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 compounds X and Y based on the given molecular formula and the reaction conditions. Here’s a step-by-step breakdown of the solution:

### Step 1: Identify Compound X
The molecular formula of compound X is given as C₄H₈Br₂. We need to find possible structures for this compound.

**Possible Structure:**
1. **1,2-Dibromobutane:**
- Structure: CH₃-CHBr-CH₂-CH₂Br

### Step 2: Reaction with Zinc in Alcohol
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, leading to the formation of an alkene.

**Reaction:**
- The two bromine atoms (Br) are eliminated, resulting in the formation of a double bond between the first and second carbon atoms.

### Step 3: Identify Compound Y
After the reaction, the resulting compound will be an alkene.

**Resulting Alkene:**
- The product formed is But-1-ene (CH₃-CH=CH₂-CH₃).

### Step 4: Conclusion
Based on the analysis, we can conclude that:
- Compound X is **1,2-Dibromobutane**.
- Compound Y is **But-1-ene**.

### Final Answer
Compounds X and Y respectively are **1,2-Dibromobutane and But-1-ene**.