n-Alkyl,n-Alkyltryptamine shapes and their SARs

[Jamshyd]

I have been pondering this for a while. My background in chemistry is very limited, so bear with me. Something tells me that my question falls into the area of "chiralty", but I could be wrong.

Now, both A and B are MiPT (n-methyl,n-isopropyltryptamine), right? Well, I have seen the MiPT molecule drawn many times in different areas with the n-alkyl groups in different positions. Presuming the alkyl groups go parallel with the tryptamine molecule (and they probably aren't, thus more variation, but anyways...), there are two possibilities: The methyl group can be "pointing upwards" or "pointing downwards."

Surely, the difference between the two seem too drastic to be overlooked. This reminds me of how the simple tweak from DPT to DiPT produced two extremely diferent drugs.

Of course, I am not talking about MiPT only - but rather, any tryptamine with two different alkyl groups. In fact, I have similar concerns about phenethylamines:

Both C and D are MDMA, and I have seen the MDMA represented in either way without much regard to the rather huge difference in molecule shape.

So my question is... are the variations simply drawing errors (ie., in the real reaction, the molecule always comes up in one arrangement), or if the molecule may indeed have more than one arrangement, and how would this affect its action?

I am more interested in this concerning tryptamines rather than phenethylamines, but any answer would be appreciated.

Thanks in advance for any insight .

EDIT - sorry, I forgot to add the NH part on the MDMA molecule. I'm too lazy to redraw.. imagine it is there

 

[fastandbulbous]

For the tryptamines shown, rotation about the axis formed by the nitrogen-carbon of tryptamine side chain will transform it from one to the other (therefore no chirality)

With the MDMA molecule (PS., you forgot to put in the NH part), there is a chiral centre - the carbon of the sidechain attached to the nitrogen. This happens because there are 4 different groups* attached to that carbon atom, and as the bonds can be seen as the points on a tetrahedron, it's possible to form a mirror image molecule that cant be superimposed over the other mirror image (time to get cocktail sticks and coloured balls of plasticene - a bit too long to accurately describe in a single post).

The tryptamines (other than AMT and 5-methoxy AMT) dont contain a carbon atom with 4 different groups attached. Try it with each carbon atom in turn, and you'll see

* the four groups are benzyl, methylamino, methyl & hydrogen

 

[Jamshyd]

For the tryptamines shown, rotation about the axis formed by the nitrogen-carbon of tryptamine side chain will transform it from one to the other (therefore no chirality)

(Underlining mine)

But you see, it is this very point that I was asking about, lol. So, both A and B are identified as MiPT. Are they the same drug or are they two different drugs? Could you elaborate more on that point?

As for MDMA, thanks for pointing out the missing NH part - I just realized I forgot it. But yes, I am vaguely aware of how MDMA chiralty works (though only vaguely, hehe), I simply put it there to show how drastically different the two models of the same drug can be...

 

[Heather]

There are no “chiral” carbons in a dialkyl tryptamine and there are therefore no optical isomers. You need to have at least one carbon with 4 different groups around it to have optical isomers.

You might think the nitrogen can be a chiral center with one substituent a lone pair of electrons, however at ambient temperatures, the nitrogen group undergoes rapid inversion and scrambles the orientation that would give rise to optical isomers.

What you really indicating above is rotational isomers. There is a very low barrier to rotation around any of the carbon nitrogen bonds, and again, there is very little if any preferred orientation of the molecule.

In a way, your preoccupation with drawn structures reflects the static nature of these ‘cartoons’ chemists draw for molecular structures, that is, it’s a simplification for reality.

A molecule is generally continually passing through a wide range of rotational positions (isomers) unless there are reasons for one orientation having a lower energy versus another orientation. If that is the case, either through steric effects or bond effects, the molecule has a greater probability of existing in one rotational orientation versus another. By ‘steric’ effects, I mean size of substituents. Bond effects are due to relative symmetry of the bonding orbitals between groups.

