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Author: Subject: Amino alcohol via Akabori, trial run
unome2
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[*] posted on 22-5-2009 at 09:28


Solvent free, heated for x hours @ 130C, then hydrolysis of the 5-membered ring with 5% Aq AcOH

What is more interesting perhaps is the suggested reaction mechanism, which seems to suggest that the Akabori/Monotami(??) Reaction, which has caused so much angst, is merely a modified version of the Erlenmeyer Reaction, using a,n-dimethylglycine (alanine being a-methylglycine, thus n-methylalanine = a,n-dimethylglycine) to form the oxazole (see here where the oxazole [193] is formed from the reaction of alanine & benzaldehyde) alkylated with benzaldehyde to form [194] and subsequent hydrolysis to a-methyl-threo-B-phenylserine [195].

The decarboxylation product of a-methyl-threo-B-phenylserine would be d,l-Norephedrine if IIRC?:cool:

This opens some doors, removes the fog and allows us to work out how this sucker works. The yields with unsubstituted benzaldehydes will suck as 130C is kinda high for reacting them given their BP. Maybe we could form the aldimine/ketimine with some sort of chiral reagent, this "MIGHT" allow for better EE's (@ the 2-carbon) than are otherwise attainable.

While the formation of n-methylalanine should be simple enough - via amination of pyruvic acid / 2-bromopropionic acid with methylamine or even n-methylation of alanine, the fact remains that the yields will still be lowish. This might offer a decent route to norephedrine/norpseudoephedrine that can be converted to give P2P however. That might be useful to someone, especially if the MD-norephedrine from piperonal & alanine (apparently 87% yield according to the Japanese paper) could undergo said dehydration/hydrolysis in good yield to give MDP2P.

The only real question is whether that MD bridge is going to stand up to hydrolysis with 80% H2SO4... If it does, then this is a potentially VERY nice route to MDP2P starting from piperonal - 80% for the condensation & 80% for the hydrolysis starting from easily acquired alanine (OTC). Even if the piperonal has to be made (from vanillin), this is still viable (the worst yields are in the conversion of vanillin to protocatechualdehyde to piperonal). This step would be done prior to fucking around with the alanine. The condensation itself should be similar to that of vanillin and creatinine (done in the melt).;)

Also, if one wanted to avoid wastage of piperonal (made from vanillin which is cheap), then consider preforming the aldimine/oxazole from alanine & vanillin first, this will mean you save one equivalent of piperonal (thus avoiding oxidation of the same). Also, if the MD ring can stand up to H2SO4, then we could skip the AcOH hydrolysis of the alkylated oxazole - simply perform the double hydrolysis (of the oxazole then of the norephedrine) in one pot distilling the MDP2P straight out of the pot. This would cause wastage of vanillin and any unreacted piperonal, but it is an option perhaps?

[Edited on 22-5-2009 by no1uwant2no]
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[*] posted on 22-5-2009 at 16:20


More good stuff no1uwant2no,

But, "Solvent free, heated for x hours @ 130C, then hydrolysis of the 5-membered ring with 5% Aq AcOH" - is basically the normal method, so where does the 48% yeilds come from?
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[*] posted on 22-5-2009 at 18:32


Normal method according to whom? The method originally used by Akabori was done in Pyridine apparently.
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[*] posted on 22-5-2009 at 21:36


i mean the normal method as most people are using on this forum, and in this thread.
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[*] posted on 23-5-2009 at 00:39


Quote: Originally posted by no1uwant2no  

What is more interesting perhaps is the suggested reaction mechanism, which seems to suggest that the Akabori/Monotami(??) Reaction, which has caused so much angst, is merely a modified version of the Erlenmeyer Reaction, using a,n-dimethylglycine (alanine being a-methylglycine, thus n-methylalanine = a,n-dimethylglycine) to form the oxazole (see here where the oxazole [193] is formed from the reaction of alanine & benzaldehyde) alkylated with benzaldehyde to form [194] and subsequent hydrolysis to a-methyl-threo-B-phenylserine [195].

