Which one of the following compounds will give a yellow precipitate when reacted with $\mathrm{I}_2 / \mathrm{NaOH}$ ?

The iodoform test is given by compounds containing the $\mathrm{CH}_3\mathrm{CO}-$ group or the $\mathrm{CH}_3\mathrm{CH(OH)}-$ group. These groups are oxidized to the methyl ketone group, which then reacts with $\mathrm{I}_2 / \mathrm{NaOH}$ to form a yellow precipitate of iodoform ($\mathrm{CHI}_3$).

Analyzing the options:
1. $\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_2 \mathrm{OH}$ is a primary alcohol and does not contain the required $\mathrm{CH}_3\mathrm{CH(OH)}-$ structural unit.
2. $(\mathrm{CH}_3)_3 \mathrm{C}-\mathrm{OH}$ is a tertiary alcohol and cannot be oxidized to a methyl ketone.
3. $\mathrm{C}_6 \mathrm{H}_5 \mathrm{OH}$ (phenol) does not contain the $\mathrm{CH}_3\mathrm{CH(OH)}-$ or $\mathrm{CH}_3\mathrm{CO}-$ group.
4. $\mathrm{C}_6 \mathrm{H}_5 \mathrm{CHOH}-\mathrm{CH}_3$ is a secondary alcohol containing the $\mathrm{CH}_3\mathrm{CH(OH)}-$ group attached to a phenyl ring. Upon oxidation, it forms acetophenone ($\mathrm{C}_6 \mathrm{H}_5 \mathrm{COCH}_3$), which contains the $\mathrm{CH}_3\mathrm{CO}-$ group and thus gives a positive iodoform test.

Answer: $\mathrm{C}_6 \mathrm{H}_5 \mathrm{CHOH}-\mathrm{CH}_3$