Q. Among the following ethers, which one will produce methyl alcohol on treatment with hot concentrated HI?
A) CH₃–CH₂–CH₂–CH₂–O–CH₃
B) CH₃–CH₂–CH(CH₃)–O–CH₃
C) (CH₃)₃C–O–CH₃
D) (CH₃)₂CH–CH₂–O–CH₃
Explanation
[Solution from source] With tert-butyl methyl ether, HI cleavage goes by SN1 through the stable tert-butyl carbocation, giving tert-butyl iodide and methyl alcohol. For the primary and secondary alkyl methyl ethers the reaction is SN2, where I⁻ attacks the less hindered methyl group, giving methyl iodide and the other alcohol. Answer (3).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.