On the contrary. The authors propose a completely different mechanism in the paper you posted. You are confusing oxazol-4(5H)-ones (which are active methylene compounds) with oxazolidines (which are not active methylene compounds). But anyway, the mechanism the authors propose is also wrong since it does not survive the Occam's razor. That is because it is generally not allowed to propose a mechanism where the pKa difference between an intermediate (a carboanion in this case) and the other species or reaction media is more than 7 or 8 units. The scheme is only correct if interpreted as formalism, but not as a mechanism.
Quote:
But, "Solvent free, heated for x hours @ 130C, then hydrolysis of the 5-membered ring with 5% Aq AcOH" - is basically the normal method, so where does the 48% yeilds come from?

The yield based on benzaldehyde accounting the stoichiometry from this later paper is two times higher than the one calculated using a 1:1 stoichiometry assumed previously. According to this later paper one benzaldehyde is lost in the condensation with the aminoalcohol to give the hydroliticaly labile oxazolidine, hence the stoichiometry of the reaction is 2:1 (PhCHO vs. alanine).




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[*] posted on 23-5-2009 at 09:47


no1uwant2no, you may want to have a look at US patent 5346828. The Erlenmeyer phenylserine synthesis is applied to benzaldehyde and alanine to form the corresponding substituted phenylserine.

Quote:
Sodium hydroxide (6 g, 150 mmoles) and 8.9 g (100 mmoles) of D,L-alanine are dissolved in 25 mL of water and the solution cooled to around 5° C. while stirring 5 under a nitrogen atmosphere. To the solution is added 21.2 g (200 mmoles) benzaldehyde and the mixture stirred at 5° C. for approximately one hour. The mixture is then warmed to room temperature and maintained for approximately 20 hours. Concentrated hydrochloric 10 acid then is added to bring the reaction mixture to pH 2.0. After stirring the acidic solution for two hours, the aqueous phase is separated, extracted with ethyl acetate and evaporated under vacuum. The resulting material is twice extracted with 80 mL hot absolute ethanol, followed by evaporation of the ethanol. The resulting material is again extracted with 40 mL of absolute ethanol, followed by removal of the ethanol. The extract then is dissolved in at least a 1:1 methanol water solution and absorbed on polymethacrylate column. The column then is eluted with the same 1:1 methanol water solution and the 30 mL fractions are combined and evaporated. Purification on the polymethacrylate column then is repeated using 95% ethanol to yield racemic 2-amino-2-methyl-3-hydroxy-3-phenylpropionic acid.
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[*] posted on 23-5-2009 at 15:56


Nicodem, I don't pretend to understand how it works... It is just that seeing the rearrangement of the expected system to a 5-membered ring made me look into the Erlenmeyer-Plochl amino acid synthesis (which uses glycine and benzaldehyde to form amino acids for those who may be interested).

As the initial product formed between the alanine & an equivalent of benzaldehyde would be the a-methyl variant of the same oxazolone formed in the Erlenmeyer-Plochl, the fact that this decarboxylates prior to alkylation does not alter that, does it?

Why the alkylation takes place sans-active methylene is beyond me... But if the authors are to be believed (well, their claims to having found said 5-membered ring mainly), the rearranged oxazoleidine (rearrangement taking place after decarboxylation) undergoes alkylation at the SAME site (the 2-carbon of the alanine) with decarboxylation. That is intriguing all by itself.
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[*] posted on 24-5-2009 at 10:28


You seem to believe that some kind of rearrangement to an oxazolidine happens and also that the reaction proceeds via an alkylation of an oxazoline?
I don't know where you saw this, but the schemes in the paper you yourself uploaded, particularly the scheme 3, clearly states this is not the case (though I have no idea what the text says). First of all you need to know that you can not call a reaction rearrangement, unless it is intramolecular and actually involves a structural rearrangement. For example, there are no rearrangements in the Akabori reaction. You can help yourself understanding this if you draw the reaction and number each carbon atom. If all connectivities between them remain in the same connectivity order as in the original reactants, then the reaction involves no rearrangement.
The formation of the oxazolidine is just a normal condensation between a beta-aminoalcohol and benzaldehyde (see the last step in the scheme 3; such a condensation is pretty much the most common way of forming oxazolidines). As far as I can see the oxazolidine formation has no major impact on the Akabori reaction itself except in that it changes its stoichiometry, because one benzaldehyde equivalent is lost due to this condensation with the all the (N-methyl)-1-phenyl-2-aminopropanol isomers formed.
There is however an important, even though mostly formalistic, consequence (unless the oxazolidine forms only when using N-methylalanine but not with alanine, which is however unlikely). Namely, the Akabori reaction gives a much better yield when the correct stoichiometry is accounted for. Much of the fame of its low yields is thus due to using the wrong stoichiometry in calculating the yield and/or the belief of having used alanine as the limiting reagent. For example, CycloKnight always used substoichiometric amounts of benzaldehyde in his experiments, even in those cases where he believed of having used an excess of it (unless I missed some example while rapidly skimming trough the start of the thread). So it is yet to see what the yields are when using alanine as the limiting reagent (maybe this is explained in that paper, but since I don't understand Japonese…).

