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yanzi
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[*] posted on 18-1-2016 at 18:12
synthesis of methyl iodide from methanol, HCl and sodium iodide?


What's the drawback to using a combination of concentrated 37 wt% HCl and sodium iodide (perhaps with a phase transfer catalyst like TBAB) to turn alcohols into alkyl halides?

In fact, what's the drawback to using HCl + NaI as an HI substitute for any reaction where HI is needed? The only drawback I think is the solubility of NaI in organic solvents versus the lipophilic nature of HI. But what if you toss in a little PTC (I have 500g of TBAB) to help with the solubility of sodium iodide in the alcohol?

From the way I see it the key transition state I--CH3(H2O+) would still be able to occur.

[Edited on 19-1-2016 by yanzi]
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[*] posted on 18-1-2016 at 18:30


I believe the main drawback is partial oxidation of the iodide to iodine, which is why H3PO4 is generally preferred as the displacing acid.

There's also this alternative:

http://www.orgsyn.org/demo.aspx?prep=CV2P0399

[Edited on 19-1-2016 by blogfast25]




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yanzi
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[*] posted on 18-1-2016 at 20:30


lol I don't want to deal with elementary phosphorus. I have a pretty good supply of sodium borohydride. Surely this can reduce any elemental iodine that forms? The only issue is that I don't want my sodium borohydride reacting with my protonated alcohol, but AFAIK sodium borohydride only reacts with with tertiary (sometimes benzyl, allyl, certain secondary) halides in a highly polar solvent like DMSO or DMF. The protic nature of my alcoholic solvent (MeOH or EtOH, depending on my target alkyl halide) should reduce the nucleophilicity of my borohydride and make it just a "mop-up" agent for I2?

[Edited on 19-1-2016 by yanzi]
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DrMethyl
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[*] posted on 19-1-2016 at 06:23


I don't understand why you need BH4 to make MeI ???

You don't need to work with elemental P, MeI can be made with iodide salt, methanol and phosphoric acid like blogfast suggested. I did the reaction twice, one time it worked well : I distilled MeI from the mixture and the orther time it didn't work out for any reason ...
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[*] posted on 19-1-2016 at 06:39


Blogfast, I generally prepare iodine by treating potassium iodide with concentrated HCl and then by oxidizing the HI in equilibrium to elemental iodine. There doesn't seem to be much of an appreciable production of iodine without an oxidizing agent present as only a mild brownish color evolves and no precipitation accompanies the acidification.

One drawback I can see in the production of methyl iodide is the tendency of methyl iodide to hydrolyze, which I imagine wouldn't be helped by the presence of so much water in your HCl. That being said, I have managed to prepare methyl iodide in pretty wet conditions simply using iodine, methanol, and aluminium powder. The yields, however, was a dismal 25% or so; not much to write home about. With water present, a lot of the iodine ends up as hydrated aluminium triiodide, from which it can be recovered with moderate difficulty.

If you're not looking to purchase concentrated phosphoric acid to use, I might recommend treating tricalcium phosphate with fairly concentrated sulfuric acid and just vacuum filtering off any precipitated solids. Your filtrate should be sufficient for preparing MeI.




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yanzi
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[*] posted on 7-2-2016 at 10:52


Quote: Originally posted by DrMethyl  
I don't understand why you need BH4 to make MeI ???

You don't need to work with elemental P, MeI can be made with iodide salt, methanol and phosphoric acid like blogfast suggested. I did the reaction twice, one time it worked well : I distilled MeI from the mixture and the orther time it didn't work out for any reason ...


The BH4 would reduce any generated iodine back into iodide. Yield control.

I am considering ordering some calcium sulfate, which is a high-efficiency low-capacity dehydrating agent (it can reduce water levels all the way to 0.004 g/L), I probably plan to use a mixture of magnesium sulfate (which is higher capacity, but lower efficiency at 2.8 g/L) and calcium sulfate.

How much water is allowed to be present in the reaction mixture when making MeI from MeOH? If I start with 50 mL of methanol, would having 200 microlitres of water in the reaction mixture at any one time (by using the more economical magnesium sulfate instead of calcium sulfate) be a big issue?

