An optically active compound is

Options

A. 1 - bromobutane
B. `beta` - bromobutyric acid
C. 2 - bromo - 2- methylpropane
D. 1- bromo - 2 - methylpropane

Answer

To determine which of the given compounds is optically active, we need to follow these steps:

### Step 1: Understand the Definition of Optically Active Compounds
An optically active compound is one that has a chiral center, typically a carbon atom that is sp3 hybridized and bonded to four different groups. This results in non-superimposable mirror images known as enantiomers.

### Step 2: Analyze Each Compound
We will analyze each of the four given compounds to check for chirality.

#### Option A: Bromobutane
1. **Structure**: CH3-CH2-CHBr-CH3 (Bromobutane)
2. **Check for Chirality**: The carbon with the bromine (C3) is attached to two hydrogen atoms and two methyl groups (CH3). Since it has two identical groups (H), it is not chiral.
3. **Conclusion**: Not optically active.

#### Option B: Beta-Bromobutyric Acid
1. **Structure**: CH3-CH(Br)-CH2-COOH (Beta-Bromobutyric Acid)
2. **Check for Chirality**: The highlighted carbon (C2) has four different groups: CH3, H, Br, and CH2COOH.
3. **Conclusion**: This carbon is chiral and thus the compound is optically active.

#### Option C: 2-Bromo-2-methylpropane
1. **Structure**: (CH3)2C(Br)CH3
2. **Check for Chirality**: The carbon with bromine (C2) is bonded to three identical methyl groups (CH3) and one bromine. Since it has three identical groups, it is not chiral.
3. **Conclusion**: Not optically active.

#### Option D: 1-Bromo-2-methylpropane
1. **Structure**: CH3-CH(Br)-CH(CH3)-CH2
2. **Check for Chirality**: The carbon with bromine (C2) is attached to two identical methyl groups (CH3) and two other groups (Br and H). Since it has two identical groups, it is not chiral.
3. **Conclusion**: Not optically active.

### Step 3: Final Conclusion
After analyzing all options, the only compound that is optically active is **Option B: Beta-Bromobutyric Acid**.