Which one of the following, with HBr will give a phenol?

Solution

Chemistry ma’am · 3 steps

A phenol needs oxygen sitting directly on the ring, so look for O bonded straight to a benzene carbon.

1. How HBr cleaves ethers. HBr first protonates the ether oxygen. Then Br⁻ breaks one C–O bond. It either attacks the less hindered carbon (S_N2) or waits for a stable carbocation to form (S_N1, tertiary or benzylic).
2. Aryl C–O survives. The ring carbon is sp², and resonance gives the O–C(aryl) bond partial double-bond character. Br⁻ cannot attack it, so the oxygen stays on the ring and ends up as phenol.

   Option	What links O to the ring	Bond that breaks	Products
   (a) C₆H₅-O-CH₃	nothing, O is on the ring carbon	O-CH₃	phenol and CH₃Br
   (b) C₆H₅CH₂CH(OCH₃)₂	a CH₂-CH chain, two carbons	the acetal C–O bonds	an aldehyde, C₆H₅CH₂CHO, and methanol
   (c) C₆H₅CH₂-O-C(CH₃)₃	a benzylic CH₂	O-C(CH₃)₃, via the tertiary cation	benzyl alcohol and tert-butyl bromide
   (d) C₆H₅CH₂-O-CH₃	a benzylic CH₂	C₆H₅CH₂-O, via the benzylic cation	benzyl bromide and methanol

3. So. Only anisole, option (a), has oxygen bonded directly to the ring. HBr removes the methyl group as CH₃Br and leaves phenol, C₆H₅OH.

The idea to keep. In HBr cleavage of an alkyl aryl ether, the aryl–O bond never breaks, so you get phenol and an alkyl bromide. A CH₂ between O and the ring gives benzyl compounds, not phenol.

Answer from the NTA final answer key. Solution by PrepWiser.