Which of the following is formed when 2 butene is treated with alkaline `KMnO_(4)` higher temperature

Options

A. `CH_(3) - underset(OH)underset(|) (CH) - underset(OH)underset(|) (CH) - CH_(3)`
B. one molecule of `CH_(3)CHO` and one molecule of `CH_(3)COOH`
C. `2` molecules of `CH_(3)COOH`
D. `2` molecules of `CH_(3)CHO`

Answer

To solve the problem of what is formed when 2-butene is treated with alkaline KMnO4 at high temperature, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Structure of 2-Butene**:
- The molecular formula of 2-butene is C4H8. Its structure can be represented as:
\[ \text{CH}_3-\text{CH}=\text{CH}-\text{CH}_3 \]
- This shows that there is a double bond between the second and third carbon atoms.

2. **Understand the Reaction Conditions**:
- The reagent is alkaline KMnO4, which is a strong oxidizing agent. The reaction is carried out at high temperature.

3. **Reaction Mechanism**:
- The KMnO4 will insert an oxygen atom into the double bond of 2-butene. This is known as syn-dihydroxylation.
- The reaction can be illustrated as:
\[ \text{CH}_3-\text{CH}=\text{CH}-\text{CH}_3 \xrightarrow{\text{KMnO}_4} \text{CH}_3-\text{C(OH)}-\text{C(OH)}-\text{CH}_3 \]
- This results in the formation of a diol (a compound with two hydroxyl groups).

4. **Rearrangement and Oxidation**:
- Under the conditions of high temperature and the presence of KMnO4, the diol can undergo further oxidation.
- The hydroxyl groups can be oxidized to form carbonyl groups (aldehydes), leading to the formation of acetic acid (CH3COOH) when further oxidized.

5. **Final Products**:
- The final products of the reaction are two molecules of acetic acid:
\[ 2 \text{CH}_3\text{COOH} \]

### Conclusion:
The correct answer is that when 2-butene is treated with alkaline KMnO4 at high temperature, it forms two molecules of acetic acid (CH3COOH).