Why do MDMA and 4-FA both release 5-HT? And methamphetamine only releases 1/10 of 5HT

[InterestingFACT]

Another oddity which does not seem to fit with the data. Amphetamine and phenmetrazine are known monoamine releasers, with high selectivity for DA and NE. Para fluorine substitution on amphetamine (4-FA) drastically increases SERT affinity. Yet, meta fluorine substitution doesn't increase SERT affinity as expected, but instead slightly reduces it compared to amphetamine. Both phenmetrazine and 3-F-phenmetrazine are very selective for DA and NE but the former compound (ec50 SERT ~8000nM) experiences a 3-4 fold increase from this substitution (3FPM ec50 SERT = 2500nM). What can explain these observations, especially the amphetamine substitution one. SERT tolerates an oxygen meta to the ring (MDMA, 6-APB). Somehow the fluorine is creating slightly negative interactions in 3-FA where as in 4-FA it seems to be binding strongly. The article does suggest a hydrogen bond between one of the fluorines in fenfluramine and the threonine residue, and although the distance will be shorter in fenfluramine, if this is so surely there should be a weaker H-bond in 3-FA and thus would slightly increase affinity rather than decrease it. Any insights are appreciated.

My grasp of chemistry is much looser than many of the folks in this thread, but wouldn't this just have to do with the difference between para and meta substitutions on a benzyl moiety? Halogens are a weird exception in that they're a deactivating group, but they attach ortho/para, against the expected resonance--because they want to dump their lone pair but also end up pulling s electrons from the ring. So a meta halogen is dropping its lone pair on the ring in contrary to the alkyl group's directing, and this would change the resonance therefore either the conformation of the ring, or the partial charge/configuration of the alkyl tail--Which could mean that 3-fa just wouldn't have the right shape.

 

[aced126]

How would the phenyl ring change conformation? It is flat and has very little degree of movement. The effect of the substitution changing the energy minimised conformation of the molecule seems legit but does can anyone please confirm how it does this is more detail.

 

[belligerent drunk]

So 3-FA is less selective for SERT than amphetamine whereas 3-Fphenmetrazine is more selective for SERT than phenmetrazine? Honestly at first glance I'm at loss for ideas; the ether (of morpholine ring) moiety shouldn't affect the electron density at the fluorine much. If there is HB involved as the article about fenfluramine suggests, then 3-Fphenmetrazine should instead be relatively less selective than 3-FA because of the weak inductive effect of the oxygen. So, sterical problems? Doesn't explain why meta oxygen behaves well.

Halogen and alkyl substitutions behave the same way in resonance contrary to what IFACT suggests (if I read correctly), so in 3-F substituted amphetamines both the fluorine and the alkyl "direct" into the same spots aka the 2, 4 and 6 positions. And obviously the benzene ring won't change conformation - it's planar and will stay that way. I haven't heard of substitutions making the ring change conformation anyway.

May well be that something obvious has evaded my eyes, but I honestly don't see any reason why the morpholine ring should affect the way meta-fluoro substitution affects the molecule's binding at SERT. Except maybe the oxygen takes part in binding to something as well and both the fluorine and oxygen bind in a way that creates a rather stable "polydentate" type of complex which wouldn't be possible without the fluorine (or the oxygen in amphetamine's case).

 

[InterestingFACT]

How would the phenyl ring change conformation? It is flat and has very little degree of movement. The effect of the substitution changing the energy minimised conformation of the molecule seems legit but does can anyone please confirm how it does this is more detail.

Mmmmm you're absolutely right, benzene is always planar unless it's bridged. Silly me. I would still be interested in whether this would affect

So 3-FA is less selective for SERT than amphetamine whereas 3-Fphenmetrazine is more selective for SERT than phenmetrazine? Honestly at first glance I'm at loss for ideas; the ether (of morpholine ring) moiety shouldn't affect the electron density at the fluorine much. If there is HB involved as the article about fenfluramine suggests, then 3-Fphenmetrazine should instead be relatively less selective than 3-FA because of the weak inductive effect of the oxygen. So, sterical problems? Doesn't explain why meta oxygen behaves well.

Halogen and alkyl substitutions behave the same way in resonance contrary to what IFACT suggests (if I read correctly), so in 3-F substituted amphetamines both the fluorine and the alkyl "direct" into the same spots aka the 2, 4 and 6 positions. And obviously the benzene ring won't change conformation - it's planar and will stay that way. I haven't heard of substitutions making the ring change conformation anyway.

May well be that something obvious has evaded my eyes, but I honestly don't see any reason why the morpholine ring should affect the way meta-fluoro substitution affects the molecule's binding at SERT. Except maybe the oxygen takes part in binding to something as well and both the fluorine and oxygen bind in a way that creates a rather stable "polydentate" type of complex which wouldn't be possible without the fluorine (or the oxygen in amphetamine's case).

