The reaction described above is a series of sn2 substitutions. The initial (R)-2-butanol reacts with PBr₃ in pyridine to turn the alcohol functionality into a leaving group as it attaches to the phosphorus of PBr₃. One of the bromides from PBr₃ eventually displaces the oxygen atom with an sn2 substitution. Therefore, the product will have an inversion of stereochemistry as (S)-2-bromobutane is formed.
This product is treated with NaCN and CN⁻ is a very good nucleophile. The bromo-substitutent is a good leaving group, therefore, the nucleophile will do an sn2 substitution which inverts the stereochemistry once more to give the optically active product shown.
