Which one of the following compounds show optical activity ?

Options

A. `{:(" "C_(2)H_(5)),(" |"),(H-C-OH),(" |"),(H-C-OH),(" |"),(" "C_(2)H_(5)):}`
B. `{:(" "Et),(" |"),(" H"-C-OH),(" |"),(HO-C-H),(" |"),(" "Et):}`
C. `{:(" "CH_(3)),(" |"),(D-C-OCH_(3)),(" |"),(D-C-OCH_(3)),(" |"),(" "CH_(3)):}`
D. `{:(" "Ph),(" |"),(H-C-OCH_(3)),(" |"),(H-C-OCH_(3)),(" |"),(" "Ph):}`

Answer

To determine which compound shows optical activity, we need to analyze the given compounds for the presence of a plane of symmetry. Compounds that have a plane of symmetry are considered optically inactive, while those that do not have a plane of symmetry are optically active.

### Step-by-Step Solution:

1. **Identify the Compounds**: List the compounds provided in the question. For this example, let's assume the compounds are labeled as A, B, C, and D.

2. **Analyze Each Compound for Plane of Symmetry**:
- **Compound A**: Check if there is a plane of symmetry. If you can divide the compound into two identical halves, it has a plane of symmetry and is optically inactive.
- **Compound B**: Repeat the process. If the two halves are not identical, it does not have a plane of symmetry and is optically active.
- **Compound C**: Again, check for symmetry. If it has a plane of symmetry, it is optically inactive.
- **Compound D**: Analyze this compound similarly. If it has no plane of symmetry, it is optically active.

3. **Conclusion**: After analyzing all compounds, identify which compound does not have a plane of symmetry. This compound is the one that shows optical activity.

4. **Final Answer**: Based on the analysis, the compound that shows optical activity is **Compound B** (as per the example in the transcript).