Methyl Iodide synthesis I was trying to determine whether I could synthesize methyl iodide by combining methanol, potassium iodide and sulphuric acid together... The way I see it, the sulphuric acid will protonate the OH instead of the I forming a very stable leaving group of water. Then the iodine ions would donate an electron pair to the carbon. You would be left with the salt Potassium bisulfate.... Then to remove my methyl iodide, just distil. 

Will this work?
So what I said in the comment still holds: it could work (particularly in combination with a Dean-Stark apparatus to remove excess water), but there is no way to know for sure unless someone tries it out.

The more usual (and early) method is described in Organic Syntheses and utilizes the reaction of methanol with iodine in the presence of phosphorous



I don't know the exact reaction mechanism (if anyone has a clue, please feel free to edit!) but what happens is that $\ce{PI_3}$ is generated in situ from the elemental iodine and phosphorous and reacts with methanol to yield the methyl iodide and phosphorous acid

$\ce{3 CH3OH + PI3 -&gt; 3 CH3I + H3PO3}$

Phosphorous acid does not act as nucleophile and so the problem of a competing backward reaction, which you would have in your proposed synthesis, is largely prevented.

I'd also like to add some caution: the reaction described in Organic Syntheses is not suited for beginners or in amateur setting. The chemicals involved are nasty: highly toxic, highly flammable, alkylating and pyrophoric. Don't try this at home :)