In which of the following pairs of compounds, will second structure have more contribution to resonance hybrid than first ?

Options

A.
B.
C. `F-underset(NH_(2))underset(|)overset(o+)(C)-OH " & " overset(o+)(F)-underset(NH_(2))underset(|)(C)-OH`
D.

Answer

To determine which of the given pairs of compounds has a second structure that contributes more to the resonance hybrid than the first, we need to analyze the resonance structures based on the placement of charges and the electronegativity of the atoms involved. Here’s a step-by-step solution:

### Step 1: Understand Resonance Structures
Resonance structures are different ways of drawing the same molecule that show the delocalization of electrons. The more stable the resonance structure, the more it contributes to the resonance hybrid. Stability is often influenced by the placement of charges, particularly the electronegativity of the atoms carrying those charges.

### Step 2: Analyze Each Pair of Compounds
1. **First Pair**:
- **Compound 1**: Positive charge on carbon.
- **Compound 2**: Positive charge on electronegative oxygen.
- **Conclusion**: Compound 1 is more stable because positive charge on carbon (less electronegative) is preferred over positive charge on oxygen (more electronegative). Thus, the first compound contributes more to the resonance hybrid.

2. **Second Pair**:
- **Compound 1**: Lone pair on nitrogen.
- **Compound 2**: Positive charge on electronegative nitrogen.
- **Conclusion**: Similar to the first pair, the positive charge on nitrogen (more electronegative) makes compound 2 less stable. Therefore, compound 1 contributes more to the resonance hybrid.

3. **Third Pair**:
- **Compound 1**: Negative charge on electronegative fluorine and positive charge on carbon.
- **Compound 2**: Positive charge on electronegative fluorine and negative charge on carbon.
- **Conclusion**: The second compound is less stable due to the positive charge on fluorine. Thus, compound 1 contributes more to the resonance hybrid.

4. **Fourth Pair**:
- **Compound 1**: Positive charge on electronegative fluorine.
- **Compound 2**: Positive charge on carbon.
- **Conclusion**: Here, the positive charge on carbon (less electronegative) is more stable than on fluorine (more electronegative). Thus, compound 2 contributes more to the resonance hybrid.

### Step 3: Identify the Correct Answer
From the analysis, we find that:
- In pairs A, B, and C, the first compound has more contribution to the resonance hybrid.
- In pair D, the second compound has more contribution to the resonance hybrid.

Thus, the correct answer is **Option D**.