Ethers - Acidic Cleavage

Cleavage of unsymmetrical ethers by HI begins with protonation of the ether oxygen, after which iodide attacks one carbon of the oxygen-carbon framework. The mechanism that operates depends on the structure of the carbon attached to oxygen. For tert-butyl methyl ether, the tertiary centre readily ionises, so the reaction proceeds by an SN1-like route in which iodide is too weak a nucleophile to attack the crowded tertiary carbon; instead, the methyl group is attacked by iodide via SN2 only when the carbon is methyl. In practice the protonated ether breaks so that the stable tert-butyl cation forms and combines with water giving tert-butanol, while the methyl portion ends up as methyl iodide. Therefore tert-butyl iodide and methanol is wrong because iodide does not bond to the tertiary carbon, and the answer with two iodides or with methane is impossible with one equivalent of acid. This selectivity is highlighted in NCERT ether chemistry and is a favourite JEE Advanced mechanism question. The contrast with a dimethyl or diethyl ether is instructive, because there both carbons are primary and iodide attacks by SN2 to give two molecules of the alkyl iodide, whereas a branched centre switches the mechanism to SN1 and changes which fragment retains oxygen. With limited acid, only one cleavage event occurs per ether molecule. As a check, the more stable carbocation always determines which fragment keeps the oxygen, consistent with tert-butanol forming.