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Author: Subject: Preparation of weak hydroiodic acid(for making NaI)
B.D.E
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[*] posted on 29-12-2019 at 08:01
Preparation of weak hydroiodic acid(for making NaI)


Hey there, so I need some sodium iodide for the "Finkelstein reaction".

This is what I have:
Iodine, KI, phosphoric acid(85%), NaOH, KOH, HCl(~30%).

Because NaI is waaay more soluble than KI, NaOH and KOH(in literary every solvent I could find), I aim for a preparation without any ionic side products.

The obvious way is to mix the KI with the phosphoric acid and distill in order to obtain HI(aq)(which can later be neutralize with NaOH to produce NaI and water). The problem is that I don't have the equipment needed in order to reach ~400C.


Is there a way to make HI solution(no matter the concentration) at temperatures below ~230C? preferably with decent yields(in regards to the KI).


Any attempts to help would be much appreciated.
Ben :)

[Edited on 29-12-2019 by B.D.E]
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[*] posted on 29-12-2019 at 13:45


Well you could also make iodine and reduce that using H2S or red P, hydrazine hydrate should also work. But why do you want the destillation to reach 400°C? The HI/Water azeotrope boils at 127°C, so quite below 400°C. Maybe you'll need a reaction temperature higher than the 127°C, but I don't think so. But I'm pretty shure you won't need to reach even nearly 400°C ;)
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[*] posted on 29-12-2019 at 19:56


You could also just try the Finkelstein reaction with KI. It might not work as well as NaI (since the solubility difference wouldn't be as favorable) but I think it's worth a try.



As below, so above.
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[*] posted on 29-12-2019 at 21:05


It is possible to convert iodine directly to an iodide salt via aqueous redox chemistry. This is described in the attached patent. There may be other references to be found with a bit of searching.

KI can be used in a Finkelstein reaction. It has limited solubility in acetone or MEK so the reaction will be slow. It does react efficiently in DMF but that may not be of much help if you cannot find DMF or your reactants are incompatible with DMF.

AvB

Attachment: Iodine to Iodide US1918622.pdf (407kB)
This file has been downloaded 36 times

[Edited on 30-12-2019 by AvBaeyer]
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[*] posted on 30-12-2019 at 00:11


Another option: react elemental iodine with magnesium metal in water, filter off excess magnesium, then add sodium hydroxide or carbonate and filter again. You can remove excess Na2CO3 by dissolving the iodide in acetone or ethanol.
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[*] posted on 30-12-2019 at 01:26


Iodine+phosphorus or hypophosphorus acid equals HI which can be distilled then neutralised with naoh
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[*] posted on 30-12-2019 at 08:02


You wrote that you have iodine and NaOH. If the NaIO3 does not interfere then just dissolve I2 in NaOH and warm up. If NaIO3 causes problems then you can separate NaI/NaIO3 from the mixture using ethanol where NaI soluble, NaIO3 insoluble.
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[*] posted on 1-1-2020 at 13:18


Quote: Originally posted by Tellurium  
Well you could also make iodine and reduce that using H2S or red P, hydrazine hydrate should also work. But why do you want the destillation to reach 400°C? The HI/Water azeotrope boils at 127°C, so quite below 400°C. Maybe you'll need a reaction temperature higher than the 127°C, but I don't think so. But I'm pretty shure you won't need to reach even nearly 400°C ;)

I've read the "argox" guide for making HI(from the wiki):
"The reaction is not over when all the brown acid has boiled out of the
reactor, it has just begun. Keep the heat on high and watch in
amazement as more and more and more acid keeps forming in the
condenser. Donít worry about the thermometer at the still head going
above 127?C, it is still HI(aq) coming over, just the temp inside the
reactor is getting HOT. At 400?C both HI(aq) and HI will come over.
Lots of HI at a larger scale, so be prepared for it. About 10% of the total
acid production will be in the form of HI that must be collected in the
water trap/receiver." page 9/20

I think I'll give it a try anyway and see for myself with H3PO4. After all I'm doing it in a much smaller scale the he did.
Quote: Originally posted by Metacelsus  
You could also just try the Finkelstein reaction with KI. It might not work as well as NaI (since the solubility difference wouldn't be as favorable) but I think it's worth a try.

The reaction I'm trying to do is very low yielding. The paper I've read state they've only got 13% yield after 2 hours, using NaI as a catalyst, at a pressure of 10 atm. I'm planing leaving it for a day or so at atmospheric pressure hoping for the best. I agree it's possible that KI would also work, but I prefer first trying it with NaI.
Quote: Originally posted by AvBaeyer  
It is possible to convert iodine directly to an iodide salt via aqueous redox chemistry. This is described in the attached patent. There may be other references to be found with a bit of searching.

KI can be used in a Finkelstein reaction. It has limited solubility in acetone or MEK so the reaction will be slow. It does react efficiently in DMF but that may not be of much help if you cannot find DMF or your reactants are incompatible with DMF.

