During the preparation of alkanes by
Kolbe's electrolytic decarboxylatio of `CH_3CH_2COONa` ,
which of the following products are formed

Options

A. `CH_3CH_2CH_2CH_3`
B. `CH_2=CH_2`
C. `CH_4`
D. `CH_3CH_2OH`

Explanation

To solve the question regarding the products formed during the Kolbe's electrolytic decarboxylation of sodium propionate (CH3CH2COONa), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Reaction**:
- The Kolbe electrolytic decarboxylation involves the electrolysis of a carboxylate salt. In this case, we have sodium propionate (CH3CH2COONa).

2. **Oxidation at the Anode**:
- During the electrolysis, the carboxylate ion (CH3CH2COO-) is oxidized at the anode. This oxidation leads to the formation of a radical.
- The reaction can be represented as:
\[ CH3CH2COO^- \rightarrow CH3CH2COO^{\cdot} + e^- \]
- Here, CH3CH2COO^{\cdot} is the propionate radical.

3. **Decarboxylation**:
- The propionate radical (CH3CH2COO^{\cdot}) can lose a carbon dioxide molecule (CO2) to form a free radical:
\[ CH3CH2COO^{\cdot} \rightarrow CH3CH2^{\cdot} + CO2 \]
- This results in the formation of the ethyl radical (CH3CH2^{\cdot}).

4. **Possible Reactions of the Ethyl Radical**:
- The ethyl radical (CH3CH2^{\cdot}) can undergo two main reactions:
- **Dimerization**: Two ethyl radicals can combine to form butane (C4H10):
\[ 2 CH3CH2^{\cdot} \rightarrow C4H10 \]
- **Disproportionation**: The ethyl radical can also disproportionate to form ethane (C2H6) and ethylene (C2H4):
\[ CH3CH2^{\cdot} \rightarrow C2H6 + C2H4 \]

5. **Final Products**:
- Therefore, the products formed during the Kolbe's electrolytic decarboxylation of sodium propionate are:
- Butane (C4H10)
- Ethane (C2H6)
- Ethylene (C2H4)

### Conclusion:
- Based on the above reactions, the correct products formed during the Kolbe's electrolytic decarboxylation of CH3CH2COONa are butane and ethane.