• N&PD Moderators: Skorpio | thegreenhand

Metabolisation of smoked methamphetamine hydrochloride (Ice) and ambiguous scientific research into this subject.

Of course methamphetamine being a base (pKa~10.4) at normal physiological pH (7.4) the equilibrium is shifted to the left. So, most of it exists in the plasma in the form of protonated METH (with the counterion being chloride or other anions in the body). Actually pretty much all of it is in the form of protonated: for 1 non-protonated (free base) molecule you’ll have ~1000 molecules in protonated form (3 log units difference between pKa of meth and plasma pH). So essentially there is pretty much no free base METH ("pure METH") once absorbed. This is true for the free base as well. Once absorbed (by eating or vaping) it will quickly get converted to the protonated form. That’s why you’ll see scientists referring to METH HCl (pure crystal) and METH only once in solution. Correct way will be to call it methamphetaminium (protonated form of methamphetamine with Cl counterion or others anions in the body like phosphate, sulfate, carboxylate...etc but mostly Cl as it is the most abundant).

As for binding/blocking of the Monoamines transporters in the brain which is the psychoactivity mechanism of stims, it is the protonated METH that binds and blocks them as you could imagine. Else, if that was the free base ("pure" METH) that is psychoactive, then you’ll have to smoke ~1000x more METH to feel the same effects (ie, you’ll need smoke 50grams of meth!!!!!??? to feel effects of 50 mg). And that is because concentration of free base is so low relative to the salt in the body/brain. Here is a pic of METH (protonated form) bound to the dopamine transporter DAT (notice the counterion is now a carboxylate part of the DAT protein: protonoted METH in green with NH2+ in blue and DAT carboxylate COO- in blue-red). Same things with the other 2 (NET and SERT).

Eh, maybe I missed something here, but I'm not quite sure how you arrived at the notion that there's almost no free based meth in serum once ingested, or that you'd need 50g of meth for the free form to be psychoactive?

Given the pKa of meth is 10.1 and normal blood pH around 7.4, it's pretty easy to calculate that around 6.3% of what's ingested should thus be expected to exist in its non-ionised form in serum at any given moment. That will vary somewhat over time as it distributes around the body (eg parts of the CNS, a few organs, lipid tissue and so on, which can maintain higher concentrations), but it's a pretty useful general indicator and not in any way insignificant.

As for the protonated forms being the main psychoactive ones, I think that's a bit of a stretch given they're not exactly lipophilic. How are they supposed to even reach CNS binding sites like that?
 
Eh, maybe I missed something here, but I'm not quite sure how you arrived at the notion that there's almost no free based meth in serum once ingested, or that you'd need 50g of meth for the free form to be psychoactive?

Given the pKa of meth is 10.1 and normal blood pH around 7.4, it's pretty easy to calculate that around 6.3% of what's ingested should thus be expected to exist in its non-ionised form in serum at any given moment. That will vary somewhat over time as it distributes around the body (eg parts of the CNS, a few organs, lipid tissue and so on, which can maintain higher concentrations), but it's a pretty useful general indicator and not in any way insignificant.
Hendersen-Hasselback equation..check it out on wiki: To get the species distribution of a base B (free v protonated) with a pKa at a given pH:
the ratio protonated BH+ versus free base B = 10^(pKa-pH).
So if pKa=pH, ratio is 10^0=1 ie 50:50 BH+ versus free base B
if pKa is one unit above pH, ratio is 10:1 (ie 10^(1))
2 units above ratio is 100:1; 3 units above ratio is 1000:1...etc etc.

pKa (meth) = 10.1, pH (plasma) = 7.4, difference = 2.7, ratio METH+ v METH free base = 10^2.7 = 500 in other words you’ll have 500 ionized METH for 1 free base METH in blood (pH7.4 ) or 0.2% free base and 99.80% protonated or ~2 free base for 1000 charged...0.2% not 6.3%! how do you come up with 6.3%? (note earlier I used predicted METH pKa of 10.4 for simplicity..kind of lazy to go get experimental value) but it doesn’t change the point!
There are bunch of freeware where you just plug in pKa and pH it will give you % ratio of free v protonated!

.. As for the protonated forms being the main psychoactive ones, I think that's a bit of a stretch given they're not exactly lipophilic. How are they supposed to even reach CNS binding sites like that?
No it is not.. so far as the psychoactivity is mediated by interaction with Dopamine/NET/SERT/VMAT transporters, the free base doesn’t interact with the target transporters (DAT, NET SERT VMAT) but the ionized, charged form of the drug (METH-Hydrochloride for a lack of better way to put it) does. Here is a picture of methamphetamine (in green-blue-white) being transported/blocking the DA transporter:

{have no idea how to post images on here..used to be simpler ??} but here the links:


The transporters used a carboxylate group (negatively charged) to recognize the substrate (either natural dopamine or METH, AMPH, cocaine..etc). You can easily see that the bound METH molecule is charged positively and the ionized amine group (NH2+ in blue-white) interacting with a DAT carboxylate (COO- in red-blue). It makes sense actually: if the free base is the one interacting with the transporter and thus psychoactive, then any neutral molecule will be as good as METH for example replace METH nitrogen with a carbon or oxygen or sulfur which are neutral at pH7.4, it will give you stims as potent as METH. But that is not the case. Else, I’d be extremely rich!!!

That’s why I am saying, if the free base was the psychoactive form, then you’d have to use 500 times the dose (eg 50 mg x 500= 25 grams! to get effects of 50 mg) simply because the concentration of free base is 500X lower than the ionized form (assuming same affinity to target transporters)...Good’day everybody
 
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