A compound has the formula `C_2HCl_2Br.` The number of non - identical structures that are possible is

Options

A. one
B. two
C. three
D. four

Answer

To determine the number of non-identical structures for the compound with the formula \( C_2HCl_2Br \), we will follow these steps:

### Step 1: Calculate the Degree of Unsaturation
The degree of unsaturation (DU) can be calculated using the formula:
\[ DU = C - \frac{H + X}{2} + 1 \]
where:
- \( C \) = number of carbon atoms
- \( H \) = number of hydrogen atoms
- \( X \) = number of halogen atoms (Cl, Br)

For \( C_2HCl_2Br \):
- \( C = 2 \)
- \( H = 2 \)
- \( X = 2 \) (2 Cl)

Substituting the values:
\[ DU = 2 - \frac{2 + 2}{2} + 1 = 2 - 2 + 1 = 1 \]

### Step 2: Interpret the Degree of Unsaturation
A degree of unsaturation of 1 indicates that there is either a double bond or a ring in the structure.

### Step 3: Construct Possible Structures
1. **First Structure**: A double bond between the two carbon atoms with both chlorine atoms on one carbon.
- Structure: \( C=C(Cl)(Cl) \) with \( H \) and \( Br \) on the other carbon.

2. **Second Structure**: A double bond with one chlorine on each carbon.
- Structure: \( C=C(Cl)(H) \) with \( Cl \) on one carbon and \( Br \) on the other carbon.

3. **Third Structure**: A double bond with the two chlorine atoms on opposite sides (trans configuration).
- Structure: \( C=C(Cl)(Br) \) with \( H \) on one carbon and \( Cl \) on the other carbon.

### Step 4: Identify Non-identical Structures
- The first structure has both chlorines on the same carbon, making it unique.
- The second structure has one chlorine on each carbon, which is also unique.
- The third structure has the chlorines on opposite sides, making it a distinct geometric isomer.

### Conclusion
The total number of non-identical structures possible for the compound \( C_2HCl_2Br \) is **3**.