How does sodium hydroxide react with chlorobenzene?

Electron withdrawing groups activate the benzene ring to nucleophilic attack.

NaOH does react with chlorobenzene, but only under extreme conditions.

Aryl halides cannot undergo an #"S"_"N"2# reaction. The C–Cl bond is in the plane of the ring and, to attack from the back, the nucleophile would have to appear inside the benzene ring. This is not possible.

An #"S"_"N"1# reaction is possible but unfavourable. It would involve the unaided loss of the leaving group and the formation of an aryl cation.

Electron withdrawing groups ortho and para to the leaving group activate the ring to nucleophilic attack. The more electron withdrawing groups you have, the faster the reaction becomes.

For example, the relative rates for reaction with NaOH are:

Chlorobenzene: 1
4-Nitrochlorobenzene: 7 × 10¹⁰
2,4-Dinitrochlorobenzene: 2.4 × 10¹⁵
2,4,6-Trinitrochlorobenzene: Too fast to measure

The reaction goes by a two-step #"S"_"N""Ar"# (substitution, nucleophilic, aromatic) mechanism.

Step 1 is an addition. Step 2 is an elimination reaction. So this is an addition-elimination reaction.

#"S"_"N""Ar"# Nucleophilic Aromatic Substitution

Step 1: Addition

Consider the reaction of 2,4-dinitrochlorobenzene with aqueous NaOH.

OH⁻ attacks the ipso carbon (the carbon bearing the leaving group). This destroys the aromaticity of the ring. But it forms a resonance-stabilized σ-complex in which the negative charge is delocalized around the ring and the O atoms of the nitro groups.

Step 2: Loss of the leaving group

The Cl⁻ leaves, regenerating the aromatic π system.

The video below describes the nucleophilic aromatic substitution mechanism.