5.) 1H+13C NMR Spectra + Mass Analysis of 

PHOTOOXIDIZED DMT.

Sample:

The same Spice as from 2.). This time it was laying in DIRECT SUN outside for 2 days.

Actually I have to add that it was some time partially covered with a standard kitchen plastic foil as I was afraid of rain. But I think we may still say it was exposed to ~ 2 days of direct sun.

Now before showing the NMRs I have a new NMR Prediction. This time: N,N-Dimethyltryptamine-N-Oxide. So I did not know what could possibly happen while this sample is standing in the sun, but a fairly senseful reaction would be just the addition of an Oxygen out of O2 from the air onto that lone electron pair of the Dimetyhl-N.

Therefore this is the calculated spectrum of the N-Oxide, this time calculated with Chemdraw Ultra instead of Mestrenova (doesn’t look so nice, but its smaller in size):

Where comes the difference in the NMR between the standard molecule and the N-Oxide? I can tell you afterwards if you want. Now let’s have a look if this is indeed forming ...

Here is the 1H-NMR of that sample:

Observation:

Hell yes! Maybe it is not visible in the first sight, but watch your eyes, you can PERFECTLY see a smaller Set of

-    The BIG Methyl Signal

-    Both of the 2 CH2 Groups

with a shift to the left – as predicted. Well in this case the shift told by Mestrenova is not exactly what we see here, but that does not need to worry us, it’s just a prediction and this IS the N-Oxide.

And now what the REALLY interesting part is: There is NEARLY NO other stuff contained here, all other foreign signals were practially also visible in the unoxidized sample.

This means: Upon UV-exposure the N-Oxide Formation is the ONLY relevant reaction (or at least within 2 days).

Let’s do some calculations in the Conclusion section. Side info: it was wrong to normalize the aromatic Signals to 5 as they also contain that ones of the N-Oxide. I rather should have normalized the Dimethyl-Signals to 6. But that really doesn’t matter, as the important thing is the comparison of the DMT Signals to the N-Oxide Signals and for this its totally unimportant. =P

Here is the 13C-NMR of that sample:

Observation:

Well to be shorter here: We get the exact same thing. We have the regular DMT Signals, but we also have the SHIFTED Signals. One is shifted slightly to the right instead of left, but that should be fine. Also no relevant other Signals, except DMT and the N-Oxide. Cool!

Here is the LC-MS of that sample, where we should be able to also observe the formation of the N-Oxide. DMT itself has a molecular weight of M = 188,269 u so as like we saw in 4.) its peak will be at 189 u in the MS. As the N-Oxide has an additional Oxygen, which weighs 16 u, we will have 189 + 16 =205 u IF we can observe any N-Oxide. So let’s have a look.

Observation:

Wow! What a nice result. We got 2 very cool aspects here. The first: We perfectly see the DMT peak, which is by far the main peak in this complete spectrum. And directly beneath there is the Peak of the N-Oxide. If you take a look below: You can see in the spectrum of the positive charged Ions we see 189 for DMT and that other Peak has 205.

Side info: Even though DMT is less polar than the charged N-Oxide, it is eluted faster. The reason: In the eluent is also Trifluoro-acetic acid, which creates DMT*H+ and this compound is more polar overall than DMT+O-. This is the reason why the N-Oxide seems less polar than the unpolar Freebase.

Well let’s do some hardcore math: 205 – 189 = 16 !

Hooray, that other Peak is produced by a Substance which is DMT + 1 Oxygen ... N-Oxide!

And so we can say also based on this LC-MS:

Upon the exposure of UV-Light to DMT we NEARLY ONLY observe the formation of N-Oxide. No other reaction or any new products in a relevant amount are observed. (I know there are other peaks, but they are quite small and are possibly just the signal noise ...)

Conclusion:

Ok what can we say now? Based on the NMR + the MS we can identify the formation of the DMT-N-Oxide and nearly no other relevant products. Therefore I may conclude: Upon exposure of 2 days of direct sun DMT partially gets converted into the N-Oxide and not quite much else.

But what is the cool part now?! We can even do a really nice math, which I guess has not been done before. Calculating, how much DMT gets destroyed/converted within that time.

For this we just divide the Integral of the DMT-Signals with the Integral of the N-Oxide Signals. For this we should only use the Dimethyl-/CH2-Signals.

Here we get (1,05 / 8,96) * 100 = 11,72 %.

So now I can finally say: Upon UV Exposure of 2 days, 11,72 % of the DMT will be converted into the N-Oxide. = D

I’m doing a longterm investigation sooner to check if this is a continuating process or if it stopps after some time.