Which of the following reactions will not give four-membered cyclic compound ?

Options

A. a.
B. b.
C.
D.

Answer

To determine which of the given reactions will not yield a four-membered cyclic compound, we will analyze each option step by step.

### Step 1: Analyze Option A
- The reaction involves **BrCH2(CH2)4Br** treated with **Mg**.
- This reaction leads to the formation of a Grignard reagent: **MgBr(CH2)4MgBr**.
- Since the Grignard reagent does not involve the formation of a cyclic compound, **Option A does not produce a four-membered cyclic compound**.

### Step 2: Analyze Option B
- The reaction involves a compound that undergoes **homolytic bond fission** when exposed to light.
- This results in the formation of free radicals, which subsequently combine to form a bond.
- When treated with **H2** in the presence of **nickel**, a four-membered cyclic compound is formed.
- Therefore, **Option B does produce a four-membered cyclic compound**.

### Step 3: Analyze Option C
- The reaction involves **two molecules of ethene**.
- When exposed to light, these molecules undergo **homolytic bond fission**, forming free radicals.
- The free radicals from both ethene molecules combine to form a bond, resulting in a cyclobutane ring.
- Thus, **Option C also produces a four-membered cyclic compound**.

### Step 4: Analyze Option D
- The reaction is a **Wurtz reaction**, where **Br + 2Na** leads to the formation of a dimer.
- In this case, the reaction does not yield a cyclic compound but rather a linear alkane.
- Therefore, **Option D does not produce a four-membered cyclic compound**.

### Conclusion
After analyzing all options:
- **Option A** and **Option D** do not yield four-membered cyclic compounds.
- The correct answer to the question is **Option A**.