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 aromatic substitution due to the strong activating $+M$ effect of the $-OH$ group. The solvent plays a critical role in controlling the extent of bromination.

Step 2: Detailed Explanation:
1. Bromine water: In aqueous medium, phenol ionizes to form the phenoxide ion, which is even more reactive. This leads to poly-substitution, giving 2,4,6-tribromophenol (white precipitate).
2. Low polarity solvents ($CS_2, CHCl_3, CCl_4$): At low temperatures ($273\text{K}$), the ionizing power of the solvent is low. Phenol stays mostly in its non-ionized form. This limits the reaction to mono-substitution.
3. The $-OH$ group is ortho-para directing. Due to steric hindrance at the ortho position, the para-isomer ($p$-bromophenol) becomes the major product when reacted with $Br_2$ in $CS_2$ at low temperature.

Step 3: Final Answer:
The reagent is $Br_2$ in $CS_2$ at $273\text{K}$.