Question

One mole of an alkene reacts with acidic $KMnO_{4}$ to give two moles of ethanoic acid. What is the alkene?

• A. 2-Methylpropene
• B. 1-Butene
• C. 2-Pentene
• D. 2-Butene
• E. 2-Methyl-2-butene

Hint:

Terminal alkenes ($R-CH=CH_2$) yield $CO_2$ and a carboxylic acid under these conditions.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
This question involves the oxidative cleavage of an alkene using acidic potassium permanganate (KMnO\(_4\)). This strong oxidizing agent breaks the carbon-carbon double bond of the alkene, and the nature of the products (ketones, carboxylic acids, or CO\(_2\)) depends on the substitution pattern of the double bond.
Step 2: Key Formula or Approach:
- The reaction is oxidative cleavage. The C=C double bond is broken completely. - Each carbon atom of the original double bond becomes the carbonyl carbon of a new molecule. - If a carbon of the double bond is bonded to at least one hydrogen (\(=CHR\)), it is oxidized to a carboxylic acid (\(-COOH\)). - If a carbon of the double bond is bonded to two alkyl groups (\(=CR_2\)), it is oxidized to a ketone (\(C=O\)). - If a carbon of the double bond is bonded to two hydrogens (\(=CH_2\)), it is oxidized to carbon dioxide (CO\(_2\)) and water. The problem states that the product is two moles of ethanoic acid (CH\(_3\)COOH). To find the original alkene, we can work backward from the products.
Step 3: Detailed Explanation:
The product is two molecules of ethanoic acid: \[ \text{CH}_3\text{COOH} \quad + \quad \text{HOOCCH}_3 \] To reconstruct the alkene, we remove the two oxygen atoms from the carboxyl groups and join the two carbonyl carbons with a double bond. \[ \text{CH}_3-\text{C}(=\text{O})\text{OH} \quad + \quad \text{HO}(\text{O}=)\text{C}-\text{CH}_3 \] Removing the oxygens and joining the carbons gives: \[ \text{CH}_3-\text{CH}=\text{CH}-\text{CH}_3 \] This molecule is but-2-ene (or 2-Butene). Let's verify this by writing the forward reaction: \[ \underset{\text{2-Butene}}{\text{CH}_3-\text{CH}=\text{CH}-\text{CH}_3} \xrightarrow{\text{acidic KMnO}_4, \Delta} \underset{\text{Ethanoic acid}}{2\text{CH}_3\text{COOH}} \] Each of the double-bonded carbons (=CH-CH\(_3\)) has one hydrogen and one methyl group attached. Therefore, upon oxidative cleavage, each part becomes a carboxylic acid with two carbons, which is ethanoic acid. Since the alkene is symmetric, we get two identical molecules.
Step 4: Final Answer:
The alkene is 2-Butene.