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Waffles SS
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What is best ratio of I2,NaOH (or KOH),acetone for making iodoform?
Woelen suggest: 1gr Iodine + 1cc(~0.7gr)Acetone+0.5gr NaOH for making 0.5gr Iodoform
I think this ratio is not true because in excess alkaline Iodoform decompose(probably to Diiodocarbene)
According to Stichometry:
3I2 + CH3COCH3 + 4NaOH -> CHI3 + CH3COONa + 3NaI + 3H2O
761.4 of iodine react with only 58gr acetone for making Triiodoacetone and finally Haloform reaction complete with 160gr of NaOH
Compare 0.5 gram of NaOH for 1gram I2 with Stichometry
1 gr Iodine really need 0.0761 gr Acetone and 0.210gr NaOH
am i wrong?
[Edited on 21-4-2014 by Waffles SS]
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woelen
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The precise ratio is not really important. You want all iodine to be used up, so use excess of acetone and NaOH. You need to do the reaction in dilute
solutions and the reaction only runs at high pH, so you need some excess NaOH.
Iodoform slowly decomposes in alkaline solution, but this reaction is very slow. At the amounts I used, this effect is not strong and if you isolate
the CHI3 quickly after it is formed, then losses, due to decomposition can be neglected.
By using 1 cc of acetone and 0.5 grams of NaOH you assure that all iodine (if you use a gram) is used up, even if you have no means of weighing
accurately. E.g. if you take 0.4 grams of NaOH or 0.6 grams of NaOH and you take 0.9 ml of acetone or 1.1 ml of acetone, it still works. My "recipe"
is quite robust.
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Waffles SS
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Thanks,
Are sure about Yield?~0.5 gram from 1 gr Iodine?
(last time i tried 10gr Iodine + 10ml Acetone +100ml H2O and 45.5ml 10% KOH drop by drop but Yield was 3gr)
I want to make 100gr Idoform what is your suggested ratio?(Multiple 100x?)
I use KOH instead of NaOH,what is your suggested method for separating KI from remaining solution(I dont think only evaporation to dryness work well
because solution contain Iodoacetone and KOH)
[Edited on 22-4-2014 by Waffles SS]
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woelen
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Yield is appr. 0.5 grams per 1 gram of I2 (half of the iodine ends up as iodide ion, the rest goes into CHI3 and because CHI3 is in terms of weight
nearly 100% iodine, I say just a little over 0.5 grams).
Keep in mind that CHI3 is soluble somewhat in water and this may lead to losses and that may be the reason that you only obtained 3 grams instead of 5
grams from iodine.
One way to obtain better yields is extracting all CHI3 into diethyl ether and then let the ether evaporate.
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Boffis
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If its not a rude question, may I ask what's the great interest in iodoform? After all, as far as I know you can't make either a substance to blow you
mind or blow your hand off from it, or can you? Or am I being cynical.
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IrC
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Woelen, you said "Dissolve KI in dilute hydrochloric acid and add a slight excess of hydrogen peroxide."
Over the years I see details such as this but they leave me in the dark. One of the reasons I wish I had focused on chemistry instead of electronics.
No doubt all the real chemists here already understand exactly what you mean. Mind you I'm not complaining, you are the best chemist I have ever seen
online. But what does it mean exactly? What percent is 'dilute' Hydrochloric acid? What percent is the peroxide? Does it mean store bought 3 percent?
On the acid, if say one had 37 percent from Home Depot used to clean concrete or whatever, does dilute mean 9 parts water for one acid? How does one
know the dilution?
I am not making Iodoform I have no need for it but this thread conveniently mentions two things I read often yet never know exactly what is meant
(dilute acid and un-named percent H2O2). Long ago I learned to not keep chemicals unless there is a need or future possible use, after the hassle I
had getting rid of many I collected over the years no one would take. Anyway if you could clarify this it would be helpful, since I have wondered over
the years many times what they meant when reading statements such as yours quoted above.
"Science is the belief in the ignorance of the experts" Richard Feynman
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woelen
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The statement, such as the one you quoted, means that concentrations are not critical at all.
In this context, dilute hydrochloric acid can be anything from let's say 3% HCl to 15% HCl, but even these borders are not hard borders. The
concentration of the hydrogen peroxide also does not matter, you just have to add enough and then a little extra.
In practice, if I would make iodine from KI, HCl and H2O2 I would proceed as follows.
Suppose I have 5 grams of KI for this experiment.
I would take 10 ml of conc. HCl and add 20 ... 25 ml of water (no precise measurement needed, just roughly) and swirl well. Then I would dissolve the
KI in this. This will be easy.
Next, I would carefully add H2O2 (e.g. common 6% stuff, but 3% material will do the job also) until iodine forms a flocculent solid in the liquid and
then I would add a little extra.
The resulting liquid then will be brown, but not very strongly brown. The I2 will be present as solid.
Next, I would add a lot of water, allow the iodine to settle, and decant as much as possible of the brown liquid, without pouring away solid iodine.
Adding water again, allowing the iodine to settle, and decant again removes nearly all acid. This product is suitable as a start point for making
CHI3. No need to isolate the iodine in dry form.
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chornedsnorkack
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So, let´s review the conditions for iodoform reaction... starting from, say, ethanol.
The steps should be:
1) CH3CH2OH+I2+2NaOH->CH3CHO+2NaI+2H2O
2) CH3CHO+I2+NaOH->CH2ICHO+NaI+H2O
3) CH2ICHO+I2+NaOH->CHI2CHO+NaI+H2O
4) CHI2CHO+I2+NaOH->CI3CHO+NaI+H2O
5) CI3CHO+NaOH->CHI3+HCOONa
Is that correct?
This would work if iodine and base are both in excess. But what if they are not? Is Step 1) the rate controlling step such that a limited amount of
iodine is converted quantitatively to iodoform and iodide, or does the reaction stop at some step?
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