5) Step 5 - pulling with hot heptane

The solution was spiked with NaCl and brought to near boiling. 6 pulls were performed with generous amounts of heptane (using both the nice whirlwind via magnetic stirring at 1500 rpm and some manual swirling).

Most of the heptane was evaporated. The solvent never became cloudy, hinting that not much DMT would be found there.

Freeze-precipitation resulted in no crystals - just a thin layer of pale yellow "dust" that did not smell like freebase at all.

6) Conclusion

What a letdown Here are my top 3 guesses on where the loss may have occurred:

1. The 30% hydrogen peroxide over 5 days may have over-oxidized or polymerized the DMT beyond recovery.

2. The zinc reduction technique may have been wrong. Perhaps the zinc quickly reacted with the acid to produce zinc chloride and hydrogen, and never really acted on much of the dissolved compound?

3. The amount of zinc may have been insufficient.

At this point I need to give up and see if someone else gets more lucky.

Yes, I made sure the reduction was done in strongly acidic conditions (+50ml 2M HCl, pH ~ 1 tested with a pH paper) before the final basification and pulling.

I added H2O2 to freebase DMT once, to oxidize it. After evaporating it, what remained was inactive.

Can't remember the concentration, but it was probably 3%.

ahh ok. that makes sense

Technically its the other way around, hydrogenations are reduction reactions. Chemical reactions are described by the movement of electrons i.e. bonds, vs the movement or change in atoms or molecules.

Reductions as well as hydrogenations can certainly happen in alkaline conditions, acids provide a source of protons (H+), which is not the same thing as a source of hydrogen. H+. Acids are not a source of hydrogen but a source of protons. Protons are a more common form of hydrogen due to the low electronegativity of the atom (2.1), meaning it holds its electrons very weakly, and can stabilize a positive charge effectively due to it's small size and spherical orbital shape.

The only purpose of having protons present in the reduction, is so that after your substrate (harmaline) picks up the electron from the reducing agent (zinc) it will exist transiently as a highly basic charged carbanion, it will pick up a proton from the solvent to be neutralized, as long as it is capable of giving one. Any polar protic solvent is sufficient, methanol, ethanol, acetic acid, water, even alkaline water.

H+ is not a reducing agent but actually an oxidant. It possesses no electrons and is capable of gaining some, carrying a charge makes it favorable to do so. The result is a bond, usually form of a dative bond, accepting a pair of electrons from a negatively charged species into it's empty s orbital. This however is neither oxidation or reduction, since the electron count remains the same, it just participates in energy lowering interaction (hydrogen bonding).

When H+ gets reduced by zinc you generate H2 gas, and that is all the bubbling you will see during the reaction. This is actually a bad thing, because you want the zinc to react with your harmaline, not with protons. This is why you need a very large excess of zinc metal to actually complete the reaction. If you wanted to optimize this, you would change the conditions so that the redox potential of harmaline is as low as possible, and the redox potential of H+ is as high as possible. This can be altered by changing the pH or the solvent.

Dissolving metal reductions as this one are by the same principles in how batteries work. The electrolyte in batteries can be acidic, like your lead-acid car battery, or they can be basic, like your good ol' alkaline AA batteries.

Redox potential is directly related to electronegativity, and the electronic orbital structure of the elements.

http://hyperphysics.phy-...ase/Tables/electpot.html

The table is measured in relative difference in the reduction half reaction potentials

The reduction of H+ to generate molecular hydrogen is actually used as the universal reference, standard hydrogen electrode (SHE), and so is set as the reference point (0.00 V) the absolute is something like 4.44 V ircc

2H+(aq) + 2e- -> H2(g) 0.00 V

For zinc:

Zn2+(aq) + 2e- -> Zn(s) -0.76 V

These tables are always written in their reduction form (Gaining electrons), but since we use zinc as the reducing agent, we want to flip the equation so we see the redox potential when zinc is oxidized.

