Thank me later Δ8THC via ptsa

Hempire Building Delta 8
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In every CBD → THC setup, the fundamentals are always the same. The yield of D8 or D9 is a function of the reaction conditions.

Δ9 THC is the kinetic product
Δ8 THC is the thermodynamic product

cctc201901589-fig-0002-m
cctc201901589-fig-0002-m646×622 43 KB

Under kinetic conditions you can imagine your CBD molecules as a lazy hiking group. They start in the valley of CBD, looking left and right on the hills TS1 and TS2 and decide to climb the smaller hill TS1, down into the valley of Δ9. This is not the deepest valley, but yeah the lazy boys don’t really care.

If your setup is under thermodynamic conditions, your hiking group is diligent. They will start to climb both hills to look which valley is the deepest. Once they know which valley is the deepest, they will go there and stay overnight. Because it is well known that the best buds are in the deepest valleys.

If your setup has energy in abundance, you will have a hiking group on meth. They will go everywhere all the time, restless and without sleep.

And a bit more scientific:

With low energy in your system, the average energy of the molecules is low and more CBD molecules have enough sufficient energy to cross activation energy Ea1 than Ea2. Therefore the reaction preferentially proceeds along the blue path to Δ9 THC. The reaction is not reversible since the molecules lack sufficient energy to reverse to CBD.

With some slightly more energy, reaction 1 will become reversible when sufficient molecules have enough energy to cross the reverse reaction barrier for reaction 1, while reaction 2 remains irreversible. So although Δ9 THC may form initially, over time it will revert to CBD and react to give the more stable Δ8 THC.

At high levels of energy, both reaction 1 and 2 are reversible and the product ratio of the reaction is dictated by the equilibrium constants for D8 and D9.

After 5 month of studying this reaction, I find this model explains the best what is going on inside our flasks/vessels. This has cost a lot of headaches.
Hope this helps with your problem of Δ8 going back to Δ9. Slightly reducing the energy should be the key. I am working on pathway 1 (Δ9 THC), so this is just a guess based on this theory.

Knowledge is the only thing that grows when shared

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I don’t think cbd is intermediate between d8 and d9. I’m also note sure if both derive directly from CBD. But the transition between d8 and d9 may be direct. Moreover, they seem to form an equilibrium with two other iso d8 isomers.

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You aren’t staining to visualize the spots? Or is that stained with Iodine?

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I am not sure that is the way to think about it?
we are talking about reactions concerning catalyst…

are you proposing that a single catalyst lowers the energy of activation
to two different levels? You can propose that the energy of activation
is different for CBD to D9 vs CBD to D8…but how do you know?
Certainly specific catalyst plus CBD yields a mixture of both and other isomers…but the reaction pathways are questionable. Catalysts by their very nature can be stereoselective… and therefore might complicate
interpretation of multi pathway product composition.
Since catalyst only lowers the energy a activation, can the isomerization
be carried out by heat alone? Perhaps use a Parr reactor allowing
pressure containment with heating. We know from ubiqutous distillation
proceedures that CBD is very happy in vapor state / inert gas/ partial vacuum at or above boiling temp. Moreover, we do not see rearrangements in the injection port or on column in GC.
But if one were to explore the higher temp range in a ration manner
you might observe the apperance of D9 or D8 first with out catalyst.

perhaps someone doing GC-MS can just turn up the injection port temp
a few hundred degrees (!) and let us know …hoping the products are stable? The observation might give a frame work to think in.
Sort of a “whose on first” question.

Is there not an entire science of head space analysis as you ramp the
temp to decomposition…and also analyze the decomp products?

I am not sure for all cases but “thermodynamic” stabilization
often involves entropic considerations, not necessarily delta H.

continued curiosity…

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mk?
the spots? do you mean the ones posted by
someone else? TLC plate of Phosphoric rxn?

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I am trying to understand the fundamentals, otherwise it will just be try & error and with an approach like this the only thing I need is luck. I am pleased about everyone who is helping to understand the process.