Also, the Erlenmeyer-Plochl synthesis is a reaction quite different from the Akabori reaction. It starts with N-acyl-alpha-amino acids (usually hippuric acid) since only these can cyclisize to the corresponding oxazol-4-one when treated with acetic anhydride. The formation of this oxazolone which has an active methylene group (a -CH2- group that is acidic enough to form a carboanion) is a prerequisite for the condensation with aldehydes. The products of this reaction are 5-alkylidene substituted oxazol-4-ones (these can be further transformed to a number of different products such as alpha-amino acids, pyruvic acids, etc.). As you see the reaction has nothing in common with the Akabori reaction: the starting compound and the product are different, as well as the mechanism and the reaction conditions. Not even the patent example posted by Manimal above uses the Erlenmeyer-Plochl synthesis since it uses completely different reaction conditions and starts with an alpha-amino acid instead of its N-acyl derivative. It is however similar to the Akabori reaction conducted at conditions where no decarboxylation can occur (if that patent claim is true at all!).

[Edited on 24/5/2009 by Nicodem]




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[*] posted on 24-5-2009 at 13:42


Have a look at the schematic in this paper, the initial product of the reaction between benzaldehyde and alanine is an azlactone[193], essentially the same type (albeit the a-methyl variant) which is formed first during the reaction between glycine and one equivalent of benzaldehyde.

Now, somehow this must rearrange if a different, decarboxylated 5-membered ring was found by the Japanese researchers after alkylation with another equivalent of benzaldehyde, provided the same original azlactone formed when the n-methylalanine was treated with benzaldehyde. But if we start with a 5-membered ring and end with a different 5-membered ring, then there has to have been a rearrangement at some point - to my mind at least (granted, I am arguing from the position that what I have cited is gospel - not from such huge font of knowledge as to how azlactones, etc. work).

Someone asked why the yields vary between the benzaldehydes. The variation in the reported yields in that paper, with the various benzaldehydes, might be attributable to the nature of the benzaldehydes themselves (some of which are solid and the reaction would take place in the melt). These might
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[*] posted on 25-5-2009 at 04:35


Quote: Originally posted by no1uwant2no  
Have a look at the schematic in this paper, the initial product of the reaction between benzaldehyde and alanine is an azlactone[193], essentially the same type (albeit the a-methyl variant) which is formed first during the reaction between glycine and one equivalent of benzaldehyde.

I think you misunderstood that reaction scheme. The oxazolinone 193 is not formed from alanine and benzaldehyde, it is instead formed by the usual method by treating N-benzoyl-alanine with acetic anhydride. It is the 194 that is formed by the addition of 193 on benzaldehyde (as explained in the text). If you would have checked the oxidation states in the compound 193 you would have noticed that its synthesis from alanine via a condensation reaction can not be done with benzaldehyde but with a benzoic acid derivative (note that the carbon at oxazole position 2 has a double bond with nitrogen!). You should read the chapter 7.3.1.1 where the preparation of oxazol-4-ones from N-acyl-amino acids is described, that should clear up your confusion.
Quote:
Now, somehow this must rearrange if a different, decarboxylated 5-membered ring was found by the Japanese researchers after alkylation with another equivalent of benzaldehyde, provided the same original azlactone formed when the n-methylalanine was treated with benzaldehyde. But if we start with a 5-membered ring and end with a different 5-membered ring, then there has to have been a rearrangement at some point - to my mind at least (granted, I am arguing from the position that what I have cited is gospel - not from such huge font of knowledge as to how azlactones, etc. work).

