An alkyl chloride produces a single alkene on reaction with sodium ethoxide and ethanol. The alkene further undergoes hydrogenation to yield 2-methylbutane. Identify the alkyl chloride from amongst the following :

Options

A. `ClCH_2CH(CH_3)CH_2CH_3`
B. `ClCH_2CH_2CH_2CH_3`
C. `ClCH_2CH(CH_3)CH_2CH_3`
D. `CH_3C(Cl)(CH_3)CH_2CH_3`

Explanation

To solve the problem, we need to identify the alkyl chloride that produces a single alkene when reacted with sodium ethoxide and ethanol, and that alkene must further yield 2-methylbutane upon hydrogenation.

### Step-by-Step Solution:

1. **Understanding the Reaction**:
- The reaction involves an alkyl chloride reacting with sodium ethoxide (a strong base) in ethanol. This reaction typically leads to dehydrohalogenation, which results in the formation of an alkene.

2. **Identifying the Alkene**:
- The alkene formed must be such that it can undergo hydrogenation to yield 2-methylbutane. The structure of 2-methylbutane is:
```
CH3
|
CH3-CH-CH2-CH3
```

3. **Analyzing the Alkyl Chlorides**:
- We need to analyze the given options for alkyl chlorides. The options are:
1. ClCH2CH(CH3)CH2CH3
2. (another option not specified)
3. ClCH2CH(CH3)CH2CH3 (same as option 1)
4. (another option not specified)

4. **Reaction of Option 1**:
- For the first option, ClCH2CH(CH3)CH2CH3, when sodium ethoxide reacts, it can lead to the removal of HCl and form the alkene:
- The reaction can be represented as:
```
ClCH2-CH(CH3)-CH2-CH3 + NaOEt → CH2=CH(CH3)-CH2-CH3 + NaCl + EtOH
```
- This results in the formation of 2-methylbutene.

5. **Hydrogenation of the Alkene**:
- The alkene (2-methylbutene) can be hydrogenated:
```
CH2=CH(CH3)-CH2-CH3 + H2 → CH3-CH(CH3)-CH2-CH3
```
- This confirms that the product is indeed 2-methylbutane.

6. **Conclusion**:
- Therefore, the alkyl chloride that produces a single alkene (2-methylbutene) upon reaction with sodium ethoxide and ethanol, which further yields 2-methylbutane upon hydrogenation, is:
- **1-chloro-2-methylbutane** (from option 1).

### Final Answer:
The alkyl chloride is **1-chloro-2-methylbutane**.