Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: sun725 on December 06, 2016, 03:50:37 PM
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Hi,
At the moment I have a project regarding: Differences in solubility of diasteromeric salts and isolating one enantiomer from a racemic mixture. We are working with (RS)-2(4'-isobutylphenyl)propionic acid (ibuprofen)
However, there is a question I have really hard time figuring out and it says:
How does the number equivalents og KOH(potassium hydroxide) influence the outcome of this experiment.
Hoping for an explenation on this. Thanks a lot.
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It is a forum rule that you must show your attempt or give your thoughts before we can help you. I'll give you a small hint, however. Don't overthink this problem.
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How does the number equivalents og KOH(potassium hydroxide) influence the outcome of this experiment.
In addition to the post above, you should post the details of the experiment so that we know what it was.
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Thanks for answering.
The exercise is called: Separation of enantiomers by formation of a diasteromic salt.
To explore differences in solubility of diasteromeric salts and use it to isolate one enantiomer from a racemic mixture. The racemic mixture here is
RS)-2(4'-isobutylphenyl)propionic acid (ibuprofen).
And the question I was asked: How does the number equivalents og KOH(potassium hydroxide) influence the outcome of this experiment
What I think is that adding KOG would make the mixture more soluble, also that the relationship between should be 2:1 other wise it wouldn't be possible to isolate.
That's my attempt and I'd love to know more... Thanks again.
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We are still missing a key piece of information; there has to be something forming the salt besides ibuprofen.
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Adding KOH will just give you the potassium salt of ibuprofen and you still have the mixture of enantiomers, not a mixture of diastereoisomers.
So as Babcock asked, what have you forgotten to include in your question?
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Hi,
I think what I'm missing here is adding a base which in this situation is
(S)-(a)-Phenylethylamine...?
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OK, now we are getting somewhere.
So what happens if you use KOH, I assume after the separation of the diastereoisomeric salts?
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I would interpret the question to mean what if you included 0, 1, 2, ... equivalents of KOH during the separation. However, I am not sure.
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Sun725, until you post a full experimental procedure, it is not clear what the question means and people are having to guess. Post the procedure.
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@Dan,
I posted every single detail written on the assignment therefor I don't know what to add anymore.
The questions directly stated: Discuss how the number equivalents of KOH influences the outcome of this experiment.
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I would interpret the question to mean what if you included 0, 1, 2, ... equivalents of KOH during the separation. However, I am not sure.
It might be right ...
As stated before they directly say: Discuss how the number equivalents of KOH influences the outcome of this experiment.
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In optimized procedure 1 mole of KOH is used for 2 moles of ibuprophen and a bit more than 1 mole of phenylethylamine. It gives food for thought.
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In optimized procedure 1 mole of KOH is used for 2 moles of ibuprophen and a bit more than 1 mole of phenylethylamine. It gives food for thought.
Okay, I was close to that earlier, however I missed adding the phenylethylamine.
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@Dan,
I posted every single detail written on the assignment therefor I don't know what to add anymore.
The questions directly stated: Discuss how the number equivalents of KOH influences the outcome of this experiment.
What is missing is the experimental procedure - i.e. what you actually did, which reagents were mixed, in what ratios, and in what order etc.
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Racemic ibuprofen 16 mmol is dissolved in 0,24 M KOH 7,92 mmol. A magnet is added and the mixture is heated to 75-85 degrees. When everything is dissolved,
(S)-(-)-1-phenylethylamine(1,0 mL) is added and we had to keep stirring for an hour.
Afterwards the Recrystallization procedure occurs ...
The next day we used the crystals and added 2M sulfuric acid(25 mL), ethyl acetate (15 mL) and a magnet etc .
We ended with a thick oil and had to measure the melting point.
I understand why (S)-(-)-1-phenylethylamine is added, however stille not sure on the KOH and can't answer the question yet. I wrote earlier that adding KOH would make the process more soluble.
Also, It's also possible to add for example (R)-(-)-1-phenylethylamine and it would have turned out the same, right?
