The REIMER-TIEMANN reaction converts phenol to salicylaldehyde. The reagents are

1. NCERT §7.7 (Phenol reactions): Reimer-Tiemann introduces a $\mathrm{-CHO}$ group at the ORTHO position of phenol. 2. REAGENTS: chloroform ($\mathrm{CHCl_3}$) and aqueous NaOH, followed by acid workup. 3. Mechanism: NaOH deprotonates $\mathrm{CHCl_3}$ to $\mathrm{CCl_3^-}$, which loses a chloride to generate a DICHLOROCARBENE ($\mathrm{:CCl_2}$). The carbene attacks the ortho position of the phenoxide ion. Hydrolysis of the resulting dichloro group gives the aldehyde. 4. Product: ortho-hydroxybenzaldehyde (SALICYLALDEHYDE). 5. Option A would be a Mannich-style attempt. Option C is the KOLBE reaction (which gives salicylic ACID, not salicylaldehyde). Option D is chlorination of phenol (different product). _Source: NCERT Class 12 Chemistry Part 2, Ch 7, §7.7 (Phenol reactions — Reimer-Tiemann), p. 18–19._