1 - Identify the reaction type: Ph—C≡CH with MeO⁻/MeOH is a nucleophilic addition of methoxide to a phenylacetylene. The phenyl group stabilizes negative charge at the alpha carbon (adjacent to Ph), making the terminal carbon (beta to Ph) electrophilic enough for nucleophilic attack under basic conditions. Step 2 - Regioselectivity: MeO⁻ attacks the terminal carbon (the less substituted end, CH), since the resulting vinyl carbanion is stabilized by the adjacent phenyl group (benzylic/vinylic carbanion stabilization). This gives the vinyl anion Ph—C⁻=CH(OMe), which is then protonated by MeOH. Step 3 - Stereochemistry: The addition of MeO⁻ to alkynes under these conditions proceeds predominantly in an anti fashion (trans addition), giving the (E)-configuration where Ph and OMe are on opposite sides, i.e., Ph and H are on one carbon and OMe and H are on the other carbon with trans geometry. This corresponds to option (b): Ph and H on the left sp2 carbon, OMe and H on the right sp2 carbon. Step 4 - Why other options fail: - Option (a): This would be styrene or a reduced product without OMe incorporation; MeO⁻ is a nucleophile, not a reducing agent. - Option (c): Ph—C≡C—OMe would require substitution at the terminal alkynyl C–H bond (replacing H with OMe), which requires different conditions (e.g., strong base + electrophilic oxygen source), not simple MeO⁻/MeOH addition. - Option (d): Ph—C(OMe)=CH₂ would result from Markovnikov-type addition (OMe to internal carbon), which is not favored under these nucleophilic addition conditions; nucleophile attacks the terminal carbon. Thus the major product is the vinyl ether with Ph and H on one carbon and OMe and H on the other (option b), formed by anti-addition of MeOH across the triple bond with nucleophilic regioselectivity. Therefore, the correct answer is B.