Question

When aqueous HI reacts with methoxyethane, methyl iodide and ethanol are formed, but when aqueous HI reacts with 2-methoxy-2-methylpropane, a mixture of methanol and t-butyl iodide is formed. Explain ?

Answer 1

`HI+underset("Methoxyethane")(CH_(3)-O-C_(2)H_(5))rarrunderset("Methyl iodide")(CH_(3)I)+underset("Ethanol")(CH_(3)CH_(2))OH`
In this case, the reaction follows `S_(N^(1))` path `I^(-)` ion (nucleophile ) attacks protonated ether from the less stearically hindere side containing `CH_(3)` group

In the second case, the reaction proced as follows : `HI+CH_(3)O-underset(2-"Methoxy-2-methylpropane")underset(CH_(3))underset(|)overset(CH_(3))overset(|)C-CH_(3)rarrCH_(3)OH +I- underset("t-butyliodide")underset(CH_(3))underset(|)overset(CH_(3))overset(|)C-CH_(3)`
The reaction followes `S_(N^(1))` path. protonated ether initially formed cleaves to give methanol and a tertiary carbocation in s slow step. The carbocation reacts with `I^(-)` ion to from butyl iodide.