[Edited on 7-2-2016 by yanzi]
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[*] posted on 7-2-2016 at 11:14


I actually would rather not deal with concentrated phosphoric acid as I hear that long-term exposure of borosilicate glass to phosphoric acid will eventually corrode the glass (not as much as hydrofluoric acid, but still). Probably from the reaction of the calcium/lime in the glass with fluoride and phosphate anions...

Here's my scheme so far for a pilot reaction:
50 mL methanol --> 1.2 mol MeOH
25 mL concentrated 12M HCl -> 1.05 mol water, 0.3 mol HCl

So my theoretical maximum yield is 25% or 0.3 mol MeI. This is OK, considering the excess methanol is cheaper than concentrated HCl or sulfuric acid and can probably be recycled.

Heat reaction mixture to 45C.

Once the mixture is well-mixed, I am going to add 19 grams (0.16 mol) of magnesium sulfate, which should be able to absorb 1 mol of water (1 mol of magnesium sulfate can absorb 7 mol of water).

Then add 45g (0.3 mol) sodium iodide -- which I don't know if I should add piecewise or gradually, or all at once.

Distill MeI, which boils at 42.5 C.

I am aiming for around 0.15 mol MeI, or 15% yield from methanol. This yield sounds horrible, but probably aiming for 50% yield in terms of amount of NaI consumed. Plus 0.15 mol MeI is a *lot* in terms of equivalents for reacting with nucleophiles.

[Edited on 7-2-2016 by yanzi]
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[*] posted on 7-2-2016 at 12:27


Quote: Originally posted by yanzi  
How much water is allowed to be present in the reaction mixture when making MeI from MeOH?


I think you're supposed to be aiming for none, so I'd try that first.




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[*] posted on 7-2-2016 at 20:27


What do you mean by none? Is the efficiency of magnesium sulfate (2.8 g/L) etc. OK for this procedure or will I eventually need to add calcium sulfate (with a dehydrating efficiency of 0.004 g/L) after most of the water is gone?

Are there any complications I should envision from mixing 1 eq. methanol + 12M HCl (0.25 equivalent HCl) + magnesium sulfate? I guess I'm hoping the HCl stays dissolved in the methanol after the magnesium sulfate absorbs all that water, rather than outgassing...

There's no harm in using a large excess of magnesium sulfate, right? (i.e. some of it will be undissolved) This is because I need to dehydrate both the HCl and the water generated from the protonated -OH leaving group.

[Edited on 8-2-2016 by yanzi]

[Edited on 8-2-2016 by yanzi]

[Edited on 8-2-2016 by yanzi]
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S.C. Wack
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[*] posted on 8-2-2016 at 13:33


Quote: Originally posted by yanzi  
What do you mean by none?


No water, as detailed in the preparation of iodomethane thread.

Trivial bonus trivia: the dude who published this would be 203 if he was still alive (AFAIK it was neither the alkyl iodides or nitroglycerin that killed him suddenly at 85).




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[*] posted on 8-2-2016 at 13:52


Quote: Originally posted by yanzi  
What's the drawback to using a combination of concentrated 37 wt% HCl and sodium iodide (perhaps with a phase transfer catalyst like TBAB) to turn alcohols into alkyl halides?



What makes you even think you'd get CH3I instead of (the even lower boiling) CH3Cl?

Primary alcohols can't be halogenated with HX in the presence of water.

[Edited on 8-2-2016 by blogfast25]




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[*] posted on 9-2-2016 at 04:45


Quote: Originally posted by blogfast25  

Primary alcohols can't be halogenated with HX in the presence of water.


Please do not spread misinformation. Just because you are unaware of such instances does not mean it is not possible. Please find attached a pertinent reference by Norris et al.

In addition, reactions may be facilitated and yields are frequently improved by using additives such as zinc chloride (with aq. HCl) or sulfuric acid (with aq. HBr). Despite the additive, these reactions are still conducted in the presence of water.

The conditions tend to be substrate dependant, seeing how the reactivity of alcohols spans quite a wide range. Not only do you have to consider whether the alcohol is 1*/2*/3*, but also whether it is allylic/benzylic, and electronic effects of other nearby substituents etc. For example, it is well known that benzyl alcohol reacts with concentrated aqueous HCl to give a good yield of the chloride despite being a 1* alcohol. Methyl chloride on the other hand requires passing HCl gas through a solution of zinc chloride in methanol.

