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

[aced126]

It seems contradictory as one has electron donating groups through resonance and the other has an electron withdrawing group through induction. They both have to bind to SERT and unbind from SERT once inside the cell. If the phenyl ring in both said compounds is pi stacking with tyr, phe or trp then surely the more donated ring i.e the one with the methylenedioxy bridge on it should bind with the strongest interactions but as a result be poor at unbinding once exposed inside. 4-FA, with a deactivated ring, should reach the neuron in the highest concentrations, then meth with a neutral ring and finally MDMA. Is it just that the latter compound is a superb agonist at TAAR1? Or maybe a better VMAT2 substrate? Does TAAR and VMAT have exactly the same aa sequence in dopamine and serotonin neurons? Or will they be slightly different and have different binding targets as a result?

 

[serotonin2A]

All three compounds act in the same basic manner but the big difference is their selectivity. (+)-METH releases DA and NE with EC50 of 10-20 nM, but the concentration required to release 5-HT is almost 2-orders of magnitude higher. So the effects on DA and NE max out before METH can alter 5-HT.

By contrast, MDMA is slighly more potent a releaser of 5-HT compared with DA and NE. So it will tend to release all three with some preference for 5-HT.

4-fluoro-AMPH is closer to METH in it's selectivity, but compared to METH it is much less selective for DA vs 5-HT. So at doses that release DA it will also have moderate effects on 5-HT release.

There is no difference in the structure of VMAT of TAAR1 in different neurons.

 

[aced126]

If VMAT and TAAR1 are the same, that means the only difference in selectivities of all these compounds would be getting into their respective neurons and getting into their respective neurons only?

So why is it that 3 compounds all of the same structure, one with a neutral ring being least selective for SERT, then one with an electron deficient ring having higher affinity and finally an electron rich ring having the highest.

neutral<electron deficient<electron rich for SERT. Why is that so? Wouldn't the order deficient<neutral<rich or vice versa be expected? Obviously I guess lipophilicity could play a factor in this but any other reasons as to why this trend is observed?

 

[serotonin2A]

One of the oxygens in the methylenedioxy ring in MDMA may also contribute to binding to SERT, possibly via H-bonding. Van der Wasls interactions/pi stacking are probably not the only type of interaction that is occuring. Contrary to your assertion, the three molecules have different structures, and that alters their relative affinity. It is also possible that SERT requires more lipophilic ring substituents, hence alkoxy > F > H.

Other differences are that (a) they have different relative affinities for VMAT and TAAR1; (b) MDMA is also a 5-HT2B agonist, which may faciliitate its ability to release 5-HT; and (c) fluorine can potentially form a hydrogen bond.

 

[aced126]

One of the oxygens in the methylenedioxy ring in MDMA also contributes to binding to SERT. Van der Wasls interactions are not the only type of interaction that is occuring. Contrary to your assertion, the three molecules have different structures, and that alters their relative affinity.

The other differences are that (a) they have different relative affinities for VMAT and TAAR1. MDMA is also a 5-HT2B agonist, which may faciliitate its ability to release 5-HT.

Sorry, I meant same structures apart from ring substituents. And with that I'd like to change 4-FA to 4-FMA I guess.

 

[Dresden]

I think it has something to do with the fact that F and O are the two most electronegative elements. You seem to be making this way too complicated.

 

[aced126]

I think it has something to do with the fact that F and O are the two most electronegative elements. You seem to be making this way too complicated.

Yes those elements are very electronegative, but they are substituents on a ring. The effect on the ring is that the ring becomes more nucleophilic in the case of MDMA. 4-FA will make the ring (slightly) less nucleophilic than benzene.

 

[aced126]

One of the oxygens in the methylenedioxy ring in MDMA may also contribute to binding to SERT, possibly via H-bonding. Van der Wasls interactions/pi stacking are probably not the only type of interaction that is occuring. Contrary to your assertion, the three molecules have different structures, and that alters their relative affinity. It is also possible that SERT requires more lipophilic ring substituents, hence alkoxy > F > H.

Other differences are that (a) they have different relative affinities for VMAT and TAAR1; (b) MDMA is also a 5-HT2B agonist, which may faciliitate its ability to release 5-HT; and (c) fluorine can potentially form a hydrogen bond.

If lipophilicity was needed at SERT, then para-chloro-AMPH (and bromine derivative to a slightly greater extent) would be observed to be much better 5-HT releasers than they are because halogen substituents on a ring raise lipophilicity significantly (except fluorine which increases it slightly). In fact I'd argue that the alkoxy group actually brings down the lipophilicity of the overall molecule to lower than that of METH.

 

[serotonin2A]

If lipophilicity was needed at SERT, then para-chloro-AMPH (and bromine derivative to a slightly greater extent) would be observed to be much better 5-HT releasers than they are because halogen substituents on a ring raise lipophilicity significantly (except fluorine which increases it slightly). In fact I'd argue that the alkoxy group actually brings down the lipophilicity of the overall molecule to lower than that of METH.