But there is no azlactone involved in the Akabori reaction. I still do not get it where you got the impression this was so. There is nothing of this kind in the schemes of that Japonese paper. The only connection with the oxazoles is in that these researchers found out that one equivalent of benzaldehyde is lost due to the condensation with the product thus forming the oxazolidine (see the last step in scheme 3).
Besides, if no rearrangement happens, then you can not call a reaction as rearrangement, and obviously no rearrangement happens in the Akabori reaction - all the structural fragments are accounted for exactly in the same connectivity as in the starting compounds. (see http://en.wikipedia.org/wiki/Rearrangement_reaction)

PS: Please start using the upper caps for N when this indicates the heteroatom position. It is annoying to always read things like " n-methylalanine" since this means something other than N-methylalanine. The prefix "n-" means normal isomer (as for example in n-butanol, etc.) which is completely of topic here.




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[*] posted on 25-5-2009 at 05:51


Quote: Originally posted by no1uwant2no  
Have a look at the schematic in this paper, the initial product of the reaction between benzaldehyde and alanine is an azlactone[193], essentially the same type (albeit the a-methyl variant) which is formed first during the reaction between glycine and one equivalent of benzaldehyde.

Unfortunately, that paper cannot be read. Someone with the right access please download and post it here.
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[*] posted on 25-5-2009 at 15:36


There is no rearrangement in what is described in the Akabori reaction papers, I concede that certainly - but if alanine reacts with one mol of benzaldehyde to give an azlactone - that is what happens in the cited article (which is from Google books so I cannot download it JohnWW - here is the citation but: "The Chemistry of Heterocyclic Compounds, Oxazoles" David C. Palmer, pp.171-2)

But the issue I have is that if the benzaldehyde and the a-amino acid are common to both, what is so difficult in suggesting that the reaction can go either way? Put simply, same reagents - slightly different conditions (although the acid anhydride might be simply acting as a dehydration agent, which I suspect the heat of the Akabori-type reaction might do just as effectively), bizzarely similar products - both going through a 5-membered ring...

PS As to the oxidation state of the benzylidene - why would it be precluded from having a double-bond on Nitrogen? Not being argumentative, I'm trying to get my head around it.

[Edited on 26-5-2009 by no1uwant2no]
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[*] posted on 25-5-2009 at 23:26


Quote: Originally posted by no1uwant2no  
There is no rearrangement in what is described in the Akabori reaction papers, I concede that certainly - but if alanine reacts with one mol of benzaldehyde to give an azlactone - that is what happens in the cited article (which is from Google books so I cannot download it JohnWW - here is the citation but: "The Chemistry of Heterocyclic Compounds, Oxazoles" David C. Palmer, pp.171-2)

Seems like I have to say everything three times before you actually get to read it. I must say that your old habits die hard. So I'll try one more time, more concisely, before I give up.
You say "if alanine reacts with one mol of benzaldehyde to give an azlactone" and I say: benzaldehyde and alanine do not react to give an azlactone!
Please read the text concerning the scheme where that compound 193 is mentioned and more so the chapter on the synthesis of azlactones from N-acyl-amino acids. I already told you that once you read it you should clear up your confusion.
Besides, why don't you just try to draw the condenstation between alanine and benzaldehyde so that you can see it can not give any azlactones unless the reaction involves an oxidation (always check the oxidation states of the left and right side of the equation!).
Quote:
But the issue I have is that if the benzaldehyde and the a-amino acid are common to both, what is so difficult in suggesting that the reaction can go either way?

But they are not common to both! I already explained it more than once that the syntheses based on the condensation or alkylation reactions on oxazol-4-ones (azlactones), like is the case with the Erlenmeyer reaction, always start with N-acyl-alpha-amino acids (obviously, since only these can cyclisize into azlactones). Benzaldehyde has nothing to do with the Erlenmeyer or related reactions, utmost it can be used as an aldehyde in a condensation reaction to prepare the 5-benzylidene derivatives, but not for the synthesis of the azlactone itself.

Quote:
Put simply, same reagents - slightly different conditions (although the acid anhydride might be simply acting as a dehydration agent, which I suspect the heat of the Akabori-type reaction might do just as effectively), bizzarely similar products - both going through a 5-membered ring...

