A single alkene is produced when an alkyl bromide reacts with sodium ethoxide and ethanol. This alkene undergeoes hydrogenation and produces 2 - methyl butane . What is the identity of the alkyl bromide ?

Options

A. 1 - bromo - 2, 2 - dimethylpropane
B. 1 - bromobutane
C. 1 - bromo - 2, methylbutane
D. 2 - bromo - 2, methylbutane

Answer

To solve the problem, we need to identify the alkyl bromide that, when reacted with sodium ethoxide and ethanol, produces a single alkene that upon hydrogenation yields 2-methylbutane. Let's go through the options step by step.

### Step 1: Analyze the Reaction
The reaction involves an alkyl bromide reacting with sodium ethoxide in ethanol. Sodium ethoxide acts as a base, promoting elimination reactions (E2) if beta hydrogens are present.

### Step 2: Evaluate Each Option

**Option 1: 1-bromo-2,2-dimethylpropane**
- Structure:
```
CH3
|
CH3-C-CH2-Br
|
CH3
```
- This compound has no beta hydrogens available for elimination. Therefore, it cannot form an alkene and will undergo substitution instead (SN2), producing an ether.
- **Conclusion**: This option is incorrect.

**Option 2: 1-bromobutane**
- Structure:
```
CH3-CH2-CH2-CH2-Br
```
- This compound has beta hydrogens. The reaction with sodium ethoxide will lead to elimination, producing butene:
```
CH2=CH-CH2-CH3 (butene)
```
- Hydrogenation of butene yields butane, not 2-methylbutane.
- **Conclusion**: This option is incorrect.

**Option 3: 1-bromo-2-methylbutane**
- Structure:
```
CH3
|
CH3-CH-CH2-Br
|
CH3
```
- This compound has beta hydrogens. The reaction with sodium ethoxide will lead to elimination, producing:
```
CH3-CH=CH-CH3 (2-methylbutene)
```
- Hydrogenation of 2-methylbutene yields 2-methylbutane.
- **Conclusion**: This option is correct.

**Option 4: 2-bromo-2-methylbutane**
- Structure:
```
CH3
|
CH3-C-CH2-Br
|
CH3
```
- This compound has beta hydrogens. The reaction with sodium ethoxide can lead to two possible elimination products:
- One alkene from the elimination of one beta hydrogen.
- Another alkene from the elimination of the other beta hydrogen.
- This results in two different alkenes, which contradicts the requirement for a single alkene.
- **Conclusion**: This option is incorrect.

### Final Answer
The alkyl bromide that produces a single alkene upon reaction with sodium ethoxide and ethanol, which then yields 2-methylbutane upon hydrogenation, is **1-bromo-2-methylbutane**.