Match the reactants in column $-I$ with the reaction conditions in column $-II$ to find the corresponding products in column $-III$.

Column $-I$ (Reactant)	Column $-II$ (Reaction Condition)	Column $-III$ (Product)
$A$. $CH_3CN + C_6H_5MgBr$	$i$. $Ether, H_3O^+$	$p$. Benzaldehyde
$B$. $Benzene + CH_3COCl$	$ii$. $Hg^{2+}, H_2SO_4, H_2O$	$q$. Acetophenone
$C$. $Toluene$	$iii$. $Anhydrous AlCl_3$	$r$. Propiophenone
$D$. $Phenyl ethyne$	$iv$. $x. Cl_2, hv$; $y. H_2O, 373 \ K$	$s$. Acetaldehyde

(A-I-Q, B-III-Q, C-IV-P, D-II-Q) The correct matches are:
$A$ $\rightarrow i$ $\rightarrow q$: $CH_3CN + C_6H_5MgBr$ $\xrightarrow{Ether} CH_3C(NMgBr)C_6H_5$ $\xrightarrow{H_3O^+} CH_3COC_6H_5$ (Acetophenone).
$B$ $\rightarrow iii$ $\rightarrow q$: Friedel-Crafts acylation of benzene with $CH_3COCl$ in the presence of anhydrous $AlCl_3$ gives Acetophenone.
$C$ $\rightarrow iv$ $\rightarrow p$: Side-chain chlorination of toluene followed by hydrolysis gives Benzaldehyde.
$D$ $\rightarrow ii$ $\rightarrow q$: Hydration of phenyl ethyne in the presence of $Hg^{2+}$ and $H_2SO_4$ gives Acetophenone.