Concept:Chemical tests allow differentiation between saturated, double-bonded, and terminal triple-bonded hydrocarbons.
Solution:- Baeyer's Test (Alkaline \( \text{KMnO}_4 \)): Decolorization indicates the presence of a pi bond (unsaturation). Both alkenes and alkynes decolorize \( \text{KMnO}_4 \).
- Tollens' Test (Ammoniacal \( \text{AgNO}_3 \)): Formation of a precipitate indicates the presence of an acidic terminal alkyne (like ethyne).
- The unknown gas decolorizes \( \text{KMnO}_4 \) (meaning it is unsaturated) but does NOT form a precipitate with \( \text{AgNO}_3 \) (meaning it is not a terminal alkyne).
- Ethylene (ethene) is an alkene. It is unsaturated (passes Baeyer's test) but lacks acidic protons (fails Tollens' test).
Why other options are incorrect:- Methane and Ethane are saturated alkanes; they fail Baeyer's test entirely (will not decolorize \( \text{KMnO}_4 \)).
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