Question

Write the mechanism of the reaction of HI with methoxymethane.

Answer 1

Methoxymethane and equimolar amounts of HI form a mixture of methyl alcohol and methyl iodide. The mechanism of the reaction is as follows –

Step 1: Generation of oxonium ion:

\[\begin{array}{cc}
\phantom{..............................}\ce{H}\\
\phantom{..............................}|\\
\ce{\underset{Methoxymethane}{CH3 - \overset{\bullet\bullet}{\underset{\bullet\bullet}{O}} - CH3 + H - I} ->[Protonation][rapid] \underset{Dimethyl oxonium ion}{CH3 - \overset{+}{\underset{\bullet\bullet}{O}} - CH3 + I^-}}
\end{array}\]

Step 2: Nucleophilic attack on oxonium ion:

\[\begin{array}{cc}\\
\ce{H}\phantom{......................}\\
|\phantom{......................}\\
\ce{I^- + CH3 - O^+ - CH3 ->[S_{N}2][dim] \underset{Methyl iodide}{CH3 - I} + \underset{Methanol}{CH3OH}}
\end{array}\]

If excess of HI is used then methanol formed in the second step gets converted into methyl iodide by the following mechanism:

\[\begin{array}{cc}
\phantom{............................}\ce{H}\\
\phantom{............................}|\\
\ce{CH3 - \overset{\bullet\bullet}{\underset{\bullet\bullet}{O}} - H + H - I ->[Protonation][dim] \underset{Protonated Methanol}{CH3 - \underset{\bullet\bullet\phantom{...}}{O^+} - H + I^-}}
\end{array}\]

\[\begin{array}{cc}
\ce{H}\phantom{.................}\\
|\phantom{.................}\\
\ce{I^- + CH3 - \underset{\bullet\bullet\phantom{...}}{O^+} - H ->[S_{N}2][dim] \underset{Methyl iodide}{CH3 - I} + H2O}
\end{array}\]