The cleavage of methyl tertbutyl ether with HI gives

The correct answer is Option (1) → methanol and tertbutyl iodide

Ethers are cleaved by concentrated HI. The acid first protonates the ether oxygen; then the C–O bond breaks in the way that is easier for the groups present. With primary or secondary groups, iodide attacks the smaller alkyl group by S_N2. But when one group is tertiary, the protonated ether loses an alcohol molecule to give a stable tertiary carbocation, and iodide joins that carbocation (S_N1). So the halide formed is the tertiary halide.

For methyl tert-butyl ether:

    \( \displaystyle \ce{(CH3)3C-O-CH3 + HI <=> (CH3)3C-O^+(H)-CH3 + I-} \)

    \( \displaystyle \ce{(CH3)3C-O^+(H)-CH3 -> (CH3)3C+ + CH3OH} \)

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

The net reaction is:

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

So the products are methanol and tert-butyl iodide.

Why the other options are incorrect

• Option (2) methyl iodide and tertbutanol — Incorrect. These would form by S_N2 attack of iodide on the methyl group, which is the path for ethers with primary or secondary groups; here the tertiary carbocation forms first.

• Option (3) tertbutyl iodide — Incorrect. This is only one of the two products; the methyl group leaves as methanol.

• Option (4) methanol — Incorrect. This too is only one product; the tert-butyl part ends up as the iodide.

NCERT (2026-27 edition):
Cleavage of C–O Bond in Ethers – “… However, when one of the alkyl group is a tertiary group, the halide formed is a tertiary halide.”

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