Acid, dragonflies and the 5HT2A receptor

[Morninggloryseed]

I dunno. Just is. Seems from PIHKAL, the drug is very inconsistent. And then my own personal experienced proved to be one of marked dissociation…very different than any of the other PEAs I have tried. I guess I consider it an oddball because it is not one of the ‘magic half-dozen’ and because of the wildly varied responses to this material.

 

[kidamnesiac]

I really enjoyed reading this thread when it was first posted, had some ideas similar to this and came back to look and have a problem with f&b's conjugation therom-
as smyth pointed out, keto-enol tautomerism does not work the way proposed on pg. 1 by f&b. think of a protein, the bond length of the c-o bond is 1.23 angstroms while the c-n bond is 1.33. This implies the bonds are of similar length and that the same type of phenomenon, what organic chem students are taught to draw as "resonance", is occuring here as would occur at a terminal carboxylate. The electron density is shared almost equally between the N and O which results in a very rigid structure, the basis for those ramachandran plots all biochem students soak in. but the important thing to get out of this is that the bond is NOT a keto-enol structure and cannot be drawn as such in the manner it is drawn. Smyth brought this up but it was effectively ignored as far as I can tell.
I also think it is seriously pushing the envelope to say that the pi orbitals that are in the c ring of the ergoloids interacts in an H-bond fashion. Dogma has it that only N, O and Cl bonded H's are able to H-bond. This isnt to say there isn't a region of electronegativity there, but, as far as my limited knowledge stretches, I can think of no examples of a pi system participating in H bonding, especially in with such a weakly proton donating (wrong terminology there but it escapes me) atom as is found in a serine residue. Particularly a delocalized system that while inevitably be less electron rich in that particular area.
I think there are some good ideas here, but in general the generalization is off. The animated gif showing the overlaying PEA shows the methoxy groups sticking out into space with no overlap in the ergoloid nucleus, suggesting the binding pocket must be rather large in that direction as well AND contain activatable residues that would correspond to the catecholamine binding that the PEAs/As would present to the pocket. Unless you want to make the argument that their effect is based at the D sites, in which case, how do the ergoloid/tryptamines get in there? I feel their binding must be similar in both pockets.

Sorry for dragging this up but I think some more theorizing with fresh perspectives couldn't hurt. And of course, I have nothing of significance to contribute, only criticism

 

[fastandbulbous]

The animated gif showing the overlaying PEA shows the methoxy groups sticking out into space with no overlap in the ergoloid nucleus

That's because it's a 5-HT receptor and corresponds to where the OH group of serotonin would go and interact with the protein (this corresponds to the 5-methoxy group of phenethylamines in terms of binding sites). This isn't present on egolines bwcause that charge is interacting with the C=O bond. Without the keto-enol type of model it becomes a lot more difficult to explain the planar conformation of LSD when the receptor is active(not lowest energy state - see Nichols paper about the 2,4-dimethylaziridine derivative)? It can only do this if it has a 'rigid' backbone that is conjugation all the way up from the benzene ring. It is the planar nature (through being fully aromatic) that gives the benzodifurans their high receptor affinity.

The pi electrons were the weakest part, but there must be some sort of charge related issue at work or else it's impossible to explain the increased potencies of BOD, BOB and other beta-methoxylated derivatives over the parent 2C-X compound in light of the data about the beta methyl derivatives. (It could be interacting with another nearby aromatic system of an amino acid in the protein structure or a region of charge)

 

[kidamnesiac]

The pi electrons were the weakest part, but there must be some sort of charge related issue at work or else it's impossible to explain the increased potencies of BOD, BOB and other beta-methoxylated derivatives over the parent 2C-X compound in light of the data about the beta methyl derivatives. (It could be interacting with another nearby aromatic system of an amino acid in the protein structure or a region of charge)

