Statement-1: Butan-2-ol is optically active.
Statement-2. Its mirror image is non-superimposable on it.

Options

A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation of statement-1.
B. Statement-1 is true, statement-2 is true, Statement-2 is not a correct explanation of statement-1.
C. Statement-1 is true, statement-2 is false.
D. Statement-1 is Fasle, Statement-2 is True.

Explanation

To analyze the statements provided in the question, we will evaluate each statement regarding butan-2-ol and its properties.

### Step 1: Understanding Butan-2-ol
Butan-2-ol is a four-carbon alcohol with the hydroxyl (-OH) group attached to the second carbon. The structure can be represented as follows:

```
CH3
|
CH3-CH-CH2-OH
|
H
```

### Step 2: Identifying Chiral Centers
A chiral center (or chiral carbon) is a carbon atom that is bonded to four different substituents. In butan-2-ol, the second carbon (C2) is bonded to:
- A hydroxyl group (-OH)
- A methyl group (CH3)
- An ethyl group (CH2CH3)
- A hydrogen atom (H)

Since C2 is bonded to four different groups, it is indeed a chiral center.

### Step 3: Optical Activity
A compound is optically active if it has at least one chiral center. Since butan-2-ol has a chiral center, it is optically active. This confirms Statement-1: "Butan-2-ol is optically active."

### Step 4: Mirror Image and Non-Superimposability
The mirror image of a chiral compound is non-superimposable, meaning that you cannot align the mirror image with the original compound perfectly. For butan-2-ol, its mirror image will have the same connectivity of atoms but will be arranged in a way that cannot be superimposed onto the original molecule. This confirms Statement-2: "Its mirror image is non-superimposable on it."

### Conclusion
Both statements are true:
- Statement-1 is correct because butan-2-ol is optically active due to the presence of a chiral center.
- Statement-2 is correct because the mirror image of butan-2-ol is non-superimposable.

### Final Answer
Both Statement-1 and Statement-2 are true.

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