The Result of an Experiment
Optically pure (R)-2-bromooctane was made to react with sodium hydroxide (a nucleophile) in DMF (dimethylformamide) as solvent.
The thorough analysis of the reaction mixture tells us that the result is the major production (80%) of an optically active alcohol,
together with a series of optically inactive alkenes. It is reasonable to think that the nucleophile (HO−) attacks the electron-deficient carbon of 2-bromooctane, the one bonded to the bromine. Consequently, the alcohol should be 2-octanol. Let's check a chemical catalog, where we can find the following data:
The comparison between the measured and tabulated optical rotations tells us that the alcohol is mainly (not exclusively) the S enantiomer. The olefins have been separated and analyzed. The global result is summarized in the following scheme:
These results lead to the following conclusions:
- The halogen was mostly substituted by an OH group.
- Alcohol configuration is mostly S, contrary to that of the starting haloalkane.
- Therefore, the substitution reaction mostly happened with INVERSION of configuration.
- The production of olefins shows that the halogen was in part ELIMINATED.
- The elimination process must be complex because a number of different olefins are formed.
- A small amount of the R alcohol was formed, suggesting that a small amount of RACEMIZATION took place.
- The reaction rate depends on the two reagents, haloalkane and nucleophile, thus suggesting that the two of them must intervene in the rate-limiting step of the reaction.
These logical deductions should allow us to propose a reaction mechanism, i.e. a reasonable sequence of steps describing precisely how and when the bonds are broken and formed.