Has any one got any views on the reductive methylation of amino acids that selectively acheives monomethylation or at least a high percent without the
usual difficulties with seperating other subsituted amines.
thanks azostoichiometric_steve - 30-4-2008 at 04:10
Quote:
Originally posted by azo
Has any one got any views on the reductive methylation of amino acids that selectively acheives monomethylation
take 1eq amino acid, 1eq paraformaldehyde, and 2-5eq oxalic acid dihydrate, heat, workup. theres a paper in here.
"eschweiler-clarke methylation something"MagicJigPipe - 30-4-2008 at 06:41
That seems like a roundabout way to Eschweiler-Clarke methylation. I wonder if it works as well as the "standard" excess formic acid and formaldehyde
reaction.
This is a pretty good explanation of the "original" (best one I found searching with google. Wikipedia wins this time).
[Edited on 4-30-2008 by MagicJigPipe]azo - 30-4-2008 at 18:19
Thanks a lot magic and steve
I have looked at the eschweiler_clark reaction the only problem is it doesn't stop at the secondary amine stage and will almost definitely will
produce some tertiary amine ? But is there any way of stopping this from happening.
ex. ph control , time of reaction , temp reaction or other ways of stearic hindering the formation of tertiary products .
regards azo.stoichiometric_steve - 1-5-2008 at 07:34
Quote:
Originally posted by azo the only problem is it doesn't stop at the secondary amine stage
read the paper supplied by solo, it explains the monomethylation procedure.azo - 1-5-2008 at 16:44
Thanks steve .
If i am correct i think you mean the amine is methylated when treated with aqueous formaldehyde and a phosphate buffer to form the imine and then
reduced with zinc dust.
I have some alanine lying around and i will see if i can acheive a seperation of a mono methylated product and will post results .
regards azo.MagicJigPipe - 1-5-2008 at 17:50
Just out of curiosity, why do you want to monomethylate the amino group on alanine? What is this compound used for?
I can't find any information on 2-methylamino-propionic acid. That's why I ask.jon - 1-5-2008 at 18:57
akabori reactionMagicJigPipe - 1-5-2008 at 19:20
So you're going through all this trouble to get acetaldehyde (or the amino aldehyde)? Would (N-methyl)-alanine yield something different? In the
second Akabori reaction (reduction) it should just yield 2-methylamino-propionaldehyde. What is THAT used for?
Oxidation of alanine (Akabori):
(N-methyl)alanine ----> acetaldehyde + CO2 + NH3(methylamine?)
I still don't understand what you are getting at here. What is your ultimate goal? Is it really acetaldehyde? And if it is 2-methylamino-propanal,
what will you use that for? Seems like it might be an interesting compound.
[Edited on 5-1-2008 by MagicJigPipe]smuv - 1-5-2008 at 20:57
2-methylamino-1-phenylpropan-1-ol. I am not so sure about the stereospecificty (need to look up mechanism).
Edit: It is not stereoselective because the first step, is the formation of an imine
[Edited on 1-5-2008 by smuv]MagicJigPipe - 1-5-2008 at 21:45
Quote:
I have some alanine lying around and i will see if i can acheive a seperation of a mono methylated product and will post results .
Where would the phenyl group come from, smuv? Did azo mean phenylalanine in which case he is trying to obtain benzaldehyde instead
of acetaldehyde?
[Edited on 5-1-2008 by MagicJigPipe]azo - 1-5-2008 at 22:53
good work smuv.
It is not stereospecific !
When i was talking about imine formation i meant between formaldehyde and aminoacid and then reduced to N methyl alanine ? Does this answer your
question.stoichiometric_steve - 2-5-2008 at 01:24
Quote:
Originally posted by azo
Thanks steve .
If i am correct
you are not
they use paraformaldehyde and oxalic acid and thats when you end up with the reduced n-methylamino acid. nothing else. wtf!!!azo - 2-5-2008 at 03:03
azo, if you had simply read what stoichiometric_steve said and/or the reference MagicJigPipe gave, you could have saved all that typing about
a different although related reaction.
The paraformaldehyde and oxalic acid method does give a bit of di-Me, or if slightly too little is used leaves a bit of amine unmethylated. With a
little care you can keep the unwanted products to a few percent, just practice a bit to calibrate. The relative base strengths of R2NH and R3N can
help with purifying the product.
If you want really specific mono-Me and nothing else, SFAIK you'll have to go through a route where you isolate and purify an intermediate and then
further react it.MagicJigPipe - 2-5-2008 at 09:13
Azo, one last time. Are you trying to methylate alanine and then reduce or oxidize it? What is your goal?
