Question

Identify product 'B' in following reaction:
$\text{Cumene} \xrightarrow{\text{KMnO}_4,\ \text{KOH},\ \Delta} \text{A} \xrightarrow{\text{H}_3\text{O}^+} \text{B}$

• A. Benzoic acid
• B. Benzophenone
• C. Phenol
• D. Benzaldehyde

Hint:

Do not confuse this side-chain oxidation with the industrial "Cumene Process" used to manufacture phenol and acetone, which explicitly requires oxygen ($\text{O}_2$) gas and a weak acid catalyst rather than a harsh alkaline $\text{KMnO}_4$ oxidant.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The question requires us to trace the products of a two-stage chemical reaction starting with cumene (isopropylbenzene) under powerful oxidizing conditions.

Step 2: Key Formula or Approach:
Alkyl side chains attached to a benzene ring undergo vigorous oxidation when treated with hot alkaline potassium permanganate ($\text{KMnO}_4$), provided the benzylic carbon has at least one benzylic hydrogen atom. Regardless of the alkyl chain length, it is entirely oxidized to a carboxylic acid group ($-\text{COOH}$).

Step 3: Detailed Explanation:
Cumene is structurally isopropylbenzene, $\text{C}_6\text{H}_5\text{CH}(\text{CH}_3)_2$. The central carbon linked to the ring contains exactly one hydrogen atom (benzylic hydrogen), which makes it reactive towards oxidation. When heated with alkaline $\text{KMnO}_4$, the entire isopropyl alkyl side group is degraded and oxidized to form the potassium salt of benzoic acid, potassium benzoate ($\text{C}_6\text{H}_5\text{COOK}$), which represents intermediate 'A'. In the subsequent step, treating intermediate 'A' with hydronium ions ($\text{H}_3\text{O}^+$) protonates the carboxylate ion, yielding

benzoic acid ($\text{C}_6\text{H}_5\text{COOH}$) as the final stable product 'B'.

Step 4: Final Answer:
Product 'B' is benzoic acid, matching option (A).