Sciencemadness Discussion Board

Amino alcohol via Akabori, trial run

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swip2 - 12-2-2008 at 22:12

Swip2 dident notice wether diffrent solvents can be used, And whitch ones work better.

Swip2 assumes an avarige non polar solvent would do for the post reaction work up :(

manimal - 17-2-2008 at 12:58

Sodium carbonate can be used to base ephedrine or norephedrine, but isn't strong enough to base benzylamine because benzylamine can form a carbonate salt whereas ppa cannot, no? So wouldn't the addition of a carbonate seperate the ppa as a freebase, but not the benzylamine, which forms a carbonate?

PPA Sulphate or Citrate maybee

swip2 - 17-2-2008 at 18:50

Quote:
Originally posted by IOC
Try alanine/benz, heat directly, when CO2 is done add equal amount of toulene then add equal amount of H2O and acidify to PH 4, seperate water phase and 3 x methylene chloride to clean.

Boil down H2O pahse to a brown paste then add dry acetone to leave clean white powder.

Havnt had time to qualify MP, next time though

IOC......out


So how about a PPA Sulphate or a citrate salt.
Swip2 just isent sure weather a sulphate salt would reduce to the desired amine

[Edited on 18-2-2008 by swip2]

Nicodem - 19-2-2008 at 11:16

From here on no more schizophrenic third person writing or use of acronyms of the SWIM-type will be allowed. Any other methods of posing or pretending of being involved in some criminal activities, or similar childish attempts at being kewl, are also not desired.

Any post not respecting the above will be deleted.

This warning is valid in general and not just this thread.

This used to be a very nice and informative thread about an interesting reaction. Do not deteriorate it! If you have nothing useful to add, then rather think twice before posting.

Ephoton - 19-2-2008 at 23:41

sorry tar.gz was ment to mean tar ball :) I was in a hurry and should have thought more about my post.

stoichiometric_steve - 20-2-2008 at 03:22

i was about to complain, too. maybe some cleanup should be done here. from page 2 it's mainly bullshit being posted.

LSD25 - 24-2-2008 at 15:52

Quote:
Originally posted by Nicodem
From here on no more schizophrenic third person writing or use of acronyms of the SWIM-type will be allowed. Any other methods of posing or pretending of being involved in some criminal activities, or similar childish attempts at being kewl, are also not desired.

Any post not respecting the above will be deleted.

This warning is valid in general and not just this thread.

This used to be a very nice and informative thread about an interesting reaction. Do not deteriorate it! If you have nothing useful to add, then rather think twice before posting.


Done,

All non-informative and cookery related material that I have access to is now deleted and will not return.

This should please Sauron & Polverone no fucking end.

Out

Nicodem - 24-2-2008 at 23:23

I don't know what you are up to, but your posts were one of the few here that did not fit the criteria for removal. Besides, the warning was set for 19/2/2008 and forward and not for already posted material. Take notion that his is not a thread where you can demonstrate your political opinions by performing irrational acts!

Edit: Now that I think of it, you are actually trying to be kewl if the definition of kewlness is stretched a bit. Maybe your above post actually breaks the anti-kewls rule after all. I'll have to think about it and perhaps remove it.

[Edited on 25/2/2008 by Nicodem]

akabori

azo - 7-3-2008 at 14:37

I was looking at the akabori reaction and come up with the idea of reacting phenylalinine with formilin the reaction seems to follow the same mechanism as the benz method only that the end product would be 1 phenyl 2 amino propane 3 ol . The only thing that i thought would be getting the temperature high enough to decarboxylate the phenylalinine due to the boiling point of water
and the reactivity of the formaldehyde which is a lot more reactive than benzaldehyde
if it would work i thought it would give a better yield due to the solubility of phenylalinine and formilin and easier for the workup with less tar.
has anyone seen this reaction before and do you think it is possable

regard azo

Maja - 28-4-2008 at 10:11

Retried Akabori reaction with Benzaldehyde and Alanine... This time I used overhead stirrer and kept temperature 150-165* for about 2hours. Some benzylamine and other products distilled over 2hours, but not a lot of. OK ... This is phase were I faced some problem.. Tried to extract with toluene... And guess what ? No layers ... Just brown solution with no layers. Emulsion ? I don't think so... Maybe someone knows how to solve this problem ?

