Which can show geometrical isomer ?

Options

A.
B.
C.
D.

Answer

To determine which compound can show geometrical isomerism, we need to analyze the structures of the given compounds. Geometrical isomerism typically occurs in alkenes where there is a double bond, and it is characterized by the presence of different groups attached to the carbon atoms involved in the double bond.

### Step-by-Step Solution:

1. **Identify the Compounds**: The compounds given are:
- 4-methylpent-2-ene
- 4-methylpent-3-ene
- 3-methylpent-2-ene
- 2-methylpent-2-ene

2. **Draw the Structures**:
- **4-methylpent-2-ene**:
- Structure: CH3-CH(CH3)-CH=CH2
- The double bond is between the second and third carbon atoms.
- Both sides of the double bond have the same groups (H and CH3), so it cannot show geometrical isomerism.

- **4-methylpent-3-ene**:
- Structure: CH3-CH2-CH(CH3)=CH2
- The double bond is between the third and fourth carbon atoms.
- The groups attached to the double bond are different (H and CH3 on one side, H and CH2CH3 on the other), allowing for cis and trans isomers.

- **3-methylpent-2-ene**:
- Structure: CH3-CH(CH3)=CH-CH2-CH3
- The double bond is between the second and third carbon atoms.
- The groups on either side of the double bond are not different (both sides have H), so it cannot show geometrical isomerism.

- **2-methylpent-2-ene**:
- Structure: CH3-C(CH3)=C(CH3)-CH3
- The double bond is between the first and second carbon atoms.
- Both sides of the double bond have the same groups (CH3), so it cannot show geometrical isomerism.

3. **Conclusion**:
- The only compound that can show geometrical isomerism is **4-methylpent-3-ene**. It can exist as cis and trans isomers due to the different groups attached to the double bond.

### Final Answer:
The compound that can show geometrical isomerism is **4-methylpent-3-ene**.