This may have been hashed over before in this thread, but why not simply that they are a difficult substrate for MAO?
Has the methylone equivalent of MDA ever been tested (nonylone?) While it is a large bag of worms when invoking MDMA, one could postulate that the need for increased doses in methylone is that the ketone actually interferes with the binding of the molecule in the pocket and thus lowers its intrisic activity. Of course, having ketone there also makes it a much nicer target for other xenobiotic metabolizers that can simply take a whack at the carbonyl and make the mdphenylacetic acid in a quick step. Thus reducing the bioavailability. But it certainly is not more potent.
I also still have trouble believing Nichols theory about the H-bonding ethers. Think about diethyl ether. what's its water solubility? rather low, it's obviously not participating in much H-bonding. A furan system *might* be rationalizable through some pi interactions that I am at a loss to rationalize now. A aromatic-O-aliphatic ether? maaaaaybe. But an aliphatic-O-aliphatic ether? Especially one that would be donating into the chain to compensate for the withdrawing amine? It's a stretch in my feeble mind.
I have to look over the rest again before addressing it. Thanks for humoring me

 

[hussness]

A search about cathinone dimerization will answer your question about nonylone, by the way.

 

[fastandbulbous]

one could postulate that the need for increased doses in methylone is that the ketone actually interferes with the binding of the molecule in the pocket and thus lowers its intrisic activity.

MDMA & methylone don't exert their activity via the 5HT2a receptor (most active isomer is the S form, opposite to psychedelic amphetamines where it's the R isomer). Methylone is a more polar molecule and more easily filtered out by the kidneys etc (why beta-keto derivatives are always shorter acting that their respective amphetamine). Also the keto group reduces it's penetration of the blood-brain barrier when compared with a -CH2- of an amphetamine because of it's increased polarity

 

[kidamnesiac]

This isn't present on egolines bwcause that charge is interacting with the C=O bond. Without the keto-enol type of model it becomes a lot more difficult to explain the planar conformation of LSD when the receptor is active(That's because it's a 5-HT receptor and corresponds to where the OH group of serotonin would go and interact with the protein (this corresponds to the 5-methoxy group of phenethylamines in terms of binding sites).

Ok I finally see where you're going with this. And I still don't buy it. Theres minimal chance that the carbonyl O is interacting with the same residue as the 5-MeO oxygen is. If you're viewing the LSD molecule as essentially planar (see below), the the distance between them in real space is going to be Minimum 3 angstroms (imagine a benzene ring sitting in the middle of the a and d rings), far to large for h-bonding with the same residue. it looks like it could be closer to 8 angstroms if it really is totally planar.

A while ago I built a model of lsd with my model kit. It showed me why all lewis structures of LSD look so damned funny. you HAVE to bend the first carbon off of the indole ring and the stereogenenic tertiary carbon connected to the amine to allow for the proper sp3 bonding in those carbons. and while this may not have *much* bearing on the nichols/f+b theory (your the last name in 5-ht2a activity, how do you like that? ) of planarity in the conjugated backbone (this part can still be viewed as mostly planar), I feel it should be noted that any model or 2d approximation of LSD clearly shows that the c+d rings are forced "downward" into the plane of the surface by ~~~~~15-25 degrees. Ever notice how the ergoline structures in TIHKAL look so damned goofy compared to the other tryptamines? You really have to bend the indole nucleus to make it fit on the plane of the paper.

Finally, my reductionist tendencies catch up with me again and I must bring up the monkey wrench that is ibogaine. A quick look @ the structure *clearly* shows it would not be planar in the c and d rings (correct nomenclature???). No eight membered ring can be, particularly when there are bridged rings involved. Yet, according to this paper, http://www.nevapress.com/cnsdr/full/5/1/27.pdf , it has a 3 fold lower affinity than LSD for the 5-ht2A receptor, which is still significant. There is no way I've found to bend that ring to overlap with the ergoloid structure. How does it get in there?

And I'm still not clear on this idea-I look @ the 4 position as a "bump" that pushes a "button" on the receptor in the 2,5 series. Different substituents result in different ways of pushing the button but the size and electronegativity is what is causing the majority of differences in the effects of the 2c and DO series. Get rid of the bump, no psychedelic activity. but neither LSD nor the IEAs have anything there to push the button in your model. I don't think this has really be resolved in this thread so I'm just using it as more evidence at this point.

So there ya go, my critique. If you want to continue this in PM you can send me one as I don't think I can yet...

 

[kidamnesiac]

From the horses mouth...

