A compound decolourises Baeyer's reagent and gives a mixture of propanoic acid and ethanoic acid when treated with a hot and conc. solution of `KMnO_4` . The compound is

Options

A. pent-1-ene
B. pent-2-ene
C. 2-methylbut-1-ene
D. 2-methylbut-2-ene

Answer

To solve the problem, we need to identify the compound that decolorizes Baeyer's reagent and produces a mixture of propanoic acid and ethanoic acid when treated with hot and concentrated KMnO4. Let's break down the solution step by step:

### Step 1: Understand the Reaction with Baeyer's Reagent
Baeyer's reagent is an alkaline solution of potassium permanganate (KMnO4) that is used to test for unsaturation in organic compounds. When an unsaturated compound (like alkenes) reacts with Baeyer's reagent, the pink color of the reagent disappears, indicating the presence of a double bond.

**Hint:** Look for an alkene that can react with Baeyer's reagent.

### Step 2: Analyze the Products
The problem states that the reaction produces a mixture of propanoic acid (CH3CH2COOH) and ethanoic acid (CH3COOH). This suggests that the original compound must have a structure that can be oxidized to yield these two acids.

**Hint:** Consider how the structure of the original compound can lead to the formation of these two acids upon oxidation.

### Step 3: Identify the Possible Structures
The options given are:
1. Pent-1-ene
2. Pent-2-ene
3. 2-Methylbut-1-ene
4. 2-Methylbut-2-ene

We need to determine which of these alkenes can be oxidized to form both ethanoic acid and propanoic acid.

**Hint:** Focus on the carbon skeleton of each option and how they can be oxidized.

### Step 4: Oxidation of Each Compound
- **Pent-1-ene (CH2=CH-CH2-CH2-CH3)**: Oxidation would not yield both acids.
- **Pent-2-ene (CH3-CH=CH-CH2-CH3)**: Oxidation can yield CH3COOH (ethanoic acid) and CH3CH2COOH (propanoic acid).
- **2-Methylbut-1-ene (CH2=C(CH3)-CH2-CH3)**: Oxidation would not yield both acids.
- **2-Methylbut-2-ene (C(CH3)=C(CH3)-CH2-CH3)**: Oxidation would not yield both acids.

**Hint:** Consider the structure of pent-2-ene and how it can be oxidized to yield the required products.

### Step 5: Conclusion
After analyzing the options, we find that pent-2-ene is the only compound that, when oxidized with hot and concentrated KMnO4, can produce both propanoic acid and ethanoic acid. Therefore, the correct answer is:

**The compound is Pent-2-ene.**

### Final Answer
**Pent-2-ene** is the compound that decolorizes Baeyer's reagent and gives a mixture of propanoic acid and ethanoic acid when treated with hot and concentrated KMnO4.

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