So far we have a lot of theories and great ideas on this board from forward thinkers. But none of those ideas really matched my experiences (with pTSA, phosphoric acid and ZnCl2)
I started to dive into this theory and found this paper. This is another context, but they labeled d8 as thermodynamic and d9 as kinetic product, too. (Synthetic pathways to tetrahydrocannabinol (THC): an overview - Organic & Biomolecular Chemistry (RSC Publishing) DOI:10.1039/D0OB00464B)

For example, phosphoric acid. Our observations can be explained with this model.
If you pour phosphoric acid into boiling solvent with CBD. Within a few minutes you will reach your maximum of D9 (kinetic product, which forms the fastest). Since your system has energy in excess, your pathway 1 is reversible and therefore your D9 will start to decrease over time. This is not like a gear box where you shift from gear 1 into gear 2. This happens simultaneity. That explains why your timing is so important on regular reflux setups when you aim for high d9. At the same time it is nearly impossible to reach very high d9 numbers in this setup.

Rogue is playing around with pressure and made some amazing yields with it. Why? This can be explained with this model, too. He is raising the boiling point and therefore increasing the energy on this setup. This means: pathway 1 is for sure reversible, but it also increases the speed of the kinetic path. This way you are able to convert the biggest part of CBD into D9, before D8 is formed in big amounts.
My TLC plate above is from a 8 bar pressure run, too.

@moronnabis Thanks for pointing out stereoselectivity! This was not on my radar. So this is the explanation, why our results vary when switching solvents and/or catalysts. It has an influence on the difference between the activation energies for Δ8 and Δ9. So you have solvent/catalyst mix A that has a bigger ΔEa than solvent/catalyst mix B. I do believe that all Δ8/Δ9 catalyst can be either turned into high Δ8 or high Δ9, based on setup conditions. I can confirm that you are able to produce high Δ9 AND high Δ8 with pTSA/toluene, but high Δ9 is much more challenging for me. For sure there are catalyst that perform better or worse and ND Δ9/Δ8 won’t be achievable with many.

Does this maybe also relate to the formation of isomers?
Compare the results of pTSA with different solvents at RT in this paper.
Hexane is the only solvent, which led to the formation of Δ8-iso-THC.

So there seem to be other pathways, which get influenced by the choice of solvent/catalyst.

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Sorry, my bad. Wrong reply button =)

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You aren’t staining to visualize the spots? Or is that stained with Iodine? Thx

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Just iodine in a jar :slight_smile:

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I think this is where you get it wrong.
The term “kinetic-” and “thermodynamic-” products are in fact a simplification of language.
Thermodynamics tell you where the equilibrium is, and kinetics tell you how fast it gets there.
The parameters of the reaction (temperature, pressure, pH, relative concentrations…) can have an influence on both aspects.

D9 is kinetically favored, it forms faster than d8. One can see d9 increasing concomitently as CBD decreases. D9 reaches a maximum, and then decreases. But CBD keep on decreasing in parallel. D9 is actually further converted (mainly) to d8.

One question is whether if d8 can be formed from d9 or if it can it be form from CBD as well.
My take is that if forms only from d9.

The main driving energy is actually brought by the equilibrium created when one mix CBD with the catalyst. From there, the system will react anyway until it reaches the thermodynamic equilibrium back. If one brings heat to this system, one accelerates all the steps. D9 forms faster, but so does d8. The point where this equilibrium is actually depends on the parameter, but also strongly on the nature of the catalyst.

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Beware. The results shown in this paper should be taken with quite some circumspection.
The underlying analytics does not seem really on point. All the described solvents give similar isomers, just in different proportions or a different rates.

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How would you optimize for D8 in the reaction? Keep heating until it goes to completion and hope it is favorable toward a high D8 percentage?

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According to all (consistent) reports here, this reaction never goes to 100% conversion of d9 to d8.
Heat brings you faster to high d8, but also favors other isomers.
The highest number seems to be reached at low temperature, but longer reaction time.

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:call_me_hand:t4::call_me_hand:t4::call_me_hand:t4:

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Just a reminder for preparation on these reactions & steps to help with higher successful conversion rates. If anyone has any questions I’m always available. :call_me_hand:t3::call_me_hand:t3:

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I think this is definitely the case. For the intermediate of d8-CBD is never found.

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Interesting point since none of us have identified d8-CBD on a chromatograph. Has anyone attempted to make d8-CBD to test this? I think this could give us an opportunity to understand a possible intermediate and see how it behaves under analytics.

On a side note… doesn’t CBD obey different IUPAC rules so it wouldn’t technically be called d8-CBD? More like d6-CBD?

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Probably a stupid question, why is iso always starting material and not a broad spec distillate?

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Cause with a more pure input you won’t have as many side reactions or a low TAC overall. Broad spectrum will always create different peaks & unknowns as it will have “impurities”. Hope that helps clarify

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Start reading here:

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