A compound 'X', molecular formula `C_4H_8Br_2` on treatment with zinc in an alcoholic solution, forms an alkene Y. Compounds X and Y respectively are

Options

A. 1, 2-dibromobutane and but-2-ene
B. 1, 3-dibromobutane and but-1-ene
C. 1, 2-dibromobutane and but-1-ene
D. 1, 4-dibromobutane and but-2-ene

Answer

To solve the problem, we need to identify the compound 'X' with the molecular formula C₄H₈Br₂ and determine the alkene 'Y' that is formed when it is treated with zinc in an alcoholic solution.

### Step-by-Step Solution:

1. **Identify Compound X**:
- The molecular formula C₄H₈Br₂ suggests that the compound has four carbon atoms and two bromine atoms.
- A common structure for such a compound is 1,2-dibromobutane (1,2-di-bromo-butane), where the bromine atoms are attached to adjacent carbon atoms (vicinal dibromide).

2. **Reaction with Zinc**:
- When 1,2-dibromobutane is treated with zinc in an alcoholic solution, a dehalogenation reaction occurs. This means that the bromine atoms will be removed from the molecule.
- The removal of the two bromine atoms from the adjacent carbon atoms leads to the formation of a double bond between those two carbon atoms.

3. **Formation of Alkene Y**:
- The product formed after the removal of the bromine atoms is but-1-ene. This is because the double bond will form between the first and second carbon atoms, resulting in the structure CH₂=CH-CH₂-CH₃.

4. **Final Identification**:
- Therefore, the compounds X and Y can be identified as:
- Compound X: 1,2-dibromobutane (1,2-di-bromo-butane)
- Compound Y: But-1-ene

### Conclusion:
The compounds X and Y respectively are:
- Compound X: 1,2-dibromobutane
- Compound Y: But-1-ene