Ease of ionization to produce carbocation and bromide ion under the treatment of `Ag^(o+)` will be maximum in which of the following compounds?

Options

A.
B.
C.
D.

Answer

To determine which compound will have the maximum ease of ionization to produce a carbocation and bromide ion when treated with `Ag^(o+)`, we need to consider the stability of the resulting carbocation formed after the ionization process.

### Step-by-Step Solution:

1. **Identify the Compounds**: We need to analyze the given compounds (let's denote them as A, B, C, and D) to see how they will react with `Ag^(o+)` and what kind of carbocation will be formed.

2. **Understanding Ionization**: When `Ag^(o+)` is added, it will interact with bromide ions (Br^-) in the compounds. The ionization process will lead to the formation of a carbocation. The stability of this carbocation is crucial for determining which compound ionizes more easily.

3. **Analyzing Compound A**: In compound A, if we add `Ag^(o+)`, it will lead to the formation of a carbocation. We need to check the stability of this carbocation. If it has stabilizing groups (like alkyl groups), it will be more stable.

4. **Analyzing Compound C**: In compound C, there is a nitrogen atom involved. When `Ag^(o+)` is added, the lone pair on nitrogen can stabilize the carbocation formed. However, the presence of a phenyl (pH) ring can affect the stability due to resonance effects.

5. **Analyzing Compound D**: In compound D, the nitrogen is attached to a methyl group (CH3). The methyl group can donate electron density through the inductive effect, which can stabilize the carbocation formed.

6. **Comparing Stability**:
- Compound A: The stability of the carbocation depends on the groups attached. If they are not strong electron-donating groups, it may not be very stable.
- Compound C: The lone pair on nitrogen can stabilize the carbocation, but the phenyl ring may hinder the stabilization due to resonance.
- Compound D: The methyl group provides electron density, enhancing the stability of the carbocation.

7. **Conclusion**: After analyzing the stability of the carbocations formed from each compound, we find that compound D has the most stable carbocation due to the +I effect of the methyl group, which makes it easier to ionize. Therefore, the ease of ionization to produce a carbocation and bromide ion under the treatment of `Ag^(o+)` will be maximum in **compound D**.

### Final Answer:
The ease of ionization to produce carbocation and bromide ion under the treatment of `Ag^(o+)` will be maximum in **compound D**.