Which of the following compounds is optically active?

Options

A. 1-butanol
B. Isopropyl alcohol
C. Acetaldehyde
D. 2-butanol

Explanation

To determine which of the given compounds is optically active, we need to identify if any of the compounds contain a chiral carbon atom. A chiral carbon is one that is bonded to four different groups or atoms. Let's analyze each compound step-by-step:

### Step 1: Identify the Compounds
We have four compounds to analyze:
1. **1-butanol** (Compound A)
2. **Isopropyl alcohol** (Compound B)
3. **Acetaldehyde** (Compound C)
4. **2-butanol** (Compound D)

### Step 2: Draw the Structures
- **1-butanol**: CH3-CH2-CH2-CH2OH
- **Isopropyl alcohol**: (CH3)2CHOH
- **Acetaldehyde**: CH3CHO
- **2-butanol**: CH3-CH(OH)-CH2-CH3

### Step 3: Analyze Each Compound for Chirality
1. **1-butanol**:
- The carbon atoms in 1-butanol are either bonded to two hydrogens or are part of a straight chain. None of the carbons are bonded to four different groups. Thus, it is **not optically active**.

2. **Isopropyl alcohol**:
- The carbon atoms in isopropyl alcohol are also bonded to either two hydrogens or are part of a branched structure. There is no carbon bonded to four different groups. Thus, it is **not optically active**.

3. **Acetaldehyde**:
- The carbonyl carbon in acetaldehyde is bonded to a methyl group (CH3), a hydrogen (H), and an oxygen (O). However, it is also bonded to another carbon (the carbonyl group), making it planar and not chiral. Thus, it is **not optically active**.

4. **2-butanol**:
- The second carbon in 2-butanol is bonded to four different groups: a methyl group (CH3), an ethyl group (CH2-CH3), a hydroxyl group (OH), and a hydrogen (H). This carbon is chiral, making 2-butanol **optically active**.

### Conclusion
Based on the analysis, the compound that is optically active is **2-butanol** (Compound D).