Which of the following compounds will be optically active?

Options

A.
B.
C.
D.

Explanation

To determine which of the given compounds is optically active, we need to follow a systematic approach. Here’s a step-by-step solution:

### Step 1: Understand the Conditions for Optical Activity
For a compound to be optically active, it must meet two conditions:
1. **Presence of a Chiral Carbon**: A chiral carbon is typically an sp³ hybridized carbon atom that is bonded to four different groups (or atoms).
2. **Absence of a Plane of Symmetry**: The molecule must not have any plane of symmetry, which would allow it to be superimposed on its mirror image.

### Step 2: Analyze Each Compound
Let’s analyze the compounds one by one:

#### Compound 1:
- **Observation**: Contains a double bond (C=C).
- **Analysis**: The presence of a double bond indicates that at least one carbon is sp² hybridized, which cannot be chiral. Therefore, this compound does not have a chiral carbon.
- **Conclusion**: Not optically active.

#### Compound 2:
- **Observation**: Contains a nitrogen atom and carbon atoms with similar substituents.
- **Analysis**: One of the carbons is bonded to two hydrogen atoms, which means it does not have four different groups. Thus, it does not meet the criteria for chirality.
- **Conclusion**: Not optically active.

#### Compound 3:
- **Observation**: Contains a carbon atom bonded to four different groups.
- **Analysis**: This carbon has four different substituents (e.g., CH3, H, and two other distinct groups). Since there is no plane of symmetry (the groups are arranged in such a way that they cannot be superimposed on their mirror image), this compound is chiral.
- **Conclusion**: Optically active.

#### Compound 4:
- **Observation**: Contains similar groups on the same side.
- **Analysis**: This compound has a plane of symmetry because the groups can be divided into two identical halves. Therefore, it does not meet the criteria for optical activity.
- **Conclusion**: Not optically active.

### Final Answer
Based on the analysis, **Compound 3** is the only compound that is optically active.

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