Q1 · 2026
Given below are two statements :
Statement I : Phenol on treatment with /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
Answer: D
- The reaction of phenol with /aq. KOH is the classic Reimer-Tiemann reaction, which introduces a -CHO group onto the aromatic ring, predominantly at the ortho position relative to the -OH group, with para-substitution as the minor pathway. [IMAGE: Reimer-Tiemann reaction mechanism showing formation of ortho-hydroxybenzaldehyde as major product and para-hydroxybenzaldehyde as minor product]
- Since the reaction actually favors the ortho product as major (not the para product as claimed in Statement I), Statement I is false — the major/minor assignment given in the statement is reversed from the actual outcome.
- Regarding Statement II: the ortho-isomer (o-hydroxybenzaldehyde) has intramolecular hydrogen bonding between its -OH and -CHO groups, making it more volatile and steam-distillable, while the para-isomer has intermolecular hydrogen bonding (leading to higher boiling point and lower volatility). This difference in volatility allows the two isomers to be effectively separated via steam distillation, confirming Statement II is true. Hence, Statement I is false but Statement II is true, so the answer is option D.