Many of our earlier attempts to produce a rigid analogue that would mimic the bioactive conformation of phenethylamines have been based on the idea that the side chain might adopt a conformation where it was coplanar with the aromatic ring, as exemplified in the ergolines (e.g., see Monte et al.8). Yet extremely potent analogues have been synthesized where the aromatic methoxy groups of the phenethylamines are constrained into a conformation coplanar with the aromatic ring.4,10 These benzodifuran analogues, through peri interactions between the additional methylene groups and the side chain, show an energetic penalty for the side chain to adopt an in-plane orientation. Thus, it seems more reasonable to assume that the side chain might lie in a plane perpendicular to the aromatic ring plane. Consistent with that reasoning, NMR spectroscopic data have shown that the aqueous solvated conformation of amphetamine is antiperiplanar.11 This finding strengthens the argument for an out-of-plane orientation but is somewhat confounded by nonbonded steric interactions provided by the methoxy groups.

et. al, Nichols J. Med. Chem.; 2006; 49(19) pp 5794 - 5803

for some reason google scholar doesnt find it...

On the other hand-
If you buy my argument that the c+d rings are bent down into the plane of the surface and map your overlay with the cycloalkane moiety, you see that the potent R isomer actually would point "down" into the plane of the surface. same with all of the other pea derivatives (besides MDMA like you mention). But I still don't like that model...

Once I can get back to a copy of chemdraw I'm thinking about posting a totally different idea. I ran across several computer model (yea, i dont trusts em neither) papers showing the indolic N interacting with the same residue as the d ring N in LSD. This would mean the alkylamine would most likely be interacting in the same area as the amide function in LSD-read, lots of electronegativity in LSD, lots of potency. (what about some thio-esters as LSD analogues...) This explains why the DEA structure of LSD is so important-same reason it is in the simple IEA's-its the money spot, defining how tightly a ligand can bind. Ever wonder why DET is the only potent, truly psychedelic orally active IEA?? (makes me wonder what LSDiPT would be like...). this could be developed to fit the 2,4 somethin-izidine paper nichols just put out...
At any rate a quick pubmed would find the papers showing the models as I forgot to save them in my meanderings.
This model is also flexible at precisely the right position-as you say, the lysergic acid ring structure is pretty planar. But the amide region is NOT. as pointed out earlier in the thread, the alpha carbon to the carbonyl would allow for full rotation of the bond (alright, not full rotation but any 3-d conformation), one of which would push the carbonyl O down, one up (periplanar). This would put either the N or the O into close proximity with the money spot. Whether it is the O or N depends on which one prefers to be down as this would represent the same conformation as R in the PEAs.
The nice thing about this model? Do another overlay with the 2-MeO group overlaying the indolic N as you did for the 5 group (basically just reverse your logic and make the wing of the 2 position overlap with the natural wing of the pyrrole ring) Ok, that looks pretty but...WHATS THAT!? now the 5-MeO group is in the same position for both the IEA and the PEA. so is the 4 position. neato! and the alkyl group is pointing in the same direction as before, albeit one carbon shorter on the PEA (more corresponding to the position of the carbonyl O than the amide N). K...
So take this little moiety and put the indolic N overlaying the LSD d-ring N. You have to invert it from our normal perspective about the 4-position on PEAs. But look where that 4 position is-Right where the indolic N on LSD is...The 5 position also fits into the indolic region and could perhaps also be overlayed with the N. the important feature is that, in the LSD struture, there is a large, stearicly bulky, electron-rich BUT NOT basic OR H-bonding region in the same position as the money spots on the PEAs and IEAs.
veeeety interesting

Can anyone point out any holes (or even follow?) this logic before I go and chemdraw it up?

and how to test this?? build some amide analogues of the common IEA's and see if they hold water via IV or intercranial injection (they wouldn't cross the BBB would they...). (I3AA is quite easy to get....)

Lastly-this model holds water for ibogaine-the c-d ring system is certainly out of plane with the indole nucleus and the harmala-like constrained N (same as the IEA N of course) is stuck in a not impossible but unfavourable position to hit the money spot.

 

[HeaT]

thats amazing. Someone should try halogenating LSD at the same place that Bromo dfly is halogenated, or adding the ethylamine and closeby atoms onto a 'dragonfly' compound

 

[kidamnesiac]

thats amazing. Someone should try halogenating LSD at the same place that Bromo dfly is halogenated

I think you missed the point...

