We took 6g of racemate = 3 g of D-amph+3 g of L-amph
We got
2.63g yield = x%
3.0g theory = 100%
2.63*100/3=87.7%
We got
2.63g yield = x%
3.0g theory = 100%
2.63*100/3=87.7%
- By Pororo
Thanks for an interesting study. But I'm surprised with your tartaric acid proportions.
6g of amph sulphate is 3.49 g of racemic amph base. Then why do you use 2.45 g of d-tartaric? It shall come in molar proportions 1:4, i.e. 0.97g of d-tartaric is enough???
6g of amph sulphate is 3.49 g of racemic amph base. Then why do you use 2.45 g of d-tartaric? It shall come in molar proportions 1:4, i.e. 0.97g of d-tartaric is enough???
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- By Pororo
Have you done this part by yourself?
I did. I was pretty surprised, with result, then I checked properties..
It works the same as sulphuric: one molecule of d or l tartaric acid bonds 2 molecules of base.
I guess there is the same mistake in your dextrometh explanation. Or excess of acid is added for some other technical reason, (reduce solubility for example)
I did. I was pretty surprised, with result, then I checked properties..
It works the same as sulphuric: one molecule of d or l tartaric acid bonds 2 molecules of base.
I guess there is the same mistake in your dextrometh explanation. Or excess of acid is added for some other technical reason, (reduce solubility for example)
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- By Pororo
In certain solvents you see it clear, when your add acid salt starts to precipitate immediately.
You know the weight, so you know how much acid you need to add.
If you add a part of the whole acid, you may filter solids and add more, and see if salt precipitates or nor.
Salt stops precipitate approx as you reach molar 1:4.
Yield is around 95.
Try to do it, and you will see. Or better check tartaric acid spec.
You know the weight, so you know how much acid you need to add.
If you add a part of the whole acid, you may filter solids and add more, and see if salt precipitates or nor.
Salt stops precipitate approx as you reach molar 1:4.
Yield is around 95.
Try to do it, and you will see. Or better check tartaric acid spec.
- By G.Patton
Hi. I read the source of this method again [attached to the post] and found that I was incorrect in my opinion but you also incorrect in yours, dear @Pororo.
As author said, two moles of phenylethylamine react with one mole of d-tartaric acid and form the solution of neutral salt d,l-phenylethylamine d-tartrate (as you said, 2 moles of amine and 1 mole of acid). D-phenylethylamine salt is dissolved in the solution while l-isomer is precipitated. So that my proportions are correct but meaning of my words in comments above was not.
I really appreciate for you notes and comments, you've helped to find mistake.
As author said, two moles of phenylethylamine react with one mole of d-tartaric acid and form the solution of neutral salt d,l-phenylethylamine d-tartrate (as you said, 2 moles of amine and 1 mole of acid). D-phenylethylamine salt is dissolved in the solution while l-isomer is precipitated. So that my proportions are correct but meaning of my words in comments above was not.
I really appreciate for you notes and comments, you've helped to find mistake.
- By Pororo
I don't know how you count, but of course there is NO ANY mistake in my calculations.
If you want to acidify the whole 1 mole of base with DL-tartaric, you need to use 0.5 mole of this acid.
If you want to extract only D or L from 1 mole of base with D or L-tartaric, you need to use 0.25 mole of D or L acid accordingly.
If you want to acidify the whole 1 mole of base with DL-tartaric, you need to use 0.5 mole of this acid.
If you want to extract only D or L from 1 mole of base with D or L-tartaric, you need to use 0.25 mole of D or L acid accordingly.
- By G.Patton
I sent you above the quotation of Nabenhauer method. Please, read full article carefully. I tell the same.
In accordance with this article, if you take the proportion like you said (1moles amine with 0.25 moles of d-tartaric acid), you'll get 0.2-0.25 moles d-amine d-bitartrate in the precipitate from 1 moles of amine free base.
