Sciencemadness Discussion Board

Preparation of Methyl Iodide Using HCl Gas

neddy17 - 17-12-2017 at 23:15

I ran a reaction I'd been intending to try for a while today, producing a moderate yield of methyl iodide. I'm curious as to what you all think, although having only done this once.
I dissolved potassium iodide in 50 mL of methanol, about 8g dissolved. I put some excess in thinking it may help force the equilibrium. With stirring, HCl gas was bubbled in over 20 to 30 minutes from a NaCl/H2SO4 generator, probably at least a mole based on NaCl, but I didn't measure this. Over the course of the gas introduction the mixture went from clear to cloudy, progressing from faint yellow to fairly red. After gas addition, I let it stir for another 30 mins and distilled. I collected two fractions, the first up until about 50 C, which had a density of 1.62 g/mL, 2.2g total. The second, up until about 58 C had a density of 1.25 g/ml, 3.5 g total. Ran an NMR on the first, with a peak at 2.2 ppm integrating to 40% of total. Pretty clean spectra with no water. Other than methanol, just a broad singlet around 2.6 which I have yet to identify.
I did this with ebay KI and hardware store acid and methanol, didn't dry anything at all. I was actually thinking catalytic water may help with acid dissociation, not sure how well HI dissociated in MeOH. But this seems moderately successful, and I think worth scaling up. Next time, I might reflux before distilling, but I'm not sure if this would cause me to lose too much HI. I might also distill to dryness and then fractionally distill the distillate. Any input?

KiWiki - 19-12-2017 at 08:29

1) Source?
2) A strong acid will release a weaker acid. Since HI(aq) is the stronger acid nothing will happen:

KI(aq) + HCl(g) -> KI(aq) + HCl(g) (equilibrium)

3) There is no way to get good yields (over 10%) by the reaction of methanol to methyl iodide by solid KI or NaI with combined with a stronger acid.

Prepare PI3 instead (P and iodine in DCM) and let PI3 react with MeOH

4) For the entire procedure please: UTFSE

DJF90 - 19-12-2017 at 08:40

Actually KiWiki, theres a nice write up in the Prepublications section on the preparation of MeI using KI and H3PO4, by Arrhenius, where yields in excess of 80% were reported.

Whilst you're right about the equilibrium between HCl and HI, hydrogen iodide does not need to be formed in situ - it is the methanol that needs to be protonated, and the iodide can then displace the water.

Melgar - 19-12-2017 at 09:09

I was going back and forth over whether this would work or not. My hunch is that it would work, just not very well, and that phosphoric acid would work a lot better as the acid, since it'd make for an easier workup.

Niter of Potash - 19-12-2017 at 09:43

I guess that first MeCl is formed, and then Cl in methyl cholride gets swapped for I from KI, forming MeI + KCl. KCl should then fall out from solution driving reaction forward.

NurdRage used similar "trick" in his pyrimethamine synthesis, step 5, when he turned p-chlorobenzyl chloride into p-chlorobenzyl iodide, only using acetone as solvent in that case.

Melgar - 19-12-2017 at 10:09

Quote: Originally posted by Niter of Potash  
I guess that first MeCl is formed, and then Cl in methyl cholride gets swapped for I from KI, forming MeI + KCl. KCl should then fall out from solution driving reaction forward.

NurdRage used similar "trick" in his pyrimethamine synthesis, step 5, when he turned p-chlorobenzyl chloride into p-chlorobenzyl iodide, only using acetone as solvent in that case.

Nope, the reaction you're referring to is the Finkelstein reaction, and works due to the fact that sodium iodide has a anomalously high solubility in acetone compared to sodium chloride.

This reaction works because iodide ions have an easier time donating electrons than chloride ions do, since its valence electrons are further away from the nucleus. And because this is an aqueous solution, it makes more sense to think of it as a bunch of K+, H+, Cl-, and I- ions floating around, none of which is permanently attached to any other one. Thus, for this reaction the K+ and Cl- ions are nonparticipants.

JJay - 19-12-2017 at 10:59

Quote: Originally posted by Melgar  
I was going back and forth over whether this would work or not. My hunch is that it would work, just not very well, and that phosphoric acid would work a lot better as the acid, since it'd make for an easier workup.


Nicoderm posted an experiment where he stored methanol, reagent grade HCl, and sodium iodide for a week in a bottle, and the results were pretty convincing and the yields were well over 50% despite the lack of optimization. I am pretty sure it would work with potassium iodide unless there are solubility issues (I didn't see any but didn't spend a great deal of time looking).

sykronizer - 27-12-2017 at 22:38

Is that some smoking cessation patch you are referring to there, is it ? .......

JJay - 28-12-2017 at 02:17

*Nicodem

He's a highly respected moderator who has made many contributions.

See: http://www.sciencemadness.org/talk/viewthread.php?tid=65075

[Edited on 28-12-2017 by JJay]

Melgar - 18-1-2018 at 21:23

Quote: Originally posted by JJay  
*Nicodem

He's a highly respected moderator who has made many contributions.

He also seems to be in his element when he's grouchy. Some of his best posts are calling out people who don't know what they're talking about, but pretend to. I get the feeling he's dabbled in the dark arts in his younger days, and doesn't see anything wrong with that, as long as you're doing something new and interesting.