An alkyl bromide (X) reacts with sodium in ether to form 4,5-diethyloctane. The compound 'X' is

Options

A. `CH_(3)(CH_(2))_(3)Br`
B. `CH_(3)(CH_(2))_(5)Br`
C. `CH_(2)(CH_(2))_(3)CH(Br)CH_(3)`
D. `CH_(3)-(CH_(2))_(2)-CH(Br)-CH_(2)-CH_(3)`

Explanation

To solve the problem, we need to identify the alkyl bromide (X) that reacts with sodium in ether to form 4,5-diethyloctane. This reaction is a Wurtz reaction, which involves the coupling of two alkyl halides in the presence of sodium.

### Step-by-Step Solution:

1. **Identify the Product:**
The product given is 4,5-diethyloctane. This indicates that the final compound has a total of 8 carbon atoms (octane) and two ethyl groups attached to the 4th and 5th carbon atoms.

2. **Determine the Structure of the Product:**
The structure of 4,5-diethyloctane can be represented as:
```
CH3
|
CH3-CH-CH-CH-CH2-CH2-CH3
|
CH2-CH3
```
This shows that the compound has a straight chain of 8 carbon atoms with ethyl groups on the 4th and 5th carbon.

3. **Determine the Alkyl Bromide (X):**
In a Wurtz reaction, the alkyl bromide (X) must have a structure that allows it to provide the necessary carbon atoms when two molecules of X react with sodium. Since the product has 8 carbon atoms and two ethyl groups, we can deduce that the alkyl bromide should be a compound that can yield these groups.

4. **Construct the Alkyl Bromide:**
To form 4,5-diethyloctane, the alkyl bromide must be:
```
CH3-CH2-CH2-CH2-CH2-CH2-Br
```
This is 1-bromo-hexane (C6H13Br), which has 6 carbon atoms in the straight chain. When two molecules of 1-bromo-hexane react with sodium, they will couple to form the desired product.

5. **Final Answer:**
Therefore, the compound 'X' is **1-bromo-hexane**.