Past exam of the chemistry B course of the University of Cambridge / 2025 / Paper 2 / B8 / b

Write $L^{-}$ for the cyclopentadienyl ligand bearing its $\mathrm{C}{\mathrm{H}}_2$–2-pyridyl arm; ${\eta }^5:\kappa N$ means that its ring and pyridyl nitrogen both bind the same metal. $D=[\mathrm{Mo}({\eta }^5:\kappa N\text{-}L)(\mathrm{CO}{)}_2(\mathrm{NCMe}){]}^{+}$. The base removes the pyridinium N–H proton in $C$, restoring a pyridine lone pair. Intramolecular nitrogen coordination then replaces one acetonitrile ligand and closes a chelate ring. Both CO ligands are retained; the molybdenum remains Mo(II), $d^4$.

This is an proton transfer followed by chelation, with metal electron count $4+6+2_N+4_{\mathrm{CO}}+2_{\mathrm{NCMe}}=18$. The chelate effect, favorable formation of the Mo–N bond, and release of a solvent molecule explain the driving force. Merely invoking a 16-to-18-electron change would be misleading because $C$ is already 18-electron. In the preceding acid treatment, protonation of the ${\eta }^3$ allyl ligand allows release of propene; the other proton resides on the pendant pyridine, accounting for $C$'s overall $2+$ charge. One proton is removed here; the cyclopentadienyl tether is not deprotonated or detached.

$E=\mathrm{Mo}({\eta }^5:\kappa N\text{-}L)(\mathrm{CO}{)}_2\mathrm{Cl}$, with chloride replacing the remaining acetonitrile. The pendant pyridine remains attached. The substitution removes the positive electric charge on $D$ but does not change the Mo(II) oxidation state. In ionic electron counting, both neutral acetonitrile and anionic chloride ligand donate two electrons, giving $4+6+2_N+4_{\mathrm{CO}}+2_{\mathrm{Cl}}=18$ in $E$.

acetonitrile is the remaining weak, exchangeable monodentate donor. Its displacement forms a favorable Mo–Cl bond at a cationic center, retains the intramolecular Mo–N bond, and restores an 18-electron product after substitution. The following formulas specify the coordination changes directly:Tetramethylammonium is a spectator counterion; it is not a carbon donor to Mo.

The readily displaced acetonitrile sites have both been exhausted. A second chloride would have to displace a strongly bound carbonyl ligand, open the chelate ring, or associate to an already 18-electron center to make an unfavorable crowded 20-electron species. The product is neutral, so further chloride uptake also lacks the charge-neutralization advantage of $D\to E$ and would generate an anionic complex if substitution occurred. The chelate effect disfavors loss of the tethered pyridine, while strong pi backbonding helps retain CO under the stated conditions. These combined kinetic and bonding arguments explain the absence of further reaction. The 18-electron rule alone would not forbid replacing CO by chloride, since that replacement is isoelectronic; it therefore cannot be the whole explanation.

For an intact complex in the usual fast-conformational-averaging regime, counting distinct chemical shifts rather than individual split lines gives $C:9$, $D:11$, $E:10$ signals.

In $C$, the uncoordinated pyridinium arm permits the paired positions of a monosubstituted cyclopentadienyl ligand to average: its four ring hydrogens give two sets, $2+2$. The linker $\mathrm{C}{\mathrm{H}}_2$ gives one set, the four inequivalent pyridinium C–H positions give four, its N–H gives one, and the two symmetry-related bound acetonitrile methyl groups give one. Hence $2+1+4+1+1=9$. The exchangeable N–H resonance may be broadened or absent experimentally; nine is the structural count before exchange removes it.

In the rigid cis-dicarbonyl chelates $D$ and $E$, the different pyridyl-N and fourth-leg donors destroy the operation that related the two sides of the substituted ring. Each of the four Cp hydrogens is distinct, and the two linker hydrogens are diastereotopic. Together with four pyridine C–H sites, this gives $4+2+4=10$. The remaining bound acetonitrile adds one methyl signal to $D$, giving 11; $E$ has none, giving 10. The two enantiomers of a racemate have identical spectra in an achiral solvent and do not double these counts. This prediction assumes resolved resonances and no rapid chelate opening that averages inequivalent sites; variable-temperature proton nuclear magnetic resonance would test that assumption.