`CH_3-CH=underset(Br)underset(|)(C)-underset(Cl)underset(|)(C)-CH-CH_3`
How many geometrical isomers are possible for this compound?

Options

A. 2
B. 3
C. 4
D. 6

Explanation

To determine the number of geometrical isomers for the given compound, we need to analyze its structure and the presence of a double bond. Here’s a step-by-step solution:

### Step 1: Identify the Structure
The compound given is:
\[ CH_3-CH=Br-C(Cl)-C-CH_3 \]

### Step 2: Locate the Double Bond
The double bond is between the second and third carbon atoms (C2 and C3). This is crucial because geometrical isomerism (cis-trans isomerism) occurs around double bonds.

### Step 3: Determine Substituents on the Double Bond
For geometrical isomerism to occur, each carbon in the double bond must have different substituents. In this case:
- C2 has a bromine (Br) and a hydrogen (H).
- C3 has a chlorine (Cl) and a carbon chain (C with CH3).

### Step 4: Check for Cis and Trans Isomers
- **Cis Isomer**: The higher priority groups (Br and Cl) are on the same side of the double bond.
- **Trans Isomer**: The higher priority groups (Br and Cl) are on opposite sides of the double bond.

### Step 5: Count the Geometrical Isomers
Since we have one double bond with different substituents on each carbon, we can form two geometrical isomers:
1. **Cis Isomer**: Br and Cl on the same side.
2. **Trans Isomer**: Br and Cl on opposite sides.

### Conclusion
Thus, the total number of geometrical isomers for the given compound is **2**.

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