Hilski - 28-4-2008 at 19:28

Quote:
Originally posted by Maja
Retried Akabori reaction with Benzaldehyde and Alanine... This time I used overhead stirrer and kept temperature 150-165* for about 2hours. Some benzylamine and other products distilled over 2hours, but not a lot of. OK ... This is phase were I faced some problem.. Tried to extract with toluene... And guess what ? No layers ... Just brown solution with no layers. Emulsion ? I don't think so... Maybe someone knows how to solve this problem ?


There really aren't any layers. The toluene is mostly to thin things out a bit so it will be easier to filter out the unreacted alanine and maybe some tar etc. The last time I tried this (quite a while back) I just filtered the thinned solution a few times to clean it up as much as possible and then added water and HCL to extract the reaction product. That's really the only time I saw any kind of layers form. I never really had much of a tar layer at the bottom.
Once the water has been evaporated, the amine salt is left behind as a dirty paste (Just like in the photos) and is kind of a bitch to clean up, but it can be done.

As an aside, I seem to remember having VERY strong ammonia-like vapors coming from the reaction solution. I'm not sure if this was from a lot of benzylamine (or other side products) forming, or if it was just from the amine that that was the desired product of the reaction.


**I almost forgot...After the aqueous layer has been separated, be sure to wash it several times as well as you can with a non-polar solvent. DCM definitely works best for this if you have it. Otherwise it'll take a ton of ethyl acetate to clean up the paste at the end.
Pretty much everything Cycloknight put in his write up works the way he says it does if you follow his procedures.

[Edited on 29-4-2008 by Hilski]

azo - 28-4-2008 at 20:00

? what isomer is produced by the akabori is it norephedrine
or norpseudoephedrine.
I haven't heard it mentioned

thanks azo.:)

Maja - 29-4-2008 at 11:23

Oh, I see... So why does he state that you need to extract with toluene ? :] I will try what you said. I have DCM. I don't actually remember from the past if there was layers or no ..

I think it's norpseudoephedrine if IIRC :]

[Edited on 29-4-2008 by Maja]

Hilski - 29-4-2008 at 16:01

Quote:
Oh, I see... So why does he state that you need to extract with toluene ? :] I will try what you said. I have DCM. I don't actually remember from the past if there was layers or no ..

I think it's norpseudoephedrine if IIRC :]


I'm not really sure why he used the term 'extract' when referring to the procedure right after the reaction completes. I believe it's just semantics, and I THINK what he was intending to convey was that one just needs to add toluene to the final reaction mixture. Then filter out the unreacted alanine and wash the alanine with some clean toluene to make sure no product is left behind. Then pool all the toluene together to do the the extraction on the toluene with aqueous HCL.

If I'm not mistaken, the reaction product is a mixture of d- and l- norephedrine, a.k.a phenylpropanolamine. There might also be some norephedrine in there too, but I don't think so.

jon - 1-5-2008 at 19:03

benzylamine is only formed about 1-3% it is soluble in h2o completely not an issue. diphenylethanolamine is insoluble in alcohol as a base per a scientist at sigma aldrich .6g/100ml with heavy sonification. this is the major impurity.

Hilski - 17-5-2008 at 20:44

So Maja, any luck? I'm just curious, because I don't remember reading any posts by anyone other than Cycloknight who actually tried the experiment and got it to work . IIRC I ended up with %15-%20 yields (based on benzaldehyde) when I did it last.

HeilHamas - 11-11-2008 at 19:37

Quote:
Originally posted by jon
I've given the serine route a thought but after some research I concluded it may or may not be possble because of the steric factors involved with alanine the carbon bearing the alpha hydrogen is substituted in alanine in the case of glycine it is not.

[Edited on 23-12-2007 by jon]


It's been demonstrated to work for alanine and nitrobenzaldehyde in an uncatalyzed reaction. Nitrobenzaldehyde is more reactive in such reactions, but if steric hinderance doesnt hinder alanine and nitrobenzaldehyde, why would it hinder alanine and benzaldehyde? It should be tried in a standard base-catalyzed reaction.

chochu3 - 14-11-2008 at 20:41

PPA*HCl
very sol. H2O
sol. Alcohol
insol. ether, benzene, chloroform

PPA freebase
sol. ether

Alanine
sol. H2O
slightly sol. Alcohol, pyridine
insol. Acetone, ether

Best way to workup this reaction would be to do the extraction with ether so it will solvate the PPA and leave the alanine behind. Referenced from Handbook of Chemistry and Physics 85th edition except the solubilty of the phenylpropanolamine freebase which was read from a patent online here

hector2000 - 4-12-2008 at 08:03

there is no way for convert ppa to pesudoehedrine?
i think if we use methyle alanine then we will have pesudoephedrine

hector2000 - 21-12-2008 at 10:14

why final product has tart and bitter taste?

l-alanine to ppa

sk13m11 - 26-3-2009 at 07:39

can i use beta-alanine instead of l-alanine

Siddy - 27-3-2009 at 03:55

@ sk13m11 - You can use l-alanine, d-alanine or dl-alanine. NOT BETA (B)!

