Sciencemadness Discussion Board - Morpholine derivate from secondary amine/alcohol

Morpholine derivate from secondary amine/alcohol

Dr.SAM - 5-9-2008 at 06:45

Hello folks, this is my first post.
I apolagize in advance if it's against the rules to open a topic at the first post.

I need little help with a reaction i theorized.

Basically i want to form a morpholine derivate from a secondary alcohol and amine.

The reaction I came up with was:

R1-CHOH-CH2-NH-R2 + Br-CH2-CH2-Br -->

R1-Morpholine-R2 + Br2

Is it better to use dibromoethane or diiodoethane?
Is this reaction possible?

Could someone help me on this please.

Thanks

[Edited on 5-9-2008 by Dr.SAM]

ScienceSquirrel - 5-9-2008 at 06:55

That should be 2HBr.

In principle it may work.
Your biggest problem may be the much greater nucleophilicty of an amine group compared with an alcohol.
The size of your R1 and R2 groups could pose problems as well due to steric hindrance.

smuv - 5-9-2008 at 07:16

I have vaguely seen it refrenced as working with EtX2; but no reaction conditions were given. The more classic method is via ethlene oxide to form the corresponding ethanolamine which can be cyclicised with acid catalysis.

If you try with EtX2 post the results as I think many would be interested.

P.S. Don't worry about aziridine formation it doesn't work under 'various conditions' as one paper stated. I guess the major conceivable side products would be substituted ethylene diamines or over-alkylation of the amine.

[Edited on 9-5-2008 by smuv]

Dr.SAM - 5-9-2008 at 10:18

Thanks for replying.

I was planning to dissolve the compound as a free base in absolute ethanol and then gently add EtBr2 while heating at 50 'C. Would it be good? Should I add some acid?

Would Diodoethane work better than EtBr2?

What sideproducts might this reaction give?

smuv - 5-9-2008 at 10:38

I covered side products in my above post...

I would stick with whatever is cheaper (probably dibromoethane). You might have to experiment with the reaction conditions. Definitely acid would not help; you would want a non-nucleophilic base. Alkylating the amine is easy but I am not sure how easily the intramolecular cyclization will occur; it is entropically favored of course but I do not know if it would still require a reagent that can quantitativly deprotonate the alcohol (as in a Williamson ether synth) or if it is so favorable that a weakish base would work.

I would experiment with things; I am not sure enough about what conditions are required to make a good suggestion. Does anyone else on these forums have an idea/experience?

Dr.SAM - 5-9-2008 at 15:33

After some reserch I came out with a promising synthesis.

The reaction involves making and ether of a secondary alcohol and alkylating a secondary amine. In order to complete the ether bond, according to the williamson ether synthesis, a polar aprotic solvent should be used to speed the reaction and get higher yields.
So i would chose the very common acetone as a solvent.
the reaction takes place under basic conditions but unfortunately sodium hydroxide is insoluble in acetone, so is sodium carbonate i think. on the other hand amines are soluble in organic solvents and the molecule we're speaking of has got already an amine group, so i don't think its necessary to add another organic base, plus i dont really have any available organic base. Adding a concentrated solution of ammonia would make the solvent more protic so i dont know if i should consider it a wise thing to do.
A soluble iodide could be used as a catalyst in this reaction.
Alkyl iodides have practically no polarity, they are good leaving groups but they are unable to polarize the carbon to C+ so i think dibromoethane would be more effective.
The reaction should be carried to 60-100 'C.

Am I doing anything wrong?

smuv - 5-9-2008 at 16:28

Yes. If you don't add any base some of your unreacted substrate will be converted to the non-nucleophilic hydrobromide (or iodide depending on your alkylating agent) and kill your yields.

I dont really understand what you are trying to describe; why not post the paper.

