A dibromo derivative of an alkane reacts with sodium metal to form an alicyclic hydrocarbon. The derivative is

Options

A. 1,1-dibromopropane
B. 2,2-dibromobutane
C. 1,2-dibromoethane
D. 1,4-dibromobutane

Explanation

To determine the dibromo derivative of an alkane that reacts with sodium metal to form an alicyclic hydrocarbon, we can follow these steps:

### Step 1: Understand the Reaction
The reaction in question involves a dibromo derivative of an alkane reacting with sodium metal. This type of reaction is known as the Wurtz reaction, where two alkyl halides react with sodium to form a new carbon-carbon bond, leading to the formation of a cyclic compound.

### Step 2: Identify the Structure of the Dibromo Derivative
A common dibromo derivative that can undergo this reaction is 1,4-dibromobutane. Its structure can be represented as:
```
Br Br
| |
H2C-CH2-CH2-CH2
```
This compound has bromine atoms attached to the first and fourth carbon atoms of a four-carbon chain.

### Step 3: Predict the Product
When 1,4-dibromobutane reacts with sodium, it will lead to the formation of cyclobutane. The sodium will facilitate the removal of the bromine atoms, allowing the carbon atoms to bond together and form a ring structure.

### Step 4: Write the Reaction
The reaction can be summarized as follows:
\[ \text{2 CH}_2\text{Br-CH}_2\text{CH}_2\text{CH}_2\text{Br} + 2 \text{Na} \rightarrow \text{C}_4\text{H}_8 + 2 \text{NaBr} \]
Where \( \text{C}_4\text{H}_8 \) is cyclobutane.

### Conclusion
Thus, the dibromo derivative of an alkane that reacts with sodium metal to form an alicyclic hydrocarbon is 1,4-dibromobutane.

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