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Distilling hexane during mini a/b?

chr0nic

Esteemed member
Hello chemists,

I have been doing some theroy research on distillation, and had a question if someone wouldn't mind indulging me!

Say you ran a mini acid/base, and the final step left you with your two layers... one had DMT freebase in the NPS (in this example I'm going to say hexane), the other layer being basified water (in this case, say it is sodium hydroxide). The goodies are currently in the hexane, and we're at the point where one would use a separatory funnel or pippette to perform extraction.

If you were to run a basic distillation on the 'soup' at this point, what would happen here?

Hexane has a boiling point of approximately 70c, so that would begin to vaporise and move across... does it "drop" the DMT? I assume it wouldn't carry across, does the freebase migrate back into the basified water?

I chose hexane as an example due to its low boiling point relative to water, sodium hydroxide, and DMT inself.

I love learning, and this puzzled me. I don't have a chemistry background, so would be grateful for the knowledge. Thanks for helping with my curiosity!
 
Hello chemists,

I have been doing some theroy research on distillation, and had a question if someone wouldn't mind indulging me!

Say you ran a mini acid/base, and the final step left you with your two layers... one had DMT freebase in the NPS (in this example I'm going to say hexane), the other layer being basified water (in this case, say it is sodium hydroxide). The goodies are currently in the hexane, and we're at the point where one would use a separatory funnel or pippette to perform extraction.

If you were to run a basic distillation on the 'soup' at this point, what would happen here?

Hexane has a boiling point of approximately 70c, so that would begin to vaporise and move across... does it "drop" the DMT? I assume it wouldn't carry across, does the freebase migrate back into the basified water?

I chose hexane as an example due to its low boiling point relative to water, sodium hydroxide, and DMT inself.

I love learning, and this puzzled me. I don't have a chemistry background, so would be grateful for the knowledge. Thanks for helping with my curiosity!
DMT will not co-distill with hexane, but it might with much heavier liquid hydrocarbons like hexadecane. @1ce has attempted co-distillation with cyclohexanol among other things some while back.
 
I have a related question. Could some cases of smelly DMT be from molecules being carried into the air with the volatile solvent?

I have this jar of gooey DMT that smelled so very strongly shortly after the solvent appeared to evaporate. But upon microscope inspection the crystals are coated with a liquid that is not drying after weeks. At first I thought solvent traces, but maybe it's just the 'goo' form of DMT. Still, I wonder if there is solvent trapped in the goo, and if the solvent vapor pressure is increasing the vapor pressure of DMT? if not physically carrying DMT molecules into the air.

Or what does cause a stronger smell in some cases? For my jar, the smell has gone down significantly, but there seemed to be a peak where it no longer smelled like solvent but smelled very strongly of what is typically associated with the scent of DMT. The best crystals I have don't smell at all though so I'm confused if DMT is naturally scentless and the typical smell is related to something else?
 
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@chr0nic check this thread out
 
@chr0nic check this thread out

Heya, that's pretty cool! Made for a good read! Thanks for linking me in 😊

DMT will not co-distill with hexane, but it might with much heavier liquid hydrocarbons like hexadecane. @1ce has attempted co-distillation with cyclohexanol among other things some while back.

Thanks, Tranform! If DMT will not co-distill with hexane, what will happen to it if the solvent reaches vaporisation boiling point whilst holding the DMT?
 
Thanks, Tranform! If DMT will not co-distill with hexane, what will happen to it if the solvent reaches vaporisation boiling point whilst holding the DMT?
Theory indicates that the vapour pressure of DMT at that temperature (~69°C in the case of hexane) will add to the vapour pressure of the hexane, meaning the mole fraction of DMT - assuming ideal behaviour - will be equal to the ratio of its vapour pressure to the sum of the vapour pressures of DMT and hexane, or as an equation:
QDMT = PDMT/(PDMT+Phexane)
In reality, one or other of the components may decrease the vapour pressure of the other. I'll spare the details on that for now. The upshot is the amount of DMT that co-distills with hexane will be more than zero, but not necessarily anything like a useful amount.

Another couple of things that affect the amount of carry-over are the degree of ebullition (harder boiling carries DMT in spray droplets from vigorous bubbling) and the distillation path length, whereby a longer path (such as a fractionation column) will cause DMT to recondense and drip back into the boiling flask. In this latter case, a full separation of hexane from DMT should occur.
 
Theory indicates that the vapour pressure of DMT at that temperature (~69°C in the case of hexane) will add to the vapour pressure of the hexane, meaning the mole fraction of DMT - assuming ideal behaviour - will be equal to the ratio of its vapour pressure to the sum of the vapour pressures of DMT and hexane, or as an equation:
QDMT = PDMT/(PDMT+Phexane)
In reality, one or other of the components may decrease the vapour pressure of the other. I'll spare the details on that for now. The upshot is the amount of DMT that co-distills with hexane will be more than zero, but not necessarily anything like a useful amount.

Another couple of things that affect the amount of carry-over are the degree of ebullition (harder boiling carries DMT in spray droplets from vigorous bubbling) and the distillation path length, whereby a longer path (such as a fractionation column) will cause DMT to recondense and drip back into the boiling flask. In this latter case, a full separation of hexane from DMT should occur.
Very interesting! Thanks a lot for taking the time- the formula is beyond my understanding, but I do get what is happening conceptually. Cheers ☺️
 
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