when optically active (R)-2-bromoctane ([alpha] = +39.6 degrees) was allowed to react with a mixture of water and ethanol, 2-octanol with [alpha] obs of + 1.0 degrees was obtained as the major product. From the chemical literature it is known that optically pure (R)-2-octanol has a specific rotation of -10.0 degrees. What is the optical purity, or enantiomeric excess, of the 2-octanol produced in this reaction? 10% S. What is the ratio of enantiomers of 2-octanol in the product mixture? 55% (S) and 45% (R). However, I am not sure how to draw the structure of the major enantiomer produced in this reaction. Indicate the configuration of the stereogenic center using wedged and hashed lines.