Question

$p$-Bromophenol is the major product formed when phenol is treated with

• A. Bromine water
• B. $Br_2$ in acetic acid at $300\text{K}$
• C. $Br_2$ in $CCl_4$ at $300\text{K}$
• D. $Br_2$ in $CS_2$ at $273\text{K}$
• E. $Br_2$ in acetone at $273\text{K}$

Hint:

Solvent matters! Water $\to$ Tribromo (3 Br). Carbon disulfide ($CS_2$) $\to$ Monobromo (1 Br). This is a very common distinction in organic chemistry tests.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
Phenol is highly reactive towards electrophilic substitution because the -OH group is strongly activating. The choice of solvent determines whether mono-substitution or poly-substitution occurs.
Step 2: Key Formula or Approach:
- Polar solvent (H\textsubscript{2}O): Phenol ionizes to phenoxide ion, which is extremely reactive. Result: 2,4,6-tribromophenol.
- Non-polar solvent (CS\textsubscript{2}, CCl\textsubscript{4}): Ionization is suppressed. Result: Mono-substitution (ortho and para).
Step 3: Detailed Explanation:
When Bromine reacts with Phenol in a low-polarity solvent like Carbon disulphide (CS\textsubscript{2}) or Chloroform at low temperatures (273K), the activation of the ring is moderate.
As a result, only mono-bromination occurs. The para-isomer (p-bromophenol) is obtained as the major product due to less steric hindrance compared to the ortho-isomer.
Step 4: Final Answer:
The reagent is Br\textsubscript{2} in CS\textsubscript{2} at 273K.