PCA and PBA are very effective releasers, but they are also extremely neurotoxic. Where do you get that they are not effective 5-HT releasers? PCA is definitely more effective than METH:

http://pubs.acs.org/doi/pdf/10.1021/bi00144a010

https://www.google.com/url?q=http://www.maps.org/images/pdf/1992_berger_1.pdf&sa=U&ved=0CBoQFjAHOB5qFQoTCNnTybfb78gCFVDXYwodWeIB4w&usg=AFQjCNH_ecr-_9dsG5GOPGR-NEs5r85CjA

It's not the lipophilicity of the entire molecule that is going to matter. Binding doesn't work like that. There are particular residues that will be facing the ring and they may prefer a lipophilic substituent at the 4-position. http://onlinelibrary.wiley.com/doi/10.1002/minf.201300013/pdfn.

 

[aced126]

If a lipophilic substituent is needed in the 4 position then in fact an oxygen in that position would create unfavourable interactions with the supposedly lipophilic residues facing the rings, would it not?

 

[adder]

There is a substantial difference between alkoxy and hydroxy groups (alkoxy groups are only hydrogen bond acceptors while hydroxy groups are both hydrogen bond acceptors and donors). Compare water solubility of diethyl ether and ethanol or even n-butanol, and it all should be clear. The interaction between the para substituent and the binding site is likely an interaction based on a hydrogen bond or opposite (partial) charge attraction, with that being said both fluoro and alkoxy can form such an interaction. Also, hydroxy and alkoxy are both electron-donating (through resonance) and electron-withdrawing (through induction), the partial positive charge will still be on the oxygen atom when it's bound to the phenyl ring, same as with fluorine.

 

[serotonin2A]

If a lipophilic substituent is needed in the 4 position then in fact an oxygen in that position would create unfavourable interactions with the supposedly lipophilic residues facing the rings, would it not?

No, it wouldn't. As adder wrote, compare the water solubility of ethanol and ether. Alkoxy groups are widely known to be lipophilic, but are capable of accepting a hydrogen bond.

Adder, I used to assume that a H-bond was involved in the potency of MDMA, PMA, etc but the modeling/docking studies I have seen haven't identified a hydrogen bond donor in the vicinity. That also wouldn't explain the potency of 4-methylamphetamine. That still leaves unanswered the question of why you can't replace both of the oxygens in the MDMA methylenedioxy ring with carbons and retain potency at SERT but that might be a stereochemical issue.

 

[aced126]

the partial positive charge will still be on the oxygen atom when it's bound to the phenyl ring, same as with fluorine.

I don't understand why there is a partial positive charge on the oxygen? Also if increased SERT affinity comes from hydrogen bonds or charge attraction (say if there was a charged lysine residue) then how come 4-methylamphetamine (5HT Ki = 53.4nM) and mephedrone, are good 5-HT releasers. SERT takes both para charged and lipophilic substituents? What about 6-APB as well?

 

[belligerent drunk]

I'm not sure what adder means by partial positive charge. According to resonance structures, oxygen (or fluorine) should have partial positive charge, but inductive effect overrides it. Phenolic oxygen definitely has quite the negative charge - the oxygen is a hydrogen bond acceptor and its hydrogen is a donor. Given, it is still probably less negative than an aliphatic oxygen, but definitely not positive.

5-HT2A, what stereochemical issue do you mean regarding the replacement of oxygens in methylenedioxy ring?

 

[aced126]

Aren't phenolic oxygens more negative than aliphatic hydroxyls? I thought the phenyl ring stabilised the negative charge on the phenoxide ion (which is why phenol (pKa=10) is more acidic than ethanol (pKa=15.7).

 

[belligerent drunk]

It does stabilize the charged ion, because the ring "absorbs" some of the charge. If you look at the resonance structures of a phenolate anion, then you see that the negative charge of the oxygen is reduced (and given a more positive charge in the protonated form). That is what fascilitates the increase in acidity - the increased "lack of electron density" on the O-H bond in phenols. X-H is more acidic if the bond lacks electron density, as in the electron density is shifted away from it, making proton a good leaving group (if the conjugated base is stabilized in some way). The rule is usually actually opposite - the atom bonded to the H needs to be less negatively charged to increase acidity when comparing similar compounds. What I mean is not C-H vs O-H, but O-H vs O-H or Cl-H vs Cl-H with different substitutes on the electronegative atom. By the way this is also the reason alpha carbons in ketones/aldehydes et al are somewhat acidic.

You could look at it from the point of view of conjugated bases/acids. A C- ion is a very strong base, because the charge is (usually) fairly localized on the atom and given carbon's meager electronegativity it "can't really keep it to itself" making it a strong base. Now when comparing oxygen acids the electronegativity (simplistically) stays the same, but what matters is how localized the charge is. In aliphatic OH compounds its fairly localized, even more so than in H-OH. If you look at something like sulphuric acid, then it's still an O-H acid, but the electron density of OH bonds are greatly reduced due to other S=O/S-OH bonds, making it a strong acid.

