On bromination, the electron rich phenoxide ion will be attacked most readily

Options

A. on the negatively charged oxygen atom
B. on the ortho and para carbon atoms
C. on the meta carbon atom
D. on the ortho carbon atom.

Answer

To solve the question regarding the bromination of the phenoxide ion, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Phenoxide Ion**:
- The phenoxide ion is derived from phenol (C6H5OH) by removing a hydrogen ion (H+), resulting in C6H5O⁻. This ion has a negative charge on the oxygen atom, which contributes to the electron density of the aromatic ring.

**Hint**: Remember that the phenoxide ion is more reactive than phenol due to the negative charge on the oxygen.

2. **Resonance Structures**:
- The negative charge on the oxygen can participate in resonance with the aromatic ring. This means that the negative charge can delocalize over the ring, enhancing the electron density at certain positions (ortho and para) of the ring.

**Hint**: Draw resonance structures to visualize how the negative charge distributes itself across the aromatic ring.

3. **Electron Density Distribution**:
- The resonance structures show that the ortho and para positions of the phenoxide ion become electron-rich due to the delocalization of the negative charge. This makes these positions more reactive towards electrophilic substitution reactions, such as bromination.

**Hint**: Identify which positions on the ring have increased electron density; these will be the sites for electrophilic attack.

4. **Electrophilic Bromination**:
- In the presence of bromine (Br2) and a Lewis acid catalyst (like AlBr3), the electrophilic bromine (Br⁺) will preferentially attack the ortho or para positions of the phenoxide ion due to the increased electron density.

**Hint**: Recall that electrophiles are attracted to regions of high electron density.

5. **Final Products**:
- The bromination will yield bromophenol, where the bromine atom is attached either at the ortho or para position relative to the hydroxyl (OH) group. After bromination, the product can be hydrolyzed to regenerate the hydroxyl group.

**Hint**: Consider the possible products and their positions after bromination.

6. **Conclusion**:
- Therefore, we conclude that the bromination of the phenoxide ion occurs most readily at the ortho and para positions due to the electron-rich nature of these sites.

**Hint**: Summarize the key points: phenoxide ion is ortho and para directing due to resonance, leading to increased reactivity at these positions.

### Final Answer:
The electron-rich phenoxide ion will be attacked most readily at the ortho and para positions during bromination.