Hexane has a fairly high boiling point, so evaporating hexane is far away from a pleasure.
The problem is that without completely evaporating the hexane, I cannot know the exact amount of free base that was isolated.
Fortunately, a university course in physics, or chemistry, or something else, where Raoult laws were studied, comes to the rescue.
Knowing the ebullioscopic constant of hexane, the boiling point of hexane, molar weight of base, and the weight of a solution of the base in hexane at a given boiling point, it is possible to calculate the weight of the free base with fairly high accuracy.
For example, 60 of MDMA HCL was taken for base extraction at the beginning. It corresponds to base weight 50.5 gr.
Base was separated with 200 ml of hexane cas 64742-49-0.
A flask with hexane was set for T=80. As this point reached, flask weight was checked, and it was 116 gr.
Next control measure was done at T=85C. Weight was 89 gr.
Last control measure was done at T=90C. Weight was 79 gr.
All this evaporation up to 90C took around 25-30 minutes.
Now we need to calculate base weight from these data.
2nd Raoult law allows us to calculate the shift of boiling point knowing weight of solvent and solved substance and vice versa.
In fact, Raoult's second law will not work exactly for large solute contents, but it can indicate the boundaries with some accuracy.
Let's see how it works.
Msolv - solvent weight
Msub - substance weight
m - molar weight of substance
dT - shift of boiling point
E - ebullioscopic constant solvent
Hexane boiling point is 68.9C
E = 2,75
m = 193,25
Let's calculate for first step.
dT = 80 - 68.9 = 11.1
Msolv = E * Msub * 1000 / (dT * m)
Now we need to solve the following equation: Msolv + Msub = Msub + E * Msub * 1000 / (dT * m) = 116 gr.
It has a solution at Msub = 50.83.
Now do the same for T=85: Msub = 47.24
and for T=90: Msub = 47.18.
Probably freebase weight shall be somewhere around 47-50 gr.
We need 21.3 ml of HCl 36% to acidify 48 gr of freebase. So I added 100 ml of acetone and dropped 21 ml of HCl and checked PH, it was around 7. So I added more 0.5 ml of HCl to reach PH = 2.
It refers to the base weight exactly as it was calculated - 48 gr of extracted freebase!!!
It means, all my calculations work perfectly, and such method allows to determine weight of freebase in hexane solution very very precisely.
As I acidified cold base with cold HCl on ice bath, it turned to solids immediately. On the photo you can see how hexane, water from HCl and acetone separated in the beaker.
Don't forget: polypropylene is not resistant for benzine and similar substances, so, to complete crystallization, liquids were moved to the freezer in the glass beaker to avoid increasing of impurities.
While hexane was evaporated, I noticed it becomes more and more yellowish. If someone could tell me what is it, natural property of hexane to change color or base was damaged under such temperatures (up to 90C) - I would appreciate.
The problem is that without completely evaporating the hexane, I cannot know the exact amount of free base that was isolated.
Fortunately, a university course in physics, or chemistry, or something else, where Raoult laws were studied, comes to the rescue.
Knowing the ebullioscopic constant of hexane, the boiling point of hexane, molar weight of base, and the weight of a solution of the base in hexane at a given boiling point, it is possible to calculate the weight of the free base with fairly high accuracy.
For example, 60 of MDMA HCL was taken for base extraction at the beginning. It corresponds to base weight 50.5 gr.
Base was separated with 200 ml of hexane cas 64742-49-0.
A flask with hexane was set for T=80. As this point reached, flask weight was checked, and it was 116 gr.
Next control measure was done at T=85C. Weight was 89 gr.
Last control measure was done at T=90C. Weight was 79 gr.
All this evaporation up to 90C took around 25-30 minutes.
Now we need to calculate base weight from these data.
2nd Raoult law allows us to calculate the shift of boiling point knowing weight of solvent and solved substance and vice versa.
In fact, Raoult's second law will not work exactly for large solute contents, but it can indicate the boundaries with some accuracy.
Let's see how it works.
Msolv - solvent weight
Msub - substance weight
m - molar weight of substance
dT - shift of boiling point
E - ebullioscopic constant solvent
Hexane boiling point is 68.9C
E = 2,75
m = 193,25
Let's calculate for first step.
dT = 80 - 68.9 = 11.1
Msolv = E * Msub * 1000 / (dT * m)
Now we need to solve the following equation: Msolv + Msub = Msub + E * Msub * 1000 / (dT * m) = 116 gr.
It has a solution at Msub = 50.83.
Now do the same for T=85: Msub = 47.24
and for T=90: Msub = 47.18.
Probably freebase weight shall be somewhere around 47-50 gr.
We need 21.3 ml of HCl 36% to acidify 48 gr of freebase. So I added 100 ml of acetone and dropped 21 ml of HCl and checked PH, it was around 7. So I added more 0.5 ml of HCl to reach PH = 2.
It refers to the base weight exactly as it was calculated - 48 gr of extracted freebase!!!
It means, all my calculations work perfectly, and such method allows to determine weight of freebase in hexane solution very very precisely.
As I acidified cold base with cold HCl on ice bath, it turned to solids immediately. On the photo you can see how hexane, water from HCl and acetone separated in the beaker.
Don't forget: polypropylene is not resistant for benzine and similar substances, so, to complete crystallization, liquids were moved to the freezer in the glass beaker to avoid increasing of impurities.
While hexane was evaporated, I noticed it becomes more and more yellowish. If someone could tell me what is it, natural property of hexane to change color or base was damaged under such temperatures (up to 90C) - I would appreciate.
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