Question

The correct options for the products A and B of the following reactions are : 

• A. A
• B. B
• C. C
• D. D

Hint:

In polar solvents (like water), phenol undergoes trisubstitution with bromine. In non-polar solvents (like $CS_2$ or $CHCl_3$), it undergoes monosubstitution.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Electrophilic aromatic substitution on phenol is highly sensitive to the solvent. Phenol is extremely reactive due to the strong $+M$ effect of the $-OH$ group.
Step 2: Detailed Explanation:
1. Reaction with $Br_2/H_2O$:
Water is a highly polar solvent. In water, phenol ionizes to the phenoxide ion ($C_6H_5O^-$). The negative charge on oxygen increases the electron density on the ring even more than the $-OH$ group. This results in polysubstitution at all available ortho and para positions.
Product A = 2,4,6-tribromophenol (a white precipitate).
2. Reaction with $Br_2/CS_2$ at low temperature:
$CS_2$ is a non-polar solvent. Phenol does not ionize significantly in this medium. The reactivity is lower, allowing for mono-bromination. Since the $-OH$ group is ortho/para directing, a mixture of o-bromophenol and p-bromophenol is formed, with the para isomer usually being the major product due to steric hindrance.
Product B = p-bromophenol (major).
Step 3: Final Answer:
A is 2,4,6-tribromophenol and B is p-bromophenol.