Maybe a starting point would be to mix 1eq dibromoethane (or whatever you want to use), excess K2CO3 with the amino alcohol in acetone or butan-2-one and reflux it for a while. The potassium carbonate will not dissolve but it will eat up some acid and to a very small extent detonate the alcohol. I would worry about using stronger basses as it may possible to eliminate a halide from the dihaloethane to produce a vinyl halide.

Don't use nucleophilic bases like ammonia...

Dr.SAM - 6-9-2008 at 11:59

So the full yhtesys would be.

Boiling absolute ethanol in conc. sulphuric acid in a reflux apparatus to give ethene gas.
The gasses escaping the reflux column are going to be bubbled into a cold solution of sodium hydroxide to remove SO2 and ethanol vapors.
the gas will be then going through a sodium hydroxyde/copper chloride column to remove humidity and any probable alcohol traces.

In order to produce bromine gas I was planning to bubble Cl2 gas made from conc. HCL and NaOCl in a saturated solution of potassium bromide kept in an ice bath to dimish the bromine-water solubility. The bromine under cold conditions should separate easily from the water layer especially if the water is saturated by potassium chloride.

The pure bromine, still kept under ice cold conditions is ging to be poured into a bubbling apparatus.
Later I'm going bubble the previsously made ethene gas into pure bromine until the liquid becomes prefectly clear meaning that all the bromine reacted with ethene to give pure 1,2 dibromomethane.

Then I'm plnning to dissolve the hydroxyamine base in acetone and ad an excess of anhydrous potssium carbonate and then obviously the 1,2 dibromoethane.
However since the base form of the molecule is a sticky, oily liquid i would prefer dissolving the hydrochloride salt in hot acetone wich would then be neutralized back to a base by the potassium carbonate.

I'm planning to reflux the mixture at 60 to 90'C with occaional swirling. Probably i'm going to ad a tiny iodine chrystal to act as a catalyst according to the Williamson ether synthesys route.
The reflux will last 6 to 8 hours and once complete I'm goin to slowly add conc. hydrochlorid acid to turn the compound into its hydrochloride form again.
The acetone is goin to be gently evaporated and the resulting chrystals are going to be purified with an acid freebase extraction.

Am I missing anything?

[Edited on 6-9-2008 by Dr.SAM]

smuv - 6-9-2008 at 12:37

If you have to make the ethylene dibromide; you would be better off making ethylene oxide and sticking to well known procedures to making morpholine derivatives.

There are some things wrong...
-On a small scale that is a crappy way of producing bromine...you might as well just use ethylene dichloride; it could likely get the job done.

-I wouldn't bother washing the gasses. Sulfur dioxide wont be much of a problem; it may reduce some Br2 to HBr, but this is trivial. Just fractionate your product after synthesis to remove the very little amount of halohydrin and monohaloethane that may be produced. Remember with relatively cheep reagents yields can be sacrificed for an easier procedure.

-I think making the freebase in situ is a bad idea; it complicates things and because of the low solubility of potsassium carbonate it might not work so well.

not_important - 6-9-2008 at 17:19

Consider treating ethylene glycol with excess hydrobromic acid, simpler than the EtOH => ethylene, + Br2 => C2H4Br2

Dr.SAM - 9-9-2008 at 07:59

Unfortunately i dont know where to get ethylene glycol, neither hydrobromic acid.

Although i know how to get conc. sulphuric acid, absolute ethanol, and potassium bromate.
It's much easier for me to reflux absolute ethanol in conc sulphuric acid and then filter the gasses escaping from the condenser in a solution of cold NaOH.

By the way, do you guys think that the reaction i posted above is practically possible?
what could i add to increase the yield?

Are 5 hours of refluxing enough or should i wait a day or so?
Another question: could traces of water in my acetone mixture mess up my reaction?

