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Acid, dragonflies and the 5HT2A receptor

Back to LSD

Just going back to that comment you made before

I'm confused, You've got the hydrogen bond pocket that the 5-methoxy group would fit into (the top one)... but what fits into that in LSD
Click to expand...


The modified LSD molecule shown below should form a hydrogen bond in the same way that the 5-hydroxy group of serotonin (or 2-methoxy of DOM)does. It shouldn't have any steric hinderance problems, as that part of the receptor can easily accomodate a furan ring. A 5-methoxy group would be preferable to a hydroxy group, as the latter's polarity would cause a slower penetration of the blood brain barrier.


59655LSD_modification.JPG



I'm unaware of any natural compound that contains a lysergic acid group with a methoxy/hydroxy group in the position shown. It would require a total synthesis, starting from a 5-methoxyindole compound. As such, I dont think that we're likely to see it appear anytime soon
 
I discussed with a friend (doing a Ph.D. in the area of organic chemistry) my ideas as regarding the ring substituted 2-phenylbutenes, and how difficult a synthesis would be.

Afterwards, he pointed out something to me that I hadn't seen originally, and it's something that I believe is the conerstone of LSD's extremely high potency (well all hallucinogenic ergolines actually); the conjugation of the the double bond in the C ring isn't the total extent of the conjugation. LSD is actually conjugated right up to the amide function because of the keto-enol isomerism that takes place whenever a C=O functional group is present. Due to the nature of conjugated bonds/sp2 hybridized carbon atoms, all four rings will have most of their atoms held in the same plane, essentially making LSD a "flat" molecule, except for the ethyl groups of the amide function. As has been stated by Nichols, molecules which can present a flat face to the 5HT2a receptor will have a much higher affinity (the phenylbutene derivative I that suggested may be even more potent than the corresponding amphetamine because only the carbon atom corresponding to the alpha-methyl group of amphetamines wouldn't be constrained in the same plane as the benzene ring). Anything disrupting the conjugation will also disrupt the carbon atoms held in the same plane.


The diagram below shows the extent to which LSD is restrained, making it a flat molecule (in the structure on the right hand side (enol form) bonds in the same plane are in black, the others in a purple/grey colour).


59655LSD_keto-enol_isomerization.JPG
 
Why in that first structure is flat all the way around to that heterocyclic nitrogen? Wouldn't you start getting considerable wiggle around that carbon in the top ring that branches off to the nitrogen? Likewise with the two ethyl groups?
 
Oops! The colour/contrast code w.r.t the conjugated bonds was only meant to apply to the enol form isomer on the right hand side. The classic structural diagram on the left was only in there to show that the conjugated system isn't immediately obvious with the keto (C=O) version.

When you start thinking in terms of keeping the molecule as planar/rigid as possible (via conjugated bonds), it starts to broaden the scope for compounds that might possibly be 5HT2a agonists (such as ones containing the A, B and C ring of LSD).
 
Keto-enol tautomerization.

You'd expect the nitrogen lone-pair to donate its electrons in the above diagram.

I mean what we are looking at is an amide. Obviously there is going to be restricted rotation about the -C(O)N- axis because there is some pi character in the bond connecting C and N.
 
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Hmmmm, one somewhat unresolved question I have (and it could be due more to my lack of understanding than anything else), but if having a "flat face" to present to the receptor increases affinity, why does an alpha-methyl group sticking out seem to increase potency? Or is this a situation where this methyl group adds more to increased aborbtion across the BBB, and its potency is unrelated to receptor affinity? Or am I misunderstanding the shape of the methyl group at that position?

oh, and one last question: On the final molecule, the Fig 6a, did you decide to go back to a methyl group at the beta position? I thought that earlier in the paper you had discussed how adding an oxygen atom to get a methoxyl group at this position was far preferable, and was the reason why BOD was more potent than beta-methyl 2C-D? Is there a reason why the fully aromatic 4-bromo-dragonfly molecule couldn't have a beta-methoxy group?

One last thing: 4-bromo-n,n-dimethyl-alpha-methyl-beta-methoxy-dragonfly.....is this the holy grail we're lusting after? (I threw in the dialkyl groups just for symmetry, I felt it was similar enough to tryptamines that the nitrogen just looked lonely).

