1-bromo-3-chlorocyclobutane when treated with two equivalents of Na, in the presence of ether which of the following 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-by-Step Solution:

1. **Identify the Reactants**:
- The compound is 1-bromo-3-chlorocyclobutane. It has a bromine atom at the 1-position and a chlorine atom at the 3-position on the cyclobutane ring.

2. **Understand the Reaction Conditions**:
- The reaction involves treating the compound with two equivalents of sodium (Na) in the presence of ether. This suggests a coupling reaction, likely a Wurtz reaction, where sodium is used to couple two halogenated compounds.

3. **Mechanism of the Reaction**:
- Sodium will react with the halogens (bromine and chlorine) to form radicals. The sodium will remove the halogen atoms from the cyclobutane, generating radicals.
- The radicals can then couple together to form a new carbon-carbon bond.

4. **Formation of the Product**:
- In this case, the radicals generated from the removal of bromine and chlorine can couple together. Since there are two halogen atoms, the coupling can lead to the formation of a dimer.
- The product formed will be bicyclo[1.1.0]butane, which is a bicyclic compound formed by the coupling of the two radical centers.

5. **Final Product**:
- Therefore, the final product of the reaction of 1-bromo-3-chlorocyclobutane with two equivalents of sodium in ether will be bicyclo[1.1.0]butane.