1. Concept involved

A compound will react readily with dilute $\mathrm{NaOH}$ if it is sufficiently acidic to donate a proton to $\mathrm{OH^-}$.

So we compare the acidity of the given alcohols/phenol.

2. Check each option

Option A: $\mathrm{C_6H_5CH_2OH}$ (benzyl alcohol)

This is an alcohol. Ordinary alcohols are very weak acids and do not react readily with dilute $\mathrm{NaOH}$.

So, A is incorrect.

Option B: $\mathrm{C_2H_5OH}$ (ethanol)

Ethanol is also a typical alcohol and is even less acidic than phenol. It does not react readily with dilute $\mathrm{NaOH}$.

So, B is incorrect.

Option C: $\mathrm{C_6H_5OH}$ (phenol)

Phenol is appreciably more acidic than ordinary alcohols because its conjugate base, phenoxide ion, is resonance stabilized.

Reaction:

Hence phenol reacts readily with dilute $\mathrm{NaOH}$.

So, C is correct.

Option D: $\mathrm{(CH_3)_3COH}$ (tert-butyl alcohol)

This is also an alcohol and is not acidic enough to react with dilute $\mathrm{NaOH}$.

So, D is incorrect.

3. Why phenol reacts but alcohols do not

• Phenol forms phenoxide ion: $\mathrm{C_6H_5O^-}$ which is stabilized by resonance.
• Alcohols form alkoxide ions: $\mathrm{RO^-}$ which are not resonance stabilized.

Therefore, phenol is acidic enough to react with dilute $\mathrm{NaOH}$, whereas normal alcohols are not.

4. Final answer

The compound that readily reacts with dilute $\mathrm{NaOH}$ is:

So, the correct option is C.