Which of the following compounds is optically active ?

Options

A. 1-Bromobutane
B. 2-Bromobutane
C. 1-Bromo-2-methylpropane
D. 2-Bromo-2-methylpropane

Explanation

To determine which of the given compounds is optically active, we need to identify if any of the compounds have a chiral center. 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 Definition of Optical Activity
Optically active compounds are those that can rotate plane-polarized light. This property is exhibited by chiral compounds, which contain at least one chiral carbon atom.

**Hint:** Remember that a chiral carbon must have four different substituents attached to it.

### Step 2: Analyze Each Compound
We will analyze each compound provided in the question to check for chiral centers.

1. **1-Bromobutane (C4H9Br)**:
- Structure: CH3-CH2-CH2-CH2Br
- The carbon atoms in this compound do not have four different groups attached. Therefore, it is not chiral and is optically inactive.

**Hint:** Look for identical groups attached to the carbon atoms.

2. **2-Bromobutane (C4H9Br)**:
- Structure: CH3-CH(Br)-CH2-CH3
- The second carbon (C2) has four different groups: a bromine atom (Br), a methyl group (CH3), an ethyl group (CH2-CH3), and a hydrogen atom (H). Thus, this carbon is chiral.

**Hint:** Identify the groups attached to the carbon to determine chirality.

3. **1-Bromo-2-methylpropane (C5H11Br)**:
- Structure: CH3-CH(Br)-CH(CH3)-CH3
- The carbon with the bromine (C1) has two hydrogen atoms attached, making it not chiral. The second carbon (C2) has two methyl groups attached, which also makes it not chiral.

**Hint:** Check for any carbon with two identical groups.

4. **2-Bromo-2-methylpropane (C5H11Br)**:
- Structure: (CH3)3CBr
- The central carbon has three identical methyl groups and one bromine, which means it is not chiral.

**Hint:** A carbon with three identical groups cannot be chiral.

### Step 3: Conclusion
After analyzing all the compounds, we find that **2-bromobutane** is the only compound that has a chiral carbon and is therefore optically active.

**Final Answer:** The optically active compound is **2-bromobutane**.