The number of optically active compounds in the isomers of `C_(4)H_(9)Br` is

Options

A. 1
B. 2
C. 3
D. 4

Explanation

To determine the number of optically active compounds in the isomers of \( C_4H_9Br \), we will follow these steps:

### Step 1: Identify the possible structural isomers of \( C_4H_9Br \)

The molecular formula \( C_4H_9Br \) indicates that we have four carbon atoms, nine hydrogen atoms, and one bromine atom. We can have different structural isomers based on the arrangement of carbon atoms and the position of the bromine atom.

### Step 2: Generate the structural isomers

1. **1-Bromobutane**:
- Structure: \( CH_3-CH_2-CH_2-CH_2Br \)
- This compound does not have any chiral centers.

2. **2-Bromobutane**:
- Structure: \( CH_3-CH(Br)-CH_2-CH_3 \)
- This compound has one chiral center at the second carbon (attached to Br, CH3, H, and CH2), making it optically active.

3. **1-Bromo-2-methylpropane**:
- Structure: \( (CH_3)_2CH-CH_2Br \)
- This compound does not have any chiral centers.

4. **2-Bromo-2-methylpropane**:
- Structure: \( (CH_3)_3CBr \)
- This compound does not have any chiral centers.

### Step 3: Identify chiral centers

A chiral center is a carbon atom that is attached to four different substituents. In our generated isomers:

- **1-Bromobutane**: No chiral centers.
- **2-Bromobutane**: One chiral center (optically active).
- **1-Bromo-2-methylpropane**: No chiral centers.
- **2-Bromo-2-methylpropane**: No chiral centers.

### Step 4: Count the optically active compounds

From the analysis above, we find that only **2-Bromobutane** is optically active due to its chiral center.

### Conclusion

The number of optically active compounds in the isomers of \( C_4H_9Br \) is **1**.

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