Chemical Forums
Chemistry Forums for Students => Organic Chemistry Forum => Topic started by: Guacoholy on November 08, 2020, 01:06:15 PM
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Essentially, the question I am given is how many grams of the SSS diastereomer is produced if 1.00g of L-Proline is used and L-Proline is the limiting reagent. Reaction is attached.
I calculate 2.17mmol and .922g of SSS diatereomer.
(1.00g/115(g/mol L-Proline))= 8.68mmol L-Proline x (1 mole SSS diastereomer/4 mole L-Proline) = 2.17mmol x 424.44(g/mol SSS diastereomer) = .922g SSS diastereomer.
I am getting conflicting answers from my professor, the answer key and my peers, it seems simple enough to me: for every mole of a specific diastereomer, the reaction requires 4 moles of L-Proline. What am I missing here? The answer key says 2 moles l proline for specific diastereomer my peers are saying 1:1.
I have been struggling with this seemingly simple question for days, due to my answer conflicting with others, please someone explain this to me.
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for every mole of a specific diastereomer, the reaction requires 4 moles of L-Proline
For every mole of a specific enantiomer of Phenylsuccinic acid, the reaction requires 2 moles of L-Proline.
The numbers provided are correct.
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for every mole of a specific diastereomer, the reaction requires 4 moles of L-Proline
For every mole of a specific enantiomer of Phenylsuccinic acid, the reaction requires 2 moles of L-Proline.
The numbers provided are correct.
Thank you so much for your reply!
Just to confirm, my calculation of .922g of a specific diastereomer is correct? Also, yes, 2 moles of L-Proline need to react with 1 mole of a specific enantiomer. Since we use a racemic mixture I would need to use 4 moles of L-Proline to create 1 mole of a specific diastereomer, since the reaction creates both?
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1 mole of (R)-PheSuc + 1 mole of (S)-PheSuc + 4 moles of (S)-Pro = 1 mole of (S)-PheSuc·2(S)-Pro + 1 mole of (R)-PheSuc·2(S)-Pro
but racemate usually is written as (RS)-PheSuc then for 1 mole of racemate you need 2 moles of Pro and you will get half a mole of each diastereomeric salt.
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Thank you very much.