Which of the following compounds is an optically active compound

Options

A. 1-butanol
B. 2-butanol
C. 3-pentanol
D. 4-heptanol

Explanation

To determine which of the given compounds is optically active, we need to identify the presence of a chiral center in each compound. A chiral center is typically a carbon atom that is bonded to four different groups. Here’s a step-by-step solution:

### Step 1: Understand the concept of optical activity
Optically active compounds are those that can rotate plane-polarized light. This property arises from the presence of chiral centers in the molecule.

**Hint:** Remember that for a compound to be optically active, it must have at least one chiral center.

### Step 2: Identify the compounds
Assuming we have four compounds to analyze (let's denote them as A, B, C, and D), we will examine each one for chiral centers.

**Hint:** Look for carbon atoms bonded to four different substituents.

### Step 3: Analyze each compound
1. **Compound A (e.g., 2-butanol)**:
- Structure: CH₃-CH(OH)-CH₂-CH₃
- The carbon with the hydroxyl group (OH) is bonded to CH₃, OH, CH₂, and CH₃.
- This carbon is a chiral center because it has four different groups attached.

2. **Compound B (e.g., 2-pentanol)**:
- Structure: CH₃-CH(OH)-CH₂-CH₂-CH₃
- The carbon with the hydroxyl group is bonded to CH₃, OH, CH₂, and CH₂.
- This carbon is not a chiral center because two of the groups (the two CH₂ groups) are the same.

3. **Compound C (e.g., 4-heptanol)**:
- Structure: CH₃-CH₂-CH(OH)-CH₂-CH₂-CH₃
- The carbon with the hydroxyl group is bonded to CH₃, CH₂, OH, and CH₂.
- This carbon is not a chiral center because two of the groups (the two CH₂ groups) are the same.

4. **Compound D (another example)**:
- Structure: CH₃-CH(CH₃)-CH₂-CH₃
- The carbon with the hydroxyl group is bonded to CH₃, CH₃, OH, and CH₂.
- This carbon is not a chiral center because two of the groups (the two CH₃ groups) are the same.

### Step 4: Conclusion
Based on the analysis, only **Compound A (2-butanol)** has a chiral center and is therefore optically active.

**Final Answer:** The optically active compound is **Compound A (2-butanol)**.