Yes Decarboxylation is a very valuble tool when mastered because aside from all the larger amino acids there is much potential in the process for the synthesis of simple alkylamines such as Methylamine from the decarboxylation of Glycine or Ethylamine from Beta-Alanine.

Thanks for the input Naf much appreciated.

PS: I just wanted to make mention that the reaction performed with cyclohexenone is a ketone catalyzed decarboxylation and simular reactions can also be performed with the use of the easy to aquire Acetone or MethylEthylketone. The decarboxylations proceed under higher temperatures but this leaves low yeilds and side reactions but the presence of a ketone speeds this up. I will search for the references for the use of Acetone in the mean time.

"Analytical characterisation of the routes by thermolytic decarboxylation from tryptophan to tryptamine using ketone catalysts, resulting in tetrahydro-beta-carboline formation."

Yes INDEEDY that is ALL very interesting. Cant at the moment see much of a problem with the CO2 removal with such an addition. The hydrolysis step has left me wondering, but all quite interesting to say the least.

Last step of the mechanism

Now KIDDIES, Remember those whose name starts with the letter A, well you get to order/ buy your Phe on Monday, Those with names starting with B -  on a Tuesday, etc etc etc.

ps  Middle and last names DO NOT COUNT< and yes, CHECK your fucking calenders. 

Thanks, yes there was alot I wanted to add but just left out to try not to confuse the issue. If you are really interested after reading the post, hopefully you can go back and read the patents at least having a rough idea of what they are doing now(hopefully it sheds a bit of light).

@2BFrank,
"Cant at the moment see much of a problem with the CO2 removal with such an addition"

If you are referring to the hydrolysis you are on the right track, de-acylation occurs in acidic medium. De-acylation reduces the stability of the ring leaving  the vulnerable to further attack. You cant see it in the above pics but the free electrons on the nitrogen and the neighboring oxygen actually contribute to extra resonance structures as shown below.

As you can see that carbon at number 2 is a sitting duck for an electrophilic reageant (Lewis acid), thats were the ring opens with a Lewis Acid. Position 5 is also vulnerable, but not so much in this case.
And look at the hydrolysis scheme, acid removes R1 and the double bonded oxygen (leaves as an aldehyde) by de-acylating it, if you remember acylation was carried out earlier in that scheme it was used to create an acyliminium ion with a positive charge, which was attracted to the negatively charged sodium cation works like magnet (in that appraoch). And closes the ring. And now the very electrophilic R2 is removed because it is vigorously attacked by a lewis acid. With that pulled out, the ring pretty much unzips and spirings into the familiar looking end product.

If I have not explained something properly, let me know. Or you still just cant see it, dont hesitate to ask.

note; In my explaination above I said;
" It also gets tricky as the ring can form trans or cis affecting stereochemistry, so a second chiral center is introduced via the aldehyde which orientates trans formation which retains stereochemistry."

That really didnt come out right,  the acetals of the aldehyde can end up being either cis or trans to each other once the ring forms. The cis configuration is preferred because it directs the alkyl groups that we are introducing into the right position. Thats why they go on about introducing a new chiral center by adding a suitable aldehyde. Because once the original chiral center is destroyed when the ring is formed, note; that this happens when the ring forms, not schiff base formation. Without a second chiral center on the aldehyde when the first chiral center is destroyed if none remained nothing would direct the alkyl group and it would attach any way possible and a racemic product would result. Directing the alkyl group via VanDerWalls forces with a chiral center, pushes (so to speak) the alky group into the right position.

Cool Sedit, I just about to post and got a message that someone just did. I will post my response to Naf !,do what I got to do etc/.

See it, COOL - 1,2,3, That is C2, attacked by lewis acid, opening ring, R1-simple acid hydrolyisis and same to break the ether bridge, forming the AA and the aldehyde.  The initial C2 through me, so thanks for the resonance/effect reminder.

Also the alpha Carbon being methylated, then if anything, he would of thought would promote the de-carboxylation. The prep is very interesting all round.

