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Topic: REDUCTION OF AZIDES  (Read 9835 times)

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Offline staudinger

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REDUCTION OF AZIDES
« on: September 14, 2012, 06:21:17 AM »
I was trying to reduce aliphatic azides wtih Staudinger, the usual H2/Pd/C method and others. But i couldn't get the expected amine or the cyclized product by removing the mesyl LG. I always see a very polar product at the base of my TLC . I tried to characterize this product but it was not fruitful. Any one who did this reaction can share his experience in alievating this problem

Offline discodermolide

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Re: REDUCTION OF AZIDES
« Reply #1 on: September 14, 2012, 06:22:38 AM »
How did you work it up?
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Offline staudinger

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Re: REDUCTION OF AZIDES
« Reply #2 on: September 14, 2012, 06:47:56 AM »
i just added little water and extract it with dichloromethane. This is for Staudinger method. When i used hydrogenation i just filtered and submitted to column.
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Offline discodermolide

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Re: REDUCTION OF AZIDES
« Reply #3 on: September 14, 2012, 06:52:16 AM »
The amine will stick to a column. Why not make the HCl salt and crystallise it. Or did you not remove the MsOH before the hydrogenation?
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Offline Dan

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Re: REDUCTION OF AZIDES
« Reply #4 on: September 14, 2012, 08:44:25 AM »
Quote
i couldn't get the expected amine or the cyclized product by removing the mesyl LG.

So you have something like this?

N3-[chain]-OMs  :rarrow: pyrrolidine/piperidine etc.

I have done several of these reactions with H2, Pd/C and have never had a problem. They are usually very clean reactions. Purification in my cases was performed by ion exchange chromatography on acidic resins because the products were polyhydroxylated pyrrolidines and far too polar for chromatography on silica.

Can you provide any other structural information? I take it you expect it to be non-polar enough for silica gel chromatography?

I would be wary of using Staudinger reduction as the phosphine may react with the alkyl mesylate.

did you not remove the MsOH

This is an important point. The product of hydrogenation/cyclisation will be the ammonium mesylate salt of the cyclic amine, which is unlikely to be mobile on silica (is this your stationary phase for the column you did?) and may also be water soluble (you might have lost it in the workup). You will need to treat the crude salt with a base (e.g. use sat. bicarbonate in the extraction rather than water) to generate the free base form of the amine before attempting chromatography on silica. If you can't move it on a silica TLC, you can't move it on a silica column.

Did you try running TLC on triethylamine doped silica and triethylamine doped solvent? This reduces streaking problems.

For the hydrogenation method, I would advise filtering, concentrating and then getting a crude NMR before attempting any purification. This way you can tell whether the problem is the reaction or the workup. If it looks OK, take a small amount of the crude and try adding sat. bicarb and extracting with an organic solvent. Check the organic fractions by TLC to see if you can find anything and get an NMR to see if the free base of the product is in there. 

If it's there, then you can safely use this method of isolating the free base on the bulk of the crude. Find TLC conditions that give appropriate mobility and low streaking, then column.

If it's not there, this means the free base is water soluble and you will have to use an alternative method. By testing the workup on a small proportion of material, you have not lost it all. Alternative purification techniques include ion exchange chromatography and recrystallisation.

Good luck.
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Offline staudinger

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Re: REDUCTION OF AZIDES
« Reply #5 on: September 14, 2012, 09:00:58 AM »
I started from 1,2,4-hexane triols. the last precursor before i did my reduction is, the 1,2-position is protected by mesylate and the six position is taken by azide. The target synthesis was reduction of azides followed by concomitant cyclization. The NMR of the last substrate was very clear, but the NMR after attemted reduction was messy. What i saw is there are still two mesylate group. If the expected reaction was done i wouldn't see one of these mesyl group since NH2 is strong nucleophile to kick out from the substrate. I have also tried to purify it but the yield was very low, may be it sticks to the column as you suggested. I didn't try the method that you proposed. If it is working i will try it.

Offline discodermolide

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Re: REDUCTION OF AZIDES
« Reply #6 on: September 14, 2012, 09:09:45 AM »
Perhaps the reaction worked. But from what I've read you did not remove the methane sulfonic acid formed. So I guess you have a methane sulfonic acid salt of your amine product.
Make it basic and extract. This , if my scenario is correct, will remove the methane sulfonic acid as it's sodium salt. Then the amine should remain in the organic layer. Then you can make an HCl salt if necessary.
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Offline Dan

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Re: REDUCTION OF AZIDES
« Reply #7 on: September 14, 2012, 09:22:11 AM »
I started from 1,2,4-hexane triols. the last precursor before i did my reduction is, the 1,2-position is protected by mesylate and the six position is taken by azide.

So, I have attached a picture of what I think you are saying (a hexane-1,2,4-triol with Ms at O1 and O2 and an azide at C6?). Can you confirm? If not, please draw.

If that is correct there are a number of potential issues with this reaction on paper. THFs, piperidines and pyrrolidines could all form in this reaction.

Quote
What i saw is there are still two mesylate group. If the expected reaction was done i wouldn't see one of these mesyl group since NH2 is strong nucleophile to kick out from the substrate.

The crude will be an ammonium mesylate salt. The -OMs that gets displaced becomes the counterion of the ammonium salt. You will still see two mesylates unless you treat it with a base.
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Offline discodermolide

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Re: REDUCTION OF AZIDES
« Reply #8 on: September 14, 2012, 09:29:59 AM »
Dan, at least our reasoning is more or less in agreement:))
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Offline staudinger

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Re: REDUCTION OF AZIDES
« Reply #9 on: September 14, 2012, 09:51:22 AM »
Sorry, It is 1,2,6-hexanetriol not 1,2,4-triol. I replaced the 1,2-position with mesylate and the last 6,on with azide. As you said we have been also expecting the six and seven membered ring(as a byproduct).
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Offline discodermolide

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Re: REDUCTION OF AZIDES
« Reply #10 on: September 14, 2012, 09:54:44 AM »
You really need to examine your work-up procedure, as both Dan and I have suggested you may well have the mesylate salt of the amine.
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Offline Dan

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Re: REDUCTION OF AZIDES
« Reply #11 on: September 14, 2012, 10:43:29 AM »
Sorry, I forgot to attache the image before. Here is a new one. Is that correct?

I'm not surprised this is messy. You can make a piperidine or an azepane, and these can also form bicyclic aziridines quite easily. From memory (i.e. possibly wrong), Steve Davies (Oxford) demonstrated this quite nicely in a paper reporting de novo synthesis of 1-deoxynojirimycin a couple of years ago. I'm sure there are other examples, I just remember that one specifically.

Which compound do you actually want? I would probably design a more specific route if the dimesylate does not behave. With two reactive electrophilic sites, this is an ambitious approach if you want a clean product.

Normally you can also treat the crude with acetic anhydride and a base (e.g. pyridine) to form the more tractable amide. However, with your extra mesylate even then you will risk a second intramolecular cyclisation by displacement of of the second mesylate by the amide oxygen (to form an oxazoline). Intramolecular displacements like this are standard methods for making oxazolines (especially bisoxazoline ligands).

You might have better luck if you take 6-azidohexane-1,2-diol and form a cyclic sulfate. This should stop at monodisplacement. Otherwise a protecting group strategy for activation of only one hydroxyl is probably the way to go.
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