First of all the conditions are anything but similar and the products are not even closely related. Not to even mention that in the Erlenmeyer reaction benzaldehyde has no role while in the Akabori reaction one equivalent is necessary in order to form the amine/imine for the alpha-CH group activation so that this can participate in the condensation with another equivalent of benzaldehyde as well as to allow the decarboxylation of the -COOH group.
In short: for the cyclization of N-benzoyl-alanine you need zero equivalents of benzaldehyde; while for the formation of the oxazolidine end product of the Akabori reaction you need two equivalents of benzaldehyde. (your homework: draw the reactions and balance the equations)
I really can not explain better. It is your turn to read all the posts all over again, do some reaction drawing and some reading, because I really can not explain in any simpler words.




…there is a human touch of the cultist “believer” in every theorist that he must struggle against as being unworthy of the scientist. Some of the greatest men of science have publicly repudiated a theory which earlier they hotly defended. In this lies their scientific temper, not in the scientific defense of the theory. - Weston La Barre (Ghost Dance, 1972)

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[*] posted on 30-5-2009 at 00:24


Thanks Nicodem, I've been busy - always helpful:)

OK

So if the conditions are not even remotely similar, it is just happy coincidence that they both lead to 5-membered rings? I've obviously been reading into it more than I should (kind of a habit of mine:P)...

Sorry, I've not replied sooner, but I've been running around like a blue arse fly...

So is the "Akabori" with the solid (up until the rxn temps) aldehydes working through the same procedure - ie. using two equivalents of the aldehyde to form the imine first, or are they simply condensing with it (like the orgsyn procedure cited earlier)?
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[*] posted on 25-6-2009 at 20:22


whats stopping you from doing the akabori on piperonal with straight alanine then a reductive alkylation with formaldehyde.

followed up by an oxidation with a jones reagent or such to make methylone ;)

[Edited on 26-6-2009 by Ephoton]




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smile.gif posted on 25-8-2009 at 04:06
Did it with some mods


These are some of the stuff I did differently :
(I did the cooking without solvent, just pure benzal + l-alanine, for 1hour 40mins, solution is dark red then wait to cool before I added Xylene to extract. I was going to distill but since I only did 4.93gr of l-alanine and 10ml of benzaldehyde, the vapor wasn't that much and my distillation equipment is kind of big :( so when the vapor turned into liquid, it drop back into the RBF)

1. Used Xylene as pulling solvent
2. After I added NaOH to the aquaeus solution, I added Xylene
3. Half of the Xylene with ppa freebase (supposedly), is dried (still waiting as I post this), half of the Xylene is gassed with HCl. (Made the HCl gas using HCL (l) dropped into H2SO4 (l), which then transferred to CaCl2, then put out), which then I filtered... this yields yellowish color (almost like vaseline like) products.
4. The crystals within the RBF I washed with methanol. Methanol turned pale yellow, and I'm left with salt like stuff (supposedly l-alanine salts). Methanol is being dried as I post this.

My question to more senior members are, if kind enough to point me to the right direction:
1. When I gas the last Xylene solution, is it ppa.HCl? (I did the Al foil burn test and it have no left overs)
2. The cyrstals in the RBF after methanol wash, is it l-alanine? (I did another Al foil burn test :cool:, they turned into cotton like stuff) was wondering if I can reuse this salt like products as l-alanine for next batch.
3. Is there anywhere I did wrong?


Note: All the solvent are technical grade, about 95% purity.

Anyway, thanks for replying and not flaming :D
I'd post some pic but my bluetooth is not working for some reason... =\

[Edited on 25-8-2009 by WaTau]

[Edited on 25-8-2009 by WaTau]

[Edited on 25-8-2009 by WaTau]
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[*] posted on 25-8-2009 at 10:27


Has anyone actually performed a rigourous analysis of the product obtained by this reaction to determine it's unequivocal composition?

According to JCS 1953, p. 3255 the only recoverable compound is 1,2-diphenylethanolamine.