There is a fair few things going on with the ring formation/ orientation and benefit of cis/trans issues. I got to go, but will most certainly look at this in detail, as such mechanisms one can usually learn PLENTY. Also trying to think of suitable OTC lewis acid that would be useful. Would their be hindrance issues???

2b

"There is a fair few things going on with the ring formation/ orientation and benefit of cis/trans issues"

Yes I sincerely wish I hadnt posted the above on stereochemistry, I got lost in translation and what I posted came out incorrect.

An optically pure amino acid is used, say either D or L. It is reacted with an aldehyde forming and oxazolidinone ring with two stereogenic centers. The first comes from the L-phenylalanine and the second comes from the aldehyde when the ring forms(they are represented as bold black lines in the reaction schemes). When the ring closes it makes these stereogenic centers either Cis or Trans to each other, the trans isomer is the one needed. When the enolate of the oxazolidinone is prepared by basing the oxazolidinone ring its double bond ruins the original stereogenic center that came from phenylalanine as said before by changing the symmetry, thats why the second stereogenic center is needed, to retain chirality. With chirality retained alkylation of the trigonal center proceeds diastereoselectively due to the influence of the second stereogenic center.

edit; I referred several times to  the ring in the alanine/benzaldehyde akabori as an oxazolidinone, although this is correct for the first case where a ketone function is present. The 'none' ketone nomenclature is dropped in the second case where it obviously isnt a ketone, and is an oxazolidine ring.

Also just looking at that science madness thread, and they refer to they oxazolidine ring several times? But the mechanism proposed resembles an aldol addition?

And look at the hydrolysis scheme, acid removes R1 and the double bonded oxygen (leaves as an aldehyde) by de-acylating it, if you remember acylation was carried out earlier in that scheme

If you mean R1 and a double bonded oxygen = R1-CO- (acyl) leaves in an aldehyde R1COH form, that is of course not the case. It leaves as R1COOH eventually.

it was used to create an acyliminium ion with a positive charge, which was attracted to the negatively charged sodium cation works like magnet (in that appraoch).

Sodium cation is not negatively charged. Acyliminium ion is attracked to carboxylte anion COO-, sodium salt is just a form of carboxylate anion.

And now the very electrophilic R2 is removed because it is vigorously attacked by a lewis acid.

Carbon atom is not attacked by acid, only nitrogen/oxygen are.

You can see the nucleophilic aldehyde will attracted to the electrophilic zwitterion and they react in a classic aldol addition.

Just the contrary, electrophilic aldehyde and nucleophilic zwitterion

Hi again!

In theory that acyl group could leave and end up as an aldehyde, carboxylic acid or alcohol. These are by-products and there identity is not revealed in any of the papers referenced.

It seems to me, that aldehyde or alcohol (in our case R1CHO, R1CH2OH) could be formed if it is a reductive deacylation - some sort of hydrid ion H- attack at carbonyl group, for example, treatment with DIBAL-H (diisobutylaluminiumhydride, AlH(i-Bu)2) would produce an aldehyde in a high yield. Alcohol could be formed in case of some metal hydrides, but not LiAlH4 - it reduces amides directly to amines.
But, if it is a usual hydrolysis (acidic, basic), then nothing would happen to the oxydation state of the carbon atom. If in the papers referenced it is still stated that they found some aldehyde /alcohol - please tell me (i did not thoroughly look through them), but still formation of those products while hydrolysis seems doubtful for me

But how sure are you that a carboxylic acid is going to formed from an acyliminium ion in acidic medium?