You are being extremely vague, IMO.smuv - 2-5-2008 at 10:41
[Edited on 2-5-2008 by smuv]azo - 2-5-2008 at 16:45
hi magic
Sorry to be vague but what i am trying to acheive is the mono methylation of alanine without reducing the carbonyl group to the diol , when i can do
this i am going to try the akabori reaction with phenylpropionaldehyde to confirm that the low yield of just using alanine is due to the electron
cloud of the ring
being close to the alpha carbon and not just because the amine of the amino acid reacts directly with the aldehyde which won't happen when you are
using a secondry amine .
regards azo.MagicJigPipe - 2-5-2008 at 17:30
That's not azo. I'll just assume he wants to use the same aldehyde. "UTFSE" is rude. About as rude as STFU. As you can see (http://www.chempensoftware.com/reactions/RXN006.htm) there are at least 3 different types of Akabori reactions and only one of which involves any
AROMATIC ALDEHYDE. So, saying I should have known what he was doing just by "UTFSE" is bullshit.
Thanks for not being rude, azo. Now I know.
[Edited on 5-2-2008 by MagicJigPipe]azo - 2-5-2008 at 22:08
I am use to people jumping to conclusions ! But what i have learnt with chemistry is what it looks like on the surface is most often wrong until you
look in to it more or have all the relevent information.
thanks magic.Nicodem - 2-5-2008 at 22:53
Quote:
Originally posted by azo
being close to the alpha carbon and not just because the amine of the amino acid reacts directly with the aldehyde which won't happen when you are
using a secondry amine .
Secondary amines react with aldehydes just like primary amines. The only difference is that the hemiaminal eliminates H2O to form either the highly
electrophilic iminium compoound (in acidic media) or the nucleophilic enamine (in close to neutral). The enamine/iminium species equilibrate by
protonation/deprotonation and tautomerism. Hence the N-monoalkylated amino acids can react by exactly the same mechanism in the Akabori reaction. Only
the tertiary and quaternary amines can't.azo - 3-5-2008 at 01:10
I no that Nicodem thats because there is 1 spare valence electron on the nitrogen not counting the lone pair on a secondary amine.
But i think there will be less of a reaction between the amino acid and the aldehyde if it was n methylated due to the greater steric hinderance of
the n methylated amino acid .
regards azo.LSD25 - 12-5-2008 at 02:18
I think you will find that further alkylation will still take place on the amine even if it is converted to a secondary amine. There are any number of
papers where the authors use an aldehyde to convert the amine to an imine in order to protect against further alkylation of the amine - then hydrolyze
that imine at the end. What about using the imine of the amino acid and formaldehyde (for instance), then alkylating the amino acid with the relevant
aromatic aldehyde? You would then not need to remove the imine - simply reduce it to get the a,n-dimethylamino acid.azo - 13-5-2008 at 16:03
I understand that you will get a little tertiary amine . I have seen info on the use of an aldehyde to form a imine to protect against further
alkylations it would defently work but i don't no how practical it would be as i have seen no procedures for it.
I think you missed something here i am not wanting to produce an aromatic n dimethylamino acid i am wanting to produce a secondry amino acid and then
decarboxylate via akabori with a aromatic aldehyde to produce a phenyl amino alcohol.
IS THAT CLEAR ENOUGHazo - 13-5-2008 at 19:13
WRONG..........
IF YOU LOOK A BIT HARDER YOU WILL SEE THAT YOU DON'T GET
PHENYL PROPANOLAMINE FROM A SECONDRY AMINE AND IF YOU LOOK AT PRIOR POSTS YOU WILL SEE THAT I AM USING PHENYLPROPIONALDEHYDE WHICH WILL NOT GIVE
PPA.
REGARDS AZOLSD25 - 13-5-2008 at 23:43
Truly?
Seems I was wrong then - of course you won't mind providing poor lil' ol' me with the specific, detailed reaction that occurs when alkylating an amino
acid via the akabori procedure? I'd really like to read it, 'cos though I've seen it cited many, many, many times, I have yet to see anybody detail
anything which provides a detailed overview of how the reaction works - unless it is analoguous to these reactions:
You'll probably have noted that the alkylation only proceeds once the imine is formed from the amino acid and the second equivalent of the aldehyde?
That is what I was suggesting, although I don't know how well it would work with formaldehdye - what about forming the imide with formic acid?
See for instance this article - where the author's prepared phenylserine by reference to the method of Akabori:
Now as the other name for phenylserine is b-hydroxyphenylalanine and it is formed by the alkylation of the n-benzylidene-glycine with another
equivalent of benzaldehyde, it makes sense that if one used alanine (which by analogy seems to have been the method of Akabori) that one would get
a-methyl-b-hydroxy-phenylalanine which upon decarboxylation (heat will do it apparently) gives PPA.
Now if you look about, the alkylation of n-benzylidene-glycine/alanine with benzyl halides gives the appropriate a-alkylated phenylalanine (glycine
gives phenylalanine & alanine gives a-methylphenylalanine) both still require the formation of the imine by reaction of a substituent on the amine
group of the amino acid - which effectively removes it from further reaction.
