`{:(CH_2-CO_2 K ),(|),(CH_2 -CO_2 K):} overset("electrolysis ") to underset(("major"))((A))` (Kolbe electrolysis method) Product (A) of the reaction is :

Options

A. `CH_3 -CH_3`
B. `CH_2 = CH_2`
C. `CH_3 -CH =CH_2`
D. None of these

Explanation

To solve the problem regarding the Kolbe electrolysis of potassium carboxylate, we will follow these steps:

### Step 1: Understand the Kolbe Electrolysis Process
Kolbe electrolysis involves the electrolysis of a carboxylate salt (in this case, potassium carboxylate) to produce alkenes. During this process, the carboxylate ions undergo oxidation to form free radicals.

**Hint:** Recall that Kolbe electrolysis typically results in the formation of alkenes from carboxylate salts through radical intermediates.

### Step 2: Identify the Starting Material
The starting material is given as \( (CH_2 - CO_2 K) \). This represents potassium acetate, which can be written as \( CH_3COOK \).

**Hint:** Recognize that the structure of the starting material is crucial for predicting the products.

### Step 3: Electrolysis Reaction
During electrolysis, the carboxylate ions are oxidized at the anode. The reaction leads to the formation of free radicals. In this case, the potassium acetate will produce two ethyl radicals (\( CH_2 \cdot \) and \( CH_2 \cdot \)).

**Hint:** Remember that the oxidation of carboxylate ions leads to the formation of free radicals.

### Step 4: Combine Free Radicals
The two ethyl radicals combine to form a double bond, resulting in the formation of ethene (\( C_2H_4 \) or \( CH_2=CH_2 \)).

**Hint:** Think about how free radicals can react with each other to form stable products.

### Step 5: Identify the Major Product
The major product formed from the Kolbe electrolysis of potassium acetate is ethene. Therefore, the product \( A \) is ethene.

**Hint:** After determining the product, check the options provided to confirm which one corresponds to ethene.

### Conclusion
The major product \( A \) of the Kolbe electrolysis of potassium acetate is ethene (\( C_2H_4 \)).

**Final Answer:** Ethene (option 2).