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2-Picolylamine is pyridin-2-ylmethanamine, with a pyridine nitrogen and an aminomethyl nitrogen adjacent on the ring. It binds as an unsymmetrical neutral bidentate ligand, forming a five-membered chelate ring.

A pyridine bearing an amino substituent; its NH donors and ring-N acceptor permit complementary hydrogen bonding.

A compound containing two joined pyridine rings; 2,2′-bipyridine is a common bidentate nitrogen ligand.

Two molecules of the aminopyridine form a self-complementary dimer through two N–H···N contacts. In each direction, one exocyclic amino hydrogen is the donor and the pyridine ring nitrogen of the other molecule is the acceptor.

Orient the two amino/pyridine-N edges toward one another, as in the second panel of the original homodimer drawing. Each molecule donates one NH and accepts at one ring N; the second amino H remains available for additional packing contacts. The chlorine substituent is retained but is not one of the two principal donor–acceptor sites.

water lies below pyridine and cyanide ligand in the spectrochemical series. At fixed metal and dioxolene charges, replacement by water decreases ${\Delta }_o$ and favours the high-spin limits: Co(II) remains $S=3/2$; Co(III) can become $S=2$ if the splitting falls below the pairing threshold. Replacement by cyanide increases ${\Delta }_o$ and favours Co(III), $S=0$, and Co(II), $S=1/2$. The two unchanged semiquinonate ligands would then give the Co(II) form a combined independent-spin moment of $3{\mu }_B$. The ranking predicts the direction of the change; two substituted donors alone do not guarantee a Co(III) spin crossover, since all six donors determine its splitting. A shift of the valence-tautomer equilibrium is also possible and is separate from this fixed-charge comparison.

Neutral pyridine contributes no formal charge. If $n$ of the two dioxolene ligands are dianions, the combined ligand charge is $-2-n$, so the cobalt oxidation state is $2+n$. The distinct charge assignments are thereforeInterchanging the two ligand labels in the middle row does not make another charge combination. Co(IV) belongs in this unrestricted charge-balance list even though the subsequent magnetic analysis excludes it.

Treat each semiquinonate ligand as a symmetric O,O chelate, with the free radical delocalized over the dioxolene framework. The ideal octahedral donor-connectivity structures are one trans form and two mirror-related cis forms, giving three stereoisomers. Trans refers to the two pyridine N donors, not to the two O atoms of a chelate, which must remain adjacent.

In the diagram, coloured curves join the two O donors belonging to one ligand. For an unambiguous coordinate specification, the trans form has Py at $+z,-z$, with chelate pairs $(+x,+y)$ and $(-x,-y)$. A cis pair has Py at $+z,+x$, with chelates $(-z,+y)$ and $(-x,-y)$; reflecting $y\mapsto -y$ gives the other cis form. Proper rotations cannot superpose those two cis arrangements. Ring conformations and pyridine rotations are not extra donor-connectivity isomers.

The Co(III) centre in $A^1$ has $t_{2g}^6(e_g^{\ast }{)}^0$: every occupied d subset is orbitally balanced. There is no first-order metal-centred Jahn–Teller driving force in $A^1$. This does not require all six experimental bond lengths to be equal, since pyridine and dioxolene are different donors.

The Co(II) centre in $A^2$ has $t_{2g}^5(e_g^{\ast }{)}^2$. Its two $e_g^{\ast }$ orbitals each contain one electron, so that subset gives no first-order preference. The $t_{2g}$ hole does give a weaker Jahn–Teller instability. To compare tetragonal distortions, take a positive small splitting parameter $\delta$ and retain each subset's orbital energy barycentre. On axial elongation let $d_{xz},d_{yz}$ shift by $-\delta$ and $d_{xy}$ by $+2\delta$. Five electrons then giveFor equal-magnitude compression the shifts reverse, and filling the lower $d_{xy}$ orbital first givesThe two singly occupied $e_g^{\ast }$ orbitals contribute zero net shift in either case. Thus the elementary tetragonal splitting model favours elongation for high-spin $d^7$, rather than equal electronic stabilization by the two signs. In a strictly sigma-only octahedral molecular orbital model $t_{2g}$ is nonbonding, so this weak effect is absent; it arises when the electrostatic and non-sigma interactions neglected in that diagram are restored. spin-orbit coupling, unequal donors and chelate constraints can modify the actual distortion, so the argument does not establish which experimental Co–donor bonds must lengthen.

