JEE Main 2024 — Alcohols Phenols And Ethers Question with Solution
From: JEE Main 2024 (Online) 1st February Evening Shift
Question
| List I (Reactants) | List II (Product) |
|---|---|
| (A) Phenol, Zn/Δ | (I) Salicylaldehyde |
| (B) Phenol, CHCl3, NaOH, HCl | (II) Salicylic acid |
| (C) Phenol, CO2, NaOH, HCl | (III) Benzene |
| (D) Phenol, Conc. HNO3 | (IV) Picric acid |
Choose the correct answer from the options given below :
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Show full solutionCorrect option: B
Step-by-step explanation
(A) Phenol, Zn/Δ: When phenol is reacted with zinc dust (Zn) upon heating (Δ), it undergoes a reduction reaction known as the Clemmensen reduction, which results in the removal of the oxygen from the hydroxyl group (-OH) attached to the benzene ring, thus forming benzene. Hence, (A) corresponds to (III) Benzene.
(B) Phenol, CHCl3, NaOH, HCl: This reaction is known as the Reimer-Tiemann reaction. Phenol, when reacted with chloroform (CHCl3) in the presence of an aqueous base (NaOH), followed by acidification with HCl, forms salicylaldehyde. Therefore, (B) corresponds to (I) Salicylaldehyde.
(C) Phenol, CO2, NaOH, HCl: This is the Kolbe-Schmitt reaction. In this reaction, phenol reacts with carbon dioxide (CO2) under pressure at high temperatures in the presence of sodium hydroxide (NaOH), followed by acidification with HCl, to yield salicylic acid. Thus, (C) corresponds to (II) Salicylic acid.
(D) Phenol, Conc. HNO3: The nitration of phenol with concentrated nitric acid (HNO3) produces picric acid (2,4,6-trinitrophenol). So, (D) corresponds to (IV) Picric acid.
Based on the above reactions, the correct matching would be:(A)-(III) Benzene
(B)-(I) Salicylaldehyde
(C)-(II) Salicylic acid
(D)-(IV) Picric acid
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This is a previous-year question from JEE Main 2024, covering the Alcohols Phenols And Ethers chapter of Chemistry. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.