The equasion of this reaction (from quotation):
1 mole d,l-amine + 1.2 mole d,l-tartaric acid = 0.5 mol of d-amine d-bitartrate (2 parts of amine bace per 1 part of acid)
the salt is precipitated
you take
1 mole d,l-amine + 0.25 mole d-tartaric acid = 0.25 mol of d-amine d-bitartrate as I suggest. Probably, you may get d-amine d-tartrate (1:1 base:acid) salt, which is also insoluble in this solvent. It's hard to predict. I haven't seen details of your experiment. Only your words.
If you wanna prove your argument, perform your experiment with pictures in comment section or new post.
In accordance with this article, if you take the proportion like you said (1moles amine with 0.25 moles of d-tartaric acid), you'll get 0.2-0.25 moles d-amine d-bitartrate in the precipitate from 1 moles of amine free base.
The equasion of this reaction (from quotation):
1 mole d,l-amine + 1.2 mole d,l-tartaric acid = 0.5 mol of d-amine d-bitartrate (2 parts of amine bace per 1 part of acid)
the salt is precipitated
you take
1 mole d,l-amine + 0.25 mole d-tartaric acid = 0.25 mol of d-amine d-bitartrate as I suggest. Probably, you may get d-amine d-tartrate (1:1 base:acid) salt, which is also insoluble in this solvent. It's hard to predict. I haven't seen details of your experiment. Only your words.
If you wanna prove your argument, perform your experiment with pictures in comment section or new post.
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- By Pororo
I got your idea. I don't know why you believe you are going to get bitartrate. May be because of ethanol.
If I understand terms correctly, tartrate means 1:2, but bitartrate means 1:1 molar proportions!!!
But I don't use ethanol. I use other solvents.
What I know, crystallization stops as soon as 0.25 moles of acid consumed with yield 95%. Then, if 0.25 moles of opposite acid is added, precipitation will start again with yield 95%.
And now, when 0.25 of L and 0.25 of D were applied, if I try to add more D or L or DL acid to mother liquor - nothing happens. It looks there is no base, ph is also acidic.
For the prove, may be, may be, on occasion. Not sure when I will do amph separation again, end of September may be.
If I understand terms correctly, tartrate means 1:2, but bitartrate means 1:1 molar proportions!!!
But I don't use ethanol. I use other solvents.
What I know, crystallization stops as soon as 0.25 moles of acid consumed with yield 95%. Then, if 0.25 moles of opposite acid is added, precipitation will start again with yield 95%.
And now, when 0.25 of L and 0.25 of D were applied, if I try to add more D or L or DL acid to mother liquor - nothing happens. It looks there is no base, ph is also acidic.
For the prove, may be, may be, on occasion. Not sure when I will do amph separation again, end of September may be.
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- By Pororo
I think, the point is here mr. Patton, and it is very simple: sequence fucking matters!!!
1. If you have a flask with with 1 mole of tartaric acid, d or l or dl, whatever, and start adding 1 mole of corresponding base in small portions, in the result you will get mostly bitartrate and some tartrate (+ some unreacted acid probably) with salt's resulting PH = 2.
2. This is what I do! If you have a flask with with 1 mole of base and start adding 0.5 mole of tartaric acid in small portions, in the result you will get mostly tartrate and some bitartrate (+ some unreacted acid.) with salt's resulting PH = 5.0-5.5.
Or 1 mole of racemic base and 0.25 mole of d or l acid if you are targeting to separate them
1. If you have a flask with with 1 mole of tartaric acid, d or l or dl, whatever, and start adding 1 mole of corresponding base in small portions, in the result you will get mostly bitartrate and some tartrate (+ some unreacted acid probably) with salt's resulting PH = 2.
2. This is what I do! If you have a flask with with 1 mole of base and start adding 0.5 mole of tartaric acid in small portions, in the result you will get mostly tartrate and some bitartrate (+ some unreacted acid.) with salt's resulting PH = 5.0-5.5.
Or 1 mole of racemic base and 0.25 mole of d or l acid if you are targeting to separate them
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