InterviewSolution
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InterviewSolution
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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 ? |
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Answer» SOLUTION :`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. 
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