Sciencemadness Discussion Board

synthesis of methyl iodide from methanol, HCl and sodium iodide?

yanzi - 18-1-2016 at 18:12

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]

blogfast25 - 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]

yanzi - 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]

DrMethyl - 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 ...

Amos - 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.

yanzi - 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]

yanzi - 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]

S.C. Wack - 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.

yanzi - 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]

S.C. Wack - 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).

blogfast25 - 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]

DJF90 - 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 642 times


[Edited on 9-2-2016 by DJF90]

blogfast25 - 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.

DJF90 - 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]

blogfast25 - 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.

chemplayer.. - 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.

blogfast25 - 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]

chemplayer.. - 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.

blogfast25 - 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]

chemplayer.. - 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.

blogfast25 - 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... :)

Amos - 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.

AvBaeyer - 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

chemplayer.. - 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.

DJF90 - 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.

chemplayer.. - 14-2-2016 at 03:54

KI in methanol and HCl. That's also very interesting and worth a try.

Does anyone think that plain old HI in aqueous solution (a strong one mind you) with a catalyst like ZnCl2 would work?

Yes, once you've got the equipment and reagents, finding ways to end up with vaguely pure products and even half-heartedly trying to confirm their identity is to be honest >90% of the challenge. If it's something like (say) ethyl iodide where we can wash it with water and sodium bicarbonate, then dry it completely, and can measure a rough density that is within about 5-10% then that's a good enough confirmation for us and we'll then try the product out in a subsequent reaction to see if it behaves as expected. Normally we find that the best test is trying out a subsequent reaction.

To us a crude yield is really a 'maximum yield'. It means 'hooray we got something that isn't an obviously impure sludge, and it seems to match what the procedure said we'd get or the high level expected characteristics'. Which frankly for us at this point anyway is a good enough success in itself.

We don't personally consider ourselves 'amateur scientists' or 'independent experimenters'. We just do reactions for fun and to learn as neither of us are chemists and it's such an amazing and fascinating science. Obviously if your goals are different (e.g. a specific synthesis or genuine innovation) then having a much more rigorous approach is necessary. To be fair it almost certainly adds to the enjoyment and satisfaction ultimately as well - but we're still on the road to becoming that disciplined...

blogfast25 - 14-2-2016 at 07:07

@chemplayer

Well, it's not too late do a capillary BP and a density on the EtI, unless it's been ditched of course.

JJay - 14-2-2016 at 07:38

I think HI would probably work even without a catalyst, but yields might not be the greatest that way.

I wish I had some HI. I could make some, but it is covered by all sorts of legal restrictions here.

Edit: I am too tired for the Internet right now. *yawn*

[Edited on 14-2-2016 by JJay]

Amos - 14-2-2016 at 07:49

Quote: Originally posted by JJay  
I think HI would probably work even without a catalyst, but yields might be the greatest that way.

I wish I had some HI. I could make some, but it is covered by all sorts of legal restrictions here.


Make it and use it the same day, without storing it. There are a couple of youtube videos detailing its preparation.

S.C. Wack - 14-2-2016 at 10:56

The first discovery using aluminum was published in 1926, just ethyl iodide, and had an unexpected workup procedure that nonetheless led to 80% yield they said. Then a Russian in 1941 included methyl iodide, no yield given in the abstract. Particularly interesting there is also a 70% yield of methyl iodide and phenol from anisole.

Norris illustrated ethyl iodide from HI and ethanol in Experimental Organic Chemistry, but did not mention substituting methanol. He did substitute it years earlier though: 5.5 g methanol, 105 g. 57% HI, 21 g. methyl iodide.

For the methyl and ethyl iodides, it seems reasonable to assume that the washed and dried distillate from any preparation would be as pure as the alcohol it came from.
Quote: Originally posted by DJF90  
To the best of my knowledge he unfortunately did not follow up on that.


There was an earlier post.
http://www.sciencemadness.org/talk/viewthread.php?tid=12394&...

BTW the higher solubility of the NaI here may be advantageous.

[Edited on 14-2-2016 by S.C. Wack]

blogfast25 - 14-2-2016 at 12:56

Quote: Originally posted by S.C. Wack  
The first discovery using aluminum was published in 1926, just ethyl iodide, and had an unexpected workup procedure that nonetheless led to 80% yield they said.


Yes, this related patent mentions it:
Quote:
In two works, Jones and Green, J. Chem. Soc. 1926, 270; J. Chem. Soc. 1927, 928, report the reaction of aluminum with three atoms of iodine, with hydrolysis of the aluminum tri-iodide in the presence of an alcohol and water to produce the corresponding alkyl iodide. These Workers state aluminum can advantageously be used instead of phosphorus in the synthesis.


AJKOER - 19-2-2016 at 07:01

Thought of an interesting path based on CH4 (and not CH3OH), HCl to produce Cl2 from say NaOCl, NaI, acetone and sunlight to create an impure CH3I.

