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

Solution

Chemistry ma’am · 3 steps

Ask where the nitro group sits: only ortho or para can soak up the negative charge, meta cannot help at all.

1. How the reaction runs. OH⁻ adds to the carbon carrying Cl, which makes a negatively charged intermediate (the Meisenheimer complex). Then Cl⁻ leaves. The faster that negative charge is stabilised, the faster the substitution.
2. Where the charge goes. In the resonance structures, the negative charge lands on the carbons ortho and para to the carbon carrying Cl. A NO₂ group on one of those carbons can pull the charge onto its oxygen atoms. A NO₂ group at the meta carbon never carries that charge, so it gives only a weak inductive pull.

   Option	Position of NO₂ from Cl	Resonance help
   (a) 2-chloronitrobenzene	ortho	yes, one group
   (b) 3-chloronitrobenzene	meta	none, inductive pull only
   (c) 4-chloronitrobenzene	para	yes, one group
   (d) 2,4-dinitrochlorobenzene	ortho and para	yes, two groups (fastest)

3. So. Only in (b) can the nitro group not stabilise the intermediate by resonance, so 3-chloronitrobenzene reacts with OH⁻ the slowest. The answer is (b).

The idea to keep. In nucleophilic aromatic substitution, electron-withdrawing groups speed it up only from the ortho or para position. More such groups means a faster reaction. A group at meta gives almost no help.

Answer from the NTA final answer key. Solution by PrepWiser.