In a reaction ${\mathrm{CH}}_2=\mathrm{C}{\mathrm{H}}_2\underset{\mathrm{acid}}{\overset{\mathrm{Hypochlorous}}{\to }}\mathrm{M}\stackrel{\mathrm{R}}{\to }\stackrel{{\mathrm{CH}}_2\mathrm{OH}}{\underset{{\mathrm{CH}}_2\mathrm{OH}}{|}}$ Here, M= Molecule, R = Reagent. M and R are respectively

${\text{CH}}_2=\mathrm{C}{\mathrm{H}}_2\underset{\mathrm{acid}}{\overset{\mathrm{Hypochlorous}}{\to }}\mathrm{M}\stackrel{\mathrm{R}}{\to }\stackrel{{\mathrm{CH}}_2\mathrm{OH}}{\underset{{\mathrm{CH}}_2\mathrm{OH}}{|}}$

Here, M= Molecule, R = Reagent. M and R are respectively.

Explanation: A molecule of hypochlorous acid (HOCl) reacts with an ethylene C=C double bond to form ${\mathrm{CH}}_2\mathrm{Cl}-{\mathrm{CH}}_2\mathrm{OH} \left(\mathrm{moleculeM}\right).\mathrm{Ethylene} \mathrm{glycol} \mathrm{is} \mathrm{formed} \mathrm{by} \mathrm{hydrolysis} \mathrm{with} \mathrm{aq} {\mathrm{NaHCO}}_3\left(\mathrm{reagentR}\right)$.

$\underset{\mathrm{ethylene}}{{\mathrm{H}}_2\mathrm{C}={\mathrm{CH}}_2}\stackrel{\mathrm{HOCl}}{\to }\underset{\mathrm{molecule} \mathrm{M}}{{\mathrm{CH}}_2\mathrm{Cl}-{\mathrm{CH}}_2\mathrm{OH}}\stackrel{\text{ aq }{\mathrm{NaHCO}}_3}{\to }\underset{\mathrm{ethylene} \mathrm{glycol}}{{\mathrm{CH}}_2\mathrm{OH}-{\mathrm{CH}}_2\mathrm{OH}}$

Therefore, the correct option is (B) ${\mathrm{CH}}_2\mathrm{Cl}·{\mathrm{CH}}_2\mathrm{OH}\text{ and aq. }{\mathrm{Na}}_2{\mathrm{CO}}_3$.