Here's another point regarding the BOX series-
If you have the indole N in the same position as the 2-MeO Oxygen as I suggest, you end up with the beta carbon on the ethylamine chain resembling the 3 Carbon in indole. what happens when you put the methoxy group there? pi electrons in the same place as where the double bond on the pyrolle ring is normally shown. I know this is simplified for a resonance structure in an indole lewis structure. however, by putting those pi electrons there with the BO series, you can generate an extended conjugated pi system VERY similar in nature to the indolic system. see it?
Here's an idea for a new dragonfly series-make an unsaturated pyran ring connecting the two position to the Beta position. Just like LSD, only a heterocycle. Aromaticity might be nice or maybe not needed. I get to name it though-the hummingbird series

 

[kidamnesiac]

the antithesis

So *most* of my pictures have finally been approved...

Here are 3 compounds lined up as I am going to present them overlapping. DOB, one of the strongest non-fly PEAs, 5-MeO-DET, one of the most potent IEA's in TIHKAL, and good ol' LSD-25.

Now here is DOB overlaying 5-MeO-DET.

EWWWW you're thinking. And that's what I thought too. But all I am doing it switching the overlay orientation that F+B proposed. Why? you ask...

Notice how this explains the 4 position far better than the F+B model (I promise I'm not trying to pick on ya buddy). It also explains the 2,4,6 pattern. Considering that lengthening the N-Methyl group to Ethyl or Propyl retains activity in LSD, it also allows for the extreme potency of the FLY series.
The terminal N winds up close to the carbonyl O. This area is obviously significant for the binding of LSD to the pocket. But it must be an amide. A group of broadly distributed negative charge (not point like a tertiary amine or ketone). This might be the key to why N-substituted PEA's and amphetamines are worthless. The electron donating ability of the methyl groups could create too strong a negative region for good interactions. Yes, this needs to be developed further, but this point is also wide open in the other model as the overlaying N is tertiary in LSD and primary in the PEAs.

As soon as the next picture is approved, I will show the overlay of the 5-MeO-DET. It also fits nicely in there and does a good deal to explain why the 4 and 5 positions are hot in the IEAs.

Someone is sure to say: "But KA, what about the 5,6 Methylenedioxy indole compounds that are found to be less potent than the 5-MeO counterparts?"
To that I have two answers-first, the methylenedioxy bridge is quite unstable (why you get oxidative damage from MDXX series) and MDXX compounds require massive doses relative to the unsubstitued amphetamine homologues. And we all should know that MDA acts is a classic psychedelic and this model makes much better explanation for that.
Of course, the 5,6 MDo series still could be active at higher levels and a receptor affinity assay could prove this. Even so, I'd still really like to see a pyrolle group across the 5,6 positions. or a halogen in the 6 spot.

So what is the glaring error in my logic here? Does this even make sense?

In best Lisa voice:
Look at me! Grade me! Evaluate and rank me! I'm good, good, good and oh so smart!

 

[fastandbulbous]

Notice how this explains the 4 position far better than the F+B model

No, and you've got the aromatic ring of a PEA overlapping a non aromatic ring of LSD. The model also has the btomine atom overlaying part of the aromatic ring, and an amide doesn't have anything like the electronegativity of an amine (which has been shown to bind to a carboxylic acid group of one of the trans membrane proteins). It also does nothing to explain the increased potency & binding affinity of the BOx compounds.

The whole thing also does nothing to explain why the stereochemistry of all the most active isomers of the chiral PEAs & tryptamines have the same absolute configuration as the 5 carbon of LSD - that is the biggie that your model falls down on; the iso isomers of lysergides are inactive but the S-isomers of amphetamines (or R isomers of alpha alkyltryptamines) are merely reduced in affinity/potency

To that I have two answers-first, the methylenedioxy bridge is quite unstable (why you get oxidative damage from MDXX series) and MDXX compounds require massive doses relative to the unsubstitued amphetamine homologues. And we all should know that MDA acts is a classic psychedelic and this model makes much better explanation for that.

