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Imidazolium salts for Ru catalysts

  1. Feb 19, 2010 #1
    Posts: 1

    Re: Grubbs olefin metathesis
    Hi! I'm working on synthesis of assimetric imidazolium chlorides but I'm having problems in the last step. I'm following exactly the steps in the general procedure of Waltman and Grubbs (Organometallics, 23, 2004, 3105). I'm trying to synthesize the imidazolium salts combining mesityl with methyl and cyclohexyl group. The reduction of the oxalamides with BH3.THF is complete but in the last step I never get to close the ring with HC(EtO)3. After leaving the reaction mixture heating at 120 ºC and refluxing overnight, the next day I always find some very dark solution (almost black, like carbonized) and when I filtered the solid that appeared once in that solution I only could see in H-NMR and C-NMR that no peak of carbon at about 160 ppm was present (the one closing the ring, linked to two N). I also distilated EtOH formed during the reaction to force it to the right side and added more HCl to increase acidic conditions...

    Could someone help me, please?
  2. jcsd
  3. Feb 19, 2010 #2


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    Isn't ammonium chloride used instead of HCl? Did you add 37% HCl (63% water) or dry HCl as a gas?
  4. Feb 22, 2010 #3
    I used 37% HCl solution. As I have the hydrocloride formed previously I decided to continue using the same acid. This has been the procedure followed step by step:

    To a flask charged with N-mesityl-N’-cyclohexyl-oxalamide (3.0 g, 10.4 mmol), was added BH3.THF (1 M in THF, 63 mL, 6 equiv). The resulting mixture was refluxed overnight while the solution turned clear (or even heating for 24 h to get full reduction). The mixture was then cooled to room temperature and MeOH was added dropwise till all bubbling ceased. Conc. HCl (12 M, 2.5 mL) was added and the solvent was removed by evaporation. The resulting solid was redissolved in MeOH and the solvent was again evaporated to remove the boron as B(OMe)3. MeOH was added and removed in this way twice more. The remaining white solid was washed with hexane (50 mL). The dichloride salt was transferred into a flask followed by addition of HC(OEt)3 (30 mL). The solution was heated at 120 ºC overnight.

    Could be possible that the black color of the solution was due to some borane left? It could be affecting to the reaction? I'm repeating the reaction heating it at 120ºC and distilating EtOH but no EtOH seems to be formed...

    Thanx a lot for your help.

    Best regards, Ana
  5. Feb 22, 2010 #4


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    Did you try to reduce an HCl salt?
  6. Feb 23, 2010 #5
    No, I didn't. I reduced the oxalamide and after reduction I added the HCl to form the salt. It's in the procedure...
  7. Feb 24, 2010 #6


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    After that much methanol and the HCl there is absolutely no borane left, you can be sure of that. You might try free basing the diamine and spotting it to make sure it is pure. After that, it is fairly straightforward to remake the HCl salt (bubble HCl gas into a solution of the amine in hexanes using NaCl and H2SO4). Purifying intermediates is the usual way to track down these problems. Don't waste your time trying to identify the black byproduct, just be sure that you are pure at every step where a solid is isolated.

    Multistep syntheses where no characterization/purification is performed between steps is always a suspect methodology. I have about as much success as failure when I try to repeat literature procedures described in that way. First step is always repeating it with characterization and purification at every step. It sounds like more work for you but as you have already seen for yourself, you are wasting time and effort by ignoring the importance of characterizing what you have made at every step and purifying it before proceeding. For example, the hexane wash of the solid diamine hydrochloride only removes greasy things soluble in hexanes. It wouldn't remove, for example, any boric acid impurities and might not successfully remove a partially reduced amide - amine hydrochloride which won't close of course. That is, if the cyclohexyl amide is too sterically blocked to rapidly reduce with BH3:THF, you might have an amine amide. Adding the HCl will form a solid but the next step won't go.
  8. Feb 26, 2010 #7
    Thank you very much for all the tips, chemisttree.

    Finallly I got the desired product distilating EtOH for only one hour and filtrating the solid that appeared after cooling the solution. Still the first yellow, then orange that turned to red colour to finish as black colour in the solution. But at least I could isolate my product, the NMR of the liquid is completely different product or products...(I don't mind anymore)

    Thanks again.
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