Match the reactants in Column-$I$ with their products in Column-$II$.

Column-$I$ (Reactants)	Column-$II$ (Products)
$(A)$ $p$-Nitrochlorobenzene $\xrightarrow[(ii) H^+]{(i) NaOH, 443 \ K}$	$(i)$ Phenol
$(B)$ Chlorobenzene $\xrightarrow[(ii) H^+]{(i) NaOH, 623 \ K, 300 \ atm}$	$(ii)$ $2,4,6$-Trinitrophenol
$(C)$ $2,4$-Dinitrochlorobenzene $\xrightarrow[(ii) H^+]{(i) NaOH, 368 \ K}$	$(iii)$ $2,4$-Dinitrophenol
$(D)$ $2,4,6$-Trinitrochlorobenzene $\xrightarrow{Warm \ H_2O}$	$(iv)$ $p$-Nitrophenol

(A-IV, B-I, C-III, D-II) The reaction of haloarenes with nucleophiles depends on the presence of electron-withdrawing groups (like $-NO_2$) on the benzene ring. These groups activate the ring towards nucleophilic substitution.
$(A)$ $p$-Nitrochlorobenzene reacts with $NaOH$ at $443 \ K$ to give $p$-nitrophenol. Thus,$(A \rightarrow iv)$.
$(B)$ Chlorobenzene is least reactive and requires harsh conditions $(623 \ K, 300 \ atm)$ to form phenol. Thus,$(B \rightarrow i)$.
$(C)$ $2,4$-Dinitrochlorobenzene is more reactive than $p$-nitrochlorobenzene and reacts at $368 \ K$ to give $2,4$-dinitrophenol. Thus,$(C \rightarrow iii)$.
$(D)$ $2,4,6$-Trinitrochlorobenzene is highly reactive and can be hydrolyzed by warm water to give $2,4,6$-trinitrophenol (picric acid). Thus,$(D \rightarrow ii)$.
The correct matching is: $(A$ $\rightarrow iv, B$ $\rightarrow i, C$ $\rightarrow iii, D$ $\rightarrow ii)$.