Given below are two statements Phenol on treatment with CHCl3 aq KOH
Q. Given below are two statements :

Statement I : Phenol on treatment with CHCl3/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 true but Statement II is false
B. Both Statement I and Statement II are true
C. Both Statement I and Statement II are false
D. Statement I is false but Statement II is true
Correct Answer: Statement I is false but Statement II is true

Explanation

This question is based on the Reimer–Tiemann reaction, which is a very important topic for JEE Main, JEE Advanced and IIT JEE.

When phenol is treated with CHCl3 and aqueous KOH followed by acidification, the reaction introduces a –CHO group mainly at the ortho position.

Due to intramolecular hydrogen bonding, o-hydroxybenzaldehyde is the major product, while p-hydroxybenzaldehyde is formed in smaller amount.

Hence, Statement I is false.

The o-hydroxybenzaldehyde forms intramolecular hydrogen bonding and is steam volatile, whereas p-hydroxybenzaldehyde forms intermolecular hydrogen bonding and is not steam volatile.

Therefore, the two compounds can be easily separated by steam distillation.

Hence, Statement II is true.

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