1. Identify the reaction

Phenol treated with chloroform ($\mathrm{CHCl_3}$) in the presence of sodium hydroxide ($\mathrm{NaOH}$), followed by hydrolysis/acidification, is the Reimer–Tiemann reaction.

2. What happens in Reimer–Tiemann reaction?

In this reaction, phenol first forms phenoxide ion in basic medium:

$\mathrm{C_6H_5OH + NaOH \rightarrow C_6H_5O^-Na^+ + H_2O}$

Chloroform in strong base generates dichlorocarbene ($: \mathrm{CCl_2}$), which acts as the electrophile.

The formyl group is introduced mainly at the ortho position of phenol, and after hydrolysis gives o-hydroxybenzaldehyde.

3. Main product formed

Thus, phenol gives:

$\mathrm{C_6H_5OH \xrightarrow[NaOH]{CHCl_3} o ext{-}HO{-}C_6H_4{-}CHO}$

This compound is 2-hydroxybenzaldehyde (also called salicylaldehyde).

4. Check the options

• A: Salicylic acid $\rightarrow$ This is formed in the Kolbe reaction using $\mathrm{CO_2/NaOH}$, not here. Incorrect.
• B: Benzene-1,2-diol $\rightarrow$ Catechol, not formed in Reimer–Tiemann reaction. Incorrect.
• C: 2-Hydroxybenzaldehyde $\rightarrow$ Correct.
• D: Benzene-1,3-diol $\rightarrow$ Resorcinol, not formed here. Incorrect.

5. Final answer

The product obtained is 2-hydroxybenzaldehyde.