AvB

[Edited on 30-12-2019 by AvBaeyer]

It's good to know about the DMF solubility, but unfortunately I have no access to DMF.
Quote: Originally posted by TGSpecialist1  
Another option: react elemental iodine with magnesium metal in water, filter off excess magnesium, then add sodium hydroxide or carbonate and filter again. You can remove excess Na2CO3 by dissolving the iodide in acetone or ethanol.
Okay, I do have some magnesium turnings. I also have acetone and heptane that can be used as solvents for the iodine.
I'll definitely read more into it. Thanks!

Quote: Originally posted by draculic acid69  
Iodine+phosphorus or hypophosphorus acid equals HI which can be distilled then neutralised with naoh

Unfortunately I don't have any phosphorous (:
Quote: Originally posted by Fery  
You wrote that you have iodine and NaOH. If the NaIO3 does not interfere then just dissolve I2 in NaOH and warm up. If NaIO3 causes problems then you can separate NaI/NaIO3 from the mixture using ethanol where NaI soluble, NaIO3 insoluble.

I don't think it will work.
Due to the high solubility of NaI, the equilibrium might shift towards the NaOH. I can use excess water, but eventually I'll need to purify it(and the final product of my planed reaction boil at >200C). You should also keep in mind that even small amounts of NaOH would be problematic for the reaction, as they will react with the alkyl halide and hydrolyze it.


Thanks everyone :)

I ain't planning on doing anything until the weekend, so if anyone has more ideas or suggestions please share.


Ben.
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[*] posted on 2-1-2020 at 03:07


Quote: Originally posted by B.D.E  

Quote: Originally posted by TGSpecialist1  
Another option: react elemental iodine with magnesium metal in water, filter off excess magnesium, then add sodium hydroxide or carbonate and filter again. You can remove excess Na2CO3 by dissolving the iodide in acetone or ethanol.
Okay, I do have some magnesium turnings. I also have acetone and heptane that can be used as solvents for the iodine.
I'll definitely read more into it. Thanks!

I guess you mean iodide, not iodine. Also iodide doesn't dissolve in heptane.

For the iodine you don't need an extra solvent, as it dissolves in water quite well.

Quote: Originally posted by B.D.E  

Quote: Originally posted by Fery  
You wrote that you have iodine and NaOH. If the NaIO3 does not interfere then just dissolve I2 in NaOH and warm up. If NaIO3 causes problems then you can separate NaI/NaIO3 from the mixture using ethanol where NaI soluble, NaIO3 insoluble.

I don't think it will work.
Due to the high solubility of NaI, the equilibrium might shift towards the NaOH. I can use excess water, but eventually I'll need to purify it(and the final product of my planed reaction boil at >200C). You should also keep in mind that even small amounts of NaOH would be problematic for the reaction, as they will react with the alkyl halide and hydrolyze it.


I would go for this route, as I wouldn't mind having some NaIO3 on hand. NaI and NaOH are both insanely soluble, so don't worry about that, also this reaction is pushed too completion by other mechanisms.

I2 and NaOH react to form NaOI which is very unstable and decomposes into iodide and iodate. This reaction is not an equilibrium.

you will lose a bit of I2 in the form of iodate but that is a nice to have I would say. At elevated temperatures the disproportionation is fast. Boil the mixture down and dry at above 65 degrees to get rid of the water and extract with absolute ethanol. You could redissolve in water and neutralize any NaOH with HCl or HI if you worry about the NaOH.

You only have to dissolve iodine in hot NaOH untill the purple color doesn't disappear anymore, then add a bit more NaOH untill color disappears, boil down, dry, extract with ethanol, dry, redissolve, neutralize NaOH, dry.

[Edited on 2-1-2020 by Tsjerk]
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[*] posted on 2-1-2020 at 04:19


NaOH and I2 react quickly and irreversibly to NaI and NaIO, the latter also fairly quickly reacts to NaIO3 and NaI, especially if heated somewhat. Add I2 to your hot solution of NaOH (no need to have it boilingly hot, 60 to 70 C is OK) and keep adding I2, until the solution remains brown. Then you have a slight excess of I2. You could add a single granule of NaOH to get rid of the brown color. Important is to keep the liquid hot all the time, otherwise it is hard to see whether you have excess I2 or not. The ion IO(-) is yellowish, IO3(-) ion is colorless, and if temperature is kept high, then the IO(-) only exists for a short time.

Next, concentrating the liquid by boiling down and letting stand, slowly colling down, will cause most of the NaIO3 to crystallize, while NaI remains in solution. When the liquid is cold, then decant the clear liquid above the white solid. This is your NaI-solution, which now can be boiled down. The solid is fairly pure NaIO3. A single recrystallization from distilled water makes this also very pure and is a nice by-product.




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[*] posted on 3-1-2020 at 11:52


Tsjerk & woelen, thank you for the explanation. I tried this route and I'm pretty sure it worked.
The final yield was only 7gr(~35%), probably due to some sloppiness on the purification steps.

