Past exam of the chemistry course of the year 3 of the natural sciences course of the University of Cambridge / 2022 / Paper 3 / 37 / b / i / Solution

A is the N-bound linkage isomer, $[(\mathrm{N}{\mathrm{H}}_3{)}_5\mathrm{Co}(\mathrm{NCS}){]}^{2+}$: five ammines plus NCS account for $\mathrm{CoC}{\mathrm{H}}_{15}{\mathrm{N}}_6\mathrm{S}$.

The natural route is conjugate-base substitution. Hydroxide reversibly removes an ammine proton:The amido donor promotes loss of X${}^{-}$, producing a short-lived five-coordinate $[(\mathrm{N}{\mathrm{H}}_3{)}_4(\mathrm{N}{\mathrm{H}}_2)\mathrm{Co}{]}^{2+}$ intermediate. That leaving-group-sensitive loss is the slower stage. Subsequent trapping/proton transfers are fast and no longer depend on which halide left.

Water capture followed by proton transfer gives the hydroxo complex. Competing SCN${}^{-}$ capture can occur at either end of this ambidentate ligand, producing Co-SCN and Co-NCS. Any initially captured S-bound material can undergo the base-assisted linkage rearrangement considered next. Water is overwhelmingly abundant, so the stable N-bound product can remain a minor overall product. The slow X-dependent and fast X-independent stages fit loss of X followed by chemistry of a common intermediate; a single rate-limiting direct attack by thiocyanate would not explain that separation naturally.