Which of the following compounds is optically active?

Options

A.
B.
C.
D.

Explanation

To determine which of the given compounds is optically active, we need to identify if any of them are chiral. A compound is chiral if it does not have any plane of symmetry. Here’s a step-by-step solution:

### Step 1: Understand the Concept of Chirality
- **Chirality**: A molecule is chiral if it cannot be superimposed on its mirror image. This typically occurs when a molecule has four different substituents attached to a carbon atom, leading to non-superimposable mirror images.
- **Plane of Symmetry**: A plane that divides a molecule into two mirror-image halves. If a molecule has a plane of symmetry, it is achiral.

### Step 2: Analyze Each Compound
- **Compound A**:
- It has chlorine (Cl) and bromine (Br) on wedge bonds.
- If we draw a plane that bisects the molecule, we find that the two halves are mirror images of each other.
- **Conclusion**: It has a plane of symmetry and is achiral.

- **Compound B**:
- Similar to Compound A, it has Cl on a wedge and Br on a dash.
- A plane can also be drawn that bisects this molecule, resulting in two halves that are mirror images.
- **Conclusion**: It has a plane of symmetry and is achiral.

- **Compound C**:
- This compound has two Cl atoms on wedge bonds.
- A plane can be drawn that bisects the molecule, reflecting one half onto the other.
- **Conclusion**: It has a plane of symmetry and is achiral.

- **Compound D**:
- This is a cyclobutane with Cl on a wedge and Cl on a hash.
- No plane can be drawn that divides the molecule into two mirror-image halves.
- **Conclusion**: It does not have a plane of symmetry and is chiral.

### Step 3: Identify the Optically Active Compound
- Based on the analysis, **Compound D** is the only compound that is chiral and therefore optically active.

### Final Answer
- **The correct answer is Compound D.**