Does methanol extraction avoid waxes?

How do you LLE with ethanol when it’s miscible with everything?

Dilute it silly

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This thread is confusing me, but I’m intruiged… Gonna run some heptane/methanol extraction Monday and see what happens.

It didn’t seem to affect my yields much at all with methanol. But like I said, I added a LOT of water. Like 200ml to 75ml of methanol.

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Ethanol becomes immiscible in the non-polar solvent once enough water is added.

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Hey, do you know if there’s still cannabinoids left in your methanol after you’ve been heptane washing it for a while?

“For awhile” what do u mean

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I have found almost No more cannabinoids in the hexane layer after 3 rinses with fresh 3% water methanol
But CBN thc ~ 1 % cbd 4% and CBN 10%
In ratio’s of the crude used
95% thc jumped
75% jumped of cbd
But 33% jumped of CBN
Thus My suspicion that cannabinoids each on their own have solvent prefrances
Wich i found very interesting

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He said he ran multiple heptane washes until the heptane came out clear on his insta.

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Wait, I think I misunderstood your earlier post. You’re saying you washed the cannabinoids out of the hexane with the methanol at only 3% water??? So all the cannabinoids were transferred from the hexane to the methanol.

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Yes :ok_hand:

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Interesting… I guess that makes sense given methanol’s cannabinoid solubility. Do you know what exactly stayed in the hexane?

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It’s just baffling to me that heptane is having a different reaction than hexane. But the implications might be huge for me :smiley:

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I think everyone is a little confused. LLE is usually not a black and white case of one solvent dominating the other. Rather, if you have some solvent system like hexane and methanol, where both are fair solvents for the target, there will be some partition coefficient between the two. Say you add 100 mL methanol and 100 mL hexane. Perhaps the partition is 6:4, so 60% of the THC is now in the methanol and 40% is in the hexane.

Well, you can go two ways. You can set aside the hexane and add fresh hexane and capture an additional 40%. Or you can set aside the methanol and add fresh methanol and capture an additional 60%. With enough repetitive washing, you can get more or less all the cannabinoids in either solvent you choose. There is nothing contradictory about someone extracting from hexane into methanol, and someone else extracting from methanol into hexane. One will be a little more efficient than the other, but if you use a large excess of solvent, even a weak solvent can extract from a strong one.

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All the alkanes will behave slightly differently in terms of what rhey will dissolve. Even though theyre all non polar

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So hypothetically

If I have a big ol vat of methanol semi-saturated with cannabanoids. Then added and vigorously stirred heptane at a 1:10 hep/meth ratio. How do you think the cannabinoids would concentrate into the heptane? Or would it likely just have less cannabinoids than the methanol?

@Kingofthekush420 had me wondering what was going on since the methanol and heptane were staying real nice and separated

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First of all I made that partition ratio up. I don’t know what it actually is. Maybe one of those papers people were citing have the actual number.

This article explains the math of how this works:

Interestingly, it works better to do multiple small extractions than one big one. So you will much better results extracting with 1 liter of hexane 10 times sequentially, than extracting with 10 liters once.

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Theyre inmiscible which is what you want

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If there is any way to do subscripts on this forum please tell me, because this is going to get ugly.

Example problem: suppose that the solubility of THC in methanol is twice that of hexane. We now extract 1 liter of methanol with a liter of hexane. After the first extraction:

C(meoh) = V(meOH) / (P * V(hex) + V(meOH)) = (1/(0.5*1 + 1)) = .66

So on the first wash, the hexane extracted 33% of the THC, and now 66% remains in the methanol. How many washes of this sort must we do to get 99% of the THC into hexane?

log(.01) = n * log(.66)
n = log(.01)/log(.66) ~ 11

Thus it takes 11 washes, or 11 liters of hexane to get 99% of the THC out of 1 liter of methanol, if methanol is twice as good a solvent as hexane. If we had tried to use all 11 liters at once in a single extraction, it would have been:

C(meoh) = V(meOH) / (P * V(hex) + V(meOH)) = (1/(0.5*11 + 1)) = .15

We would have only got 85%.

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I did like 8 to 10 x the ammount of heptane to methanol. My crude was dissolved 4:1 in methanol. Heptane is the best solvent for this, pentane youll need a ton more solvent. Pentane doesnt like to dissolve cannabinoids. Hexane doesnt like to seperate. Heptane wins IMO. You can use it with ethanol too

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