Which of the following does not decolourise reddish brown solution of `Br_(2)//CCl_(4)` :

Options

A.
B.
C.
D.

Answer

To determine which of the following compounds does not decolorize the reddish-brown solution of \( \text{Br}_2/\text{CCl}_4 \), we need to analyze the nature of the compounds provided in the options. The reddish-brown color of bromine in carbon tetrachloride (CCl4) is due to the presence of Br2, which is a diatomic molecule. Unsaturated compounds, such as alkenes and alkynes, can react with bromine, leading to the decolorization of the solution.

### Step-by-Step Solution:

1. **Understanding the Reaction**:
- The reddish-brown color of bromine is due to the Br-Br bond. When bromine reacts with unsaturated compounds (alkenes or alkynes), the Br-Br bond breaks, resulting in the formation of colorless products.

2. **Identifying Unsaturated Compounds**:
- Alkenes (C=C) and alkynes (C≡C) are unsaturated and will react with bromine to decolorize the solution.
- Aromatic compounds, despite having double bonds, do not react with bromine in the same way because they maintain their aromaticity.

3. **Analyzing Each Option**:
- **Option 1**: Alkene present - This will decolorize the bromine solution.
- **Option 2**: Alkene present - This will also decolorize the bromine solution.
- **Option 3**: Aromatic compound (e.g., benzene) - This does not react with bromine in CCl4 to decolorize the solution due to its stable aromatic structure.
- **Option 4**: Alkyne present - This will react with bromine and decolorize the solution.

4. **Conclusion**:
- The compound that does not decolorize the reddish-brown solution of \( \text{Br}_2/\text{CCl}_4 \) is the one that is aromatic (Option 3).

### Final Answer:
The compound that does not decolorize the reddish-brown solution of \( \text{Br}_2/\text{CCl}_4 \) is **Option 3** (the aromatic compound).