Question

1A. The product from your reaction has a sharp melting point, indication that it is either meso-hydrobenzoin or a racemic mixture of the (1R, 2R) and (1S, 2S) enantiomers. Why is it not possible that your product is either the (1R, 2R) or the (1S, 2S) pure enantiomer?

1B. What is the enantiomeric excess (ee) of a racemic mixture of enantiomers?

1C. If the racemic mixture of the hydrobenzoins were formed in the reaction of benzil with sodium borohydride, describe a set of experimental conditions for which you might get a non-racemic mixture of these two products. You can chance anything except the two starting materials.

(The reaction being described in the first part of the questions is the reation of benzil with sodium borohydride. It's Experiment 13 in Understanding the Principles of Organic Chemistry: A Laboratory Course, which is online in PDF form for free)

If you could explain your answers, I would really appreciate it. Thanks!

Answer 1

1) As the melting point is sharp, (assuming it is 137-139^oC) the product contains the majority of meso-hydrobenzoin.

If it was racemic mixture then the melting point would have been a range (i.e. not sharp).

1B) Enantiomeric excess (e.e.) is the percentage excess amount of either enantiomer in the mixture. However, in racemic mixture none-of-the enantiomer is in excess. Hence the e.e. of the racemic mixture is 0%.

1C) If we used chiral catalysts or enzymes we can obtain enantiomeric excess. The chiral catalyst including RuCl[(1S,2S)-N-p-toluenesulfonyl-1,2-diphenylethanediamine] or enzymes such as Talaromyces flavus.