@ hector2000 - The original Akabori (and MOMOTANI) paper trials n-methyl alanine (thats the whole point of the paper, semi synthetic ephedrine). They got 16% with there method which uses pyridine. The methods you read on this site increase the yield and dont use pyridine. Although since n-methyl alanine is not commercially available no one as tried it since Akabori...
You can methylate an amine (single methylation) with formaldehyde and a reducing agent.

Products via this method are racemic (4 different isomers).

[Edited on 27-3-2009 by Siddy]

sk13m11 - 31-3-2009 at 02:17

thanks for the info

unome2 - 21-5-2009 at 14:44

Sorry, under a new pseudonym (while I try and find my original activation email - I suspect it is in an account which has been heavily spammed).

Found this the other day, it is in Japanese too, but it at least shows a putative reaction scheme (via an acid labile, 5 membered ring). Interestingly, it claims that the 3,4-MD benzaldehyde (piperonal) gives up to 87% yield while plain old benzaldehyde gives only <50% (which is likely, given that it forms two sides of the 5 membered ring).

The references from this paper really ought to be rather special reading if anyone can find them too:cool:

Attachment: akabori.rewrite.2007.pdf (94kB)
This file has been downloaded 1621 times


Siddy - 21-5-2009 at 17:41

good find no1uwant2no,

You can get a ruff translation from google, and make sense of it.

But it doesnt detail the method, or say how the yield was increased.

unome2 - 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]

Siddy - 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?

unome2 - 22-5-2009 at 18:32

Normal method according to whom? The method originally used by Akabori was done in Pyridine apparently.

Siddy - 22-5-2009 at 21:36

i mean the normal method as most people are using on this forum, and in this thread.

Nicodem - 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).

manimal - 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.

unome2 - 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.

Nicodem - 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]

unome2 - 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

Nicodem - 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.

JohnWW - 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.

unome2 - 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]

Nicodem - 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.

unome2 - 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)?

Ephoton - 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]

Did it with some mods

WaTau - 25-8-2009 at 04:06

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]

manimal - 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.


Attachment: Decarboxylation 3.pdf (187kB)
This file has been downloaded 1392 times

jon - 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.

jon - 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]

entropy51 - 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."

roamingnome - 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

roamingnome - 8-12-2009 at 18:39

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

we know you got the minerals

manimal - 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.

Quantum_Dom - 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]

jon - 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]

MeSynth - 25-8-2011 at 01:10

I tried to understand the post cylco made. He claims in his second post that using no solvent is the method of choice. He says he mixed a and b then distilled at 165C and then extracts freebase PPA from reaction flask mixture (byproducts are distilled). He isn't very clear in some areas... The following is my original revision of his post. If anyone can point out screw ups or point out ways to make it more efficient feel free.

-----------------------------------------------------------

Conversion of benzaldehyde into PPA. (Akabori)

Into a suitable distillation apparatus 1 gram of l-alanine is added for every 2 ml of benzaldehyde to be used.

EXAMPLE I

A stir bar is added to a 500ml round bottom flask then 200ml of benzaldehyde and 100grams of l-alanine are added with stirring. At this point the round bottom flask is immersed in an oil bath which has been pre heated to 150-160C. It is essential to keep the stir bar spinning and the heat at 150C-159C (benzaldehyde boils in the 170s) for the duration of the reaction. The reaction is considered complete when CO2 evolution stops. [Reaction time >1.5hrs]

When reaction has stopped and the mixture has cooled enough to handle (it should still be hot) extract with toluene. All toluene extracts are combined and set aside to cool. Filter off any precipitates that form as a result of cooling.

Extract the combined toluene extracts with dilute HCl to remove the amines.

Wash the aqueous amine solution with DCM taking care not to create an emulsion.

Basify the aqueous amine solution by slowly adding granules of NaOH with stirring.

-method I
Now extract the freebase with diethyl ether. The ether extract is placed into a distillation apparatus. The ether and the free base are distilled under reduced pressure. Wash out the condenser and adapters with ether. The freshly distilled freebase/ether mixture is dried with sodium sulfate. After the sodium sulfate is filtered off and washed with additional dry ether the mixture is ready for dry hydrogen chloride gas. The beaker containing the freebase/ether is placed into an ice bath and chilled to allow for better absorption of the hydrogen chloride. After it has cooled considerably dry anhydrous hydrogen chloride gas is bubbled into the mixture until no more precipitate forms. The precipitate is then filtered off. If unsure about the completeness of the extraction repeat drying and addition of hydrogen chloride.


