Which of the following can show geometrical isomerism

Options

A. 2,3-dimethyl butane
B. 2,3-dimethyl-2-butene
C. 1,2-dimethyl cyclo hexane
D. 2-methylpentane

Explanation

To determine which of the given compounds can show geometrical isomerism (cis-trans or E-Z isomerism), we need to analyze each option based on the presence of a double bond or a ring structure, as these are typically required for geometrical isomerism to occur.

### Step-by-Step Solution:

1. **Understanding Geometrical Isomerism**:
- Geometrical isomerism occurs when there is restricted rotation around a double bond or within a cyclic structure.
- For geometrical isomers to exist, there must be two different groups attached to each carbon involved in the double bond or ring.

2. **Analyzing the Options**:
- **Option 1: 2,3-Dimethylbutane**:
- This compound is a straight-chain alkane with no double bonds or rings.
- Since it is a simple alkane, it cannot show geometrical isomerism.
- **Conclusion**: This option does not show geometrical isomerism.

- **Option 2: 2,2-Dimethylbutane**:
- This compound has a double bond between the second and third carbon atoms, but both carbons involved have identical groups attached (two methyl groups).
- For geometrical isomerism, the groups attached to the double-bonded carbons must be different.
- **Conclusion**: This option does not show geometrical isomerism.

- **Option 3: 1,2-Dimethylcyclohexane**:
- This compound is a cyclic structure with two methyl groups on adjacent carbons.
- The two methyl groups can be arranged in different spatial orientations (cis or trans).
- If both methyl groups are on the same side, it is termed "cis"; if they are on opposite sides, it is termed "trans".
- **Conclusion**: This option can show geometrical isomerism.

3. **Final Answer**:
- The compound that can show geometrical isomerism is **1,2-Dimethylcyclohexane**.