Question

Among the following ethers, which one will produce methyl alcohol on treatment with hot concentrated HI?

• A. $CH_{3}-CH_{2}-CH\left(\right.CH_{3}\left.\right)-O-CH_{3}$
• B. $\left( CH _{3}\right)_{3} C - O - CH _{3}$
• C. $\left( CH _{3}\right)_{2} CH - CH _{2}- O - CH _{3}$
• D. $CH_{3}-CH_{2}-CH_{2}-CH_{2}-O-CH_{3}$

Answer 1

Answer: (B)

In in this reaction a stable tertiary carbocation is possible so $S_{N}1$ mechanism is used and $I^{-}$ in $S_{N}1$ is taken with stable carbocation.

In rest other options $S_{N}2$ mechanism will be used as stable carbocations are not possible hence $I^{-}$ goes with $CH_{3}$ group to give $CH_{3}I$ and respective alcohols.