The final product has some yellowish tint that didn't disappeared even after two recrystallizations in acetone(in the freezer at -18C). I've read that it might be traces amounts of iodine due to the decomposition of NaI. But I doubt it will interfere the reaction so I'll just use it as is.

One interesting thing I've noticed is that the leftover product from the filtration has gone white after leaving it for a while(see attached photo).

The final product:
https://pasteboard.co/IOiBpzT.bmp

The leftovers from the filtration step:
https://pasteboard.co/IOiBIVF.bmp

And I guess that's it.
Thanks again to everyone that helped :P

Edit: Just for curiosity I rinsed the NaI with some n-heptane(which dissolve iodine, but not iodide salts).

The product after rinsing it with n-heptane:
https://pasteboard.co/IOnYUww.bmp


[Edited on 4-1-2020 by B.D.E]
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[*] posted on 4-1-2020 at 07:35


Yes, that looks like there is a bit of iodine contamination. But considering the intense color of iodine I looks minimal. I'm not an expert on this, but I think it might be possible iodine is complexing with the iodide in acetone and not in heptane.

Edit: Next time you can directly upload the pictures to Sciencemadness, that way they won't be lost when the host site you upload them to takes them of or goes down. That happened a lot in the past.
To do so you can hit "preview post" in under the field where you type your post, now you will see a field where you can upload. You also see it when you edit a post.

[Edited on 4-1-2020 by Tsjerk]
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[*] posted on 4-1-2020 at 11:03


Quote: Originally posted by B.D.E  
Edit: Just for curiosity I rinsed the NaI with some n-heptane(which dissolve iodine, but not iodide salts).


Iodine and iodide forms triiodide. I'm not sure that solvent washing would remove iodine from it.

Instead of magnesium, iron has been recommended. Followed by carbonate.




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[*] posted on 4-1-2020 at 15:42


The trouble with NaI in the Finkelstein is that NaI is strongly hygroscopic and water interferes with the reaction by solvating Cl- ions. KI is basically not hygroscopic and thus while it reacts much slower it is probably easier to ensure dry reaction conditions and therefore get a reaction.

Depending on your setup NaI may work better than KI, or not.




[Edited on 04-20-1969 by clearly_not_atara]
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[*] posted on 6-1-2020 at 10:53


Quote: Originally posted by Tsjerk  
Yes, that looks like there is a bit of iodine contamination. But considering the intense color of iodine I looks minimal. I'm not an expert on this, but I think it might be possible iodine is complexing with the iodide in acetone and not in heptane.

Edit: Next time you can directly upload the pictures to Sciencemadness, that way they won't be lost when the host site you upload them to takes them of or goes down. That happened a lot in the past.
To do so you can hit "preview post" in under the field where you type your post, now you will see a field where you can upload. You also see it when you edit a post.

[Edited on 4-1-2020 by Tsjerk]

Your idea also sounds plausible to me. And thanks for the instructions on how to upload photos directly to the post. I'll be sure to use it next time :)
Quote: Originally posted by S.C. Wack  
Quote: Originally posted by B.D.E  
Edit: Just for curiosity I rinsed the NaI with some n-heptane(which dissolve iodine, but not iodide salts).


Iodine and iodide forms triiodide. I'm not sure that solvent washing would remove iodine from it.

Instead of magnesium, iron has been recommended. Followed by carbonate.

The reaction between iodine and iodide ions is an equilibrium reaction. Assuming that the rinsing of the triiodide in heptane force any liberated iodine molecules to adopt a more permanent non-polar configuration(and thus non-reactive towards iodide ions), it is thus possible for those iodine molecules to be washed with the solvent. It is of course just wild speculations on my part, but no matter the explanation - it did worked(as evident from the pictures).

Quote: Originally posted by clearly_not_atara  
The trouble with NaI in the Finkelstein is that NaI is strongly hygroscopic and water interferes with the reaction by solvating Cl- ions. KI is basically not hygroscopic and thus while it reacts much slower it is probably easier to ensure dry reaction conditions and therefore get a reaction.

Depending on your setup NaI may work better than KI, or not.

Thanks for the information! But will the water from the hydrate really be enough to dissolve Cl- ions whilst an excess of a ketone is present?

[Edited on 6-1-2020 by B.D.E]
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[*] posted on 6-1-2020 at 11:30


Well out of curiosity I tried the reduction with hydrazine hydrate, that I mentioned in my first answer and was really surprised how fast it was. I added a few crystals of Iodine to a test tube with a bit of water. After adding a few drops of 40% Hydrazine hydrate solution it immediately started producing bubbles of nitrogen and after just around five seconds I was left with a perfectly clear solution of HI in water. It's way faster and easier than any other reaction I tried before, especially because you don't need to distill the solution, because the only reaction products should be N2 and HI. If you have any access to Hydrazine hydrate I would strongly recommend it to you. But if not you can easily prepare Hydrazine sulfate from OTC materials. :)



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