In the above reaction, product B is

1. Analysis of the Starting Material

The starting reactant is 4-(hydroxymethyl)phenol. It features a benzene ring with two distinct hydroxyl (-OH) environments:

• Phenolic -OH group: Attached directly to the aromatic ring. Due to resonance, the carbon-oxygen bond has partial double-bond character, making it highly stable and resistant to substitution.
• Aliphatic -OH group: Part of the substituent at the top, written as -CH₂OH. This benzylic alcohol is highly reactive toward nucleophilic substitution.

2. Formation of Product A

When treated with HCl and heat (Δ):

• The acid protonates the aliphatic alcohol, converting the -CH₂OH group into a good leaving group (water).
• The chloride ion (Cl⁻) acts as a nucleophile and replaces the water molecule at the benzylic position.
• The phenolic -OH group remains completely unaffected.

Therefore, Product A is 4-(chloromethyl)phenol, where the top substituent is now a -CH₂Cl group.

3. Formation of Product B (Finkelstein Reaction)

When Product A reacts with NaI (Sodium Iodide):

• This reaction proceeds via an S_N2 nucleophilic substitution mechanism, commonly known as the Finkelstein reaction.
• The iodide ion (I⁻) replaces the chloride ion (Cl⁻) at the benzylic position because sodium chloride (NaCl) precipitates out in the reaction solvent (typically acetone), driving the equilibrium forward.

Consequently, the -CH₂Cl group is converted directly into a -CH₂I group.