Phenol treated with chloroform in presence of sodium hydroxide, which further hydrolysed in presence of an acid results…

Step 1: Identify the reaction — Reimer-Tiemann reaction

Phenol + $\ce{CHCl3}$ + $\ce{NaOH}$ → followed by acid hydrolysis = Reimer-Tiemann reaction.

Step 2: Mechanism overview

1. $\ce{CHCl3 + NaOH -> :CCl2}$ (dichlorocarbene, a highly reactive electrophile)
2. Sodium phenoxide is formed in NaOH: $\ce{C6H5OH + NaOH -> C6H5ONa + H2O}$
3. Phenoxide (activated ring) undergoes electrophilic attack by $\ce{:CCl2}$ at the ortho position: → Forms a $\ce{-CHCl2}$ group at ortho position of phenoxide.
4. Hydrolysis of the $\ce{-CHCl2}$ group under basic conditions then acidification: $\ce{-CHCl2 ->[ NaOH ] -CHO}$ (aldehyde)

Product: 2-Hydroxybenzaldehyde (salicylaldehyde)

Step 3: Why the other options are wrong

• Salicylic acid (o-hydroxybenzoic acid) is the product of Kolbe's reaction (phenol + NaOH + CO2).
• Catechol and resorcinol are not formed from this reaction.

Step 4: Regiochemistry

The -OH group in phenol is ortho/para director. Carbene attacks predominantly at ortho position (steric reasons: para also gives some product, but ortho is major for intramolecular H-bond stabilized transition state).

Key Points to Remember:

• Reimer-Tiemann: phenol + CHCl3 + NaOH → salicylaldehyde (2-hydroxybenzaldehyde).
• Intermediate is dichlorocarbene (:CCl2), generated from CHCl3 + NaOH.
• The -CHCl2 intermediate is hydrolysed to -CHO.
• Kolbe's reaction (not Reimer-Tiemann) gives salicylic acid.