The compound which will have the lowest rate towards nucleophilic aromatic substitution on treatment with OH$$^-$$ is

Nucleophilic aromatic substitution is facilitated by electron-withdrawing groups (especially at ortho and para positions) which stabilize the intermediate carbanion (Meisenheimer complex).

The order of reactivity towards nucleophilic aromatic substitution:

- p-NO₂ chlorobenzene > o-NO₂ chlorobenzene > m-NO₂ chlorobenzene > chlorobenzene

NO₂ at para and ortho positions activate nucleophilic substitution by stabilizing the negative charge in the Meisenheimer complex through resonance.

NO₂ at meta position provides only inductive withdrawal, so it's less effective.

Chlorobenzene without any NO₂ group has the lowest rate of nucleophilic aromatic substitution.

The correct answer is Option 3: Chlorobenzene (no NO₂).