A gas decolourises `KMnO_4` solution but gives no precipitate with ammoniacal cuprous chloride is

Options

A. Ethane
B. Methane
C. Ethene
D. Acetylene

Explanation

To solve the problem, we need to analyze the properties of the gas that decolorizes KMnO4 solution and does not give a precipitate with ammoniacal cuprous chloride.

### Step-by-Step Solution:

1. **Understanding the Reaction with KMnO4**:
- KMnO4 (potassium permanganate) is a strong oxidizing agent and is purple in color. When a compound decolorizes KMnO4, it indicates that the compound is likely an alkene or alkyne, as these compounds can undergo oxidation reactions with KMnO4, leading to the formation of diols (alcohols) and the reduction of KMnO4, which results in the loss of its purple color.

2. **Identifying the Type of Hydrocarbon**:
- Alkenes and alkynes can both decolorize KMnO4. However, we need to determine which specific type of hydrocarbon is involved.
- Alkenes have at least one carbon-carbon double bond (C=C), while alkynes have at least one carbon-carbon triple bond (C≡C).

3. **Analyzing the Reaction with Ammoniacal Cuprous Chloride**:
- Ammoniacal cuprous chloride (Cu2Cl2 in the presence of NH4OH) is used to test for terminal alkynes. Terminal alkynes (like acetylene, C2H2) will react with this reagent to form a precipitate (copper acetylide).
- Since the gas does not give a precipitate with ammoniacal cuprous chloride, it indicates that the gas is not a terminal alkyne.

4. **Conclusion**:
- Given that the gas decolorizes KMnO4 but does not form a precipitate with ammoniacal cuprous chloride, it suggests that the gas is likely an internal alkyne (not a terminal alkyne) or an alkene.
- A common example of such a gas is ethyne (acetylene), which is a terminal alkyne, but since it does not form a precipitate, we can conclude that the gas must be an internal alkyne, such as 2-butyne or 3-heptyne.

### Final Answer:
The gas that decolorizes KMnO4 solution but gives no precipitate with ammoniacal cuprous chloride is an internal alkyne.