To determine which of the given compounds will readily react with dilute $$\mathrm{NaOH}$$, we need to understand the chemical reactivity of these compounds towards bases like sodium hydroxide ($$\mathrm{NaOH}$$).

Here's a brief overview of each compound's reactivity towards $$\mathrm{NaOH}$$:

• Option A: $$\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_2 \mathrm{OH}$$ (Benzyl Alcohol)

  The presence of a benzyl group (a phenyl group attached to a CH₂ group) adjacent to the hydroxyl group can somewhat increase the acidity of the hydroxyl hydrogen. However, benzyl alcohol is still not significantly acidic to react vigorously with a weak base like dilute $$\mathrm{NaOH}$$. Nonetheless, under certain conditions, it might undergo reactions, but not as readily as an acidic hydrogen-containing compound would.



• Option B: $$\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}$$ (Ethanol)

  Ethanol is a simple alcohol with no acidic hydrogen atoms that would react with $$\mathrm{NaOH}$$. The hydroxyl group in ethanol is not sufficiently acidic to deprotonate in the presence of a base like $$\mathrm{NaOH}$$, making it unreactive in this context.



• Option C: $$\mathrm{C}_6 \mathrm{H}_5 \mathrm{OH}$$ (Phenol)

  Phenol contains a hydroxyl group directly attached to an aromatic ring. This structural feature significantly increases the acidity of the hydroxyl hydrogen compared to alcohols. The reason behind this is the stabilization of the phenoxide ion (the conjugate base) through resonance within the aromatic ring. As a result, phenol can react with a weak base like $$\mathrm{NaOH}$$ to form sodium phenoxide and water. $$\mathrm{C}_6 \mathrm{H}_5 \mathrm{OH} + \mathrm{NaOH} \rightarrow \mathrm{C}_6 \mathrm{H}_5 \mathrm{O}^- \mathrm{Na}^+ + \mathrm{H}_2\mathrm{O}$$



• Option D: $$\left(\mathrm{CH}_3\right)_3 \mathrm{COH}$$ (tert-Butyl Alcohol)

  tert-Butyl alcohol, being a tertiary alcohol, does not have a hydrogen atom on the carbon bearing the hydroxyl group that is acidic enough to react with $$\mathrm{NaOH}$$. Tertiary alcohols, in general, are resistant to deprotonation because of the steric hindrance around the hydroxyl group and the lack of an easily removable hydrogen atom connected to the carbon with the -OH group.

Given the choices, the correct answer is Option C: $$\mathrm{C}_6 \mathrm{H}_5 \mathrm{OH}$$ (Phenol), as it will readily react with dilute $$\mathrm{NaOH}$$ due to the increased acidity of its hydroxyl group caused by the resonance stability of the phenoxide ion formed in the reaction.