From the chemical literature it is known that optically pure (R)-2-octanol has a specific rotation of -10.3degerees when optically active (s)-2-bromooctane ([alpha]=+39.6) was allowed to react with a mixture of 80% H2O and 20% ethanol.2-Octanol, with [alpha] abs of +6.0 degrees, was obtained as the major product.
1. Illustrate the structure of the major enantiomer produced in this reaction. Indicate the configuration of the stereogenic center using the wedged and the dashed lines
2. Determine the ratio of (R/S) enantiomers of 2-octanol in the product mixture?
3. Discuss what type of substitution mechanism would account for the stereo chemical outcome of this reaction.