Attachment: The reaction between alcohols and aqueous solutions of hydrochloric and hydrobromic acids.pdf (200kB)
This file has been downloaded 634 times


[Edited on 9-2-2016 by DJF90]
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blogfast25
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[*] posted on 9-2-2016 at 08:39


@DJF90:

Fair points.

My comment was made in the light OP's rather naive plan to synthesise CH3I.

There are exceptions to what I wrote but quite specific ones Even sec. alcohols (bar exceptions) can't be turned into alkyl chlorides w/o catalyst (often ZnCl2).

Thanks for the link.




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[*] posted on 9-2-2016 at 09:40


For the benefit of the OP here is a link to the mentioned Iodomethane thread in Prepublications:

http://www.sciencemadness.org/talk/viewthread.php?tid=12475

The use of an acid/iodide system relies on removing the volatile MeI as an entropic driving force. Use of a non-volatile, non-oxidising acid such as H3PO4 is ideal. You may have some (limited) success with HCl but why bother when you can use a verified and high yielding procedure? The required phosphoric acid should not be too difficult to aquire.

@Blogfast - There are always exceptions but I had a problem with the generality of your statement, seeing how incorrect it was. You mention that there are issues with generating 2* alkyl chlorides without catalyst but even so that occurs in the presence of water. You specified aqueous HX as the issue, yet there are many preparations where such acid is used for the preparation of alkyl halides (whether X = Cl, Br, I). In fact I have found it to be less common to use anhydrous HX to prepare alkyl halides from the alcohols, and in such cases an alternative reagent (system) is often used to advantage. These alternative conditions are often required for the mild/chemoselective transformation of ROH to RX in the presence of other (sensitive) functional groups.

[Edited on 9-2-2016 by DJF90]
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[*] posted on 9-2-2016 at 10:06


Quote: Originally posted by DJF90  
You mention that there are issues with generating 2* alkyl chlorides without catalyst but even so that occurs in the presence of water.


Of course, I've done it myself (isopropyl chloride). ZnCl2 wouldn't be much good (in this context) w/o some solvent.




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[*] posted on 12-2-2016 at 23:21


Another option if you have elemental iodine but no red phosphorus is to use aluminium foil instead!

I know it sounds crazy but we randomly tried it last year (video here: https://www.youtube.com/watch?v=71airqaAvPQ) using ethanol and it ended up working really well. So unless someone knows better I can't see why it wouldn't also work with methanol as well.

The downsides are that it does take a long time for the reaction to work and for distillation to complete (although if you're doing it properly with phosphorus and want a good yield it also takes a while) so you're going to need patience. The amount of heat generated (though over a period of time) by the iodine and aluminium reacting is also quite scary. The mixture stays hot for literally hours! So be wary about this on a large scale - iodine and aluminium reacting you DON'T want to get out of control.




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[*] posted on 13-2-2016 at 07:27


Quote: Originally posted by chemplayer..  
Another option if you have elemental iodine but no red phosphorus is to use aluminium foil instead!

I know it sounds crazy but we randomly tried it last year (video here: https://www.youtube.com/watch?v=71airqaAvPQ) using ethanol and it ended up working really well. So unless someone knows better I can't see why it wouldn't also work with methanol as well.



That's quite remarkable indeed! Would be well worth trying with MeOH.

Well done.

I think the final yellowish colours is due to I2: EtI isn't very stable, I think. That would explain why it was colourless after the wash with bisulphite, then picked up colour again.
<hr>

In the case of MeOH to MeI with Al/I2 you might want to consider reactive fractionation: with a fractionation column between the RBF and final condenser and considering the BPs of MeOH and MeI are sufficiently different, you might be able to distil off relatively pure MeI during reaction, thereby also pulling the equilibrium further to the right.

[Edited on 13-2-2016 by blogfast25]




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[*] posted on 13-2-2016 at 16:42


EtI is a nightmare. It seems to be a lot less stable than MeI and turns yellow in a week and brown in a month. It also has a very annoying tendency to escape any vessel you put it in (even pro quality Duran with PTFE inner-seals) over time. At least that's what we found - we also found that adding copper wire as a stabiliser didn't work to prevent this.

Now it may well be that it's an impurity in the product causing / catalysing the decomposition but we found this after re-distilling and carefully drying the product from ANY of the usual methods of producing (KI+H3PO4+EtOH / P+I+EtOH / Al+I+EtOH) - no difference, the stuff still slightly decomposed and then escaped!

