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JEE Main 202622 JanuaryMorning Shift
Exam
JEE Main
Year
2026
Date
22 January
Shift
Morning Shift
Subject
Chemistry
Chapter
Alcohols Phenols and Ethers

JEE Main 2026 · 22 January · Morning Shift · Chemistry · Alcohols Phenols and Ethers

Given below are two statements: Statement I : Phenol on treatment...

Given below are two statements:
Statement I: Phenol on treatment with CHCl3\mathrm{CHCl}_{3} /aq. KOH under refluxing condition, followed by acidification produces pp-hydroxy benzaldehyde as the major product and oo-hydroxy benzaldehyde as the minor product.
Statement II: The mixture of pp-hydroxybenzaldehyde and oo hydroxybenzaldehyde can be easily separated through steam distillation.
In the light of the above statements, choose the correct answer from the options given below

Options

  1. AStatement I is true but Statement II is false
  2. BStatement I is false but Statement II is true
  3. CBoth Statement I and Statement II are true
  4. DBoth Statement I and Statement II are false

Correct answer

B. Statement I is false but Statement II is true

Step-by-step solution

Statement I: The reaction of phenol with CHCl3CHCl_3 and aqueous KOHKOH is the Reimer-Tiemann reaction. In this reaction, the major product is oo-hydroxybenzaldehyde (salicylaldehyde) due to the formation of a stable six-membered chelate ring via intramolecular hydrogen bonding. pp-hydroxybenzaldehyde is formed as the minor product. Thus, Statement I is false.

Statement II: oo-hydroxybenzaldehyde (salicylaldehyde) has intramolecular hydrogen bonding, making it more volatile (steam volatile). pp-hydroxybenzaldehyde has intermolecular hydrogen bonding, which leads to association between molecules and a higher boiling point, making it non-steam volatile. Therefore, the mixture can be easily separated by steam distillation. Thus, Statement II is true.

Conclusion: Statement I is false but Statement II is true.
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