Concept:

To differentiate two classes of compounds visually, a reagent must cause a distinct, observable change in one but not the other.

Formula:

$$ \text{C}_6\text{H}_5\text{OH} + 3\text{Br}_{2(aq)} \longrightarrow \text{2,4,6-Tribromophenol}\downarrow \text{(white)} $$

Solution:

• Halogenation using bromine water is a classic diagnostic test for highly activated aromatic rings like phenol.



• Phenol rapidly decolorizes orange bromine water, instantly forming a highly visible white precipitate of 2,4,6-tribromophenol.



• Aliphatic alcohols lack this activated ring and do not react with bromine water, showing no color change or precipitation.

Why other options are incorrect:

Lucas test is for classifying alcohols among themselves (1°, 2°, 3°). Nitration requires harsh conditions and isn't a quick visual bench test. Iodoform only works for specific alcohols (like ethanol), not a general class differentiator for all alcohols vs phenols.