Past exam of the chemistry B course of the University of Cambridge / 2022 / Paper 2 / B7 / d / Solution

The anionic nitrogen donor replaces iodide to giveHere Cp denotes ${\eta }^5$ cyclopentadienyl and $\kappa N$ specifies attachment through the ring nitrogen of the pyrrolyl anion, ${\mathrm{C}}_4{\mathrm{H}}_4{\mathrm{N}}^{-}$. Fe(II), $d^6$, receives six electrons from Cp, four from the two CO ligands and two from an N-bound pyrrolyl ligand: $6+6+4+2=18$.

Two coupled CO stretching modes, approximately symmetric and antisymmetric combinations, account for the two strong carbonyl-region absorptions. The bent two-CO arrangement of this piano-stool complex permits both combinations to be infrared active. They are not evidence for two different chemical products.

Heating expels two CO molecules and changes the binding of the pyrrolyl ligand from ${\eta }^1$ to ${\eta }^5$, giving the sandwich azaferrocene,The four electrons lost with CO are replaced by the four additional electrons donated on changing the hapticity of the pyrrolyl ligand. Thus $B$ is also Fe(II), $d^6$, and 18-electron: $6+6+6=18$. Each complex has one time-averaged Cp proton environment, two pairs of protons from the pyrrolyl ligand, one adjacent to N and one farther from N. Hence three proton environments, with integrals $5:2:2$, fit both structures. For proton nuclear magnetic resonance, three signals means three chemical shift environments; the individual resonances can still show coupling.