Not really as the strain on the 5-membered ring is not that great and the reason you get oxidative damage from the MDXX compounds is because of the amount of dopamine that gets metabolized producing free radicles, not due to the ring opening. If it were the methylenedioxy ring that caused the problems then you'd see damage with the likes of MDAI (5,6-methylenedioxy-2-aminoindan) which you don't; you'd also get it with other drugs containing the methylenedioxyphenyl group, such as paroxetine (again which doesn't happen).

but this point is also wide open in the other model as the overlaying N is tertiary in LSD and primary in the PEAs.

That would appear to have something to do with the pKa of primary, secondary & tertiary amines. The fact that secondary amines are the most basic implies that there is some competing repulsion with another like charge in the protein chain (it also fits with the inactivity of nor-lysergic acid diethylamide and the likes of N-methyltryptamine and the monoalkylated tryptamines aslso lacking activity, even in the presence of an MAOI)

 

[kidamnesiac]

No, and you've got the aromatic ring of a PEA overlapping a non aromatic ring of LSD.

as you've said yourself, this ring is damned planar (chem3d shows this quite nicely, the pucker occurs in the d ring) It's stuck, juxtaposed between, against and half composed of a very planar system that is the indole ring.

The model also has the btomine atom overlaying part of the aromatic ring,

sure do, it represents a large region of non-h-bonding negative charge. which would love to be near a cation (lys, arg).
do a search for "cation quadrapole binding" in google or pubmed, etc. daugherty did some great work on it. the face of a aromatic is DAMNED similar to a large, covalent halogen in that it's big, electronegative, doesn't have a true h-bonding lone-pair and therefore is non-specific.

and an amide doesn't have anything like the electronegativity of an amine (which has been shown to bind to a carboxylic acid group of one of the trans membrane proteins).

No, not exactly, but there is a lone pair there where in your model there is not. why should the nature of the amide substituents matter so much in LSD if the PEA's and IEA's are able to bind with almost equal affinity?

It also does nothing to explain the increased potency & binding affinity of the BOx compounds.

Why not? the beta carbon is in almost the same spot in your model as it is in mine. and if you invoke the IEA 4-OH group as interacting with the same region as the double bond, I would say the beta position on the PEA's is closer as I have shown it (you're talking a full C-C bond away)

The whole thing also does nothing to explain why the stereochemistry of all the most active isomers of the chiral PEAs & tryptamines have the same absolute configuration as the 5 carbon of LSD - that is the biggie that your model falls down on; the iso isomers of lysergides are inactive but the S-isomers of amphetamines (or R isomers of alpha alkyltryptamines) are merely reduced in affinity/potency

but your model proposed a planar ergoline core? stereo chemistry should not matter in this case.
You've almost got me here, but I can propose something simple-the site that the carbonyl O interacts with is above the plane of the ergoloid ring. An R PEA hits this. An S interacts (but very weakly and from far away) with where the amide N would reside in LSD.

Not really as the strain on the 5-membered ring is not that great and the reason you get oxidative damage from the MDXX compounds is because of the amount of dopamine that gets metabolized producing free radicles, not due to the ring opening. If it were the methylenedioxy ring that caused the problems then you'd see damage with the likes of MDAI (5,6-methylenedioxy-2-aminoindan) which you don't; you'd also get it with other drugs containing the methylenedioxyphenyl group, such as paroxetine (again which doesn't happen).

I had the feeling I was wrong about that
I was thinking of the metabolism of the benzenedioxoles:

MDMA and MDA are O-demethylenated to N-methyl-alpha-methyldopamine (N-Me-alpha-MeDA) and alpha-methyldopamine (alpha-MeDA), respectively, both of which are catechols that can undergo oxidation to the corresponding ortho-quinones.

http://www.mdma.net/metabolites/toxic.html

That would appear to have something to do with the pKa of primary, secondary & tertiary amines. The fact that secondary amines are the most basic implies that there is some competing repulsion with another like charge in the protein chain (it also fits with the inactivity of nor-lysergic acid diethylamide and the likes of N-methyltryptamine and the monoalkylated tryptamines aslso lacking activity, even in the presence of an MAOI)

overlay the 5-MeO-compound. notice how what you're saying about mono alkylated amines holds true for LSD analogues and their potency. I think the activity of lysergamide amine has everything to do with it being held planar, lone pair orbital sticking above and below the plane, just like the fly series and the indolic N.

What about ibogaine?

and, looking @ your model, the 6 position on the IEAs corresponds to the 4 position of the PEA's. So why no 5-MeO, 6-X super-potent IEAs?

