Which of the following correctly ranks the halides in increasing order of reactivity towards `CH_(3)O^(ө)Na^(o+) ` in `CH_(3)OH`?

Options

A.
B.
C.
D. all are correct

Answer

To determine the increasing order of reactivity of halides towards `CH3O^-Na^+` in `CH3OH`, we need to analyze the electronic effects of the substituents on the aromatic ring. The reactivity of the halides will depend on the electron-withdrawing or electron-donating nature of the groups attached to the aromatic ring.

### Step-by-Step Solution:

1. **Identify the Halides**: Let's assume we have the following halides for consideration:
- A: `F` (Fluorine)
- B: `Cl` (Chlorine)
- C: `Br` (Bromine)
- D: `I` (Iodine)

2. **Understand the Role of Electron-Withdrawing Groups**: Electron-withdrawing groups (EWGs) decrease the electron density on the aromatic ring, making it more susceptible to nucleophilic attack. Conversely, electron-donating groups (EDGs) increase electron density, making the ring less reactive towards nucleophiles.

3. **Assess the Halides**:
- Fluorine is a strong electron-withdrawing group due to its high electronegativity.
- Chlorine is also an electron-withdrawing group but less so than fluorine.
- Bromine is less electron-withdrawing than chlorine.
- Iodine is the least electron-withdrawing among these halogens.

4. **Rank the Halides**: Based on their electron-withdrawing ability, we can rank the halides in terms of their reactivity towards `CH3O^-`:
- **Most reactive**: `F` (Fluorine) - strongest EWG
- `Cl` (Chlorine) - moderate EWG
- `Br` (Bromine) - weaker EWG
- **Least reactive**: `I` (Iodine) - weakest EWG

5. **Final Order of Reactivity**: Therefore, the increasing order of reactivity towards `CH3O^-Na^+` in `CH3OH` is:
- `I < Br < Cl < F`