 

[adder]

I'm not sure what adder means by partial positive charge. According to resonance structures, oxygen (or fluorine) should have partial positive charge, but inductive effect overrides it. Phenolic oxygen definitely has quite the negative charge - the oxygen is a hydrogen bond acceptor and its hydrogen is a donor. Given, it is still probably less negative than an aliphatic oxygen, but definitely not positive.

5-HT2A, what stereochemical issue do you mean regarding the replacement of oxygens in methylenedioxy ring?

I meant partial negative charge.

I don't understand why there is a partial positive charge on the oxygen? Also if increased SERT affinity comes from hydrogen bonds or charge attraction (say if there was a charged lysine residue) then how come 4-methylamphetamine (5HT Ki = 53.4nM) and mephedrone, are good 5-HT releasers. SERT takes both para charged and lipophilic substituents? What about 6-APB as well?

Even if there is an opposite charge attraction, there are many different residues around present and they will impact binding, if some bulk makes a molecule not fit into the cleft, it won't bind even though it might have charged parts at right distances, right? As for SERT, it seems that generally some kind of bulk at para and/or meta positions is desirable for whatever reason. Anyway, if there is no hydrogen bond happening between the para substituent and SERT, I imagine apart from aminoacids with aromatic rings there may be some aminoacid residues around with branched alkyl chains like valine or leucine and/or polar uncharged chains like serine, so apart from pi stacking between aromatic rings there may be an interaction based on van der Waals forces (dipole-dipole or dipole-induced dipole force). This could explain why 4-methyl can be as effective as 4-chloro or 4-methoxy as in PMA. Those forces are considered weak, but hey, if you look at solvents like hexane or toluene, they're making molecules stick together, so they might be good enough as a secondary interaction between a ligand and a receptor. I'll try to look into some ligand binding structures for SERT this week if I don't forget, it's something I wondered about too but never decided to check what's in the literature. I suppose some binding studies has been done with drugs like fluoxetine.

 

[serotonin2A]

5-HT2A, what stereochemical issue do you mean regarding the replacement of oxygens in methylenedioxy ring?

I'm comparing MDA and indanylaminopropane (IAP). The shape of the methylenedioxy ring in MDA and the cyclopentyl ring in IAP are slighly different, which may explain why IAP is less potent than MDA at SERT. That part of the binding site in the transporter may be sterically constrained and IAP may not "fit" as well as MDA.

 

[belligerent drunk]

I'm comparing MDA and indanylaminopropane (IAP). The shape of the methylenedioxy ring in MDA and the cyclopentyl ring in IAP are slighly different, which may explain why IAP is less potent than MDA at SERT. That part of the binding site in the transporter may be sterically constrained and IAP may not "fit" as well as MDA.

I just recently (thanks to adder) learned that the methylenedioxy ring in MDA is quite out of plane of the benzene ring. What does the indanyl version look like then? Somehow my gut feeling tells that the difference can't be drastic enough to warrant such difference in binding affinity. I would still go with the explanation that the oxygens' electron pairs and/or negative charge have something to do with it.

 

[belligerent drunk]

Those forces are considered weak, but hey, if you look at solvents like hexane or toluene, they're making molecules stick together, so they might be good enough as a secondary interaction between a ligand and a receptor.

There is a common misconception that vdW forces are weak. That is actually not true at all. Surprisingly, London dispersion force is among the strongest forces for most organic molecules that keeps them together. For example, about 78% percent of all forces that keep butanone molecules together are London dispersion forces; that's quite a lot for a quite small, relatively non-polarizeable and polar molecule. Some more:

-86% HCl-HCl; 96% HBr - HBr - that is quite surprising to me personally, as one could assume that the dipole moment in HCl is big enough to be a major factor.
-57% for NH3 - NH3
And the winner is, of course, the magical substance that is water with a whopping 24% of all interactions accounted to London dispersion forces.

When you consider the strength of H-bonds and dipole-dipole interactions between polar molecules like water or ammonia, such percentages are quite astonishing. Obviously this is not to say that HB is absolutely useless and should be discarded. It's just a reminder that London dispersion force is a strong force despite the fact that it is usually portrayed as a "minor" force. Also, this is just my opinion, but I think that since this force is absolutely non-selective, then in terms of binding of a substrate it kind of loses its meaning unlike HB donor/acceptor which is fairly specific. But in some cases it might explain the indifference of -methoxy or -methyl as a substrate, because both are fairly similar in terms of dispersion forces.

A good piece of reading on this subject is Reichardt's "Solvents and Solvent Effects in Organic Chemistry" http://onlinelibrary.wiley.com/book/10.1002/3527601791