What happens when C_6H_5-O-R is treated with HX?

The correct answer is Option (1) → RX and \(C_6H_5OH\) are formed

Ethers are cleaved by hydrogen halides. In an alkyl aryl ether the bond that breaks is the alkyl–oxygen bond. The aryl–oxygen bond is stronger because it has partial double bond character (the lone pair on oxygen is in resonance with the ring). After protonation of the oxygen, the halide ion attacks the alkyl carbon, never the ring carbon:

    \( \displaystyle \ce{C6H5-O-R + HX -> C6H5OH + R-X} \)

For example, anisole and HI give phenol and methyl iodide:

    \( \displaystyle \ce{C6H5-O-CH3 + HI -> C6H5OH + CH3I} \)

So the products are RX and phenol. Phenol does not react further to give a halobenzene, because its sp² ring carbon cannot undergo nucleophilic substitution.

Why the other options are incorrect

• Option (2) ROH and \(C_6H_5X\) are formed — Incorrect. This would need the stronger aryl–oxygen bond to break and a halide to attack a ring carbon, which does not happen.

• Option (3) \(C_6H_4X_2\) and ROH are formed — Incorrect. Cleavage of an ether does not put halogen atoms on the ring.

• Option (4) RX and \(C_6H_5X\) are formed — Incorrect. Two halides form only from a dialkyl ether with excess HX; phenol does not change into a halobenzene.

NCERT (2026-27 edition):
Cleavage of C–O Bond in Ethers – …