How many isomeric halogen derivatives including stereoisomers are possible for `C_(2)H_(2)Br_(2)` ?

Options

A. 2
B. 3
C. 4
D. 5

Answer

To determine the number of isomeric halogen derivatives, including stereoisomers, for the compound \( C_2H_2Br_2 \), we can follow these steps:

### Step 1: Identify the Structure
The molecular formula \( C_2H_2Br_2 \) indicates that we have two carbon atoms, two hydrogen atoms, and two bromine atoms. The structure can be either an alkene or a haloalkane. Given that we have only two carbon atoms, we can form a double bond between them.

### Step 2: Determine Possible Structures
1. **Alkene Structure**: The compound can exist as an alkene with a double bond between the two carbon atoms. The general structure would be:
- \( CH_2=CHBr \) (bromine on one carbon)
- \( CHBr=CHBr \) (bromine on both carbons)

2. **Haloalkane Structures**: We can also consider the saturated haloalkane structures:
- \( CH_2Br-CH_2Br \) (both bromines on the same carbon)
- \( CH_2Br-CH_3 \) (one bromine on one carbon, one hydrogen on the other)

### Step 3: Count Isomers
Now we will count the distinct isomers, including stereoisomers:
1. **For the alkene**:
- **Cis and Trans Isomers**:
- \( CH_2=CHBr \) can exist in two forms (cis and trans) depending on the arrangement of the bromine atoms.

2. **For the haloalkane**:
- **1,2-Dibromopropane**:
- \( CH_2Br-CH_2Br \) (both bromines on the same carbon)
- \( CHBr-CH_3 \) (one bromine on one carbon and one hydrogen on the other)

### Step 4: Total Isomers
1. **Alkene Isomers**: 2 (cis and trans)
2. **Haloalkane Isomers**: 2

### Final Count
Adding these gives us a total of 4 isomers for \( C_2H_2Br_2 \).

### Conclusion
Thus, the total number of isomeric halogen derivatives, including stereoisomers, for \( C_2H_2Br_2 \) is **4**.

---