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 will follow these steps:

### Step 1: Identify the Reactants
We have 1-bromo-3-chlorocyclobutane as our starting material. This compound contains both bromine (Br) and chlorine (Cl) substituents on the cyclobutane ring.

### Step 2: Understand the Reaction Conditions
The reaction is carried out using sodium (Na) in the presence of ether. Sodium is a strong reducing agent and can facilitate the removal of halogens through a process known as the Wurtz reaction.

### Step 3: Mechanism of the Reaction
In the presence of sodium, the halogen atoms (Br and Cl) can be removed from the cyclobutane ring. The sodium will react with the halogens to form sodium halides (NaBr and NaCl).

### Step 4: Formation of the Product
Since we have two equivalents of sodium, both halogens can be removed. The removal of both Br and Cl from 1-bromo-3-chlorocyclobutane will lead to the formation of a cyclobutane dimer.

### Step 5: Final Product
The product formed will be bicyclo[1.1.0]butane, which is a bicyclic compound resulting from the coupling of the two ends of the cyclobutane ring after the halogens have been eliminated.

### Summary of the Reaction
1-bromo-3-chlorocyclobutane + 2 Na → bicyclo[1.1.0]butane + 2 NaX (where X is either Br or Cl)

### Final Answer
The final product formed is bicyclo[1.1.0]butane.

---