In which of the following compounds is there the least delocalization of the lone pair of electrons on the nitrogen atom?

Options

A.
B.
C.
D.

Explanation

To determine which compound has the least delocalization of the lone pair of electrons on the nitrogen atom, we will analyze the options provided.

### Step-by-Step Solution:

1. **Identify the Compounds**:
We have four options to consider, which are different types of amines:
- Option A: 1° amine
- Option B: 2° amine
- Option C: 3° amine
- Option D: 1° amine (this appears to be a repetition of option A)

2. **Understand Delocalization**:
Delocalization refers to the spreading out of electrons across multiple atoms or bonds. In the case of nitrogen in amines, the lone pair of electrons can participate in resonance with adjacent π systems (double bonds or aromatic rings).

3. **Assess Resonance and Steric Hindrance**:
- **1° Amines (Options A and D)**: The lone pair on nitrogen can participate in resonance effectively since there are fewer bulky groups around the nitrogen.
- **2° Amines (Option B)**: The lone pair can still participate in resonance, but there is slightly more steric hindrance due to the presence of two alkyl groups.
- **3° Amines (Option C)**: The lone pair on nitrogen is hindered by three bulky alkyl groups. This steric hindrance causes the nitrogen atom to adopt a non-planar geometry, which reduces the ability of the lone pair to participate in resonance.

4. **Conclusion**:
The compound with the least delocalization of the lone pair of electrons on the nitrogen atom is the **3° amine (Option C)**. This is due to the significant steric hindrance that prevents the lone pair from being involved in resonance effectively.

### Final Answer:
The compound with the least delocalization of the lone pair of electrons on the nitrogen atom is **Option C (3° amine)**.