The number of optical isomers of the compound `CH_(3)CHBrCHBrCOOH` is

Options

A. `0`
B. `1`
C. `3`
D. `4`

Explanation

To determine the number of optical isomers of the compound \( CH_3CHBrCHBrCOOH \), we can follow these steps:

### Step 1: Identify the Structure
The compound can be represented structurally as follows:
- The compound has a chain of four carbon atoms.
- There are two bromine (Br) substituents on the second and third carbon atoms.
- There is a carboxylic acid group (-COOH) on the last carbon.

### Step 2: Determine Chirality
A carbon atom is considered chiral if it has four different substituents. In this compound:
- The second carbon (C2) has the substituents: H, Br, CH3, and the carbon chain (C3).
- The third carbon (C3) has the substituents: H, Br, C2, and the carboxylic acid group (COOH).

Both C2 and C3 are chiral centers because they each have four different groups attached to them.

### Step 3: Count the Chiral Centers
We have identified that there are 2 chiral centers in the compound (C2 and C3).

### Step 4: Calculate the Number of Optical Isomers
The formula to calculate the number of optical isomers is given by:
\[ \text{Number of optical isomers} = 2^n \]
where \( n \) is the number of chiral centers.

In this case, \( n = 2 \):
\[ \text{Number of optical isomers} = 2^2 = 4 \]

### Conclusion
Thus, the total number of optical isomers for the compound \( CH_3CHBrCHBrCOOH \) is **4**.

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