Question

What is the product of the reaction between n-butyl benzene and alcoholic KMnO4?

• A. Benzoic acid
• B. 2-phenylacetic acid
• C. 3-phenylpropanoic acid
• D. Cinnamic acid

Hint:

Alkaline KMnO4 oxidizes any benzylic side chain all the way down to one carbon.

Answer 1

The Correct Option is A

Step 1: Alkaline (alcoholic) KMnO4 is a strong oxidizing agent that oxidizes any alkyl side chain on an aromatic ring, regardless of its length, as long as the benzylic carbon bears at least one hydrogen. Step 2: n-Butylbenzene has the structure $C_6H_5-CH_2-CH_2-CH_2-CH_3$. The benzylic carbon, the $CH_2$ attached to the ring, has hydrogens available for oxidation. Step 3: Vigorous oxidation cleaves the side chain completely, breaking every C-C bond beyond the ring-attached carbon and oxidizing that carbon straight to a carboxylic acid, regardless of how many carbons were originally in the chain. \[ C_6H_5-CH_2CH_2CH_2CH_3 \xrightarrow{KMnO_4,\ \Delta} C_6H_5COOH \] Step 4: The three extra carbons of the butyl chain are lost as carbon dioxide and water during this exhaustive oxidation; they do not survive as a shorter chain such as an acetic or propanoic acid derivative. Step 5: The only organic product retained on the ring is the carboxylic acid directly bonded to the benzene ring, benzoic acid. \[\boxed{\text{Benzoic acid}}\]