SolutionsThe Result of an Experiment
1. Enantiomeric excess
The 2-octanol obtained contains 94% of the S enantiomer and 6% of the R. Calculate its enantiomeric excess and the expected specific rotation, given that pure (S)-2-octanol has +9.5°.
ee = 94 − 6 = 88%. [α] = 0.88 × (+9.5°) ≈ +8.4°, very close to the measured value (+9°).
2. The kinetics
Doubling the concentration of HO− or that of the halide doubles the rate of alcohol formation. What does this tell us?
The reaction is first order in each reactant and second order overall: v = k[RBr][HO−]. Both reactants take part in the rate-limiting step: the reaction is bimolecular.
3. Optically inactive alkenes
Why are the alkenes formed optically inactive?
When the double bond forms, carbon 2 stops being a stereocentre (it becomes sp2). 1-Octene and the 2-octenes are achiral.
4. The equation
Write the equation of the major substitution reaction, indicating the configuration of reactant and product.
(R)-CH3CHBr(CH2)5CH3 + NaOH → (S)-CH3CH(OH)(CH2)5CH3 + NaBr