When optically active (S)-2-bromooctane is treated with aqueous NaOH, the reaction proceeds exclusively via an S _ N 2 mechanism. What will be the stereochemical nature of the major product formed and the optical activity of the resulting reaction mixture?

Step-by-step solution

The S _ N 2 reaction involves the attack of the incoming nucleophile ( OH ⁻ ) from the side opposite to the leaving group ( Br ⁻ ). This back-side attack leads to a complete inversion of configuration at the chiral center. Therefore, the reactant (S)-2-bromooctane will be converted entirely into (R)-octan-2-ol. Because only a single enantiomer is formed as the product, the resulting reaction mixture will be optically active. A racemic mixture would only form if the reaction proceeded via an S _ N 1 mechanism. A mes