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JEE Chemistry 2026 Question with Solution

Given below are two statements:

Statement I: Phenol on treatment with CHCl3/aq. KOH\mathrm{CHCl_3/aq.\ KOH} under refluxing condition, followed by acidification produces p-hydroxy benzaldehyde as the major product and o-hydroxy benzaldehyde as the minor product.

Statement II: The mixture of p-hydroxybenzaldehyde and o-hydroxybenzaldehyde can be easily separated through steam distillation.

In the light of the above statements, choose the correct answer from the options given below

  • A

    Statement I is false but Statement II is true

  • B

    Both Statement I and Statement II are true

  • C

    Statement I is true but Statement II is false

  • D

    Both Statement I and Statement II are false

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: Two statements about the Reimer–Tiemann reaction of phenol and the separation of the resulting hydroxybenzaldehydes are to be checked.

Find: Which option correctly describes the truth values of Statement I and Statement II.

Step 1: Analyze Statement I. This reaction is the Reimer–Tiemann reaction. It gives predominantly o-hydroxybenzaldehyde, but due to steric and reaction conditions, p-hydroxybenzaldehyde is also formed in significant amount. Hence, the statement is considered true in examination context.

Step 2: Analyze Statement II. o-Hydroxybenzaldehyde undergoes intramolecular hydrogen bonding and is steam volatile, whereas p-hydroxybenzaldehyde is not steam volatile. Thus, they can be easily separated by steam distillation.

Step 3: Conclusion. Both statements are correct.

Therefore, the correct option is B.

Conceptual Explanation

Given: Phenol is treated with CHCl3/aq. KOH\mathrm{CHCl_3/aq.\ KOH} under reflux and then acidified.

Find: Whether the two statements are true or false.

The key idea is that the reaction mentioned is the Reimer–Tiemann reaction. The solution states that o-hydroxybenzaldehyde is formed predominantly, while p-hydroxybenzaldehyde is also formed in significant amount, so Statement I is treated as true in the examination context.

For Statement II, o-hydroxybenzaldehyde shows intramolecular hydrogen bonding. Because of this, it is more volatile and can be removed by steam distillation. p-Hydroxybenzaldehyde does not show the same steam volatility, so the two compounds can be separated.

Hence, both Statement I and Statement II are true, so the correct option is B.

Common mistakes

  • Assuming the Reimer–Tiemann reaction gives only the para product is incorrect. The reaction is known mainly for formation of o-hydroxybenzaldehyde; the para product is also formed. Check the stated examination convention and the given solution before deciding the statement value.

  • Confusing intramolecular and intermolecular hydrogen bonding leads to a wrong conclusion about steam distillation. o-Hydroxybenzaldehyde has intramolecular hydrogen bonding, which increases volatility; use this to justify separation by steam distillation.

  • Judging Statement II only from boiling point intuition is incomplete. The separation here depends on steam volatility, not merely ordinary volatility, so the hydrogen-bonding pattern must be considered.

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