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You convert half of ibuprophen into potassium salt and the rest into phenyethylamine salt. Since one of the ibuprophen optical isomer form less soluble diasteromeric salt then only this salt crystallizes (nearby pure). The rest of the ibuprophen (the second optical isomer) remains as potassium salt in solution. More than 1 mole of KOH to 2 moles of ibuprophen causes decomposition of the diastereomeric salt. Using only two moles of amine instead KOH and amine (1:1) causes troubles (dissolution of ibuprophen needs more time, higher cost of pure amine enantiomer, worser yield of crystals and probably less enantiomeric purity of the diastereomeric salt since occlusion in the crystal of the second diastereomeric salt).
I advice recrystalization of salt before decomposition and removing organic solvent under low pressure.
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Sorry AWK, I don't think this is quite correct. "You convert half of ibuprophen into potassium salt and the rest into phenyethylamine salt. Since one of the ibuprophen optical isomer form less soluble diasteromeric salt then only this salt crystallizes (nearby pure). The rest of the ibuprophen (the second optical isomer) remains as potassium salt in solution. "
The potassium salt is still an enantiomeric mixture.
If you use too much KOH you will just form the potassium salt and that will not react with phenylethylamine.
Now ibuprofen is insoluble in water, is it possible that what the OP used was a salt? Therefore the purpose of the KOH is to re-generate the free carboxylic acid which is organic soluble.
Otherwise I see absolutely no requirement for KOH in this experiment.
I just read from the WIKI page In some countries, ibuprofen lysine (the lysine salt of ibuprofen, sometimes called "ibuprofen lysinate") is licensed for treatment of the same conditions as ibuprofen; the lysine salt is used because it is more water-soluble.[15].
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The potassium salt is still an enantiomeric mixture.
Right, but from solution crystallizes only less soluble salt.
https://infohost.nmt.edu/~jaltig/Ibuprofen.pdf
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The potassium salt is a mixture of enantiomers not diastereoisomers, therefore you cannot resolve the potassium salt by fractional crystallisation as the solubility of both enantiomers is the same in any given solvent.
The link you provided does not utilise KOH, it describes a classical resolution exploiting solubility differences between diastereoisomeric salts.
As I said I see no role for KOH in the OP's methodology.
my mistake, sorry
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This link use KOH in amount exactly given by sun725
"To a 100 mL round bottom flask, add a stir bar, 3.0g of racemic Ibuprofen and 30 mL of 0.24 M
KOH. Clamp the flask into a heating mantle and insert a thermometer. Initiate stirring. Heat the "
"Next, add 0.9 mL of (S)-(-)- Phenethylamine dropwise and slowly to the "
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The potassium salt is a mixture of enantiomers not diastereoisomers, therefore you cannot resolve the potassium salt by fractional crystallisation as the solubility of both enantiomers is the same in any given solvent.
I've often wondered this - it's probably prohibitively expensive but maybe if you did reactions in e.g. R-butan-2-ol would that be stereoselective? Not really appropriate for this reaction mind. I know I've tried resolving a racemate with differential crystallisation before but that was on a small scale (milligrams).
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The potassium salt is a mixture of enantiomers not diastereoisomers, therefore you cannot resolve the potassium salt by fractional crystallisation as the solubility of both enantiomers is the same in any given solvent.
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As I said I see no role for KOH in the OP's methodology.
I did not say that the potassium salt is a mixture of diastereoisomers. In solution exist ions. Only one of their combination with the lowest solubility crystallizes. That is the salt of needed enantiomer of ibuprophen with enantiomer of phenylethylamine (used for optical resolution). The rest is soluble potassium salt of the other enantiomer of ibuprophen. For resolution we use ibuprophen racemate, and the same amount of moles of the mixture of KOH and phenyethylamine (approx. 1:1).
Using only phenyethylamine in resolution of ibuprophen we should use a time consuming fractional crystallization.
Concerning resolution with L-lysine. Lysine is a limiting amino acid with daily requirement 12 mg/kg of body weight (adult), hence we do not need decomposition of diasteromeric salt (one 342 mg tablet of Ibuprophen Lysine contains about 100 mg of lysine).
Probably the same "trick" with KOH is using in resolving procedure.
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So this is a type of DKR.
The lysine salt is used to enhance the solubility of ibuprofen in water.