Quote:
EXPERIMENTAL
All the decarboxylations were carried out under similar conditions, which are described for the typical case of +/- alanine.
Decarboxylation of DL-Alanine in Benzaldehyde.-DL-Alanine (10 g., 0.11 mole) and freshly distilled benzaldehyde (100 c.c., 1 mole) were heated in a 200-C.C. Claisen flask (thermometer in liquid), fitted so that gases evolved were passed through a trap cooled by solid carbon dioxide and alcohol into a test-tube containing lime-water. At about 120” the alanine dissolved with effervescence, and heating was continued so as to distil off water formed but as far as possible retain the benzaldehyde (2-5” below the b. p.). After 5-10 min. evolution of carbon dioxide practically ceased; in the trap were found ice, acetaldehyde, and a little benzaldehyde. The cooled deep red reaction mixture was shaken with 6N-hydrochlonc acid
(200 c.c.) and distilled in steam to remove all the benzaldehyde. The hot acid solution was boiled with 2 g. of animal charcoal, then filtered, and the boiling filtrate was cautiously made alkaline with pellets of sodium hydroxide. This treatment was necessary in order to ensure the precipitation of the bases in a crystalline and filtrable form. After several hours the solid was collected, washed with water, and dried to give 2 g. of crude product (8.4%). The two racemic forms of 8-hydroxy-1 : 2-diphenylethylamine were separated from the crude product by shaking with ether (150 c.c.), in which the is0 is much more soluble than the normal isomer. The residue, recrystallised from benzene, had m. p. 163-164” (Found : C, ‘78.8; H, 6-8; N, 6-7. Calc. for C,,H,,ON : C, 78.8; H, 7.0; N, 6-6%), unaltered by admixture with an authentic specimen of normal 2-hydroxy-1 : 2 diphenylethylamine (Weijlard, Pfister, Swanezy, Robinson, and Tishler, J. Amer. Chem. SOL, 1951, 73, 1216). The N-acetyl derivative (from ethanol) had m. p. 196’ ; Soderbaum (Bey., 1896, 29, 1210) gives m. p. 196-197O.


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[*] posted on 9-11-2009 at 16:11


they are using 1/1 stochiometry there that's why geez i'm retarded and i can figure that, out unlike nicodem geez, that guy will work circles around us myself included.
wonder if nicodem has a scheme to rule the, world dude could have been bill gates.
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[*] posted on 8-12-2009 at 16:57


u know there have been longwinded arguments about oxazolidones and such on other forums ,and americans seem to overtrust the japanesse; but, i'm with nicodem and i believe occam's razor applies in this case.
think about it ever try to work on a japanesse car? very tight tolerances, they must still be pissed about the war.
i would'nt be suprized if that spews over into thier english translations of chemical literature.

so jon figures he's to devise a workup stratagem seeing how 1,2diphenylethanolamine is the major product, he dutifully searched chem abs to no avail on properties.
a scientist at sigma alldick (pun intended) was consulted on it's solubilty properties and was again reffered to you guessed it. a chemical abstract search.
well he explains this has been attempted and the scientist was kind enough to give this detail:
.6g/100ml ethanol upon sonication in other words it just crashes out with the rest of the garbage.


[Edited on 9-12-2009 by jon]

[Edited on 9-12-2009 by jon]

[Edited on 9-12-2009 by jon]
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[*] posted on 8-12-2009 at 18:19


Quote: Originally posted by jon  
u know there have been longwinded arguments about oxazolidones and such on other forums ,and americans seem to overtrust the japanesse; but, i'm with nicodem and i believe occam's razor applies in this case.
think about it ever try to work on a japanesse car? very tight tolerances, they must still be pissed about the war.
i would'nt be suprized if that spews over into thier english translations of chemical literature.

so jon figures he's to devise a workup stratagem seeing how 1,2diphenylethanolamine is the major product, he dutifully searched chem abs to no avail on properties.
a scientist at sigma alldick (pun intended) was consulted on it's solubilty properties and was again reffered to you guessed it. a chemical abstract search.
well he explains this has been attempted and the scientist was kind enough to give this detail:
.6g/100ml ethanol upon sonication in other words it just crashes out with the rest of the garbage.


[Edited on 9-12-2009 by jon]

[Edited on 9-12-2009 by jon]

[Edited on 9-12-2009 by jon]
Three edits and you can't still find the energy to depress the shift key. I think this post "crashes out with the rest of the garbage."



Better to remain silent and appear a fool than to open your mouth and remove all doubt.
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roamingnome
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[*] posted on 8-12-2009 at 18:37


that's it, its time to embrace nitroethane once and for all
while not forgetting cell lysis or l-pac

but not thinking japan is always right

http://www.journalarchive.jst.go.jp/english/top_en.php

enjoy many free pdf here from japan, i spend most of my day looking at this

also download full papers from iran...hey didnt they event acid?!

http://www.ics-ir.org/jics/archive/v3/1/article/8/index.html

but unfortunately i can not find the full version of this
which says iran
http://www3.interscience.wiley.com/journal/114261883/abstrac...

