Chemical Forums
Chemistry Forums for Students => Organic Chemistry Forum => Topic started by: herrhansen on June 16, 2011, 06:58:42 PM
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Hello again.
Currently I am doing a reductive amination with a ketone and an aliphatic amine in THF with sodium triacetoxy borohydride as the reducing agent.
When doing a TLC on a F254 silica plate I got two new spots. Maybe there are two new spots because of double addition, but what surprises me is the position. The spots move up compared to the ketone. In my opinion the spots should be positioned longer "down" the TLC plate and below the starting ketone, as the amine would have more affinity to the plate and not move long up. My eluent are EtoAc/Hexane, but that should not make a difference.
Could this be true or am I having some kind of other sideproduct? Cant see what that should be. The ketone should not be able to be reduced by the sodium triacetoxy borohydride and cant see what other sideproducts there should be able to be formed.
Thanks in advance.
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If the amine is a really greasy one it might move up... can you give details?
Also, is your starting material chiral, and is the ketonic centre prochiral? What I'm getting at here is the possibility of diastereoisomeric products (2 spots) arising from addition of hydride to either diastereotopic face of a prochiral ketone.
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Are you doing any workup of your TLC sample? I have occasionally seen boronated intermediates by TLC on triacetoxyborohydride reductions.
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Thanks for two good answers.
It is 4-cyclohexyl-1-phenylbutan-2-one which I am going to react with n-propyl amine. I havent done any work up of the reaction, so it could be the boronates as you said. I will try to do that and see if it gets in position.
Thanks alot.
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I think E/Z isomers of imine can be formed in this case.
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The amine can become much more unpolar than the carbonyl compound. I have observed it often. What about Ninhydrin spray? I can only insist everytime that people do stain their TLC´s after UV visualisation...that helps a lot.
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Thats an good offer. But I hope that the triacetoxy borohydride have made its job. The TLC was taken after 48 h, so I hope the imine has been reduced.
Thanks
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You are right. Ninhydrin could be the answer. I just presumed that the amine had formed. But sometimes Ninhydrin detection seems to be weak at detecting secondary amines?
But thanks OC.
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I am pretty sure you will get a nice purple spot. GC-MS should also be an option to detect whether your product has been formed. Seems pretty unpolar and the molecular weight fits GC-MS too.
The side product you can observe is the reduced version of your starting material, the alcohol. Happens sometimes when you have not so reactive carbonyl compounds. And the kind of acetophenone you have is not so active compared to an aldehyde. The amine competes with the triacetoxy borohydride. And in case of a weaker nucleophile, the borohydride can attack also the ketone.
So I assume your yield will be not very high.
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Thanks for answering. I think I will try Ninhydrin. A good simple indication and maybe follow on GC-MS also. Regarding reduction of the ketone, I actually thought that sodium triacetoxy borohydride would reduce imines selectively over ketones and that ketones wasnt reduced at all like with NaBH4. I am stating this because in my article they use either a one pot with triacetoxy or a stepwise with the imine forming in MeOH and reduction with NaBH4.
Thanks.
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Regarding the position of your spots its not really that surprising the position as you are putting on an aliphatic amine...
Also, it might be an idea to dump in some molsieves to drive the formation of the imine and then just dump in sodium borohydride.
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Thanks.
I tried to conduct the reaction today in anhydrous MeOH as suggested. Only little conversion was seen from the starting ketone. I am wondering if I cant see the imine, because it isnt stable when taken out for TLC, so it eliminates water to yield the ketone?
The reaction was conducted under nitrogen and the imine had 8 hours to form. So at last I just tried to dump in some NaBH4 (2 eq.). According to my reference it wasnt necessary to push the reaction foreward by using mol sieves or any other drying agent and good yields was reported without.
I also tried to stain the TLC plate with Ninhydrin as OC suggested, but I could only see the propylamine used. I will try with vanilin spray soon.
Is it really so hard to do these reductive aminations without using sodium cyanoborohydride or hydrogenations? My university only allow the parr shaker for larger portions.
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According to my reference it wasnt necessary to push the reaction foreward by using mol sieves or any other drying agent and good yields was reported without.
Is this reported for exactly the same substrate - or at least a similarly hindered ketone?
A drying agent can't hurt, and it sounds like the imine is not forming under the conditions you're using, or it is very hydrolytically unstable. Maybe Dean-Stark it. It would be worth identifying the two products from the cyanoborohydride reaction - is one of them 4-cyclohexyl-1-phenylbutan-2-ol? This would point to an unfavourable ketone <> ketimine equilibrium. You may have to be quite aggressive.
