See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Step 1: Identify the degree of unsaturation. C4H6 has molecular formula with degree of unsaturation = (2×4 + 2 - 6)/2 = (8+2-6)/2 = 4/2 = 2. This suggests either a triple bond or two double bonds. Step 2: Reaction with excess Br2. (A) reacts with excess Br2 to give C4H6Br4. Starting from C4H6 and adding Br4 (i.e., 4 bromine atoms = 2 Br2 molecules), this means 2 moles of Br2 are added. A triple bond adds 2 moles of Br2 (one across each pi bond), consistent with a compound containing a C≡C triple bond. A single double bond would add only 1 mole of Br2 to give C4H6Br2, not C4H6Br4. This rules out options (c) But-1-ene and (d) But-2-ene. Step 3: White precipitate with ammoniacal silver nitrate (Tollens' reagent for alkynes / Lucas test context). Formation of a white precipitate with ammoniacal silver nitrate solution is characteristic of a terminal alkyne (RC≡CH), which reacts with Ag(NH3)2+ to form a silver acetylide precipitate (AgC≡CR), which is white/pale yellow. But-1-yne (HC≡C-CH2-CH3) is a terminal alkyne and gives this test positive. But-2-yne (CH3-C≡C-CH3) is an internal alkyne and does NOT form a precipitate with ammoniacal silver nitrate because it has no terminal ≡C-H bond. Step 4: Conclusion. Only But-1-yne satisfies both conditions: (i) adds 2 moles of Br2 to give C4H6Br4, and (ii) forms a white precipitate with ammoniacal silver nitrate due to its terminal alkyne hydrogen. Why other options fail: - (b) But-2-yne: internal alkyne, does not give white ppt. with ammoniacal AgNO3. - (c) But-1-ene: only adds 1 Br2, gives C4H6Br2, not C4H6Br4. - (d) But-2-ene: only adds 1 Br2, gives C4H6Br2, not C4H6Br4. Therefore, the correct answer is A.