ScienceSquirrel - 9-9-2008 at 08:24

To be honest it might work, on the other hand it might not.
With a proper lab, NMR spectroscopy etc these things can be far from simple.
It is possible that a competent postgraduate could make a range of these compounds in a few months but there could be some huge pitfall in your proposed route and without the proper gear you just will not find out what is going wrong.
The old style chemists were brilliant at doing a lot with very little but the price they paid was years of painstaking work and a lot of deductive reasoning.
The structure of benzene is obvious to the modern chemist armed with instrumental techniques, it was a near impenetrable puzzle to Kekule's generation!
Making ethylene dibromide at home on a preparative scale by reacting ethylene with home made bromine is going to be far from a walk in the park and that is only the foothills of your project.

smuv - 9-9-2008 at 08:31

And considering he will be consuming the product (phenmetrazine)....better be damn sure it is the desired product. Although Mp and TLC are good ways of checking approx. purity.

IMO if you have to be spoon-fed you are not ready to successfully do it.

ScienceSquirrel - 9-9-2008 at 08:49

Now the squirrel can see the hazelnut :cool:

I would follow the classic route.

I do not know where you are going to get the phenylpropanolamine from in sufficient quantity.
But I would react it with 2-bromoethanol to form the N-hydroxyethylaminopropanolamine and then cyclise this with acid. That answers your main problem about reaction centre specificity.
The ether will be be formed in a facile manner due to the adjacent phenyl group that will promote the easy formation of a carbocation by loss of water. Nucleophilic attack by the primary alcohol will then result in the formation of the six membered ring which will be entropically preferred to intermolecular reaction.
I am sure that you will find a synthesis of 2-bromoethanol somewhere on the net.

Are Italian prisons particularly comfortable?

[Edited on 9-9-2008 by ScienceSquirrel]

kolpak - 14-9-2008 at 04:21

Once I've faced with the same problem. I needed to convert amino ethanol PhCH(NH2)CH2OH into corresponding morpholine. I find out from the literature that treatment with dibromoethane leads basically to bis products: HOCH2CH(Ph)NHCH2CH2NHCH(Ph)CH2OH - the obvious result

The simple and reliabale method I used is treatment with chloroacetylchloride - gave corresponding amide from amino-group acylation, then its cycization into correspodning 2-oxomorpholine and reduction with lithium aluminium hydride.

Ketone - 16-9-2008 at 07:41

Well, it seems pretty obvious you're going to make (Well, try to make) Phenmetrazine. Those unnamed substituents (sp??) is C6H5 and CH3 aren't they?

You should just level with the people here you know, if anything it could save you alot of frustation when they start discussing all that boring chemistry instead of just telling you "Yes, you can make drugs this way" or "no, you can't make drugs that way."

I don't think you can make it this way btw. Don't got (any) experience with "substituted morpholines" but if this reaction works..
Then I think the CH3 would be in the wrong position.
Though, it might make you high all the same..

But as I said, I don't think it will work, it just seems.. wrong. Way wrong.

Quote:

Originally posted by ScienceSquirrel
Are Italian prisons particularly comfortable?

Lol..

Quote:

Originally posted by Dr.SAM
i dont know where to get ethylene glycol

Lol.... try a gas station maybe?

[Edited on 16-9-2008 by Ketone]

ScienceSquirrel - 16-9-2008 at 07:50

I think he just made a mess of his original formula, I think he means;

PhCH(OH)CH(NH2)CH3

which would give you phenmetrazine.

Notice that he eliminates bromine in the reaction which is nonsense.

I think he has just given up now anyway.

Nicodem - 17-9-2008 at 00:30

Phenmetrazine can not be made by reacting any of the 2-amino-1-phenylpropanols with 1,2-dibromoethane as far as I know. The reaction of an primary amine with 1,2-dibromoethane would not stop at the R-NH-CH2CH2-Br stage since these compounds only exist untill the amine is protonated. In neutral and basic conditions they they form the coresponding aziridines (R-N=(CH2CH2)) which are electrophilic enough to alkylate amines. Hence the results Kolpak described above. This is essencially inevitable, hence the method used to produce phenmetrazine from 2-amino-1-phenylpropanol involves the use of chloroacetyl chloride, cyclisation and amide reduction, though there is a way simpler method used by the pharm industry which however does not start from any 2-amino-1-phenylpropanol stereoisomer.

smuv - 17-9-2008 at 07:16

Quote:

In neutral and basic conditions they they form the coresponding aziridines (R-N=(CH2CH2))

The formation of a 6 membered ring interferes with the synthesis of aziridines (as I previously posted)...

