AIIMS 2018 Chemistry Reimer-Tiemann Reaction MCQ Question

In the Reimer-Tiemann reaction, phenol reacts with chloroform (CHCl₃) in the presence of a strong base, typically sodium hydroxide (NaOH), to produce salicylaldehyde (o-hydroxybenzaldehyde) through the formation of an ortho-hydroxy ketone intermediate. This reaction specifically leads to the introduction of a formyl group (-CHO) at the ortho position relative to the hydroxyl group (-OH) on the aromatic ring, resulting in salicylaldehyde. Other options are not applicable as they do not pertain to the Reimer-Tiemann reaction or its product, which is uniquely identified as salicylaldehyde.