Selective Deprotection of Bisulfite Addition Products by FeCl3×6H2O and Fe(NO3)3×9H2O Supported on Silica Gel under Solvent-Free Conditions.

can some one get this article its seems strait forward from the abstract but bisulfite-adduct Deprotection without water is quite important
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roamingnome
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[*] posted on 8-12-2009 at 18:39


yes jon step up the integrity of those posts ....

we know you got the minerals
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manimal
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[*] posted on 15-4-2010 at 17:22


Quote: Originally posted by manimal  
no1uwant2no, you may want to have a look at US patent 5346828. The Erlenmeyer phenylserine synthesis is applied to benzaldehyde and alanine to form the corresponding substituted phenylserine.


Xtaldoc claims success: http://www.drugs-forum.com/forum/showthread.php?p=678668#pos... I wonder if he elucidated this independently.
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[*] posted on 23-7-2010 at 19:43


Quote: Originally posted by manimal  
Has anyone actually performed a rigourous analysis of the product obtained by this reaction to determine it's unequivocal composition?

According to JCS 1953, p. 3255 the only recoverable compound is 1,2-diphenylethanolamine.

Quote:
EXPERIMENTAL
All the decarboxylations were carried out under similar conditions, which are described for the typical case of +/- alanine.
Decarboxylation of DL-Alanine in Benzaldehyde.-DL-Alanine (10 g., 0.11 mole) and freshly distilled benzaldehyde (100 c.c., 1 mole) were heated in a 200-C.C. Claisen flask (thermometer in liquid), fitted so that gases evolved were passed through a trap cooled by solid carbon dioxide and alcohol into a test-tube containing lime-water. At about 120” the alanine dissolved with effervescence, and heating was continued so as to distil off water formed but as far as possible retain the benzaldehyde (2-5” below the b. p.). After 5-10 min. evolution of carbon dioxide practically ceased; in the trap were found ice, acetaldehyde, and a little benzaldehyde. The cooled deep red reaction mixture was shaken with 6N-hydrochlonc acid
(200 c.c.) and distilled in steam to remove all the benzaldehyde. The hot acid solution was boiled with 2 g. of animal charcoal, then filtered, and the boiling filtrate was cautiously made alkaline with pellets of sodium hydroxide. This treatment was necessary in order to ensure the precipitation of the bases in a crystalline and filtrable form. After several hours the solid was collected, washed with water, and dried to give 2 g. of crude product (8.4%). The two racemic forms of 8-hydroxy-1 : 2-diphenylethylamine were separated from the crude product by shaking with ether (150 c.c.), in which the is0 is much more soluble than the normal isomer. The residue, recrystallised from benzene, had m. p. 163-164” (Found : C, ‘78.8; H, 6-8; N, 6-7. Calc. for C,,H,,ON : C, 78.8; H, 7.0; N, 6-6%), unaltered by admixture with an authentic specimen of normal 2-hydroxy-1 : 2 diphenylethylamine (Weijlard, Pfister, Swanezy, Robinson, and Tishler, J. Amer. Chem. SOL, 1951, 73, 1216). The N-acetyl derivative (from ethanol) had m. p. 196’ ; Soderbaum (Bey., 1896, 29, 1210) gives m. p. 196-197O.





Lots of people are arguing that the only compound obtained in this procedure is 1,2-diphenylethanolamine (DPEA) quoting this study and affirming that no PPA is obtained at all. DPEA is indeed the major product but it is easily removed in the work-up. The reason the authors are not isolating any PPA is mainly due to

1) the very little scale theyre working on (amount of PPA is almost insignificant at this scale)

2) they are omitting to concentrate the aqueous phase extracts in the work-up (the latter contains PPA*HCl). PPA freebase being miscible in water, extraction with a non polar solvent therefore fails.

[Edited on 24-7-2010 by Quantum_Dom]
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[*] posted on 25-7-2010 at 12:38


right i spoke to a scientist at sigma aldrich 1,2-dpea is practically insol. in water and alcohol.
ppa is sol. in water but it can be salted out.
the procedure is to saturate the aqueous phase with salt, and add isopropanol until phase seperation is noted and perform extractions of ppa in this manner.

[Edited on 25-7-2010 by jon]
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Polverone
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