If you need it for a reaction then best make it fresh as you require it.




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[*] posted on 13-2-2016 at 17:20


Possible decomposition products (back-of-envelope hypothesis, nothing more):

* butane + I2
* ethane + ethene + I2

Either way the stronger electron pushing by the ethyl group would help here.

Perhaps oxygen is involved here?

The real prize here would be MeI by your method though. That would get a lot of OCers here into their labs, I imagine! :D

Also potentially interesting: the equivalent reactions with Br2, perhaps in a solvent?


[Edited on 14-2-2016 by blogfast25]




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[*] posted on 13-2-2016 at 18:37


Yes we were surprised the yield was so good (65%).

We'd played around with AlCl3 to see if it would react with acetic acid, and then tried AlCl3 and P2O5 together with acetic acid in the hope of making acetyl chloride. There is some sort of reaction but we never figured out what was going on - and it certainly doesn't make acetyl chloride.

But then we wondered if AlI3 would react with alcohols and just tried it out with ethanol. If no one tries it out within a few weeks we'll do another video using methanol and see what happens as it's pretty easy to set up and run.




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[*] posted on 13-2-2016 at 19:11


Quote: Originally posted by chemplayer..  
If no one tries it out within a few weeks we'll do another video using methanol and see what happens as it's pretty easy to set up and run.


Well, it may sound like a platitude but I look forward to these results... :)




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[*] posted on 13-2-2016 at 19:59


I mentioned it briefly in an earlier comment, but yes, Chem Player's method did actually net me some methyl iodide from following their same procedure while simply using methanol instead of ethanol. I ended up with a yield of around 25% using a similar scale, but I can definitely attest to using non-anhydrous methanol and surely some of it must have escaped my reflux condenser.



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[*] posted on 13-2-2016 at 20:38


Chemplayer:

Re.: "Yes we were surprised the yield was so good (65%)."

I watched your video and it was an interesting experiment. However your yield claim is without foundation as you present no data showing proof of composition or purity of the isolated liquid. You should distill the product to show a valid boiling point and compute the yield on the basis of the distillate weight. Crude and unverified yields do not count for much in organic chemistry.

AvB
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[*] posted on 13-2-2016 at 23:08


That's a fair comment AvBaeyer. The problem with the methyl iodide one is going to be that if it takes a few hours to run then even with an ice-cold condenser there's going to be some product inevitably lost as well.

Perhaps if we have a go at the methyl iodide equivalent then we'll include product distillation and see what we end up with.




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[*] posted on 14-2-2016 at 02:32


Quote: Originally posted by AvBaeyer  

I watched your video and it was an interesting experiment. However your yield claim is without foundation as you present no data showing proof of composition or purity of the isolated liquid. You should distill the product to show a valid boiling point and compute the yield on the basis of the distillate weight. Crude and unverified yields do not count for much in organic chemistry.
AvB


I haven't watched the video but I'm glad someone has picked up on this. It appears to be a common problem amongst the amatuer chemists here, in that they obtain a product and claim a yield of XX %th without ANY characterisation data. I'm pretty sure I've said before that although we're amatuer chemists (I prefer the term independent researchers) we don't have to be amatuer about it.

In this case the starting material and product boil close enough that the fractionation will be fairly involved unless the reaction is driven to complete conversion. Even though the product is a liquid there are several things we can do to show its the right stuff...

a) Take a density reading.
b) If you're lucky enough to have a refractometer, take a reading of the refractive index. The benefit of this is that any residual ethanol can be quantified.
c) Form a crystalline derivative and obtain a melting point. This is a very old school way to do things since the advent of modern instrumentation has made it unecessary. However as "amatuers" we do not have access to such instruments and so this is as good as it will get.

As an aside, Peach had done this reaction several years ago using iodine and aluminium, although I'm not sure if it was posted or mentioned in private communication. I had a quick search on the forum but could not find it. I did come across this gem from an old friend though:

Quote: Originally posted by entropy51  
Methyl iodide can also be made by gassing a saturated solution of KI in methanol with HCL gas.

I've not been able to find the ancient reference in which I found this method, but I have used it and it does work. I'd been thinking that I should post something on it since it does not seem to be generally known.


To the best of my knowledge he unfortunately did not follow up on that.
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