What I find fascinating is that this new molecule may have other areas for variation. Obviously we can switch things around at the 4 position (iodine, methyl, ethyl, propyl, ethylthio, propylthio, etc), as with the DO and 2C compounds, but then there's also a few modifications that could be made with various dialkyl groups up at the nitrogen position (dimethyl, dipropyl, diisopropyl, etc). And then you've got the "wings" of the dragonfly to work with, as well. The oxygens at each wing would be the 2 and 5 position if this were a phenethylamine....would adding a methyl group to that oxygen...a methoxyl group, be useful? After all, it would not only stick with the desoxy TMA-2 system of two methoxy groups that is so prevalent in the DO series, but the "top" methoxy group, at the 2 position if this were a normal phenethylamine, could also function much the way the methoxyl group works in the same position in tryptamine compounds.

This substance just gives us so much to work with, and then you add in that it may have the binding affinity and dosage scale as LSD? This is definitely fascinating (unless, of course, I've managed to get everything completely wrong on this.)
 
There are lots of examples in drug chemistry where if a bare hydroxy can be converted into the corresponding methyl or ethyl ether this leads to an increase in potency.

See the link I provided in the microwave chemistry thread for a more clear cut synthetic protocol.

It is simply a Willamson's ether synthesis and yes, this could be conducted on ephedrine.

Caution: MeI is mutagenic and volatile
 
why does an alpha-methyl group sticking out seem to increase potency

Because it inhibits metabolism to the inactive compounds (it's why 2,4,5-trimwthoxyamphetamine is active, but 2,4,5-trimethoxyphenethylamine isn't). Means that more of the drug actually makes it to the brain. It appears that the alpha methyl group is acceptable, but increasing it to an ethyl group really hinders receptor binding

On the final molecule, the Fig 6a, did you decide to go back to a methyl group at the beta position?

No, it's a methylene group. The double bond conjugates with the delocalized pi electrond of the aromatic ring. Due to the nature of sp2 hybridization of carbon atoms involved in double bonds, the benzene ring, and three of the four sidechain atoms are all in the same plane. The activity of the beta-methoxy group is (I think) due to its ability to put a negative charge (electrons from lone pair, in the phenylbutene, the delocalized pi electorns from the double bond provide the negative charge) in the correct place. plain beta-methyl derivatives cause steric hinderance, but without the benefit of a localized negative charge

Is there a reason why the fully aromatic 4-bromo-dragonfly molecule couldn't have a beta-methoxy group?

None at all, if the BOB series (and a Nichols paper about alpha-methyl BOB)
relationship holds true for all DOx and 2C-x compounds, it should double the potency over the unsubstituted amphetamine/phenethylamine.

One last thing: 4-bromo-n,n-dimethyl-alpha-methyl-beta-methoxy-dragonfly.....is this the holy grail we're lusting after?

Not really. The molecule without the N,N-dialkyl groups should be very potent though. For phenethylamine 5HT2a agonists, anything other than a primary amine seems to be quite detrimental to potency

the "top" methoxy group, at the 2 position if this were a normal phenethylamine, could also function much the way the methoxyl group works in the same position in tryptamine compounds.

By George, I think he's got it (sorry, I'm not taking the piss, it's just I've always wanted to use that phrase as a reply). Thats the reasoning behind the addition of a methoxy group to LSD providing a drug that is even more potent (one of the previous posts in this thread).
The oxygen at the 5-position correlates to the nitrogen of the indole nucleus of tryptamine. As they both have lone pair electron groups (nitrogen one, oxygen two), they can both form hydrogen bonds to the same part of the protein receptor.


Makes one wonder if the benzofuran analogues of psychoactiove tryptamines would also be psychoactive.

So many variations, so little synthesis!
 
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Here's a question for you Fast... How active do you think piperazine like phenetylamines, along the lines of this... or even with the ring closed ever fruther, over the 1,6 position...
 
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Only the one in the lower right hand corner looks mildly promising, the other two just.....they feel intuitively wrong.
 
It has been shown that PEA's generally dont favor nitrogen alkylation. The 2,5-dimethoxy-4-bromophenyl-piperazine compound is available on Rh.ws and is fully active.

Extension: p-chloroamphetamine is a potent serotonin neurotoxin. In this case the rationale is that the piperidine ring could hinder it binding to SERT yet still allow NET & DAT binding. Thus such a compound looks like a good candidate structure (if the synthetic steps are not overly absurd).

