Question

An Alkene "X" on reaction with hot acidified \( \text{KMnO}_4 \) gave a mixture of Ethanoic acid and Propanone. Identify "X".

• A. Pent-2-ene
• B. 2-Methylbut-2-ene
• C. But-2-ene
• D. 2,3-Dimethylbut-2-ene

Hint:

When an alkene reacts with hot acidified KMnO4, it undergoes oxidative cleavage, splitting the double bond and forming carbonyl compounds such as aldehydes, ketones, and carboxylic acids.

Answer 1

The Correct Option is B

Step 1: Understand the reaction with hot acidified KMnO4.
When an alkene reacts with hot acidified KMnO4 (potassium permanganate), it undergoes oxidative cleavage, breaking the double bond and forming two smaller carbonyl compounds, such as carboxylic acids and ketones.

Step 2: Identify the products formed.
We are given that the reaction produces Ethanoic acid (CH3COOH) and Propanone (CH3COCH3). Ethanoic acid is a carboxylic acid and Propanone is a ketone. Therefore, the alkene must split in such a way that one product has a methyl group and the other has a carbonyl group.

Step 3: Analyze the structure of the alkene "X".
The alkene must be one that, when cleaved by hot acidified KMnO4, results in the formation of Ethanoic acid and Propanone. A plausible candidate is 2-Methylbut-2-ene. When 2-Methylbut-2-ene undergoes oxidative cleavage, the products will be Ethanoic acid (CH3COOH) and Propanone (CH3COCH3).

Step 4: Verify with the other options.
- (A) Pent-2-ene: This would produce different products upon cleavage, not the required Ethanoic acid and Propanone.
- (C) But-2-ene: This would also produce different products, not the required Ethanoic acid and Propanone
. - (D) 2,3-Dimethylbut-2-ene: This would produce different products upon cleavage.

Step 5: Conclusion.
The alkene that gives the required products is 2-Methylbut-2-ene, which corresponds to option (B).