See: Eur. J. Org. Chem. 2007, 3281–3291
DOI: 10.1002/ejoc.200700041

[Edited on 9-17-2008 by smuv]

Nicodem - 18-9-2008 at 00:31

Quote:

Originally posted by smuv

Quote:

In neutral and basic conditions they they form the coresponding aziridines (R-N=(CH2CH2))

Aziridines generally can not be prepared using 1,2-dibromoethane since they are often more reactive than 1,2-dibromoethane itself therefore leading to R-NH-CH2CH2-NH-R products. Here aziridines are just intermediates allowing for what is called neighbouring group assisted nucleophilic substitution (though, there are exceptions where some aziridines can also be isolated). Probably I did not made this clear enough in my previous post.
I would be a bit sceptical about these claims in the paper where the formation of the morpholine ring is considered only a side note without any experimental date. Also, they do not say unambiguously that they used 1,2-dibromoethane while the "various conditions" say close to nothing. I guess one could always email to the authors and ask for more information if particularly interested.
On the other hand, 1,2-dibromoethane does react with ortho-aminophenols to yield the corresponding benzomorpholines without problems and generally in good yield, but this is expected since the phenols are easily deprotonated hence the aziridine intermediate reacts more easily intramolecularly or alternatively the reaction starts with O-alklyation since the phenoxide is a good nuclephile and gives the same product. But to simulate such a condition on aliphatic aminoalcohols one would need to react 1,2-dibromoethane with the O-deprotonated aminoalcohol (for example by using NaH on the aminoalcohol in THF followed by 1,2-dibromoethane addition). However, I never saw any such example so I have doubts about it working.

Besides, I still don't full understand the practical aspects of this thread. In which way would be using 1,2-dibromoethane more practical than the classical synthesis employing the cyclisation of N-(2-hydroxyethyl)aminoalcohols in acidic media? I understand the unattractiveness of the route employing chloroacetyl chloride since it is based on the reduction of the amide, but preparing the N-(2-hydroxyethyl)aminoalcohols is hardly a difficult task giving that just about any reagent able to form vinyl oxide in situ can be used (like 2-chloro or 2-bromoethanol).

The paper referred by Smuv is attached in case someone wants to read it more thoroughly (I just check the paragraph mentioning the discussed reaction so perhaps I missed something):

Attachment: An Approach to Enantioselective 5-endo Halo-Lactonization Reactions.pdf (216kB)
This file has been downloaded 1269 times

smuv - 18-9-2008 at 07:37

Nicodem,

You did not miss anything; reaction conditions were not given, however the writers state "all our attempts (to make the aziridine) led to the exclusive formation of the cyclic amine (morpholine)." If one were sufficiently interested in making these morpholines from 1,2-dihaloethanes some experimentation in this area could be fruitful.

Admittedly I know little about aziridine chemistry; I would expect the formation of ethylenediamines as a side reaction to be entropically unfavorable as opposed to intramolecular cyclization though. Although I suppose under the conditions I proposed kinetics would probably trump energetics... I was also hoping that this cyclization would be so strongly favored that quantitative deprotonation of the alcohol would not be required.

I agree with you though, if you don't care about experimentation and just want to make the morpholine, might as well stick to known routes involving ethylene oxide and halohydrin.

[Edited on 9-18-2008 by smuv]

sonogashira - 18-9-2008 at 14:16

Quote:

Originally posted by smuvmight as well stick to known routes involving ethylene oxide and halohydrin.

Would you mind sharing these references and papers (if you have them)?

