In the reactions
`underset(CH_2)overset(CH_2)(||)underset"acid"overset"Hypochlorous"to M oversetRtounderset(CH_2OH)overset(CH_2OH)|`
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 question, we need to identify the compounds M and R in the given reactions involving ethene and hypochlorous acid.

### Step-by-Step Solution:

1. **Identify the Starting Material:**
- The starting material is ethene, which has the molecular formula \(C_2H_4\) or \(CH_2=CH_2\).

2. **Reaction with Hypochlorous Acid:**
- Ethene reacts with hypochlorous acid (\(HOCl\)). This is an addition reaction where the hypochlorous acid adds across the double bond of ethene.
- The addition of \(HOCl\) to ethene results in the formation of ethylene chlorohydrin. The reaction can be represented as:
\[ CH_2=CH_2 + HOCl \rightarrow CH_2(OH)CH_2Cl \]
- Therefore, the product M is ethylene chlorohydrin, which can be written as \(CH_2(OH)CH_2Cl\).

3. **Identify Product M:**
- M is thus identified as ethylene chlorohydrin, with the structure \(CH_2(OH)CH_2Cl\).

4. **Reaction of M with R:**
- The next step involves the reaction of M (ethylene chlorohydrin) with aqueous sodium bicarbonate (\(NaHCO_3\)). This reaction typically leads to the conversion of the chlorohydrin to ethylene glycol.
- The reaction can be represented as:
\[ CH_2(OH)CH_2Cl + NaHCO_3 \rightarrow CH_2(OH)CH_2(OH) + NaCl + CO_2 + H_2O \]
- The product formed is ethylene glycol, which has the structure \(CH_2(OH)CH_2(OH)\).

5. **Identify Product R:**
- R is identified as aqueous sodium bicarbonate (\(NaHCO_3\)).

### Final Answer:
- M is ethylene chlorohydrin (\(CH_2(OH)CH_2Cl\)).
- R is aqueous sodium bicarbonate (\(NaHCO_3\)).