Consider incorporation of the alphamethylgroup and the nitrogen into a five ring pyrrolidine compound, with the monohalophenyl ring. Such a compound is attached down below. I think such a compound is a mild stimulant with hopefully not too many cardiovascular limitations. It is certainly nothing special though.
 

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Then again F&B's candidate structure based on reduction of phenylalanine followed by construction of the corresponding morpholine ring looks like another compound worthy of examination.

The oxygen atom in the morpholine ring is a point of metabolic degradation, hence the compound shouldn't be so irrationately long-lasting that it would cause insomnia.

What is a discrepancy is that for ritalin, incorporation of morpholine ring in the structure resulted in deterioted DAT receptor binding & inhibition of dopamine uptake.

A quick razzle on Bielstein aka cross-fire would churn out the relevant documentation but unfortunately this is a project that I (and F&B) are unable/unwilling to persue.
 
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interesting, i noticed the same overlapping of the aromatic dragonfly shown in fig.2 when i was comparing the two structures my self, hmmmm. i wonder if replacing the indolic nitrogen and adding an extra furan group to lsd would render it more active.....
 
It'd be easier to do any initial studies with the benzofuran analogues of the simple tryptamine psychedelics first, as the benzofuran analogue of LSD won't be available from a natural source, as lysegric acid is, and it looks like it'd be a bastard of a total synthesis to do.

I agree with the point you're making though; if an oxygen atom can substitute for a nitrogen when it comes to hydrogen bonding (all that's different is the number of lone pair electron groups. Even the bond angles wouldn't vary that much) to the receptor protein re dragonfly vs tryptamines, why shouldn't it substitute in the case of the the indolic nitrogen of tryptamines. After all, it's not a matter of having to form salts like the sidechain nitrogen, as it doesn't really come into play when the atom is part of the aromatic system.

Wouldn't it be an eyeopener if 3-(2-N,N-dimethylaminoethyl)benzofuran (DMT benzofuran analogue) turned out to be active? It'd open up a whole series of novel psychedelics (sits twirling moustache while emitting evil cackle!)
 
By benzofuran analog do you mean replacing the ring nitrogen in TEA's with an oxygen? If so it is expected to be atleast 4X weaker than the parent structure.
 
I'm confused, You've got the hydrogen bond pocket that the 5-methoxy group would fit into (the top one)... but what fits into that in LSD?
Click to expand...

After much searching, I've eventually got an answer that also is in keeping with LSD's high affinity; the hydrogen bonds from the oxygen atom of the 5-hydroxy group in serotonin/5-methoxy group in psychedelic drugs isn't required as the serine residue that is the other half of the hydrogen bond instead has a hydrogen bond to the oxygen of the carboxamide group of LSD when configured as the enol tautomer (that's the one that shows conjugated double bonds through all 4 rings of LSD, which also forces the molecule to adopt a totally planar shape).

I've drawn a diagram to show what I mean (the light grey bit shows the position of a 5-methoxy group on the indole nucleus and assoc. hydrogen bond). The serine amino acid is part of one of the protein groups that constitute the 5HT2a receptor.

59655LSD_hydrogen_bonding.jpg
.

It's left me wondering if the D ring is needed at all to maintain activity. just as long as the conjugated system is kept in tact.

By benzofuran analog do you mean replacing the ring nitrogen in TEA's with an oxygen? If so it is expected to be atleast 4X weaker than the parent structure.
Click to expand...

Smyth
not quite sure what TEA's are, but yes I meant a replacement of the indolic nitrogen, as opposed to the sidechain nitrogen. Have you got a ref for the 4 x weaker statement, as I've never seen any papers about benofuran analogues of indole-3-ethylamine (tryptamine) psychedelics, but I'd be really interested in seeing what research has taken place.
 
Hyperion, said!

"One last thing: 4-bromo-n,n-dimethyl-alpha-methyl-beta-methoxy-dragonfly.....is this the holy grail we're lusting after? (I threw in the dialkyl groups just for symmetry, I felt it was similar enough to tryptamines that the nitrogen just looked lonely)."

I have heard this time and time again >
"is this the holy grail we're lusting after"

What would this magical chem. consist of? or a replacement of?
Lsd, mdma, sulfate mescaline a mixture of each of these. Something in that order and lasting about 8 to 10 hrs.
something short and sweet and very colorful sounds nice!
Thanks
,
 
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