An alkyl halide with molecular formula `C_6H_(13)Br` on dehyrohalogenation gives two isomeric alkenes X and Y with molecular formula `C_6H_(12)`. On reductive ozonolysis X and Y gives four compounds `CH_3COCH_3`, `CH_3CH_2CHO` and `(CH_3)_2CHCHO`. The alkyl halide is

Options

A. 4-bromo-2-methylpentane
B. 3-bromo-2-methylpentane
C. 2-bromo-2,3-dimethylbutane
D. 2,2-dimethyl-1-bromobutane

Explanation

To solve the problem, we need to identify the alkyl halide with the molecular formula \( C_6H_{13}Br \) that can produce two isomeric alkenes \( X \) and \( Y \) with the molecular formula \( C_6H_{12} \) upon dehydrohalogenation. These alkenes, upon reductive ozonolysis, yield the compounds \( CH_3COCH_3 \), \( CH_3CH_2CHO \), and \( (CH_3)_2CHCHO \).

### Step-by-step Solution:

1. **Identify the Structure of the Alkyl Halide:**
The alkyl halide has the formula \( C_6H_{13}Br \). This indicates that it is a six-carbon compound with a bromine atom. The general structure can be represented as \( R-Br \), where \( R \) is the alkyl group.

2. **Dehydrohalogenation Reaction:**
Dehydrohalogenation involves the elimination of HBr from the alkyl halide to form alkenes. We need to find an alkyl halide that can produce two different alkenes \( X \) and \( Y \).

3. **Consider Possible Structures:**
Let's consider the structure of 3-bromo-2-methylpentane:
- It has the structure:
```
CH3-CH(CH3)-CHBr-CH2-CH3
```
- Upon dehydrohalogenation, we can remove HBr from two different positions:
- Removing H from the carbon adjacent to the bromine leads to one alkene \( X \).
- Removing H from the other adjacent carbon leads to another alkene \( Y \).

4. **Identify the Alkenes:**
For 3-bromo-2-methylpentane, the possible alkenes formed are:
- \( X: CH3-CH=CH-CH2-CH3 \) (2-methyl-2-pentene)
- \( Y: CH3-CH2-CH=CH-CH3 \) (2-pentene)

5. **Reductive Ozonolysis:**
The alkenes \( X \) and \( Y \) undergo reductive ozonolysis, which cleaves the double bonds and adds oxygen. The products formed are:
- From \( X \): \( CH3COCH3 \) (acetone)
- From \( Y \): \( CH3CH2CHO \) (propionaldehyde) and \( (CH3)_2CHCHO \) (isobutyraldehyde)

6. **Match the Products:**
The products from ozonolysis match the compounds given in the question:
- \( CH3COCH3 \)
- \( CH3CH2CHO \)
- \( (CH3)_2CHCHO \)

7. **Conclusion:**
The alkyl halide that fits all the criteria is **3-bromo-2-methylpentane**.

### Final Answer:
The alkyl halide is **3-bromo-2-methylpentane**.