Question

Given below are two statements:
Statement I: Phenol on treatment with \( \mathrm{CHCl_3/aq.\ KOH} \) 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

Hint:

Intramolecular hydrogen bonding increases volatility, making o-hydroxybenzaldehyde steam distillable.

Answer 1

The Correct Option is B

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.
Final Answer: \[ \boxed{\text{Both Statement I and Statement II are true}} \]