Zn(s) -> Zn2+(aq) + 2e- +0.76 V

Now if we add the two half reactions together, the reduction half from H+ and the oxidation half from zinc, it should be balanced. The electrons cancel out, and the charges all add up, we get:

2H+(aq) + 2e- -> H2(g) 0.00 V
Zn(s) -> Zn2+(aq) + 2e- +0.76 V

Zn(s) + 2H+(aq) -> Zn2+(aq) + H2(g) +0.76 V

So zinc metal reacts with protons to generate H2 gas and +0.76 volts, and so a spontaneous process. This is essentially a battery. The protons/acid (H+) are acting as the electron acceptor here, but any other ion, molecule, could act as well. The protons serve other purposes as well, since zinc can react with oxygen and water in the air to form Zn(OH)2 as a passivating layer. The acid with appropriate counter ion can remove the passivating layer to form soluble zinc salts (ZnCl2, ZnSO4, Zn(OAc)2), so that fresh zinc metal is exposed and continue reacting.

A good example is to use another metal that possesses a lower redox potential than zinc, like copper or silver, which gives an additive voltage output. this is a zinc-copper battery you can make for kids, and also this is what is commonly done in gold refining to precipitate or dissolve specific metals. These redox reactions are leaving out the counter-ions, which are important for solubility and side reactions.


half rxns
red: Cu2+(aq) + 2e- -> Cu(s) +0.34 V
ox: Zn(s) -> Zn2+(aq) + 2e- +0.76 V

redox: Zn(s) + Cu2+(aq) -> Zn2+(aq) + Cu(s) +1.10


Enough tangents,
Now to reiterate, and get back to the original point.

You can do hydrogenations in alkaline conditions, the half rxn reducion for water is :

2H2O(l) + 2e- -> H2(g) + 2OH-(aq) -0.83 V

This is electrolysis of water, the addition of 2 electrons generates H2(g) and hydroxide ions. it costs more energy than the reduction of H+. Which makes sense, H+ is more electrophilic. If we disregard the problem of passivation, combing the half reactions we get:


red: 2H2O(l) + 2e- -> H2(g) + 2OH-(aq) -0.83 V
ox: Zn(s) -> Zn2+(aq) + 2e- +0.76 V
redox: 2H2O(l) + Zn(s) -> H2(g) + 2OH-(aq) Zn2+(aq) -0.07 V


So this shows that the reaction actually requires a voltage input, neutral or alkaline conditions significantly lower the redox potential of H+, because it is more tightly bound to water molecules. So if passivation and Zn(OH)2 weren't problems of themselves, then raising the pH might actually be more efficient.

But again, no data here on the redox potential of harmalas, but there are some papers out there. We do know that the redox potential of zinc (0.76 V) is high enough to reduce harmaline, but not harmine.

Aluminum is a pretty damn powerful reducing agent, double the potential of zinc at 1.66 V

Al(s) -> Al3+(aq) + 3e- +1.66 V

It however suffers the same problems as zinc, it is even more susceptible to passivization, and the hydroxide salts (Al(OH)3) are god awful. It will react with protons to generate H2 much much faster than it will react with harmalas. Mercury amalgam is toxic and shuold be avoided, but historically the most effective method to use aluminum in organic reductions/hydrogenations.

For example, one of the oldest old school reductions is using sodium metal dissolved in some alcohol.
Very powerful electron donor, with a standard potential of 2.71V. It was also the first material used to reduce harmalas, and it is capable of reducing both harmaline and harmine to THH.

Na(s)-> Na(s) + e- +2.71 V

In this example, acidic conditions are not necessary at all, and the reaction conditions are extremely alkaline. It however suffers the same problem, it will preferentially react protons before it reacts with your substrate. This can be controlled by keeping a lower temperature, and choosing a suitable protic solvent. Especially with pure sodium, aqueous conditions are to be avoided entirely. Sodium will abstract a proton from water to make hydroxide, or from alcohols to make alkoxide.

Na(s) + EtOH -> Na+ + EtO- + H2

To reiterate again, the idea is to strike a balance between having enough protons that they can be incorporated into the reduced THH (hydrogenation) but not so much that the reducing agent just reacts with the protons alone to generate H2. Excess almost always needs to be used. The side reactions can be inhibited by using a bulky alcohol, like isopropanol, or tert amyl alchol, this steric bulk makes it challenging to abstract the proton. And the other option is by creating a mercury sodium amalgam. You can make a 1-10% w/w alloy of sodium and mercury, which reduces the surface area and the reactivity of the sodium. But as you can imagine, it uses a hell of a lot of mercury. This is 1900's chemistry, better methods exist. Dissolving metals is old school.


