2Na+2HCI-H,+2Naci
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The reaction described above is a series of sn2 substitutions. The initial (<em>R</em>)-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 (<em>S</em>)-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.
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.
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