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.