I had a look in the Merck index and the patents mentioned there only discussed routes starting from a-bromopropiophenone or the chloroacetyl chloride route mentioned.. But no mention of this "well known" 2-bromoethanol method?

Nicodem - 18-9-2008 at 23:42

It is the most common route to morpholines and there are lot's of references (even plain morpholine is made this way!). For the synthesis of 2-phenylmorpholines you can just check the reference given in the paper I uploaded in my previous post. I think it was the reference number 11, an old JMC paper, if I remember correctly. I would guess the only reason it was not used in the industry for the phentermines synthesis (or was it?) is simply because it is not so rational and economic as the simple process starting from propiophenone.

ScienceSquirrel - 19-9-2008 at 08:20

The original synthesis of phenmetrazine from 2-amino-1-phenylpropanol.

Otto, W. G., Angew. Chem. 68, 181 (1956)

[Edited on 19-9-2008 by ScienceSquirrel]

jon - 12-1-2009 at 22:25

right, that's what he's driving at, 2-iodoethanol, k2co3, ephedrine, and methanol reflux it.
but why bother? that's a shit compound you want phenmetrazine the routes are cloroacetylchloride to phenylpropanolamine, followed by cyclization with koh and reduction with LAH.
or oxalyl choride and ppa and lah reduction.
or you could get *creative* and react iodo ethanol with ppa in dmf with CsOH to suppress dialkylation distill the hydroxyethylamine and cyclizize with conc. h2so4 via the concerted reaction they talk about.

i don't think the haloethanols form vinylic oxides in situ how? the conditions are'nt basic enough in the literature reactions??
am i just some dumb redneck or am i missing something here?.
i've never seen iodoethanol reacted with primary amino alcohols like this in any literature examples. for the suspected reason that they react the classic sn2 way any halo alkane and amine would giving rise to predominately tertiary amines.

oh and why bother with otc haloethanol synthesis you can buy from science supply houses no questions aked

[Edited on 13-1-2009 by jon]

[Edited on 13-1-2009 by jon]

[Edited on 13-1-2009 by jon]

Sauron - 13-1-2009 at 09:51

Be extremely careful that you do not have any intermediate or product with a halogen beta to either the N or the O, because if you do you will have either a nitrogen mustard or an oxygen analog of same. You do not want to be around either.

What's this about not knowing where to get ethylene glycol?

Antifreeze. Some are EG, others are propylene glycol, some mixtures. Go to an auto store or hardware store and read labels.

Hydrobromic acid? Make your own or use conc H2SO4 + NaBr.

But heed my warning. -N-CH2-CH2-Br is a bad idea.

(BrCH2CH2)2N- is a worse idea.

smuv - 13-1-2009 at 20:46

Quote:

Originally posted by jon
right, that's what he's driving at, 2-iodoethanol, k2co3, ephedrine, and methanol reflux it.
but why bother? that's a shit compound you want phenmetrazine the routes are cloroacetylchloride to phenylpropanolamine, followed by cyclization with koh and reduction with LAH.
or oxalyl choride and ppa and lah reduction.
or you could get *creative* and react iodo ethanol with ppa in dmf with CsOH to suppress dialkylation distill the hydroxyethylamine and cyclizize with conc. h2so4 via the concerted reaction they talk about.

I don't really think the methods you proposed are any easier.

Tertiary amines do not form as readily as in other alkylation reactions because the product is a weaker nucleophile than phenylpropanolamine. This is simular to the synthesis of glycine from B-chloroacetic acid, where the electron withdrawing character of the carboxylic acid makes glycine a weaker nucleophile than ammonia, therefore mono alkylated products predominate.

Also, sterics plays a factor in this case...

[Edited on 1-14-2009 by smuv]

Sauron - 13-1-2009 at 23:21

So the substituted alkanolamines are phenylpropanolamines?

Gee, that sounds familiar and it stinks worse than morpholine.