1-Bromo-3-chlorocyclobutane is treated with two equivalents of Na, in the presence of ether. Which of the following compounds will be formed.

Options

A.
B.
C.
D.

Explanation

To solve the question regarding the reaction of 1-bromo-3-chlorocyclobutane with two equivalents of sodium in the presence of ether, we can follow these steps:

### Step 1: Draw the Structure of 1-Bromo-3-Chlorocyclobutane
1. **Identify the structure**: Cyclobutane is a four-membered carbon ring. The "1-bromo" indicates that a bromine atom is attached to the first carbon of the ring, and "3-chloro" indicates that a chlorine atom is attached to the third carbon.
2. **Draw the structure**: The structure can be represented as follows:
```
Cl
|
C1 - C2
| |
C4 - C3
|
Br
```

### Step 2: Understand the Reaction Mechanism
1. **Wurtz Reaction**: The reaction described is a Wurtz reaction, which involves the coupling of alkyl halides using sodium in dry ether.
2. **Free Radical Formation**: When 1-bromo-3-chlorocyclobutane is treated with sodium, the halide (Br or Cl) leaves, forming free radicals.

### Step 3: Formation of Free Radicals
1. **Halide Elimination**: The bromine and chlorine atoms will be replaced by sodium, forming sodium bromide (NaBr) and sodium chloride (NaCl).
2. **Free Radical Generation**: The elimination of the halides leads to the formation of two free radicals on the cyclobutane ring.

### Step 4: Coupling of Free Radicals
1. **Radical Coupling**: The two free radicals can couple together by sharing their unpaired electrons, forming a new bond.
2. **Final Product**: The resulting compound will be a dimer of cyclobutane, specifically bicyclobutane, as the two radicals combine.

### Step 5: Identify the Correct Compound
1. **Final Structure**: The final product formed from the reaction is bicyclobutane.

### Conclusion
The compound formed from the reaction of 1-bromo-3-chlorocyclobutane with two equivalents of sodium in dry ether is **bicyclobutane**.