(S)-Butan-2-ol is treated with $\mathrm{SOCl}{}_2$ in the presence of pyridine. The product is

With pyridine the intermediate chlorosulfite is attacked by free $\mathrm{Cl}{}^{-}$ in an ${\mathrm{S}}_{\mathrm{N}}2$ step, giving inversion (without pyridine the ${\mathrm{S}}_{\mathrm{N}}\mathrm{i}$ path gives retention). The priority order is the same in alcohol and chloride ($\mathrm{OH}$ or $\mathrm{Cl}>\mathrm{C}{}_2\mathrm{H}{}_5>\mathrm{CH}{}_3>\mathrm{H}$), so (S) becomes (R).