-method II (original method)
The ether is distilled out of the freebase (the freebase shouldn't go with it but if it does skip some steps). The distilled ether is then dried with sodium sulfate and filtered. To the freebase is added a few ml of the dry ether to take up the freebase. The freebase/ether is poured into a smaller flask for a more efficient distillation. Carefully extract the larger flask until no freebase remains. Carfully distill off the ether once again and then switch receiving flasks. Distil the freebase. Wash the inside of the apparatus with dry ether and combine with the distilled freebase. Add enough ether to the free base to allow for gassing with HC (about 3 times the volume of the freebase). Then dry the freebase/ether extract with anhydrous sodium sulfate and filter. Wash the sodium sulfate and the filter with some dry ether. Gas the now dry freebase/ether mixture with anhydrous hydrogen chloride and filter off the precipitate.

MeSynth - 27-8-2011 at 22:20

What would happen if you used cinnamaldehyde in place of benzaldehyde? What would the product be?

For example..

200gm cinnamaldehyde and 100gm l-alanine

questions - 28-8-2011 at 03:36

I tried this reaction about 1 month ago, I mixed one teaspoon of l-alanine with 250ml of cinnamon aldehyde and heated it to 150C. After 5 min the whole garage stunk intensly of cinnamon and the solution turned dark red. I dont know what the product was but I'm assuming their was a reaction of some sort as their was lots of fizzing

questions - 28-8-2011 at 03:44

I'm always wondering if the cinnamon will react with the n-methylalanine because the benzaldehyde i make has cinnamon still in it about 20% cinnamon actually. So when I mix the solution with n-methylalanine I'm hoping it wont stop the benzaldehyde reacting with the n-methylalanine

overload - 9-10-2011 at 09:27

Can the freebase produced via the akabori be distilled without a vacuum? In other words will the free base decompose and will there be losses?

overload - 10-10-2011 at 11:21

I don't know what to make of this. I just read that the bp of ppa free base is 108C? If the main reaction is taking place at 150C how does the free base stay in the reaction flask? Is it forming an Azeotrope with the Benzaldehyde? If so I would have to guess that using less L-Analine would increase yields in the reaction flask since the more benzaldehyde that is taken up in the reaction more and more free base will escape into the distillate. Could this be why there is such a low yield? Also why doesnt cycloknite extract with the reaction mixture with Diethyl Ether instead of toluene since this would make int easier to try to recycle the benzaldehyde (it there is any to recycle and even if you did recycle it I wouldnt use it for anything but running it through this reaction again). If this is the case maybe there is a way to take the distillate and extract the free base with benzaldehyde and distil out the by-products at 150C again then convert the benzaldehyde to the bisulfite adduct and then basify and extract the ppa with ether. I have no experience or time and money to test out this theory right now. Anyone have any clue if what I'm saying has any truth? If I ran it the way cycloknite did in his second post I would use less L-Analine but for the sake of science I would first try to replicate his results before implementing my own theories.

[Edited on 10-10-2011 by overload]

Nicodem - 10-10-2011 at 11:28

overload, questions and MeSynth: Please do not persistently crap in this thread! This is not a garbage bin.

overload - 10-10-2011 at 11:38

Quote: Originally posted by Nicodem  
overload, questions and MeSynth: Please do not persistently crap in this thread! This is not a garbage bin.


Well these are serious questions I have and you guys are my only source of answers. I could go to a college and ask the chemistry teachers but were is that going to get me.. I may not be adding anything to the thread right now but I may in the future add some great stuff so help me out and see what happens. :)

overload - 11-10-2011 at 22:29

Has anyone read this patent? It has some of the same concepts in it that I talked about earlier with the benzaldehyde azerotrope with the free base.

US4552957

I've only had the chance to read the summary but it sounds really interesting.

watson.fawkes - 12-10-2011 at 09:23

Quote: Originally posted by overload  
Well these are serious questions I have and you guys are my only source of answers.
Are they serious scientific questions or serious production help questions? Scientific questions are phrased with care, not written hastily, pay attention to notation and capitalization, avoid slang acronyms, and put effort into being understood by potential collaborators. I does not seem to me that you are treating your questions with any great degree of scientific effort.

