In a reaction
`CH_(2)=CH_(2)underset("acid")overset("Hypochlorous")rarrMoverset(R)rarrunderset(CH_(2)OH)underset("| ")(CH_(2)OH)`
Here, M = Molecule, R = Reagent, M and R are respectively

Options

A. `CH_(3)CH_(2)Cl and NaOH`
B. `CH_(2)Cl.CH_(2)OH and aq.NaHCO_(3)`
C. `CH_(3)CH_(2)OH and HCl`
D.

Answer

To solve the problem, we need to identify the molecules M and R in the given reaction involving ethene (CH₂=CH₂) and hypochlorous acid (HOCl).

### Step-by-Step Solution:

1. **Identify the Reactants**: The reaction starts with ethene (CH₂=CH₂) and hypochlorous acid (HOCl).

2. **Reaction with Hypochlorous Acid**: When ethene reacts with hypochlorous acid, the double bond opens up, and the hypochlorous acid adds across the double bond. The oxygen from HOCl will attach to one carbon, and the chlorine will attach to the other carbon.

- The product formed from this reaction is 1-chloroethanol (CH₂Cl-CH₂OH). This compound is what we will refer to as M.

3. **Identify M**: Therefore, M = CH₂Cl-CH₂OH (1-chloroethanol).

4. **Identify the Next Reaction**: The next part of the question states that M reacts with R to form CH₂OH.

5. **Determine the Reagent R**: The reagent R that can facilitate the conversion of 1-chloroethanol (M) to the final product (which is likely an alcohol) is typically an aqueous base. In this case, aqueous sodium bicarbonate (NaHCO₃) is a suitable reagent.

6. **Final Product**: When 1-chloroethanol reacts with aqueous NaHCO₃, it undergoes a reaction that leads to the formation of ethylene glycol (HO-CH₂-CH₂-OH), which is a diol.

7. **Conclusion**: Thus, we conclude that:
- M = CH₂Cl-CH₂OH (1-chloroethanol)
- R = Aqueous NaHCO₃

### Final Answer:
- M = CH₂Cl-CH₂OH (1-chloroethanol)
- R = Aqueous NaHCO₃