Aside from dissolving metals as electron source, Hydrogen as a reducing agent requires a source of molecular hydrogen, hydride, or a hydrogen radical. The hydrogen atoms must be bonded to elements it possess electrons from molecules in which it is bound to having a lower electronegativity than hydrogen, i.e. aluminum (1.5) or boron (2.0). In these molecules electron density is concentrated around H in the Al-H bonds. However the mechanism in how these reducing agents is different, and related to the structure of the group 12 elements, i.e. the empty p-orbital. AlH3 for example, possesses a Lewis acid capability (just like a H+), which can activate the substrate to facilitate hydrogen transfer. These are infact, much better agents for hydrogenation or reduction. They are highly selective to certain functional groups, double bonds, ketones, etc. Excess does not need to be used, and most of the time you can use just 1 molar equivalent. Protic solvents are not necessary, but can be used without problem in many cases.

It all depends on redox potential. For instance sodium metal is a powerful reducing agent, but sodium hydride is not. So even though you have an extremely low electronegativity of sodium (0.9) bound to hydrogen, providing an extremely polarize bond, ionic interaction with electrons localized on hydrogen.
NaH -> Na+ + H-
you end up with powerful hydride source. However, since sodium has already been ionized, it actually has very little redox potential.

So you've got your hydride, but you have no driving force to move those electrons into your desired molecule. In addition, hydride is also poorly nucleophilic, so hydride tends to act as a selective and strong base. Which is useful in its own right. It will violently react with water or other acids (Proton source H+) to form hydrogen gas and a sodium salt. It has a pKa of about 35, so you can deprotonate things like amides, pyrroles or indoles. So you are doing a reduction, reducing protons to make H2. But not really useful for hydrogenation or other types of reduction on small organic molecules.

This is a really great read. Bringing the focus back to the mysterious substance of DMT-Oxide, here are some claims I am aware of:

1) One can convert DMT to DMT-Oxide by dissolving DMT in something (for example, ethanol) and adding hydrogen peroxide.
2) DMT-Oxide is as visual as DMT is, but it causes none of the mind-shattering, panic-inducing drama (so it has great recreational potential)
3) DMT-Oxide is a yellow (orange, amber) oil that would not crystallize. In order to weigh predictable dozes for smoking, one needs to mix 1 part of DMT-Oxide with 9 parts of calcium sulfate or another inert powder.
4) DMT-Oxide can be converted back to DMT freebase by mixing it with a large excess of zinc and either acetic or hydrochloric acid (the latter being faster-acting) for 2-3 hours.
5) Nobody knows what the DMT-Oxide formula is. The theories I have heard so far are: a) it is DMT with an extra atom of oxygen bonded with the amine nitrogen b) same, but the oxygen atom is bonded with the indolic nitrogen, c) it is only a change in the oxidation state of the indolic nitrogen.

I am really curious to understand the composition, and study the chemistry, of DMT-Oxide, especially if it is indeed psychoactive on its own, and if it is indeed something that DMT can naturally degrade to.

There are 2 open problems that I see:

1) Is there a fast and reliable way to obtain DMT-Oxide from DMT, without any cyclizations, polymerizations, or any other unwanted chemical side effects? In particular, does dissolving DMT freebase in ethanol and then adding 30% H2O2 accomplish that goal? Can the ethanol be replaced with IPA or vinegar? Also, can the peroxide be evaporated safely without becoming over-concentrated and damaging the organic molecules? If not, should/could it be decomposed by adding a catalyst like NaI, KI, MnO2, Fe2Cl3, etc. and does that catalytic decomposition require non-acidic conditions?

I am betting that vinegar + peroxide, followed by NaOH and KI would work just fine.

2) Is there a quick and reliable way to convert DMT-Oxide back to DMT freebase? Do the conditions need to be acidic - and if so, is acetic acid (as a paper linked to from this site's wiki suggests) the right medium, or is hydrochloric acid (recommended on this forum as faster-acting) any better? How much zinc and acid is needed? If a large excess of zinc is used, should new acid be added to keep the pH low? Should the zinc be added gradually, or all at once? Should the DMT-Oxide solution be concentrated (to keep the DMT-Oxide close to the zinc atoms) or dilute? Is mixing going to help? Does the temperature matter?

This one strikes me as a finicky and unreliable procedure that will be hard to get right, and hard to recommend.

benzyme, if you have any spare DMT freebase lying around, would it be possible to dissolve it in 1 part of ethanol, IPA or vinegar, add 3 parts of 30% peroxide, give it a few days to act, either decompose the peroxide with KI or MnO2 (needs pH > =7) / Fe2Cl3 (any pH) or simply evaporate it, and get a MS to see what the product is made of?

