Identify A, B and C in following reaction when phenol is treated...

The given reaction is the Reimer-Tiemann reaction.

Chloroform reacts with aqueous $NaOH$ to generate dichlorocarbene ($:CCl_2$), which acts as an electrophile. Phenol is converted to the more reactive phenoxide ion by $NaOH$.

The phenoxide ion attacks the dichlorocarbene at the ortho position. After proton transfer and rearomatization, the intermediate [A] formed is sodium 2-(dichloromethyl)phenolate. Its structure has an $-ONa$ group and a $-CHCl_2$ group at ortho positions on the benzene ring.

The $-CHCl_2$ group in intermediate [A] undergoes alkaline hydrolysis with $NaOH$ to form an unstable gem-diol, which immediately loses a water molecule to form an aldehyde group ($-CHO$). This gives intermediate [B], which is sodium 2-formylphenolate (having $-ONa$ and $-CHO$ groups).

Finally, acidification of [B] with $H_3O^+$ protonates the phenoxide oxygen to yield the final product [C], salicylaldehyde (2-hydroxybenzaldehyde).

Comparing these structures with the given options, the first option correctly identifies [A], [B], and [C].

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