More pointedly, it doesn't matter to most anybody on this forum that you see it as a unique oracle in your life. The participants here are ordinary amateurs, not a paid help desk, and if you want the respect of their attention it behooves you to pay proper respect in how you ask your questions.

fanglongcan - 26-10-2011 at 00:39

Quote: Originally posted by MeSynth  
I tried to understand the post cylco made. He claims in his second post that using no solvent is the method of choice. He says he mixed a and b then distilled at 165C and then extracts freebase PPA from reaction flask mixture (byproducts are distilled). He isn't very clear in some areas... The following is my original revision of his post. If anyone can point out screw ups or point out ways to make it more efficient feel free.

-----------------------------------------------------------

Conversion of benzaldehyde into PPA. (Akabori)

Into a suitable distillation apparatus 1 gram of l-alanine is added for every 2 ml of benzaldehyde to be used.

EXAMPLE I

A stir bar is added to a 500ml round bottom flask then 200ml of benzaldehyde and 100grams of l-alanine are added with stirring. At this point the round bottom flask is immersed in an oil bath which has been pre heated to 150-160C. It is essential to keep the stir bar spinning and the heat at 150C-159C (benzaldehyde boils in the 170s) for the duration of the reaction. The reaction is considered complete when CO2 evolution stops. [Reaction time >1.5hrs]

When reaction has stopped and the mixture has cooled enough to handle (it should still be hot) extract with toluene. All toluene extracts are combined and set aside to cool. Filter off any precipitates that form as a result of cooling.

Extract the combined toluene extracts with dilute HCl to remove the amines.

Wash the aqueous amine solution with DCM taking care not to create an emulsion.

Basify the aqueous amine solution by slowly adding granules of NaOH with stirring.

-method I
Now extract the freebase with diethyl ether. The ether extract is placed into a distillation apparatus. The ether and the free base are distilled under reduced pressure. Wash out the condenser and adapters with ether. The freshly distilled freebase/ether mixture is dried with sodium sulfate. After the sodium sulfate is filtered off and washed with additional dry ether the mixture is ready for dry hydrogen chloride gas. The beaker containing the freebase/ether is placed into an ice bath and chilled to allow for better absorption of the hydrogen chloride. After it has cooled considerably dry anhydrous hydrogen chloride gas is bubbled into the mixture until no more precipitate forms. The precipitate is then filtered off. If unsure about the completeness of the extraction repeat drying and addition of hydrogen chloride.


-method II (original method)
The ether is distilled out of the freebase (the freebase shouldn't go with it but if it does skip some steps). The distilled ether is then dried with sodium sulfate and filtered. To the freebase is added a few ml of the dry ether to take up the freebase. The freebase/ether is poured into a smaller flask for a more efficient distillation. Carefully extract the larger flask until no freebase remains. Carfully distill off the ether once again and then switch receiving flasks. Distil the freebase. Wash the inside of the apparatus with dry ether and combine with the distilled freebase. Add enough ether to the free base to allow for gassing with HC (about 3 times the volume of the freebase). Then dry the freebase/ether extract with anhydrous sodium sulfate and filter. Wash the sodium sulfate and the filter with some dry ether. Gas the now dry freebase/ether mixture with anhydrous hydrogen chloride and filter off the precipitate.


The ether and the free base are distilled under reduced pressure.
Distillation temperature?

Bandwith

cal - 9-4-2012 at 08:40

Quote: Originally posted by Ramiel  
I hate to be off topic CycloKnight (it looks like there's a good chemist here), but that's 37 pictures there... omg bandwidth! Someone pays for that megabyte or so for each person who view this page. In this case, perhaps a picture is worth omitting for a few concise descriptions, eh?

If you don't have unlimited then your not with the 20 th century
program.

tadpoleannihilater - 9-10-2013 at 10:57

Appologies for bringing up an old thread but I would like to know the following.

What ph should the HCL have when using it to extract the amines from the toluene?

After reading through a few of these threads I have decided to consider the ph of the HCL as important as it may damage the amines if to strong.

Im also having a bit of trouble trying to raise the ph of the hcl. I would basically only need 5ml in 1000ml to get a slightly acidic solution. Does this seem right to you guys?

My hcl is muriatic acid passed through activated carbon a few times to remove some of the yellow goo.

So right now I guess I could run a test to see if this small amount of hcl would be sufficient.

[Edited on 10-10-2013 by tadpoleannihilater]

WChase501 - 9-10-2014 at 08:41

During theAkabori reaction wasn't a primary amine used to catalyze the main reaction?....
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