Hold up hold up. I'm starting to question the claim that the meta-F has any relationship to SERT activity at all.
https://en.wikipedia.org/wiki/Phenmetrazine
https://en.wikipedia.org/wiki/3-Fluorophenmetrazine
http://jpet.aspetjournals.org/content/320/2/627.full.pdf+html

From above sources sources:
phenmetrazine: EC50 NE: 50.4 +/- 5.4, DA: 131 +/- 11, SERT: 7765 +/- 610, DA/SERT: 0.01687
3-fpm: NE: 30, DA: 43, SERT: 2558, DA/SERT: 0.01681
3-fa: NE: 16.1 +/- 1.7, DA: 24.2 +/- 1.1, SERT: 1937 +/- 202, DA/SERT: 0.0124
amphetamine: NE: 7.2 +/- .44, DA: 24.8 +/- 3.5, SERT: 1765 +/- 94, DA/SERT: 0.0141

With those error bars, 3-fa is no less selective for SERT than amphetamine. And 3-fpm is *more potent* than phenmetrazine, but maintains the same ratio of DA/SERT effects. The only differences of note here are that the morpholines are more SERT-friendly than amphetamines, but lower potency, and that 3-fpm is generally more potent than phenmetrazine, which to me indicates a general increase in "stickiness" and not any particular improvement of fit at SERT. If this is the case, and the 3-F substitution doesn't confer SERT affinity, then the question is: What do 3-CF3, 3,4-methylenedioxy, and 4-F substitutions do that 3-F doesn't.

Does anyone know if 4-f-phenmetrazine has ever been made? I've heard of 4-methylphenmetrazine but can't remember it's binding values.

 

[roi]

Does anyone know if 4-f-phenmetrazine has ever been made?

Yeah, it's available in Sweden now where 3-FPM is banned. You'll probably find a flashback thread. tl;dr: Not very potent and sucks, similar to 4-methylphenmetrazine,

 

[Dresden]

Wouldn't these molecules be of very high affinity to SERT. They are quite lipophilic and are planar.

And what about these 2 aromatics? The nitrogen isn't even basic so it should be more lipophilic than if the lone pair of the nitrogen was localised. The n-demethylated 5 isomer is apparently a thing but I couldn't find any binding data.

5-IT is said to be a good roll, but it kills a lot more people than ecstasy on a unit per unit basis:

5-IT:

IAP was an all-around disappointment,

IAP:

but your indene methamphetamines may be an improvement over IAP. I would, however, suggest the addition of a beta keto group to activate all those greasy, indolent carbons a bit:

 

[aced126]

A beta ketone group would deactivate the whole aromatic system by conjugation

 

[adder]

So does para-fluoro, yet we don't observe a drastic drop in potency for 4-FA vs. amphetamine. Actually I believe methcathinone is more potent at DAT than methamphetamine, however increased hydrophilicity makes it cross the blood-brain barrier less readily. I guess a functional group on an aryl ring can increase or decrease affinity if the interaction between the aryl in the ligand and the aryl in the receptor is based on electron density on certain carbon atoms in both groups, I suppose this might be more prominent for tyrosine or tryptophan residues, or if the neighbourhood is crowded, but it's just guessing. If you compare 4-FA with 4-MA or amphetamines with activated rings, partial charges on the carbon atoms of the aromatic ring don't seem to play as much of a role as substituents on the ring and their positioning. Also, if you compare 3-MMC with 4-MMC, it seems that there is enough room for the ligand's molecule to move for the best interaction, aromatic rings 3-MMC and 4-MMC may not be exactly in the same position (may not perfectly overlap) when bound.

 

[Dresden]

and

would also be improvements.

As for the dreaded activating/deacitvating debate,

seems to hold its own with other DOx's despite being relatively deactivated with respect to other DOx's by the Ar-NO2 group, if PiHKAL is to be believed (and it usually is).

I would further venture that

would knock your socks off despite being highly ring deactivated by the Ar-NO2 again.

Likewise for,

and

 

[aced126]

Adder the fact that we observe really activated (MDMA) and deactivated (4-FA, flephedrone etc) rings both having potential to be good 5HT releasers suggests that the substituent itself plays a much larger role in binding than the ring itself. Hence why I don't think arguments of increased or decreased ring density works because the substituent itself will probably have a bigger role in binding. In fact in a deactivated ring one would expect less pi stacking with tyrosine and so in fact weaker binding, so we would expect methcathinone to bind weaker at DAT and SERT. The ketone itself could form a hydrogen bond with tyr in DAT or thr in SERT. I do agree that ring density will play a role but in my opinion this is less so when compared to the actual substituent itself taking part in binding interactions. If all we really needed was a really electron rich ring then we could just place multiple amino groups on the ring. Leave out pharmacokinetics for one moment, if that sort of compound was tested for monoamine release or even reuptake I don't think we'd observe the expected results given the ring density hypothesis is true.

