Final answer:
The statement is A. TRUE. R-3-bromo-3-methylhexane reacting with OH⁻ will produce a 50%/50% mixture of R-3-methyl-3-hexanol and S-3-methyl-3-hexanol due to the SN2 mechanism.
Step-by-step explanation:
The statement is TRUE.
R-3-bromo-3-methylhexane reacting with OH⁻ will indeed produce a 50%/50% mixture of R-3-methyl-3-hexanol and S-3-methyl-3-hexanol. This is because the reaction follows an SN2 mechanism, where the nucleophile (OH⁻) attacks the carbon bearing the leaving group (Br) with inversion of configuration at the stereocenter.
The reaction between R-3-bromo-3-methylhexane and OH⁻ (hydroxide ion) typically involves a substitution reaction known as an SN2 (nucleophilic substitution bimolecular) reaction, particularly if the conditions are appropriate.
In this reaction, the hydroxide ion acts as a nucleophile, attacking the carbon atom bearing the leaving group (the bromine) to replace it.
The stereochemistry of the product formed in this reaction will depend on the specifics of the starting compound (R or S configuration) and the conditions of the reaction.