Which of the following decolourises solution?

Options

A. `C Cl_4`
B. `CH_4`
C. `C_2H_6`
D. `(CH_3)_3 CH`

Explanation

To determine which of the given compounds decolorizes a solution, we need to analyze the compounds in the context of their reactions with potassium permanganate (KMnO4), a common oxidizing agent known for its purple color.

### Step-by-Step Solution:

1. **Identify the Compounds**: The compounds provided are:
- CCl4 (Carbon Tetrachloride)
- CH4 (Methane)
- C2S6 (Dithiane)
- CH3 whole thrice CH (tert-Butyl group or tert-Butane)

2. **Understanding Decolorization**: Decolorization refers to the process where a colored solution loses its color, which in this case is due to the reduction of KMnO4. KMnO4 is purple due to the presence of MnO4^- ions, which can be reduced to Mn^2+ ions, a colorless species.

3. **Reaction with KMnO4**:
- **CCl4**: This compound is inert and does not react with KMnO4. Therefore, it does not cause decolorization.
- **CH4**: Methane is also a saturated hydrocarbon and does not react with KMnO4 under normal conditions. It does not cause decolorization.
- **C2S6**: This compound is not a typical hydrocarbon and does not have the necessary reactive sites to interact with KMnO4 effectively. It is unlikely to cause decolorization.
- **CH3 whole thrice CH (tert-Butane)**: This compound can react with KMnO4 in the presence of oxygen. The reaction leads to the formation of alcohols, which are colorless. During this reaction, KMnO4 is reduced to Mn^2+, resulting in the loss of the purple color.

4. **Conclusion**: Among the given options, only tert-Butane (CH3 whole thrice CH) can decolorize the KMnO4 solution by reacting with it and reducing the manganese from +7 to +2 oxidation state.

### Final Answer:
The compound that decolorizes the solution is **CH3 whole thrice CH (tert-Butane)**.

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