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

[aced126]

Okay, first off, say in cyclopentane, if the bond lengths were all the same, one could make a reasonable claim that it would adopt a planar conformation because this would give it the shape of a regular pentagon with an angle of 108 degrees between carbons, very close to the 109.5 degrees a normal tetrahedral carbon is in. So one could argue that in the planar conformation it is least strained and so is of lower energy than other conformations which would result in more strain.

But, 2 problems. First, cyclopentane itself is not observed to be planar but adopts a lower energy "envelope" conformation. Now, the reason why this is lower energy is something I'm not sure why this is the case and any insight into this problem would be good, because it seems that with this conformation there would be more overall ring strain.
https://en.wikipedia.org/wiki/Cyclopentane#/media/File:Cyclopentane3D.png
https://en.wikipedia.org/wiki/Cyclopentane#/media/File:Cyclopentane_halfchair.svg

The second problem is that the bond lengths are not the same. In the 5 carbons of indane, the benzene carbons probably have a distance of close to 1.40 angstroms and all other carbons are closer in distance to 1.54 angstroms. This is not a regular pentagon and thus the angles would not be the same if it were to adopt a planar conformation. So there would be more ring strain if it adopted planar. It can reduce this ring strain by depressing itself so that the distance between all the carbons are kept the same but the angles are as unstrained as possible. I'm trying to figure out right now what this angle of depression would be.

Finally however, on the wikipedia page of indane, the molecule does indeed look non-planar and that image is probably from a program which energy minimises the structure first.
https://en.wikipedia.org/wiki/Indane

 

[aced126]

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

Agreed on the fact that there is some free movement and that energy lowering interactions can stabilise this movement.

 

[belligerent drunk]

Finally however, on the wikipedia page of indane, the molecule does indeed look non-planar and that image is probably from a program which energy minimises the structure first.
https://en.wikipedia.org/wiki/Indane

If you look closely, the "end-methylene" in indane's pentagon is actually directed "behind the plane/page" - look at where its hydrogens are. If it was on the same plane, its hydrogens would be similar to the other methylenes in that ring.

 

[Dresden]

I already said the methylene (as opposed to the benzene ring's) hydrogens are not in the plane of indan (6 out of plane H's) or of 1,3-BDO (2 out of plane H's). However, all the other atoms are.

 

[aced126]

I already said the methylene (as opposed to the benzene ring's) hydrogens are not in the plane of indan (6 out of plane H's) or of 1,3-BDO (2 out of plane H's). However, all the other atoms are.

https://en.wikipedia.org/wiki/Indane#/media/File:Indane_3D_ball.png

Look at that diagram closely, the methylene carbon on the edge is depressed into the page.

 

[belligerent drunk]

I already said the methylene (as opposed to the benzene ring's) hydrogens are not in the plane of indan (6 out of plane H's) or of 1,3-BDO (2 out of plane H's). However, all the other atoms are.

I think you should re-read aced's and my posts. Of course all hydrogens of the sp3 carbons are out of plane, it would make no sense for sp3 carbon to adopt an all-planar configuration akin to that of sp2 (like in benzene). The "first two" methylene carbons are in plane with the other ring, the last one though isn't. If you want some graphical proof, search for pictures of indane and cyclopentane. Cyclopentane has an out-of-plane carbon; look at the way its hydrogens are relative to it and the rest of the ring. Look at what indane looks like, compare the two molecules and the way the hydrogens are arranged. That should give you some idea, because they're extremely similar. I'm not going to look for literature that says the same, if you're this adamant that the whole molecule of indane (and cyclopentane) is planar, I'd like to see the proof, out of genuine curiosity.

 

[aced126]

Ok, figured out the theoretical co-ordinates. Assumptions: ar bond length = 1.40a, c-c alkane bond length=1.54a, ar-c angle = 120 degrees, c-c angles = 109.5 degrees. Basically I found the following result using 3D vectors: if the assumptions I made were true and to be satisfied, then the end methylene carbon has to be depressed down (or up if you wish) by 1.08 angstroms. And if this is so, the angle it makes with the aromatic carbons is 31.1 degrees. If it was depressed by this much, then all the bond lengths and angles are conformed to and so there is no ring on bond strain; thus I believe this configuration is energetically most stable.

If anyone wishes to see the proof/my procedure then request it and I'll put it up in neat.

 

[serotonin2A]

I already said the methylene (as opposed to the benzene ring's) hydrogens are not in the plane of indan (6 out of plane H's) or of 1,3-BDO (2 out of plane H's). However, all the other atoms are.

I'm sorry but you are wrong -- the methylene is out of plane.

 

[aced126]

On a different note, can anyone explain why cyclopentane is not a planar configuration?

 

[InterestingFACT]

On a different note, can anyone explain why cyclopentane is not a planar configuration?

at 109.5* bond angles, cyclopentane doesn't have significant angle strain. However in its planar conformation, the eclipsed C-C bonds cause tortional strain that can be minimized by adopting an envelope or twist conformation. Even these conformations don't totally eliminate tortional strain but they significantly reduce it (10kcal/mol-->6kcal/mol).

If you go look at a Newman projection of cyclopentane you'll see what I mean.

 

[belligerent drunk]

I'm not sure, but perhaps it has something to do with the fact that the hydrogens in cyclopentane "wish" to be as far from eachother as possible. If you look at the "first pair" and the "second pair" of identical methylenes that are on the same plane, their hydrogens are quite apart because of the fifth carbon being out of plane. If everything was on the same plane, all the hydrogens would be much closer, so it is more favorable to have some bond angle strain in order to separate them. I'm not sure because it's hydrogens in this case and they are small, so I don't know if this effect is enough; with bigger substitutes there is definitely this kind of effect. If anyone has a better explanation I'd appreciate hearing it.

EDIT:

at 109.5* bond angles, cyclopentane doesn't have significant angle strain. However in its planar conformation, the eclipsed C-C bonds cause tortional strain that can be minimized by adopting an envelope or twist conformation. Even these conformations don't totally eliminate tortional strain but they significantly reduce it (10kcal/mol-->6kcal/mol).

If you go look at a Newman projection of cyclopentane you'll see what I mean.

Essentially what I had in mind, but explained significantly better.

 

[Dresden]

Ok, so for indan, 4 of the carbons are in the same plane and the fifth is puckered.

 

[aced126]

Yeah and that's the supposed reason why it has less potency than MDMA, not because there is a saturated beta carbon lol.

 

[Dresden]

MDMA's two oxygens place a big part.
Don't believe me? Try one of your non-oxygenated indene methamphetamines sometime.

I have to at least suggest

 

[Seppi]

Just to clear up a point of confusion in this thread:
Amphetamine has one additional transporter phosphorylation mechanism (CAMKIIα) that is independent of TAAR1 and which produces neurotransmitter efflux from neurons.
TAAR1 mediates PKA and PKC signaling cascades that phosphorylate transporters, but not the CAMKIIα cascade. I'm fairly certain that the triggering mechanism for the CAMKIIα cascade has just recently been identified, but I can't be sure until I get my hands on this paper: https://www.ncbi.nlm.nih.gov/pubmed/26162812

If DAT & L-type calcium channel coupling does mediate CAMKII phosphorylation, then it is likely that transporter/channel coupling also mediates 5-HT release since SERT and L-type calcium channels have also been shown to couple -https://www.ncbi.nlm.nih.gov/pubmed/24854234 - in turn, this represents another target that accounts for variability between releasing efficacy of different compounds. Unlike PKA and PKC, CAMKII phosphorylation has no effect on neurotransmitter reuptake.