'A' and 'B' formed in the following set of reactions are:

Reaction for Product A

In the first starting material, we have two different types of $-OH$ groups:

1. Phenolic -OH (Top): Attached directly to the benzene ring.
2. Benzylic -OH (Bottom): Attached to a side-chain CH2 group.

The Mechanism:

• The Phenol is unreactive: The C-O bond in phenol has partial double-bond character due to resonance. Breaking this bond to replace it with Br is extremely difficult and does not happen under these conditions.
• The Benzylic Alcohol reacts: The side-chain -OH gets protonated by HBr to form a good leaving group H2O. It then undergoes a substitution reaction (likely SN1 because the benzylic carbocation is very stable) where the Br- replaces the -OH.
• Result A: The phenolic group remains, and the benzylic alcohol becomes a benzylic bromide -CH2Br.

Reaction for Product B