Identify Intermediates in Bromination of Phenol Derivative

In the presence of an aqueous solution of sodium hydroxide ($\text{NaOH}$), phenol ($\text{C}_6\text{H}_5\text{OH}$) is deprotonated to yield the phenoxide ion ($\text{C}_6\text{H}_5\text{O}^-$):

$\text{C}_6\text{H}_5\text{OH} + \text{OH}^- \rightleftharpoons \text{C}_6\text{H}_5\text{O}^- + \text{H}_2\text{O}$

The phenoxide ion ($-\text{O}^-$) is an extremely strong activating group due to its powerful resonance electron-donating effect ($+M$ effect), which dramatically increases the electron density at the ortho and para positions relative to the meta position.

When bromine ($\text{Br}_2$) is added to this solution, rapid electrophilic aromatic substitution (EAS) takes place in a stepwise manner:

1. First Substitution (Formation of C):
   The electrophilic attack of $\text{Br}_2$ occurs primarily at the para-position of the phenoxide ion (due to lower steric hindrance relative to the ortho-positions). Deprotonation restores aromaticity, yielding the 4-bromophenoxide ion (Option C).

2. Second Substitution (Formation of A):
   The 4-bromophenoxide ion remains highly activated by the phenoxide group. Subsequent electrophilic attack by a second equivalent of $\text{Br}_2$ occurs at one of the available ortho-positions to give the 2,4-dibromophenoxide ion (Option A).

3. Final Substitution:
   A third bromination step at the remaining vacant ortho-position yields 2,4,6-tribromophenoxide, which precipitates as 2,4,6-tribromophenol upon neutralization.

Analysis of Options:

• Option A: Represents the 2,4-dibromophenoxide intermediate, which is formed after two successive brominations at the ortho and para positions.
• Option B: Represents a non-aromatic 4,4-dibromocyclohexa-2,5-dienone species, which is not an intermediate in the standard electrophilic aromatic substitution pathway to 2,4,6-tribromophenol.
• Option C: Represents the 4-bromophenoxide intermediate, formed in the initial bromination step.
• Option D: Represents a 3-bromophenoxide (meta-substituted) ion, which does not form because the phenoxide group ($-\text{O}^-$) is strongly ortho/para-directing.

Thus, the intermediates formed in the reaction are A and C.