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 G.Patton
Hello, 6g of amph sulphate (368.5 g/mol) contain 0.03256 moles free base (look at the reaction equasion, where 2molecules of amph produce 1molecule of salt and vice versa). Briefly, one mole of sulphate salt contain two moles of free base, hence moles of amph sulphate multiply x2 (0.01628x2).
Your question is why it's written to take 2.45g of d-tartaric acid.
Answer: 2.45g of d-tartataric acid is 0.01632 moles (2.45g/150.09g/mol). Hence, this 0.01632 moles (2.45g) is half of 0.03256 moles of racemic amph free base.
You need 1/2 cuz this racemic amph free base contains 1/2 of d-amph and l-amph in theory.
Do you understand?
Your question is why it's written to take 2.45g of d-tartaric acid.
Answer: 2.45g of d-tartataric acid is 0.01632 moles (2.45g/150.09g/mol). Hence, this 0.01632 moles (2.45g) is half of 0.03256 moles of racemic amph free base.
You need 1/2 cuz this racemic amph free base contains 1/2 of d-amph and l-amph in theory.
Do you understand?
- 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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