jamit
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Iodate to iodide
I am wondering if someone could direct me to the link here at sm on converting potassium iodate to potassium iodide? I checked the search but only
find topics on how to make iodide from iodine or making potassium iodate from iodine or iodide.
Would adding a reducing agent like sodium sulfite or sodium metabisulfite to potassium iodate reduce it to potassium iodide? Or can I just heat it.
Any help would be appreciated. Thanks.
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Adas
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Probably yes. Do it in water solution for better results. Heating over 80°C would probably not be necessary. Iodate is pretty reactive.
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jamit
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Does anyone know how to write out the reaction between potassium iodate and sodium sulfite?
My guess: KIO3 + Na2SO3 ---> NaI + K2SO4
Has anyone tried this?
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Adas
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If it works with Na2SO3, then probably only with large excess. Try something like Na2S2O4 or so..
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AndersHoveland
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I do not think iodate is very oxidizing in alkaline solution, there might not be a reaction.
Chlorate does not react with sulfite unless the solution is acidified.
You might also see:
http://www.demochem.de/D-Landolt-e.htm
[Edited on 5-11-2011 by AndersHoveland]
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Adas
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I think, iodate is pretty more reactive than chlorate, isn't it? :O
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woelen
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No, iodate is probably even less reactive, certainly at high pH. It is somewhat kinetically faster than chlorate, but thermodynamically it is a much
weaker oxidizer than chlorate.
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jamit
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@woelen do you think the reaction will proceed if I mix solution of potassium iodate with sodium sulfite to produce potassium iodide?
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woelen
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That reaction only proceeds if there is acid present. Without acid the reaction does not proceed.
If you have acid, then add this to your solution of the iodate, and then carefully add the sulfite, drop by drop. First a lot of dark brown iodine is
formed and even solid iodine can be observed. At a certain point this dissolves again. Keep adding sulfite, until the liquid is just colorless. Now
you have sulfate and iodide in solution.
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AndersHoveland
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Does hydrazine hydrate reduce iodate to iodide?
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jamit
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@woelen thanks. But would you be able to separate the sulfate from the iodide by the differences in solubility?
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woelen
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Yes, it does, very violently. I have a page about this
reaction on my website:
http://woelen.homescience.net/science/chem/exps/oxohalogenat...
On the webpage a hydrazinium salt is used, but this reaction also works with hydrazine in aqueous solution. I tried that as well.
First the iodate is reduced to iodine, but if excess hydrazine is used, then the iodine is further reduced to iodide. I also tried adding solid I2 to
a dilute solution of hydrazine and that also gives a violent reaction, in which a lot of gas is produced.
[Edited on 6-11-11 by woelen]
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woelen
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Quote: Originally posted by jamit  | | @woelen thanks. But would you be able to separate the sulfate from the iodide by the differences in solubility? |
What you could try is boiling down the solution quite a lot and then allow it to cool down. Sodium sulfate is less soluble and large
crystals of sodium sulfate will settle, while the more soluble NaI remains in solution. But it most likely will be quite difficult to get really pure
NaI.
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plante1999
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Use acetone to separe Na2SO4 from NaI , NaI is solube in acetone.
I never asked for this.
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S.C. Wack
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I would try heating first, starting low to get rid of any water, all the way to the mp of iodide.
In an oven, not a soup can over a blowtorch, since atmospheric exposure would be uncool. Carbon might help.
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AndersHoveland
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Assuming that hydrazine hydrate (not a hydrazine salt) is used to reduce sodium iodate, how could the reaction possibly produce elemental iodine? I
just do not see how a cloud of violet vapor could form. What would be the formula for such a reaction?
But elemental iodine reacts with acetone (at room temperature, reaction takes around 30 seconds, much faster if the pH is lowered) to form
iodoacetone. Actually, in the complete absence of water, the hydrogen iodide that simultaneously forms will reduce another molecule of acetone to
propane gas.
C3H6O + I2 --> C3H5OI + HI
C3H6O + (2)HI --> C3H8 + H2O + I2
No doubt some 1-iodopropane will also form, depending on the reactant ratio.
[Edited on 7-11-2011 by AndersHoveland]
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woelen
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| Quote: Originally posted by AndersHoveland | | Assuming that hydrazine hydrate (not a hydrazine salt) is used to reduce sodium iodate, how could the reaction possibly produce elemental iodine? I
just do not see how a cloud of violet vapor could form. What would be the formula for such a reaction? |
This
indeed is an interesting question. Fact is that a cloud is formed, even if plain KIO3 is added to e.g. 35% hydrazine solution. No acid needed, no
hydrazinium salt needed. The reaction is very violent and exothermic.
I think that the following reaction occurs:
4 KIO3 + 5 N2H4 ----> 4 KOH + 2 I2 + 8 H2O + 5 N2
The iodine, formed in the reaction partially reduces further N2H4, yielding HI and N2, another part escapes as vapor. The HI then can neutralize part
of the hydroxide in the reaction mix. But still, some issues remain with this model. After the reaction, the liquid should be alkaline according to
the equations. The amount of iodine formed is just enough to neutralize the hydroxide formed.
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UnintentionalChaos
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| Quote: Originally posted by AndersHoveland |
Actually, in the complete absence of water, the hydrogen iodide that simultaneously forms will reduce another molecule of acetone to propane gas.
C3H6O + I2 --> C3H5OI + HI
C3H6O + (2)HI --> C3H8 + H2O + I2
No doubt some 1-iodopropane will also form, depending on the reactant ratio. |
[citation needed]
Department of Redundancy Department - Now with paperwork!
'In organic synthesis, we call decomposition products "crap", however this is not a IUPAC approved nomenclature.' -Nicodem
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ScienceSquirrel
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| Quote: Originally posted by UnintentionalChaos | | Quote: Originally posted by AndersHoveland |
Actually, in the complete absence of water, the hydrogen iodide that simultaneously forms will reduce another molecule of acetone to propane gas.
C3H6O + I2 --> C3H5OI + HI
C3H6O + (2)HI --> C3H8 + H2O + I2
No doubt some 1-iodopropane will also form, depending on the reactant ratio. |
[citation needed] |
No.
The primary reaction is alpha substitution on acetone as the dominant reaction.
Di and tri substitution is preferred over mono substitution so iodoform is a preferred product over iodoacetones as iodine concentration increases.
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AndersHoveland
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I suddenly realised how little I know about the reaction. How is it that anhydrous HI is used to reduce ketones when the iodine that forms is also
known to react with ketones?
Could only a very small quantity of iodine be used as a catalyst, in the presence of larger quantities of ketone and phosphorous?
The iodine would oxidize the phosphorous, forming PI3. the PI3 would react with the water, produced as a byproduct to form P(OH)3 and anhydrous HI.
The anhydrous HI would reduce the ketone, forming I2 and H2O as byproducts. Perhaps I am just incredibly ignorant about this reaction.
Links:
http://www.demochem.de/cassy_iod_acet-e.htm
"To our suprise, the first reaction with 11 equiv of H3PO2 and 4 equiv of HI in refluxing water not only reduced the ketone group to a methylene..."
http://pubs.acs.org/doi/abs/10.1021/ol102174w
Apparently Samarium(II) Iodide can also reduce ketones:
http://en.wikipedia.org/wiki/Reductions_with_samarium(II)_iodide
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