First, perform a photolysis of CH4 in excess and Cl2 to create some CH3Cl and other products.See, for example, reaction summary and discussion at http://www.chemguide.co.uk/mechanisms/freerad/multisubcl.htm... . Or, one can repeat the path first used to synthesized by the French chemists Jean-Baptiste Dumas and Eugene Peligot in 1835 from boiling a mixture of methanol, sulfuric acid, and sodium chloride. Source: https://en.m.wikipedia.org/wiki/Chloromethane .

Second, dissolve separately the NaI and any formed CH3Cl (as a gas and other impurities) both in acetone. Combine solutions and let the NaCl precipitate out. Source to quote from Wikipedia (https://en.m.wikipedia.org/wiki/Iodine ):

"Sodium iodide is especially useful in the Finkelstein reaction, because it is soluble in acetone, whereas potassium iodide is less so. In this reaction, an alkyl chloride is converted to an alkyl iodide. This relies on the insolubility of sodium chloride in acetone to drive the reaction:

R-Cl (acetone) + NaI (acetone) → R-I (acetone) + NaCl (s)"

[Edited on 19-2-2016 by AJKOER]

[Edited on 19-2-2016 by AJKOER]

blogfast25 - 19-2-2016 at 07:20

@AJ:

You are aware that the BP of CH3Cl is about - 24 C, right?

AJKOER - 19-2-2016 at 07:36

Thanks Blogfast for the comment.

I have since edited my thread to be clear that one is to dissolve the gaseous CH3Cl in acetone.

[Edit] The bottom line of my suggested route is that if you has a good workable path to CH3Cl, then via NaI and acetone, one may be able to produce CH3I as well.

[Edited on 19-2-2016 by AJKOER]

blogfast25 - 19-2-2016 at 08:39

Quote: Originally posted by AJKOER  


I have since edited my thread to be clear that one is to dissolve the gaseous CH3Cl in acetone.

[Edit] The bottom line of my suggested route is that if you has a good workable path to CH3Cl, then via NaI and acetone, one may be able to produce CH3I as well.



The preparation of CH3Cl from methanol, NaCl and conc. H2SO4 should be unproblematic.

But you can't prepare a solution of CH3Cl in acetone without pressure to keep the CH3Cl in there. Not with such a low BP.

One possibility would be to gas a solution of NaI in acetone with CH3Cl but that's not really for the faint of heart either...

[Edited on 19-2-2016 by blogfast25]

Nicodem - 23-2-2016 at 12:06

A lot of talk instead of doing an experiment. Like this:

To NaI (18.8 g, 125 mmol) was added methanol (8 g, 250 mmol, 2 eq.) and 37% hydrochloric acid (42 mL, 500 mmol, 4 eq.). There was a slight exotherm. The yellow suspension was left standing stoppered at room temperature. The next day there was a brown liquid present bellow the precipitated salts. After standing for one week, the salts were dissolved by adding water (40 mL) and the aqueous phase decanted. The heavy red liquid product was washed with a solution of ascorbic acid (0.2 g) in water (40 mL) and separated to give a colorless liquid (9.95 g, 56%): density 2.30 g/mL (+/- 0.15 g/mL); bp 43 °C.

blogfast25 - 24-2-2016 at 10:26

Quote: Originally posted by Nicodem  

To NaI (18.8 g, 125 mmol) was added methanol (8 g, 250 mmol, 2 eq.) and 37% hydrochloric acid (42 mL, 500 mmol, 4 eq.). There was a slight exotherm. The yellow suspension was left standing stoppered at room temperature. The next day there was a brown liquid present bellow the precipitated salts. After standing for one week, the salts were dissolved by adding water (40 mL) and the aqueous phase decanted. The heavy red liquid product was washed with a solution of ascorbic acid (0.2 g) in water (40 mL) and separated to give a colorless liquid (9.95 g, 56%): density 2.30 g/mL (+/- 0.15 g/mL); bp 43 °C.


Hmmm... works rather well.

DJF90 - 24-2-2016 at 11:17

Quote: Originally posted by Nicodem  
A lot of talk instead of doing an experiment. Like this:

To NaI (18.8 g, 125 mmol) was added methanol (8 g, 250 mmol, 2 eq.) and 37% hydrochloric acid (42 mL, 500 mmol, 4 eq.). There was a slight exotherm. The yellow suspension was left standing stoppered at room temperature. The next day there was a brown liquid present bellow the precipitated salts. After standing for one week, the salts were dissolved by adding water (40 mL) and the aqueous phase decanted. The heavy red liquid product was washed with a solution of ascorbic acid (0.2 g) in water (40 mL) and separated to give a colorless liquid (9.95 g, 56%): density 2.30 g/mL (+/- 0.15 g/mL); bp 43 °C.


