On boiling concentrated \(HBr\), phenyl ethyl ether will give:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → Phenol and Ethyl bromide
Ethers are cleaved by hot concentrated HX. In an alkyl aryl ether the bond broken is the alkyl–oxygen bond, because the aryl–oxygen bond is more stable (the O lone pair is in resonance with the ring, giving the C–O bond partial double-bond character). The ether is protonated, and \( \ce{Br^{-}} \) attacks the ethyl carbon (SN2):
\( \ce{C6H5-O-C2H5 + HBr} \xrightarrow{\Delta} \ce{C6H5OH + C2H5Br} \)
The phenol formed does not react further with HBr, because its C–O bond is attached to an sp2 carbon of the ring.
Why the other options are incorrect
Option (2) Bromobenzene and Ethanol — Incorrect. This would need the strong aryl–oxygen bond to break and \( \ce{Br^{-}} \) to attack a ring carbon, which does not happen.
Option (3) Phenol and Ethane — Incorrect. The ethyl group leaves with the bromide ion as ethyl bromide; HBr is not a reducing agent that could give ethane.
Option (4) Bromobenzene and Ethane — Incorrect. Neither product forms; the aryl–O bond stays intact and no reduction takes place.
NCERT (2026-27 edition):
Cleavage of C–O Bond in Ethers – …
