Match the reactants in Column-$I$ with the major products of their reaction in Column-$II$. Find the correct match for $(I)$ from $(II)$.

Column-$I$ (Reactants)	Column-$II$ (Products)
$(A) \ CH_3CH_2CH_2-O-CH_3 + HI \rightarrow$	$(i) \ (CH_3)_3C-I + CH_3CH_2OH$
$(B) \ C_6H_5OCH_2CH_3 + HBr \rightarrow$	$(ii) \ CH_3CH_2CH_2OH + CH_3I$
$(C) \ (CH_3)_3C-OCH_2CH_3 + HI \rightarrow$	$(iii) \ CH_3CH_2I + C_6H_5OH$
$(D) \ C_6H_5OCH_2CH_3 + \text{conc. } H_2SO_4 + \text{conc. } HNO_3 \rightarrow$	$(iv) \ p-NO_2-C_6H_4-OCH_2CH_3$

(A-II, B-III, C-I, D-IV) $(A) \ CH_3CH_2CH_2-O-CH_3 + HI \rightarrow CH_3CH_2CH_2OH + CH_3I$ (Cleavage occurs at the less hindered side,$S_N2$ mechanism). Thus,$(A \rightarrow ii)$.
$(B) \ C_6H_5OCH_2CH_3 + HBr \rightarrow C_6H_5OH + CH_3CH_2Br$ (Phenol and ethyl bromide are formed). Wait,looking at the options,$(iii)$ is $CH_3CH_2I + C_6H_5OH$. If $HI$ was used,it would be $(iii)$. Given the options,$(B \rightarrow iii)$ is the intended match.
$(C) \ (CH_3)_3C-OCH_2CH_3 + HI \rightarrow (CH_3)_3C-I + CH_3CH_2OH$ (Cleavage occurs at the more hindered side due to stable carbocation formation,$S_N1$ mechanism). Thus,$(C \rightarrow i)$.
$(D) \ C_6H_5OCH_2CH_3$ undergoes electrophilic aromatic substitution (nitration) to give $p$-nitro-phenetole. Thus,$(D \rightarrow iv)$.
Final match: $(A$ $\rightarrow ii, B$ $\rightarrow iii, C$ $\rightarrow i, D$ $\rightarrow iv)$.