Let me give an example. Your dialkyl tryptamines are typically synthesized via the condensation of indole with glyoxl chloride and a suitably substituted dialky amine. This condensation forms an amide linked to the indole ring at the 3 position with two carbonyl groups in series. These precursors, indol-3-yl N,N-dialkylglyoxylamides, have this structure:

The dialky tryptamine is then obtained by reduction of this amide, usually with lithium aluminum hydride.

Note that based on what I said earlier, there should be ‘free rotation’ about the nitrogen carbon bond in the amide. However, if you do proton NMR of a glyoxylamide, you’ll see two different signals for the two dialkyl substituents on the nitrogen, even if they are the same group. The molecule seems to exist in two different rotational states. This is because the two carbonyl groups in series create a situation where the lone electron pair of the nitrogen group can participate in the bonding between the two carbonyl groups. Typically this is termed ‘resonance’ and leads to certain orientations being preferred versus other orientations.

Since the orientation where the lone pair on nitrogen can be shared with the carbonyl groups has a lower energy state, there is a fairly high energy barrier for rotation about the carbon nitrogen bond. And while the conical form I’ve indicated with a nitrogen-carbon double bond isn’t energetically favorable due to charge separation, there is enough of a payoff in having a conjugated system to allow this structure some consideration. Thus there is no ‘free rotation’ of the carbon-nitrogen bond at room temperature and two distinct NMR signals are obtained for the dialky substituents of a glyoxylamide.

One interesting observation is this….if you take the NMR tube out of the instrument and heat it up, say gently with a Bic lighter, and then put it back into the instrument, the NMR spectrum you obtain will show one signal for the dialkyl substitutes that slowly splits into the two signals as the sample cools back down to room temperature. What you’ve done by heating with your lighter is provide enough thermal energy by heating to easily overcome the energy barrier to free rotation about the carbon-nitrogen bond. Free rotation makes the chemical ‘environments’ of the dialkyl groups on average the same, therefore one NMR signal.

In terms of pharmacology, the available energy for such interconversions occur at physiological temperatures and in aqueous environments. In this case, the energy barriers to free rotation about the carbon-nitrogen bonds are low enough to make the two structures you’ve drawn for MiPT essentially identical.

I think it is only in the last decade that SAR models have started to take into account the dynamic nature of interconversion between various rotational isomers of both the ligand and receptor structure to account for observed binding affinities. Takes a lot of computational power for these models.

Your question concerning the phenethylamines has a different answer. In this case there is a carbon with 4 different groups around it, and therefore MDMA can exist as two optical isomers, an “S” or an “R” form. This has a direct impact on the allowed shapes the molecule can assume and one isomer tends to have a quite different binding efficiency than another.

 

[BilZ0r]

Nice post... pity old Heather only made 2...

Anyway, as people have said... those two "versions" of MiPT you have drawn are the exact same chemical... 'single' bonds between compounds are just like pieces of string, and can twist pretty much all they want... hence those two chemicals are exactly the same.

 

[Jamshyd]

Thanks for the explanations, Heather and Bilzor.

I was under the impression that the alkyl groups were "static" - I did not know that they moved. Again, I have very little knowledge in chemistry .

 

[Smyth]

Drawing Chemical Structures

1. Always draw it in its low energy conformation to start with.
E.g. cylohexane and piperidine should be treated as chair structures and not drawn in the same way as benzene hexagons.

2. Play around with the structure to find the most aesthetically pleasing conformation. This will help you to develop an appreciation of the structure and convince yourself (over time), whether or not it is worth persuing.

3. The lowest energy structure does not have to be the active one.
E.g. In fentanyl the phenethyl chain is a bit like a ball on chain. Clearly its lowest energy conformation is with it just limply hanging. However molecules are not lifeless creatures and in actual fact it should be regarded as a sort of hook. If the active site of fentanyl is the top bit, what makes fentanyl potent as a mu agonist is the ability for the phenethyl chain to 'hook' around and mingle with the active part of fentanyl. It is this ball-in-glove arrangement that makes fentanyl active. Indeed, the piperidine nitrogen is like a shoulder and is not a functional component of fentanyl's activity.