Again, amine oxides are well known in the chemical literature. You can even find an example of DMT N-oxide and bufotenine N-Oxide (see attached paper). They are formed with tertiary amines, so in the case of DMT you have the nitrogen on the tail oxidized to form a nitrogen ylide. It doesn't effect the conjugation of the indole ring, DMT N-oxide is colourless when pure.

I think you are confusing speculation about DMT oxidation products and DMT N-Oxide. For instance, as DMT ages and becomes yellow/orange, its speculated that is the result of oxidation about the indole ring and around the nitrogen. The reason for this is indoles commonly oxidize under oxygen to give quinones, which are often intensely coloured blues, reds or oranges, isatins are another example. Psilocin easily oxidizes into a blue quinone due to it's hydroxy substituent on the 4 position, oxidation is simple due to resonance.

Im not sure since im not up to date with the threads but maybe some confusion around it has been due to lack of detection of these products by mass spec, which is a incredibly sensitive instrument and should detect the addition of oxygen. So some speculation that there is an oxidation elimination going on at the indole nitrogen to give an isomer with the same mass. I think products with the addition of oxygen are still likely, since they are intensely coloured only traces are needed to cause a discolouration. Aromatic amines are notorious for this, in some cases the reason is dimers or polymers

Heterocycles like pyridine can form N-oxides, because the nitrogen is again fully substituted (double bond and aromatic), pyrrolidines and indoles are not conjugated this way, so they cannot. Only way you could get something like this is if some other oxidation happens first on the 2 position of the indole, a radical process with oxygen for example. That would push arrows from the indole nitrogen lone pair and give you an iminium ion, but still that lacks the lone pair necessary to form the amine oxide. I think its likely a radical oxidation like this is happening and giving one of these indole quinone type products, or dimerization, but ill wait until I see this confirmed by mass spec.





about 10 years ago i tried this experiment, just adding some 30% H2O2 to freebase DMT dissolved in some ethanol. Preparation of amine oxides from tertiary amines this way from H2O2 is a well known reaction and free of any major side reactions, you should be able to find examples online of other tertiary amines.

If without ethanol IPA should work the same, I might have even used IPA I don't remember. I made a thread on the forum should be able to find it. Don't worry about decomposition of the peroxide, the workup is simple. It will stay in the water layer, just dilute it with at least 10x volume of water (so now its <3%) and then extract out the amine oxide with DCM or chloroform. Any traces of peroxide will have decomposed during evaporation, water and oxygen. Best to also dry the extracts with MgSO4 or NaSO4 and filter, but not entirely necessary, especially working at a small scale.

Amine oxides are very polar so you need a relatively polar extraction solvent. I ran it all in a narrow test tube so it was really easy to just suck up the organic layer with a pipette and then evaporate it. It's not soluble in naphtha or hexanes, which is also a helpful indicator that the reaction worked. It is also a useful way to remove any unreacted DMT, by washing the oil well with hot hexanes. I mixed it well with hexanes, heated it near boiling with lots of stirring, decanted off the hot hexanes and saved it for later to recover any unreacted DMT. Hardly anything was recovered, so seemed the reaction went to completion in high yield.

The product I recovered was a colourless to pale yellow oil, which did not crystallize. I didn't have any analytical equipment to test it at the time (was just out of highschool) Don't remember if I did TLC, would have been a good indicator.



No this is misinformation, although the combination of the two probably has some modified effect, albeit subtle, and the DMT will overpower it all if enough is taken. Similarly combination of NMT and DMT can give some changes, while pure NMT on its own is inactive.

IME, Vaporization by the usual method of 50mg gave no effect resembling DMT. It did give some mild psychedelic stoned feeling. It had a tryptamine vibe to it too, and tryptamine like CEV, so you could call that similar to DMT, but not really doing it justice IMO. It had some visuals but more closely with a low dose of a psychedelic or the visuals from smoking cannabis, not remotely close to the visuals that DMT produces. I think the mixed reports are from people who smoked what they thought was DMT-oxide but instead they were just smoking mostly DMT. DMT is intensely visual.



