11.83 Write the mechanism of the reaction of HI with methoxymethane.

<p><strong>11.83&nbsp;</strong></p><p>The reaction of HI with methoxymethane yields two different sets of products depending upon the initial amount of HI taken.</p><p>When equal moles of HI and methoxymethane are taken, a mixture of methyl alcohol and methyl iodide is</p><p>The mechanism is given below:</p><p>In the first step, methoxymethane reacts with hydrogen iodide to extract a proton to give the dimethyloxonium ion.</p><p>In the second step of the reaction, the Dimethyloxonium ion reacts with the iodide ion present to yield methyl iodide and methyl alcohol as the product via SN² pathway.</p><p>If an excess of HI is used the methyl alcohol formed in Step II is also converted into methyl iodide by the following mechanism : -</p><p>In the first step, methoxymethane reacts with hydrogen iodide to extract a proton to give the dimethyloxonium ion.</p><p>In the second step of the reaction the Dimethyloxonium ion reacts with the iodide ion present to yield methyl iodide and methyl alcohol as the product via SN²&nbsp; pathway.</p><p>In the third step of the reaction Methyl alcohol formed above reacts with hydrogen iodide to extract a proton to give the protonated methyl alcohol which finally reacts in the fourth step with the iodide ion via SN² pathway to give methyl iodide and water as the product.</p>

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