Enantiomeric Excess Practice Problems

A solution of a chiral molecule gives a rotation about 179.9º. Propose a way to tell if the rotation is +180º or −180º.

A mixture consists of 20% of (+)-pentan-2-ol and 80% of (−)-pentan-2-ol. Determine the specific rotation of the mixture if the specific rotation of pure (−)-pentan-2-ol is 13.0º.

A solution was prepared by dissolving 1.8 g of (−)-glyceraldehyde in 9.00 mL of water. The solution was placed in a 100-mm cell and the rotation was then measured at 25℃ using the sodium D line. If the rotation observed was −1.74º, calculate the specific rotation of (−)-glyceraldehyde.

Addition of a chiral molecule to the catalytic reduction of pentan-2-one produces a slightly optically active product. Determine the percentage of (+)-pentan-2-ol and (−)-pentan-2-ol formed in the reaction if the specific rotation of the product is +0.54º. (The rotation of pure (−)-pentan-2-ol at the same condition is 13.0º)

A. (+)-pentan-2-ol =  0.54 % and (−)-pentan-2-ol = 99.46 %

B. (+)-pentan-2-ol =  47.9 % and (−)-pentan-2-ol = 52.1 %

C. (+)-pentan-2-ol =  52.1 % and (−)-pentan-2-ol = 47.9 %

D. (+)-pentan-2-ol =  95.85 % and (−)-pentan-2-ol = 4.15 %