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Question 47

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

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Solution

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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

  • Ether Cleavage: When an aryl-alkyl ether (like anisole derivatives) reacts with concentrated $HBr$, the bond between the Oxygen and the Alkyl group (the methyl group) breaks, not the bond between the Oxygen and the Ring. The bond between the ring and oxygen is strong (resonance), while the bond to the methyl group is a standard sp3- bond. The Br- attacks the methyl group SN2, forming CH3Br and leaving behind an -OH group on the ring.
  • Result B: The ether is cleaved to form a second phenolic group, resulting in Resorcinol (benzene-1,3-diol).

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