Match the column-I (reaction) with column-II(products).

	Column- I		Column-II
A	${\text{CH}}_3-\text{CH}={\text{CH}}_2\stackrel{{\text{Cl}}_{\text{2}}/{\text{H}}_{\text{2}}\text{O}}{\to }$	P	Acetic acid &Butanoic acid
B		Q	Product contains two chiral carbons
C	2-Hexene $\underset{\left(\text{ii}\right){\text{\hspace{0.17em}\hspace{0.17em}H}}_{\text{2}}{\text{O}}_{\text{2}}}{\overset{{\text{(i)\hspace{0.17em}\hspace{0.17em}O}}_{\text{3}}}{\to }}$	R	Acetic acid &propanoic acid
D	2-pentene $\underset{{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\text{,\hspace{0.17em}Δ}}{\overset{{\text{KMnO}}_{\text{4}}}{\to }}$	S	Reaction proceeds through intermediate Halonium ion