Match List-I  with List-II .

Choose the correct answer from the options given below :

In every conversion a chloride atom on an aromatic ring is being replaced by $$\!OH$$ to give a phenol. This transformation proceeds by a nucleophilic aromatic substitution (SNAr) of the $$Cl^{-}$$ ion by the nucleophile $$OH^{-}$$.

Rate of SNAr increases sharply when the ring has strong $$-I$$ and $$-M$$ groups (such as $$NO_{2}$$) at the ortho and/or para positions. These groups stabilise the intermediate Meisenheimer complex by delocalising the negative charge.

Therefore, the more $$NO_{2}$$ groups present, the milder the conditions required for hydrolysis. Arrange the four substrates in increasing ease of hydrolysis:

chlorobenzene < p-nitrochlorobenzene < 2,4-dinitrochlorobenzene < picryl chloride (2,4,6-trinitrochlorobenzene).

Now match the reagents/conditions given in List-II with this order:

Case 1: Chlorobenzene → phenol
Very unreactive; Dow’s process is used.
Reagents: $$NaOH$$, $$623\text{ K}$$, $$300\,\text{atm}$$, then $$H_{3}O^{+}$$  ⇒  Item (IV).

Case 2: p-Nitrochlorobenzene → p-Nitrophenol
One $$NO_{2}$$ group activates the ring; moderate temperature is sufficient.
Reagents: $$NaOH$$, $$443\text{ K}$$, then $$H_{3}O^{+}$$  ⇒  Item (III).

Case 3: 2,4-Dinitrochlorobenzene → 2,4-Dinitrophenol
Two $$NO_{2}$$ groups give higher activation; still milder conditions than above.
Reagents: $$NaOH$$, $$368\text{ K}$$, then $$H_{3}O^{+}$$  ⇒  Item (II).

Case 4: Picryl chloride (2,4,6-Trinitrochlorobenzene) → Picric acid (2,4,6-Trinitrophenol)
Three $$NO_{2}$$ groups make the ring so activated that simple warming with water suffices.
Reagent/condition: warm $$H_{2}O$$  ⇒  Item (I).

Thus the correct pairing is:
(A) → (IV), (B) → (III), (C) → (II), (D) → (I).

Comparing with the options, this corresponds to Option C.