The reaction of phenol with chloroform in the presence of sodium hydroxide gives

Salicylaldehyde

The correct answer is Option (3) → Salicylaldehyde

$Phenol + CHCl_3 + NaOH$ undergoes the Reimer-Tiemann reaction, which introduces a -CHO (formyl) group at the ortho position of phenol, giving o-hydroxybenzaldehyde (salicylaldehyde) as the main product.

Option-wise Explanation

1. Salicylic acid Salicylic acid is formed in the Kolbe–Schmitt reaction (phenol + CO₂ under pressure with NaOH), not with chloroform. Here the introduced group is –COOH, whereas Reimer–Tiemann introduces –CHO.

2. Acetylsalicylic acid This is aspirin, made by acetylation of salicylic acid using acetic anhydride. No acetylating agent is present in the given reaction, so this product is not possible.

3. Salicylaldehyde In alkaline medium, CHCl₃ generates dichlorocarbene (:CCl₂), which attacks the activated phenoxide ring at the ortho position. Hydrolysis of the intermediate yields o-hydroxybenzaldehyde (salicylaldehyde) — the characteristic product.

4. Mixture of ortho and para chlorophenols This type of product forms during electrophilic chlorination of phenol with Cl₂. Chloroform under basic conditions does not act as a chlorinating agent; instead, it provides the carbene for formylation.