The cleavage of methyl tertbutyl ether with HI gives

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

Core Concept:

In unsymmetrical ethers, cleavage with HI occurs at the more stable carbocation forming side

(tertiary > secondary > primary).

Explanation:

Methyl tert-butyl ether contains:

• One primary methyl group
• One tertiary tert-butyl group

In presence of HI:

Protonation of ether oxygen occurs first.

Then cleavage follows SN1 pathway at the tertiary carbon because tert-butyl carbocation is highly stable.

Thus:

tert-butyl group forms tert-butyl iodide

Remaining group forms methanol

Option 1: Methanol and tert-butyl iodide

Correct because cleavage occurs at tertiary side via SN1 mechanism.

Option 2: Methyl iodide and tert-butanol

Incorrect because SN2 at methyl group is not favoured when tertiary carbocation can form.

Option 3: tert-butyl iodide

Incorrect because methanol is also formed.

Option 4: Methanol

Incorrect because tert-butyl iodide is also produced.