Question

In the following reactions sequence, structure of A and B respectively will be :

• A.

  

• B.

  

• C.

  

• D.

  

Hint:

HBr with heat cleaves the aryl alkyl ether to a phenol and an alkyl bromide (the aryl-O bond is not broken); then Na/ether does an intramolecular Wurtz coupling of the two C-Br bonds.

Answer 1

The Correct Option is D

Step 1: The substrate is a chroman-type aryl alkyl ether (O attached to the benzene ring and to a CH₂ of the saturated ring) with a CH₂Br group on the ring.
Step 2: HBr/heat cleaves the C(alkyl)-O bond (the C(aryl)-O bond is stronger because of resonance). The product A is a phenol with a -CH₂CH₂CH₂Br chain and the existing -CH₂Br group on adjacent ring carbons.
Step 3: Na/dry ether (Wurtz) couples the two terminal CH₂Br carbons intramolecularly. The new ring contains the two aromatic carbons, 1 CH₂ and the 3 CH₂, i.e. 6 atoms, giving a tetrahydronaphthalene (tetralin) ring.
Step 4: B is therefore 5,6,7,8-tetrahydronaphthalen-1-ol (OH on the aromatic ring next to the saturated ring), which is the pair of structures in option (D).
Answer: (D)