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Q7.21(iii):
Name the reagents used in the following reactions: Bromination of phenol to 2,4,6-tribromophenol.

Solution :

Reaction Mechanism and Reagents

The bromination of phenol to 2,4,6-tribromophenol is an electrophilic aromatic substitution reaction. Due to the strong electron-donating effect of the $-\text{OH}$ group, the benzene ring is highly activated, directing incoming electrophiles to both the ortho and para positions [Per the activating effect of the hydroxyl group in electrophilic substitution].

Step 1: Reagent Identification

To achieve polysubstitution at the 2, 4, and 6 positions, phenol is treated with Bromine water ($\text{Br}_2/\text{H}_2\text{O}$). Unlike the reaction with $\text{Br}_2$ in a non-polar solvent like $\text{CS}_2$ or $\text{CHCl}_3$ (which yields mono-brominated products), the polar solvent water ionizes phenol to the phenoxide ion ($\text{C}_6\text{H}_5\text{O}^-$).

Step 2: Theoretical Justification

  • Phenoxide Ion Formation: $\text{C}_6\text{H}_5\text{OH} + \text{H}_2\text{O} \rightleftharpoons \text{C}_6\text{H}_5\text{O}^- + \text{H}_3\text{O}^+$.
  • Activation: The phenoxide ion is significantly more reactive than phenol toward electrophilic aromatic substitution because the negative charge on the oxygen atom increases the electron density on the ring through resonance [By the resonance effect (+R effect)].
  • Reaction Equation:

$\text{C}_6\text{H}_5\text{OH} + 3\text{Br}_2 (\text{aq}) \rightarrow \text{C}_6\text{H}_2\text{Br}_3\text{OH} (\text{white precipitate}) + 3\text{HBr}$

Step 3: Visual Representation of the Transformation

OH 3Brâ‚‚, Hâ‚‚O OH Br Br Br

Final Solution: The reagent required for the bromination of phenol to 2,4,6-tribromophenol is Bromine water ($\text{Br}_2$ in aqueous solution).


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