It's worth trying propylamine, H2, Pd/C. You can often do this at atmospheric pressure with a balloon of H2, though it may be slow.
Another way around that I would consider if this imine is not behaving is as follows. It's longer but probably more reliable:
1. NaBH4 (to give 4-cyclohexyl-1-phenylbutan-2-ol)
2. Tf2O, MsCl, or TsCl (that would be my order of preference for sulfonylation)
3. NaN3 (to give 2-azido-4-cyclohexyl-1-phenylbutane)
4. Propionaldehyde (1 eq), H2, Pd/C(one pot)
It's not as long as it looks - chromatography is unlikely to be required until the end of the sequence. Steps 1,2,3 should be very clean and just require washing away the salts, and I reckon step 4 will give the target secondary amine quite cleanly because the bulk of it will probably heavily disfavour overalkylation.
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Really quick great answer. Thanks. Love this board.
I am not using a reported substrate, but the reaction has been conducted on similar sterically hindered ketones, some of them required acid catalyst though.
I did not know that it was possible to do the reaction using H2/Pd/C at atmospheric pressure. I thought pressure was needed.
I have seen references for reducing ketones to hydrocarbons using ammonium formate and Pd/C, but does it also work using hydrogen and Pd/C?
I am very excited about my reduction and tomorrow I will try to reduce the ketone with NaBH4 and compare it to my reaction mixture.
I have not worked much with imines before, so I am in doubt if they are stable enough to be dried, formed, filtered and reduced or using a dean stark and then Isolate the imine and reduce it using MeOH/NaBH4?
I know conjugated imines are stable, but I doubt about this one. But even as it is a reversible reaction I cant see why so little product is formed in anhydrous MeOH. I expected maybe 25 %, but only small dots appear on TLC.
Cant see how it can be so difficult to form. But I really appreciate your synthesis. I always think of azides as explosives, but this is not the case with NaN3, so maybe I should take the longer working way :-)
Thanks.
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What also concerns me is the two new spots above the ketone. They are formed in equally amount according to TLC. I was hoping that is could possibly be the end product from the sodium acetoxyborohydride reduction, but as far as I know enantiomers have identically Rf values and are not able seperate on HPLC?
But again it could be the boronates, since they did not stain by ninhydrin spray? But I simply have to quench it and try to spray it again :-)
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Your ketone is a kind-of acetophenone, therefore not very reactive. I have done a similar reaction. Triacetoxyborohydride failed in my case completely.
The alternative is, like Dan mentioned before Dean-Stark.
Dissolve equal amounts of ketone and amine in toluene, Dean-Stark it for several hours (oil bath 150 degrees). The imine will be visible on TLC (ninhydrin). In my case it was a bit more unpolar than the ketone. Since there is an equilibrium, you will not get full consumption of ketone. Evaporate the mixture, redissolve in ethanol, add NaBH4 (2equiv) and stir at room temperature. You will get amine and alcohol of your starting ketone which can also be reisolated (and reoxidized). Best yield in my case was 54%.
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Your ketone is a kind-of acetophenone, therefore not very reactive. I have done a similar reaction. Triacetoxyborohydride failed in my case completely.
The alternative is, like Dan mentioned before Dean-Stark.
Dissolve equal amounts of ketone and amine in toluene, Dean-Stark it for several hours (oil bath 150 degrees). The imine will be visible on TLC (ninhydrin). In my case it was a bit more unpolar than the ketone. Since there is an equilibrium, you will not get full consumption of ketone. Evaporate the mixture, redissolve in ethanol, add NaBH4 (2equiv) and stir at room temperature. You will get amine and alcohol of your starting ketone which can also be reisolated (and reoxidized). Best yield in my case was 54%.
Isn't the whole point of the Dean-Stark apparatus to remove the water and so drive the equilbrium (in this case) towards imine?
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Thanks OC.
Have I have seen your method described, but I am just concerned if the imine is stable enough to survive the evaporation of the toluene and to be isolated.
I know you said close to acetophenone, but acetophenone has its ketone at the alpha carbon to the phenyl ring, in my case I have the ketone beta to the phenyl ring. Are you thinking of the steric hindrance as unreactive?
Thanks again.
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Hey ,
What's the best stain for secondary aliphatic amines