 In aqueous medium that ion will immediately hydrolise into an amide and aldehyde(if you are interested in mechanism: addition of water to carbocation resulting in R-OH2+, H+ is leaving and revealing R-OH(not a cation now), then this R-CH(OH)(NR*COR1) is further hydrolised ). And then, a usual acidic hydrolysis of amide - resulting in carboxylic acid.
But that is only one case and one route - hydrolysis of the N-C-O forming an acyliminium ion. Other routes may include hydrolysis os an esteral O-C=O bond forming N-C-OH (which is quite like a acyliminium ion but still not the same, it is a one small step closer to the final aldehyde and amide), protonation of amide function and it's elimination resulting in CHR(+)-O-CO-.. (but that is less favorable route, nitrogen is generally a bad leaving group, even protoned), and the same 3 processes but after the acyl group is removed.
Brief explanation why the cycle is more stable if there is acyl - because it destabilizes the carbocathion in case it is accidently formed by heterolytic breaking of C-O bond. (at all the pictures you attached with electron effects, charges and double bonds, all these effects exist only after a carbocathion is formed, in our case you should have drawn the same but without oxygen(or nitrogen), resulting only in one heteroatom for stabilisation of a charge. The pics you had drawn, are reflecting a one more electron oxidized carbon, you need to abstract one hydride ion H- from  our cycle, or if we had a cycle with OH  substituent in position 2, and that is absolutely different molecule. In our case, you may have drawn arrows directed to heteroatoms, that would indicate they are taking the electron pairs to them due to their electronegativity and -I effect, and that would result in a negative charge, not positive. Positive charge they (only one of them) would have after heterolysis of one of the bonds occurs. Acid is just for protonation making another atom a better leaving group(and for acyl removement).
In common words, the more stable carbocathion would form, the more favorable the heterolysis would be. The best stabilization occurs because of neighbouring heteroatom nitrogen (because in can form C=NR2+). But if the nitrogen is conjugated with carbonyl, that means part of the electrons belong to carbonyl, and there are only few left for carbocathion stabilization. That's why acyl destabilizes that carbocathion and stabilizes the cycle.

 The ring can be opened and de-acylation can occur in basic conditions, is that what you meant?

No, i did not mean that, but yes, i suppose deacylation can occur in basic medium (usual mechanism of basic hydrolysis of amides). Acyliminium ion as well can be hydrolysed in basic media.
Ring can not be opened by basic hydrolysis of N-C-O- fragment(here acid is required), but esteral bond can probably be hydrolysed without any problems, opening the ring and so on. That depends on the rate of course, i'm not sure about basic hydrolysis is fast enough to be used for such cycles.

The conjugate acid of a base and an acyliminium ion is an acid (carboxylic acid).

Sorry if i misunderstood. Acyliminium ion is not a carboxylic acid, because it does not possess  COOH function. There is nother COO- in the molecule, that can attack the alyliminium closing the ring. But acyliminion itself can not be called an acid.

And I would also like to know what you meant by the acyl groups leaves eventually?

By that i mean, that after all the hydrolysis steps and the order in which the hydrolysis occurs(i am not sure about it), the acyl group would in all cases become a R1COOH.

But the point I was trying to stress was that the Sodium was used to create an ion, that the acyliminium ion was attracted to.

Now it weems i understood what you have meant. Yes, actually you are right, sodium salt is having free COO- ions, and they are attracked to acyliminium. You can call it electrostic attraction, some may call it making a better nucleophile, some may call it - beforehand buffering the pH not to allow much HCl to be formed becasue it can stop the reaction. But i don't think that the impact of electrostatic attraction itself is critical here. For example, oppositely charged ions in a solvent are quite a common thing, but they do not suffer from attraction much, because they are surrounded by polar molecules of solvent. In case of carboxylate, it would be the same. This is one point. Another one is, that if you take a sodium salt of a-aminosulfonic acid instead of aminoacid, then you will no way get a cycle, because sulfonate is a weak nucleophile, but the charge is the same as carboxylate has.

Point three; Now you say that only oxygen and nitrogen are attacked by a Lewis acid. You must have missed the oxazolidine resonance structures I posted earlier, here just for you;

As i explained above, those effects would play a role only after the structure you have drawn will be present, and this is a bit different from our's

So a Lewis Acid (electron acceptor) is going to avoid any positive charges and go straight for the electron rich bond. Most other times the Lewis Acid would go for the free electrons on O or N, but not this time. For all N and O's electrons are busy stabilizing the ring, and O and N have actually acquired some positive character.