Well done for taking the time and initiative to give it a go Nicodem. Its a pretty nice result although I have a question... Is there a rationale behind how you conducted the experiment (molar equivalents, lack of stirring and heating)?

Nicodem - 25-2-2016 at 08:09

Quote: Originally posted by DJF90  
Is there a rationale behind how you conducted the experiment (molar equivalents, lack of stirring and heating)?

There is no reaction related rationale for lack of stirring. It's just that I'm flooded with work and could not afford to occupy a stirrer for that long. You can see the motivation as one of those moments: "Why the f***k do we have members that ask questions that can be answered so easily by a ridiculously simple experiment?"

Heating would be pointless. I would need to set up a reflux (occupying a stirrer), and with the bp of the product, the highest achievable temperature would not be particularly helpful. Containing MeI for long enough with normal cooling water would be hard on such a small scale. Just stoppering the reaction can be a compromise between reaction time and simplicity.

Molar equivalents? You don't expect me to do a DoE model on this? Except for having NaI as the limiting reagent, the rest was an arbitrary/intuitive choice.

DJF90 - 25-2-2016 at 12:25

Quote: Originally posted by Nicodem  

It's just that I'm flooded with work and could not afford to occupy a stirrer for that long.


I figured that might be the case.
Quote: Originally posted by Nicodem  

Heating would be pointless. I would need to set up a reflux (occupying a stirrer), and with the bp of the product, the highest achievable temperature would not be particularly helpful.


I was just wondering what made you deviate from the procedure used with phosphoric acid (reactive distillation). You've answered that clearly, thanks.
Quote: Originally posted by Nicodem  

Molar equivalents? You don't expect me to do a DoE model on this? Except for having NaI as the limiting reagent, the rest was an arbitrary/intuitive choice.


I didn't expect you to do anything, although I'm glad that you gave it a go. I suspected they were just educated choices.

dicyanin - 9-5-2021 at 08:53

My apologies for waking up the dead, but I thought I might as well share my own notes here using a variation of the methylsulfuric acid process (using methanol as solvent instead of water, and NaI instead of KI).

Preparation of methyl iodide:
A solution of methylsulfuric acid is made by adding 50g concentrated sulfuric acid (pre-ban drain cleaner ~92%) dropwise with stirring to 25ml (0.6mol) methanol in a cold-water bath.

In a 500ml flask equipped with magnetic stirrer and prepared for distillation, 45g (.60mol) of sodium iodide is dissolved in 70ml methanol. This is stirred in a distillation setup fitted with a Claisen adapter and a separatory funnel, until the full dissolution of the NaI (lightyellow tinge of the solution), and then heated on an oil bath until the bath temperature measures between 80-85*C. To this solution, methylsulfuric acid is added dropwise via an addition funnel, and the methyl iodide distills almost immediately. The bath temperature is brought to 90°C towards the end of the addition. A precipitate of Na2SO4 is formed, the mixture turns darker. (As sodium bisulfate cannot exist in alcoholic media, it decomposes on forming into sodium sulfate and sulfuric acid. The Na2SO4 precipitates from solution and drives the reaction forward). The product distilling over turns milky sometimes.

After about three hours, no more methyl iodide was seen to distill. The receiving flask contained 50ml of brine (saturated NaCl solution), cooled by an ice/water bath. Methyl iodide forms the lower layer, the first thing that catches the eye is its high refractive index (for which it has some uses in microscopy). The layers are separated and the MeI is dried over CaCl2. Yield: 21.7g (51%)


Results are somewhat worse than the procedure that uses water as solvent, it seems that there is no way to suppress partial oxidation to iodine using sulfuric acid/methylsulfuric acid.

Recovery of iodine:
200ml water is added to the distilling flask and the mixture is stirred until all precipitated sodium sulfate has dissolved. The whole is transferred to a 1L beaker and with good stirring 300ml 3% H2O2 is added. The mixture turned from an oily purplish to darkbrown and a precipitate formed. Stirring was discontinued and the mixture allowed to stand for an hour.

When the dark precipitate of iodine has settled, the supernatant layer was decanted off, the slurry of iodine filtrated and washed with several portions of ice-cold distilled water. The filter-cake was pressed to remove as much moisture as possible, then dried over CaCl2. The amount of iodine recovered was 9.9 grams.

Boffis - 9-5-2021 at 09:31

No apology needed for waking the dead! Nice experiment and fairly succesful. The slightly lower yield that Nicodem's experiment with HCl is probably not significant. Interestingly, I have often found that if conc sulphuric acid is too oxidizing (eg when preparing alkyl bromides from alcohols with NaBr and sulphuric acid) the problem can often be reduced by diluting the sulphuric acid to <85%. This shouldn't greatly affect things if Nicodem can get away with 37% HCl solution so you may be able to get away with an even lower concentration.