Match the reactions in Column-$I$ with their mechanisms in Column-$II$.

Column-$I$ (Reaction)	Column-$II$ (Mechanism)
$(A)$ $OH^- + CH_3Cl \to HOCH_3 + Cl^-$	$(1)$ $S_N2$ mechanism
$(B)$ $(CH_3)_3CBr + OH^- \to (CH_3)_3COH + Br^-$	$(2)$ $S_E2$ aromatic
$(C)$ $CH_3CH_2Br \xrightarrow[ethanol]{KOH} CH_2 = CH_2$	$(3)$ $S_N1$ mechanism
$(D)$ $C_6H_6 + Br_2 \xrightarrow{Fe} C_6H_5Br + HBr$	$(4)$ $\beta$-elimination (dehydrohalogenation)

(A-1, B-3, C-4, D-2) $(A)-(1), (B)-(3), (C)-(4), (D)-(2)$
$(A)$ $OH^- + CH_3Cl \to HOCH_3 + Cl^-$: This is a primary alkyl halide reacting with a strong nucleophile,following the $S_N2$ mechanism.
$(B)$ $(CH_3)_3CBr + OH^- \to (CH_3)_3COH + Br^-$: This is a tertiary alkyl halide,which undergoes substitution via the $S_N1$ mechanism due to the formation of a stable carbocation.
$(C)$ $CH_3CH_2Br \xrightarrow[ethanol]{KOH} CH_2 = CH_2$: This is an elimination reaction (dehydrohalogenation) of an alkyl halide in the presence of a base,known as $\beta$-elimination.
$(D)$ $C_6H_6 + Br_2 \xrightarrow{Fe} C_6H_5Br + HBr$: This is an electrophilic aromatic substitution reaction ($S_E2$ aromatic) where benzene reacts with bromine in the presence of a Lewis acid catalyst.