Question

Identify the product ' B ' in the following series of reactions.
Chlorobenzene $\xrightarrow{\text{i) NaOH, 623 K / 150 atm}{\text{ii) H}_3\text{O}^+} \text{A} \xrightarrow{\text{Br}_2\text{ water}} \text{B}$}

• A. Phenol
• B. o-Bromophenol
• C. p-Bromophenol
• D. 2,4,6-tribromophenol

Hint:

Bromine in $CS_2$ gives mono-substitution (ortho/para), but Bromine in water gives poly-substitution (2,4,6) for Phenol.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
Chlorobenzene is converted to phenol via the Dow process, which is then halogenated.
Step 2: Key Formula or Approach:
1. Dow Process: Preparation of Phenol.
2. Halogenation of Phenol in aqueous medium.
Step 3: Detailed Explanation:
- Step 1: Chlorobenzene reacts with $\text{NaOH}$ at high temp/pressure followed by acidification to give Phenol (A).
- Step 2: Phenol is treated with bromine water. In water, phenol exists partly as phenoxide ion, which is highly activating. This leads to multisubstitution at all available ortho and para positions.
- Reaction: $\text{C}_6\text{H}_5\text{OH} + 3\text{Br}_2 \rightarrow \text{C}_6\text{H}_2\text{Br}_3\text{OH}$ (white precipitate).
Step 4: Final Answer:
The product ' B ' is 2,4,6-tribromophenol.