I'm trying to find a paper I saw that showed the indolic N in the same position as the lysergic nitrogen generated by computer modelling. but i really need to get back to real work

 

[Smyth]

If this hypothesis were to be tested this is what I suggest doing.

React 1,4-diMeO-benzene with EtC.O.Cl (AlCl3 cat.).

Brominate the alpha position on the ketone and the 4-position on the aroamtic ring in one pot.

React alpha-halo-ketone with potassium phthalimide.

Then you do a Wittig reaction on the keto group to get the methylene functionaility.

Finally you have to deprotect the amine using hydrazine.

See here, for proof that the olefin can survive treatment with hydrazine.

http://www.organic-chemistry.org/namedreactions/gabriel-synthesis.shtm

 

[kidamnesiac]

The whole thing also does nothing to explain why the stereochemistry of all the most active isomers of the chiral PEAs & tryptamines have the same absolute configuration as the 5 carbon of LSD - that is the biggie that your model falls down on; the iso isomers of lysergides are inactive but the S-isomers of amphetamines (or R isomers of alpha alkyltryptamines) are merely reduced in affinity/potency

I can't believe I've never noticed this before but I've got one thing to say to that-
Look @ my model, put an R isomer in there for DOB.
Now: Iso-LSD, which is inactive, corresponds to the S isomer of DOB (or another PEA).
LSD, on the other hand, corresponds EXACTLY to my model and R DOB.
same with Dr. Nichols new cyclobutane compounds, while the N still fits nicely near the lysergamide N, it also flips up very nicely to where the carbonyl O is.

 

[Dondante]

I like that you're throwing a new idea out there. Still just doesn't look intuitively right, but I'd like to see some more discussion.

sure do, it represents a large region of non-h-bonding negative charge. which would love to be near a cation (lys, arg).
do a search for "cation quadrapole binding" in google or pubmed, etc. daugherty did some great work on it. the face of a aromatic is DAMNED similar to a large, covalent halogen in that it's big, electronegative, doesn't have a true h-bonding lone-pair and therefore is non-specific.

Maybe it explains DO(halogen) compounds, but what about DOM or 2C-E?

And why don't the tryptamines undergo the same huge drop in potency that LSD does when you make the diethyl bigger or smaller?

I can't believe I've never noticed this before but I've got one thing to say to that-
Look @ my model, put an R isomer in there for DOB.
Now: Iso-LSD, which is inactive, corresponds to the S isomer of DOB (or another PEA).
LSD, on the other hand, corresponds EXACTLY to my model and R DOB.
same with Dr. Nichols new cyclobutane compounds, while the N still fits nicely near the lysergamide N, it also flips up very nicely to where the carbonyl O is.

I'm going to need a model or something to think about the position of nitrogen in DOB vs. LSD. I'm not seeing it. Are you saying that the the nitrogens are in similar positions in the active isomers, or that the nitrogen of DOB is where the carbonyl O of LSD is?

Yeah, I need a 3-D model to think about this ...

Also, F&B's proposal makes more sense with the BOx series. KidA's does seem interesting when it comes to the 2,4,6 or the 2,4,5 substitutions of PEAs as well as the 5 position of the tryptamines (although in that case you would expect 5,6 tryptamines to be potent).

I mainly don't like the fact that the amine nitrogen is all over the place (at least in the 2-D pictures).

 

[HeaT]

what could explain why bromo dragonfly is more potent than DOB? I'd say that oxygens at the 3 and 6 positions are more active than at the 2 and 5.

 

[nuke]

what could explain why bromo dragonfly is more potent than DOB? I'd say that oxygens at the 3 and 6 positions are more active than at the 2 and 5.

3 and 6 is analogalous to 2 and 5.

 

[MattPsy]

^ Yup, it's just the naming scheme. The spatial arrangement is the same.

The different in affinity is the methoxy groups being held in an ideal bonding conformation; they're tethered so that they don't suffer energy loss from the flopping around they would suffer as such, if they were untethered.

 

[Morninggloryseed]

Has anyone ever looked at the individual stereoisomers of a BOx? Is it (-) or (+) BOx that carries the most activity? I'm making 3D models of interesting atoms and don't know which way I should draw it.