JEE Main 2021 · 16 March · evening · Single correct · +4 / −1
In the above reaction, the reagent ''A'' is :
- (a) LiAlH₄
- (b) NaBH₄, H₃O⁺
- (c) HCl, Zn - Hg
- (d) Alkaline KMnO₄, H⁺
Answer and solution
Answer (d) Alkaline KMnO₄, H⁺
Solution
Chemistry ma’am · 3 steps
The methyl group became COOH, so A must be an oxidising agent, not a reducing one.
- Spot the change. The starting compound is 3-methylanisole. In the product, the ring -CH₃ has become -COOH, and the -OCH₃ group is unchanged. Going from -CH₃ to -COOH is an oxidation.
- Side-chain oxidation. Hot alkaline KMnO₄ oxidises any alkyl group on a benzene ring that has a benzylic H all the way to the carboxylate, Ar-COO⁻. Acid (H⁺) is then added to turn the salt into the free acid Ar-COOH.
Reagent What it does Can it make -CH₃ → -COOH? LiAlH₄ Reduces C=O groups to alcohols No, it is a reducing agent NaBH₄, H₃O⁺ Reduces aldehydes and ketones No, it is a reducing agent Zn-Hg, HCl Clemmensen reduction: C=O → CH₂ No, it goes the opposite way Alkaline KMnO₄, H⁺ Oxidises a side chain to -COOH Yes - So. Reagent A is alkaline KMnO₄ followed by H⁺, which gives 3-methoxybenzoic acid. The answer is (d).
The idea to keep. Compare the starting compound and the product first. If C gains bonds to O, you need an oxidant. Hot alkaline KMnO₄ then H⁺ turns any ring alkyl group with a benzylic H into -COOH.
Answer from the NTA final answer key. Solution by PrepWiser.