tert-Butyl methyl ether, (CH3)3C-O-CH3 , is heated with an equimolar…

The correct answer is Option (2) → \( \ce{(CH3)3C-I} \) and \( \ce{CH3OH} \), by S_N1 through the tertiary carbocation

• The reaction starts with protonation of the ether by HI.

• When one of the alkyl groups is tertiary, the protonated ether loses methanol to give the stable tertiary carbocation, \( \ce{(CH3)3C+} \).

• Iodide then combines with this carbocation, so the reaction follows the S_N1 path and the halide formed is a tertiary halide.

      \( \displaystyle \ce{(CH3)3C-O-CH3 + HI -> (CH3)3C-I + CH3OH} \)

Why the other options are incorrect

Option (1) \( \ce{CH3I} \) and \( \ce{(CH3)3C-OH} \), by S_N2 attack of iodide on the methyl carbon — Incorrect.

• This is what happens when both alkyl groups are primary or secondary: iodide attacks the less substituted carbon and the lower alkyl group forms the iodide.

• With a tertiary group present, the S_N1 path through the tertiary carbocation is followed instead.

Option (3) \( \ce{CH3I} \) and \( \ce{(CH3)3C-OH} \), by S_N1 through the methyl carbocation — Incorrect. A methyl carbocation is far less stable than a tertiary one, so the C–O bond that breaks is the one to the tertiary carbon, not to the methyl group.

Option (4) \( \ce{(CH3)3C-I} \) and \( \ce{CH3OH} \), by S_N2 attack of iodide on the tertiary carbon — Incorrect. The products are right, but the path is not. In S_N2 the iodide attacks the least substituted carbon; here the tertiary iodide is formed through the carbocation (S_N1).

NCERT (2026-27 edition):
7.6.3 Chemical Reactions – “Iodide is a good nucleophile. It attacks the least substituted carbon of the oxonium ion formed in step 1 and displaces an alcohol molecule by S_N2 mechanism.”

“However, when one of the alkyl group is a tertiary group, the halide formed is a tertiary halide.” … “It is because in step 2 of the reaction, the departure of leaving group (HO–CH₃) creates a more stable carbocation [(CH₃)₃C⁺], and the reaction follows S_N1 mechanism.”