See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: Terminal alkynes (alkynes with a C≡C-H group) are acidic enough to react with ammoniacal silver nitrate (Tollens' reagent for alkynes) to form a white precipitate of silver acetylide (silver alkynide), whereas internal alkynes, alkanes, and alkenes do not give this reaction. Step 1: Identify compound (3): CH3CH2C≡CH is 1-butyne, a terminal alkyne. It has a hydrogen attached directly to the sp-hybridized carbon of the triple bond (≡C-H). Step 2: Identify the other compounds: - Compound (1): CH3C≡C-CH3 is 2-butyne, an internal alkyne (no terminal ≡C-H). - Compound (2): CH3CH2-CH2-CH3 is butane, an alkane. - Compound (4): CH3CH=CH2 is propene, an alkene. Step 3: Evaluate each reagent: (a) Bromine in CCl4 - decolorizes with both alkynes (1 and 3) and alkene (4), but not with alkane (2). Does not uniquely distinguish compound (3) from compound (1). (b) Bromine in acetic acid - similar to (a), reacts with unsaturated compounds; does not uniquely distinguish terminal alkyne (3) from internal alkyne (1). (c) Alk. KMnO4 - reacts with both alkynes and alkenes (oxidizes them); does not uniquely distinguish compound (3) from (1) or (4). (d) Ammoniacal silver nitrate (AgNO3/NH3) - only terminal alkynes react to form a white/cream precipitate of silver acetylide (RC≡CAg). Compound (3) CH3CH2C≡CH has the terminal ≡C-H and will give a precipitate. Compound (1) is an internal alkyne and will not react. Compound (2) is an alkane and will not react. Compound (4) is an alkene and will not react. Step 4: Therefore, ammoniacal silver nitrate uniquely identifies compound (3) as a terminal alkyne among the four compounds, since only it produces a silver acetylide precipitate. Therefore, the correct answer is D.