The main problem I see is solubility, DMT N-oxide is insoluble in water, so Zn/HCl doesn't work very well, since all the zinc reacts with the acid before it can react with the n-oxide. I would experiment with other solvents, like methanol, or search the lit for a similar reduction procedure and use their conditions.

Does this also mean decocting plant material in acidic aqueous solutions doesn't extract the DMT n-oxide well?

If that's the case, it would explain why wasn't found in the brews in this study.

Mindlusion, thank you very much for the comprehensive response.

This wiki claims DMT-N-Oxide is soluble in water. Do you believe it is almost completely insoluble in water (despite being very polar?)

Yes the wiki appears to be incorrect or at least imprecise. DMT N-oxide is partially soluble in water, but only slightly more soluble than DMT freebase in water.
Even in the strongly acidic Zn/HCl water, hardly any DMT N-oxide dissolved. That method could be improved with addition of a co-solvent or phase-transfer catalyst.
Glacial acetic acid and zinc would definitely work, but 100% acetic acid not as OTC. Vinegar can be used as well but I imagine it will suffer the same solubility problem. If the reaction is run slowly enough I am sure it would work despite the low solubility, but requires patience and maybe a lot of zinc.

found the thread on my original experiment: https://www.dmt-nexus.me...aspx?g=posts&t=23362

I'm wondering here about the choice of KI as a means of destroying excess peroxide.

ML has already outlined it as unnecessary and, considering it produces elemental iodine, in my view this is asking for side-reactions, particularly the production of 2-iodo-DMT. And how do we know if iodide doesn't reduce amine oxides? Trimethylamine oxide (TMAO) is used as an oxidising agent in the formation of alcohols from alkylboranes following hydroboration, for example.

It's certainly conceivable that TMAO might be capable of oxidising iodide, and DMT-N'-oxide is effectively the indol-3-ylmethyl derivative of TMAO.

ML's way of not dealing with concentrated peroxide is to wash the unreacted freebase out with heptane and then retrieve the N-oxide with DCM. Perhaps it is the best way (second only to evaporating the peroxide if it happens to be OK). Here the idea was to keep everything inside the same container for as long as possible, to avoid losses due to handling.

The amount of KI added was catalytic - about 15mg. The molar ratio of KI to DMT was less than 0.5%. I assume that most of the DMT should have survived. HCl does not react with KI, and NaOH does not complicate matters either. Iodine is soluble in heptane but its solubility does not decrease much with temperature (here) so it is unlikely to freeze-precipitate in the end. I did not think of 2-I-DMT.

MnO2 was considered for the decomposition catalyst but it only seems available in the fine powder form, requiring filtration to avoid emulsions. A fun way to destroy the peroxide would be to put it under a high-intensity 254nm light, but reading this paper made me wonder if the decomposition would still take too long. In hindsight, Fe2Cl3 could work, and it does not require non-acidic conditions. Not sure why I did not try it, maybe because it is nasty and corrosive.

...Has 2-iodo-DMT been studied? Is there a chance it is psychoactive? I see this, but not much else pops up.

Thanks for the clarification. Incidentally, you can make supported MnO2 by dumping activated charcoal granules into KMnO4 solution. In a procedure described by Carpino, 24g of potassium permanganate was dissolved in 300mL of boiling water. Over the course of 10 minutes, 7.5g of activated carbon [powder] was added cautiously in order to avoid excessive frothing. The supported manganese dioxide was then filtered off, washed, etc. This form of activated MnO2 is very easy to handle.

That's a good paper you linked to. Of particular interest:


2-I-DMT may even block the effects of DMT.

Thanks! Sounds incredibly easy, and it may generalize to other substances.
Activated charcoal could absorb some freebase and/or N-oxide, but I'd take that over side reactions.

If this doesn't work with acetic acid, shouldn't the tek on the Wiki be updated?

I'm doing cold vinegar MHRB soaks on some old bark with potentially DMT-n-oxide in it. After the bark is spent in that way (i.e. most n,n-DMT removed from it), I'm thinking of the following experiment:

- Basify the 'spent' bark with Ca(OH)2
- Evaporate the water
- Pull with IPA

Then, potentially a zinc reduction could be attempted on the IPA pulls.
Alternatively, FASI could be used to precipitate DMT-n-oxide from the IPA. Or the IPA could be evaporated and the result, presumably mostly DMT-n-oxide, bioassayed.

Does any of this sound reasonable?