Hot and conc. `KMnO_4` at 373-383 K reacts with pent - 2- ene to form

Options

A. propionic acid only
B. ethonoic acid only
C. a mixture of propionic acid and ethanoic acid
D. a mixture of butanoic and formic acid

Answer

To solve the problem of what products are formed when hot and concentrated KMnO4 reacts with pent-2-ene, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Structure of Pent-2-ene**:
- Pent-2-ene has the structure: CH3-CH=CH-CH2-CH3. The double bond is located between the second and third carbon atoms.

2. **Understand the Reaction Conditions**:
- The reaction involves hot and concentrated potassium permanganate (KMnO4), which is a strong oxidizing agent. The temperature range given is 373-383 K.

3. **Reaction Mechanism**:
- When pent-2-ene reacts with hot concentrated KMnO4, it undergoes oxidative cleavage. The double bond is broken, and the carbons involved in the double bond are oxidized.

4. **Formation of Aldehydes**:
- The cleavage of the double bond results in the formation of two aldehydes:
- The left part of the double bond (CH3-CH) will form propanal (CH3-CH2-CHO).
- The right part of the double bond (CH-CH2-CH3) will form ethanal (CH3-CHO).

5. **Further Oxidation of Aldehydes**:
- Since KMnO4 is a strong oxidizing agent, both aldehydes will undergo further oxidation:
- Propanal (CH3-CH2-CHO) will be oxidized to propanoic acid (CH3-CH2-COOH).
- Ethanal (CH3-CHO) will be oxidized to ethanoic acid (CH3-COOH).

6. **Final Products**:
- The final products of the reaction are a mixture of propanoic acid and ethanoic acid.

7. **Choose the Correct Option**:
- Among the given options, the correct answer is a mixture of propanoic acid and ethanoic acid.

### Conclusion:
The reaction of hot and concentrated KMnO4 with pent-2-ene produces a mixture of propanoic acid (CH3-CH2-COOH) and ethanoic acid (CH3-COOH).