`CH_(3)-CH_(2)-CH_(2)-CH_(3)`. There is free rotation about `(C_(2)ul(sigma)C_(3))` bond. The same most stable formation is repeated after rotation of

Options

A. `60^(@)`
B. `120^(@)`
C. `240^(@)`
D. `360^(@)`

Explanation

To solve the question regarding the rotation of the n-butane molecule (CH₃-CH₂-CH₂-CH₃) around the C₂-C₃ sigma bond, we can follow these steps:

### Step 1: Understand the Structure of n-Butane
n-Butane has the following structure:
- CH₃ (methyl group) at one end
- CH₂ (methylene group) in the middle
- CH₂ (another methylene group)
- CH₃ (methyl group) at the other end

### Step 2: Identify the Sigma Bond
The question specifies the rotation about the C₂-C₃ bond. This is a sigma bond, which allows for free rotation.

### Step 3: Determine the Most Stable Conformation
In n-butane, the most stable conformation is the anti conformation, where the two CH₃ groups are opposite each other. This minimizes steric hindrance and torsional strain.

### Step 4: Calculate the Rotation Required to Achieve the Same Conformation
To return to the most stable conformation after rotation, we need to consider that:
- Each 60-degree rotation around the C₂-C₃ bond will change the arrangement of the groups.
- After a complete rotation of 360 degrees, the molecule will return to its original conformation.

### Final Answer
The most stable formation is repeated after a rotation of **360 degrees**.

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