See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Step 1 - Concept: This problem involves a sequence of reactions starting from trans-stilbene (Ph-CH=CH-Ph). Step 2 - Reaction A (Br2/CCl4): Addition of Br2 across the double bond of Ph-CH=CH-Ph gives the vicinal dibromide Ph-CHBr-CHBr-Ph (meso or dl depending on stereochemistry, but the key product is the dibromide A). Step 3 - Reaction B (2NaNH2): Treatment of the vicinal dibromide with 2 equivalents of NaNH2 (a strong base) causes double dehydrohalogenation. The first equivalent removes HBr to give a vinyl bromide (Ph-CBr=CH-Ph or Ph-CH=CBr-Ph), and the second equivalent removes another HBr to give the internal alkyne Ph-C≡C-Ph (diphenylacetylene). This is product B. Step 4 - Reaction C (H2/Pd/CaCO3): Pd/CaCO3 is the Lindlar's catalyst (partially poisoned palladium). Hydrogenation with H2 over Lindlar's catalyst performs a syn (cis) addition of H2 across the triple bond of Ph-C≡C-Ph, giving the cis-alkene (Z)-stilbene: Ph(H)C=C(H)Ph where both Ph groups are on the same side, i.e., cis configuration. This corresponds to option (b): Ph/H on one carbon and Ph/H on the other carbon in the cis arrangement. Step 5 - Why other options fail: - Option (a) would be trans-stilbene, which would result from anti addition (Birch/dissolving metal conditions), not Lindlar's catalyst. - Option (c) Ph2C=CH2 is a different connectivity entirely, not formed here. - Option (d) Ph-C≡C-Ph is product B, not C. Lindlar's catalyst gives syn addition, producing cis-stilbene, which is option (b). Therefore, the correct answer is B.