In the Newman projection formula of the lost stable staggered form of `n`-butane, which of the following reasons is the causes of its unstability?

Answer

To solve the question regarding the stability of the staggered form of n-butane in the Newman projection, we can follow these steps:

### Step 1: Understand the Structure of n-Butane
n-Butane is a straight-chain alkane with the molecular formula C4H10. Its structure can be represented as:
\[ \text{CH}_3-\text{CH}_2-\text{CH}_2-\text{CH}_3 \]

### Step 2: Draw the Newman Projection
To analyze the stability of n-butane, we need to draw its Newman projection. In the Newman projection, we look down the bond between two carbon atoms. For n-butane, we can consider the bond between the first (C1) and second (C2) carbon atoms.

- **Front Carbon (C1)**: This carbon will have three hydrogen atoms (H) and one methyl group (CH3).
- **Back Carbon (C2)**: This carbon will also have three hydrogen atoms (H) and one methyl group (CH3).

The staggered conformation will have these groups positioned as far apart as possible.

### Step 3: Identify the Staggered Form
In the staggered form, the bulky methyl groups (CH3) on C1 and C2 will be positioned opposite each other. However, if we consider the eclipsed form, the bulky groups will be aligned with each other, leading to increased steric hindrance.

### Step 4: Analyze the Stability
In the staggered form, the two bulky methyl groups are positioned 180 degrees apart, which minimizes steric hindrance. However, in the target (eclipsed) form, these bulky groups are aligned, causing significant steric strain due to their proximity.

### Step 5: Conclusion on Instability
The instability of the target form of n-butane can be attributed to **van der Waals strain**. This strain arises from the repulsion between the electron clouds of the bulky methyl groups when they are in close proximity to each other.

### Final Answer
The reason for the instability of the target Newman projection of n-butane is due to **van der Waals strain**.