Identify the substrate 'S' in the following reaction.

The given reaction is a nucleophilic aromatic substitution where a leaving group on the benzene ring is replaced by an $-\text{OH}$ group from water to form 2,4,6-trinitrophenol (picric acid).

Aryl halides are generally unreactive towards nucleophilic substitution. However, the presence of strong electron-withdrawing groups (like $-\text{NO}_2$) at the ortho and para positions with respect to the halogen atom greatly increases their reactivity.

In 2,4,6-trinitrochlorobenzene, there are three $-\text{NO}_2$ groups at the ortho and para positions to the chlorine atom. This makes the ring highly electron-deficient and the $\text{C}-\text{Cl}$ bond highly susceptible to nucleophilic attack. Because of this extreme activation, the nucleophilic substitution can occur under very mild conditions, such as simply warming with water.

Reaction:
2,4,6-trinitrochlorobenzene + warm $\text{H}_2\text{O} \rightarrow 2,4,6-trinitrophenol + \text{HCl}$

Therefore, the substrate 'S' is 2,4,6-trinitrochlorobenzene.

Answer: 2, 4, 6 - trinitrochlorobenzene