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Acetonitrile, $\mathrm{C}{\mathrm{H}}_3\mathrm{CN}$, can coordinate through nitrogen as a neutral two-electron donor. Bound acetonitrile can be displaced by a stronger donor or an intramolecular chelating group.

Photoinduced CO loss gives a 16-electron fragment. In the ester-substituted complex the pendant ester carbonyl oxygen can reach chromium and capture that vacant site. A suitable structure is $\mathrm{Cr}({\eta }^6\text{-}{\mathrm{C}}_6{\mathrm{H}}_5\mathrm{C}{\mathrm{O}}_2\mathrm{Me})(\mathrm{CO}{)}_2(\kappa O\text{-ester})$, retaining the coordinated arene and adding intramolecular coordination of its ester oxygen. The added oxygen lone-pair donation restores an 18-electron count. The gaseous by-product is carbon monoxide.

heptane is a poor donor, but intramolecular ligand trapping avoids needing a donor solvent. Benzene has no pendant donor. Although the methoxy group has oxygen lone pairs, its directly attached oxygen does not have the same reach and favourable geometry as the projecting ester carbonyl oxygen; possession of a lone pair alone is insufficient. For these two substrates, acetonitrile can trap the unsaturated fragment through nitrogen to give $\mathrm{Cr}({\eta }^6\text{-arene})(\mathrm{CO}{)}_2(\mathrm{NCMe})$. Without a suitable trap, CO recombination can reverse the initial photochemical dissociation.

A reasonable product is sodium facial trimetaphosphate–molybdenum tricarbonyl, with the anionOne terminal O from each phosphorus binds Mo; the three O donors occupy one octahedral face, while the three carbonyls occupy the opposite face. The ring bridges remain P–O–P bridges and are not replaced by Mo–P bonds. Mo remains Mo(0), $d^6$, and $6+3(2)+3(2)=18$ electrons. Overall,acetonitrile can transiently replace CO during heating, but it is not required in the final six-coordinate proposal. The same anion has been structurally characterized with a different counterion: that primary study independently supports the facial O${}_3$/CO${}_3$ connectivity, without establishing identical solubility or yield for the sodium salt.^[ref]

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.

$E=\mathrm{Mo}({\eta }^5:\kappa N\text{-}L)(\mathrm{CO}{)}_2\mathrm{Cl}$, 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.

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.

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.

Acetonitrile is suitable: it dissolves the organic reagents, is polar enough to stabilize the developing ionic products and is an aprotic, relatively unreactive medium. It does not strongly hydrogen-bond-solvate the amine lone pair or compete efficiently as the methylating nucleophile. Sufficient solubility and chemical compatibility matter as well as polarity.

An acetal retrosynthesis separates the target into trans-cyclohexane-1,2-diol and cyclohexyl methyl ketone. Both fragments can be prepared using separate portions of cyclohexene.

For the diol, epoxidation with a peroxy acid such as mCPBA gives cyclohexene oxide. Acid-catalysed opening with water occurs by backside attack on the protonated epoxide, giving the anti, hence trans, diol. A syn dihydroxylation would give the wrong relative configuration.

For the ketone, add HBr to another portion of cyclohexene to obtain bromocyclohexane, form its grignard reagent with Mg in dry ether, and add acetonitrile. Hydrolyse the resulting magnesium imine salt to ${\mathrm{C}}_6{\mathrm{H}}_{11}\mathrm{COMe}$. A nitrile is useful here because addition followed by hydrolysis stops at the ketone instead of giving the double addition associated with an ester.

Finally combine the isolated trans diol and ketone with catalytic acid and remove water to form the cyclic ketal. The ketone carbon becomes the acetal carbon bearing Me and cyclohexyl; the two original diol C–O bonds retain their relative configuration. Anti epoxide opening supplies the required trans relationship before ketal formation. ketal formation is an equilibrium and the trans-fused product may be less favourable than an unstrained acetal, so effective water removal and isolation are necessary; substituting a cis diol is not an acceptable shortcut.