Dresden, I know the Wikipedia page stated several hours of amphetamine like stimulation from DON but is this due to monoamine releasing effects? I couldn't find any binding data for DON but for all other DOX it seems to just interact with serotonin receptors while not actually directly affecting monoamine release by modulation of the transporters (which is really what this discussion is about, not 5 HT receptor binding which would have a different SAR). Obviously there are downstream effects off this agonism, one of which seems to be monoamine release. I don't think DON binds to any transporters. I might be wrong though, please correct me if I am as I didn't spend much time looking through research on this.

 

[Dresden]

The sense of "activated" that I meant was something like:

a-PVP is "activated" (deactivated?) in comparison to

prolintane.

 

[aced126]

Those are reuptake blockers anyway. How would the beta ketone make the indene compound better than its reduced counterpart?

 

[Dresden]

If you have this whole substituted amphetamine molecule that is 100% hydrocarbons (with one amine), they get weaker as they get bigger. Oxygen invigorates them. Has to do with NMR energy field shifts or Lord knows (or cares) what. That's why IAP (IndanylAminoPropane) and prolintane are so lackluster. An exception is aromatic methyl groups, which are very powerful, but even that is not an exception since Ar-Me's get metabolized into the ultra electron withdrawing (oxygenated!) Ar-CO2H in time. Remember, O is the second most electronegative element and is also bigger than F, the most electronegative element, giving it the advantage of the heavy atom effect. Electronegativity is of paramount importance in understanding chemical properties and reaction mechanisms, as I'm sure you know. Also, such molecules can't hydrogen bond with anything but their amine, but this effect may or may not be all that relevant to their psychoactivity. If a molecule like this one

is found to be weak, then adding a beta keto group will often give it some adrenergic push.

Anyway, this is the model that works for me. You are not obligated to buy into it.

This

is an even better way to achieve a similar goal (activating the ring) without resorting to the adrenergic (adrenaline sucks) beta keto substituent. Removing the N-methyl,

also helps. I have a feeling I just gave you more questions than answers.

The methoxy can also go here

if you'd rather.

 

[aced126]

The reason why IAP is a weaker serotonin releasers has already been addressed; the plane in which the methylene carbon is in is depressed relative to the plane of the ring and that causes steric clash. Prolintane might suck in comparison to A-PVP because the beta ketone in the former compound might bind to the tyrosine residue and strengthen the interaction. Using your argument methylone would have more efficacy than MDMA but in fact it seems to release only a third of 5HT. It also has 13 times less affinity at VMAT2. So a chunk of its mechanism of action comes from reuptake rather than release. That seems to be the general trend: beta ketones make reuptake inhibitors better but worsen monoamine release potential. It also reduces logP so that needs to be considered when looking at overall potency.

I still don't understand why you think an activated ring > more affinity at transporter. This is not necessarily the case. Is there any other reasons why you think methoxy groups will make the molecule better (leave out downstream effects of 5HT receptor agonism for now).

 

[Dresden]

Methylone is in fact much more adrenergic in character than MDMA. However, I agree it sucks in comparison.

But good luck designing desirable designer drugs with no oxygen(s). Also, who says IAP's two rings are not in the same plane?

 

[aced126]

Actually I think I made a mistake on the angles. I just realised that by conforming to a planar regular pentagon shape, the angles between the carbons is 108°, very close to the optimal tetrahedral carbon angle (assuming all the bond lengths are relatively the same). This would mean conforming to this shape would stabilise the molecule. But... The lengths are not the same (benzene carbon carbon slightly shorter)! Let me try and do some maths to work out the angle of depression.

 

[Dresden]

The two methylene carbons directly attached to the phenyl are, without a doubt, in the same plane as the benzene ring. Why would the remaining methylene carbon directly attached to both of those other carbons not be in their plane as well? The only atoms not in the plane are the 6 hydrogen atoms attached to those three methylene carbons. Your assertion is as ridiculous as claiming 1,3-benzodioxole is not planar save for its two methylene hydrogens.

 

[belligerent drunk]

Double-check everything! Look at cyclopentane, that's the way IAP should look like.

 

[Dresden]

You're just wrong.

 

[belligerent drunk]

Care to explain why?

EDIT: Also, it's not my assertion that 1,3-BDO DEFINITELY is non-planar, I haven't been there myself, haven't seen the molecule with my own eyes. There are articles which state that it in fact is non-planar though. You have to remember that most molecules are not static, they change conformations to a significant degree very quickly (with some exceptions, like benzene for example). Molecules like IAP and BDO have certain degrees of freedom at that five-membered ring. It would be wrong to assume that statistically it is absolutely planar.

http://www.sciencedirect.com/science/article/pii/S0009261410003830