Chemical Forums
Chemistry Forums for Students => Organic Chemistry Forum => Topic started by: AlphaScent on February 26, 2014, 03:45:11 PM
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I was handed a paper today and told to try the reaction and see if it works.
I placed my aldehyde (benzaldehyde, 5 mmol) into a soluton of Iodine (5.5 mmol), 28% ammonium hydroxide ( 30 mL) and THF 5 mL. This all stirred together for one hour. The solution should go colorless after 1 hour. It did NOT after 1 hour or 2 hours. So I continued with the solution being a slight peach color, adding 3 mL of 35% hydrogen peroxide. This should "trap" the formed nitrile and produce the amide.
Upon extracting with DCM, washing with brine, drying over sodium sulfate and removing solvent i.v., there was no amide. The GC shows product and no starting material. I am convinced that it is the nitrile, not benzamide. The product at rt is a viscous liquid, not a solid as benzamide should be.
What do you all think is the next move?
I am doing the reaction again and just going to wait until the solution goes colorless. If that means letting it stir overnight, so be it.
I will also try and add excess hydrogen peroxide. 6 mL and then 9 mL on the next trial.
After all a nice chemist once said to me, "Chemistry is an experimental science"
Cheers,
thanks in advance
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Even though 'viscous liquid' might be considered proof of nitrile formation, a little more data would have more reassuring. Just looking at the procedure, this is what concerns me. The Payne oxidation (if I have this correct) uses hydroxide, carbonate or bicarbonate, I don't remember which. Replacing it with ammonia seems risky in that in addition to it acting as a base, it could also become oxidized and deplete the peroxide for hydrolysis.
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I should have also stated that I will be getting my NMR results tomorrow. So the story will be told then. I just wanted a little anticipation.
I agree that the, excess mind you, ammonia would almost surely react with such an oxidizer as peroxide.
I wonder then why publish the paper at all. If you make the nitrile in this manner, awesome. Why publish if it doesnt work? Just say a simple hydration will then lead you to an amide if desired.
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They state the peroxide must added after the first reaction is completed due to the iodide ion reaction with peroxide and lowering the yields. I must then wait for the solution to go colorless to indicate that there is no iodide present.
It may just be the reaction is not done in an hour as they state.
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What article is it? You definitely have me curious enough to go read it.
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It seems strange that they report the reaction going clear within an hour with an excess of I2 around. Unless I'm missing something in the mechanism.
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It seems strange that they report the reaction going clear within an hour with an excess of I2 around. Unless I'm missing something in the mechanism.
What is the mechanism?
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After the iodine is consumed why not try adding LiOOH (LiOH in peroxide, water/THF)?
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I only have a hard copy, but I googled it and there it was.
http://www.audreyli.com/panli/chemistry/reference/method/jo026407z.pdf
All questions posted are answered in the article.
Disco,
What about KOH if no LiOH is available to add to the 35% HOOH?
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Unfortunately Li is the ion to use here. I'm not sure you can form KOOH, I'm sure such a thing must be there but have no experience with KOH/peroxide.
For example using LiOOH you can hydrolyse a nitrile substituted beta-lactam to the amide with no ring opening!
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Wow. That is kind of intense. Really awesome. That chemoselective, huh. I do not have any LiOH.
I may try using molar equivalents of iodine. As Unassuming pointed out it is in excess and is not used up in the reaction. I assume it wont be used until the HOOH is added.
This is for a possible lab experiment for a organic lab. If going to the amide was that important cant you just reflux the nitrile in conc. HCl of H2SO4? That should hydrate it quite easily. Again who wants undergrads mixing with refluxing, concentrated acids.... its a thought though.
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I cannot discount Disco's suggestion, but I'd just try adding NaOH. I earlier suggested this as part of a Payne oxidation which is how I came to learn of it, but in Payne's paper, he refers to this as the Radziszewski reaction. See http://www.sciencemadness.org/scipics/jo01062a004.pdf
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Orgopete,
Great reference. In my case, would the nitrile made by my initail conditions not have to be isolated first and then treated under Radziszewski conditions?
I may try to add a HOOH/NaOH/NaOOH solution to test that theory, seeing as after my initial step, the intermediate is in fact a nitrile.
I am getting some NMR today at some point. I will return with results in time.
Cheers
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I don't think the peroxide anion will oxidize any ammonia. So I don't think isolating the nitrile is necessary. I haven't been following the stoichiometry but the iodine is generating HI. The peroxide reaction requires two equivalents in formation of amide under basic conditions. The peroxyimidate reacts with a second mole of peroxide to give O2 and the amide.
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I have finally had a chance to look closely at my NMR results. There is a slight bit of the amide formed. It is indeed the nitrile that is formed.
I am doing other work this week, but will be back to this potential project and let you now how the rest turns out.