Here i doubt that one more acid moity can attack this cation, maybe i very strong one. As for double bond, in this case it is not electron rich at all(resembles alkene, i know but that only looks like)! Notice, that in both resonance structures the double bond does indicate the place where positive charge is delocalised, while the third atom(having its electron pair) is the one to be attacked with LA. Besides positive charge is carried by carbon also(carbocation), that is another important resonance structure not mentioned on the pic. So, the next lewis acid moiety would attach to either of nitrogen or oxygen, but not carbon.
Hope i made it clear. It is pity that i don't have hyperchem on this PC to make drowings, so if there is something unclear - feel free to ask

Um, I haven't read the whole thing, but has anyone addressed the carbon with five bonds in scheme 1?

Yes, now the mechanism of hydrolysis is correct(second diagram). But still, those diagram with 2 resonating structures(your previous posts) seems to be depicting something else - and i suppose i realized, that seems to be given for oxazoline (double bond between C and N). Then the protonated oxazoline would be the same structure. But if it is realy for oxazolidine, that is a gross mistake for JOC paper! I mean, there is no positive charge on heteroatoms and no resonance if it is a plain oxazolidine ring. And after it is protonated, it becomes like structure 4, while oxazoline while protonation would give those resonating structures.
As for structures A and B, it seems they were given for explanation which heteroatom would be protonated in case there is acyl and without acyl. The arrows and electron clouds seem to show -I effect of the atom, that is going to be protonated. But still that is not the only and not the main reason why acyl is stabilizing the ring. By that diagram they probably want to explain the stabilization effect of acyl group in the way that there are 2 possibilities for protonation(nitrogen and oxygen) in case of A and only one in case of B. And that leads to 2 possibilities for elimination (formation of C=O(+)- and C=NR2+, A)  in comparison to one (acyliminium, B). So 2 is better then one. That is not correct explanation. Of course, there would be 2 routes in case of A,  C=O(+) and C=NR2+ cations formation. But these 2 routes are not equal in their possibility. (:joke: what is a probability to meet a red elephant in the street? answer :50%. either you'll meet, either not).   And as we can see from second diagram, they write only one route, pecause another one much less possible (both nitrogen is a bad leaving group, and resulting carbocation is less stable then 4). So that means, the main reason why acyl is stabilizing the ring, is because carbocation 4 is much less stable  if there is acyl, and thus can not be formed in that easy way, like it happens without acyl.  

They were the two resonance structures used two explain ring opening. And if you look at the one on the left (after some research), it is actually a pretty good representation of why O-1 will be attacked and subsequently protonated in acetic acid(I know, thats what you said!). Granted, there is no acyl group on N-3 nor any electron donating/withdrawing groups on C-2.  But considering that in acetic acid the first thing that will happen is protonation of O-1 removing the bond between O-1 and C-2 leaving a transient iminium ion. The iminium ion undergoes hydrolysis into our products discussed above. How quickly the transient iminium ion undergoes hydrolysis depends on electron donating/withdrawing substituents present on the ring. The fact that there is an acyl group attached to N-3 provides more stability at that locant, making certain that protonation would first occur at O-1(albeit slowly read; 20 hours, thats why so long A; is favored because it will open quickly, whereas B; is alot slower!) leaving the well know but relatively unstable iminium ion which is very vulnerable to simple hydrolysis.

Practically right, but some clarifications:
1. Acyl group does not affect O- protonation at all. Its withdrawing effect can not pass through single bonds, only double. It affects the further elimination of protonated oxygen (as alcohol), because the resulting acyliminium cation is bit destabilized, so the elimination is not that beneficial as it is in case there is no acyl. But protonation is not affected.
2. The hydrolysis of transient ion rate depends on presence/absence of acyl, but that is not so critical here. I think it is not a rate-limiting step, like acyliminium ion formation. As you wrote in the end, it is "relatively unstable iminium ion which is very vulnerable to simple hydrolysis".

Thanks again zz, I hope you become a regular here. You are good to have around!

Always welcome It is nice that someone is interested in understanding the mechanisms of reactions. On most drug resources people like to cook and mechanisms are bit neglected. But i am always glad to explain them for those wasps who are interested.