Use ionic electron counting, counting each ordinary donor atom once and keeping ligand charge distinct from denticity. A is 16-electron, B is 14-electron and C is 16-electron, so none is an 18-electron complex under this convention.

A contains a modified, monoanionic PNP pincer: the exocyclic double bond and altered pyridine-ring bond pattern in the drawing matter. Its two phosphines and its central N donor provide six electrons altogether, while neutral N${}_2$ provides two. Neutrality then gives Co(I), $d^8$, and $8+6+2=16$. Treating the modified pincer as a neutral intact pyridine ligand would incorrectly produce Co(0) and a 17-electron count. A four-coordinate approximately square planar molecular geometry in a $d^8$ complex can leave the strongly sigma-antibonding orbital empty and be stable at 16 electrons. The underlying modified-pincer chemistry has been examined using structure, spectroscopy and electronic-structure calculations.^[ref]

B has a neutral tertiary amine donor at the centre of its tripod and three anionic amido arms. Their combined charge is $-3$, so Re is Re(III), $d^4$. The four N donors give eight electrons and N${}_2$ gives two: $4+8+2=14$. Such an electron count is not inherently unexpected in an amido-supported complex: strong amido pi donation and the imposed five-coordinate geometry alter the orbital energies, and the empty orbitals are not nine equivalent slots that must all be filled. Steric protection and chelation also help stabilize this coordinatively restricted structure. The 14-electron figure uses the usual two-electron ionic count for each amido donor; pi bonding should not be appended to it inconsistently.

C has an anionic acyl carbon donor and a neutral phosphine in the chelate, together with two neutral PMe${}_3$ donors. Thus Co(I), $d^8$, receives $2+2+4=8$ ligand electrons, giving $8+8=16$. This is again a reasonable four-coordinate $d^8$ structure. The acyl C=O is part of a carbon-bound acyl ligand, not an additional free CO donor.

C is the best geometrical match for a strongly bound 1:1 complex with E. Both have an adjacent amide donor–acceptor edge. Arrange them head-to-tail so that the N–H of E donates to the carbonyl oxygen of C, and the N–H of C donates to the carbonyl oxygen of E. This closes the familiar two-hydrogen-bond cyclic amide motif, illustrated below.

C combines a strong acceptor with a useful donor in the correct spacing. B has a stronger individual donor, but its pyridine N and OH are separated around the ring and do not reproduce the compact adjacent amide edge of E. A lacks a matching donor, and D's N–H is weak and has the wrong geometry for the same amide–amide motif. Site strength and spatial complementarity must be considered together; multiplying two isolated probe constants is not a reliable prediction of the complete association equilibrium. Self-association of C or E can compete and must be included in a quantitative binding analysis.

4-Fluorophenol probes acceptor strength because it donates its OH hydrogen. The qualitative order is C $>$ D $>$ B $>$ A: the electron-rich lactam oxygen is the strongest acceptor here, the secondary amine is next, pyridine nitrogen is intermediate, and aldehyde oxygen is weakest. Resonance donation from amide nitrogen increases the lactam oxygen's basicity relative to an ordinary aldehyde or ketone oxygen. The N-methylpyrrole site in A does not add an independent amine-like acceptor.

N-Methylpyrrolidin-2-one is a carbonyl acceptor without N–H, so it instead probes the solute's donor ability. B $>$ C $\gg$ A and D: B's phenolic OH is a strong donor, the lactam N–H of C is appreciable, A lacks a conventional donor, and D's neutral secondary-amine N–H is weak. The amine can be a good acceptor and a poor donor at the same time; the two properties are not opposites on one scale.

These are association equilibria in cyclohexane, where solvent competition for hydrogen bonds is weak. The quoted constants should be compared within each probe series, not treated as solvent-independent measures. At 298 K, as an illustrative common temperature, a tenfold increase in $K_a$ corresponds to a binding free-energy advantage $RT\ln 10\simeq 5.7\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$. This does not imply the experiment's temperature was specified, nor that every difference is the energy of one isolated hydrogen bond.

A: aldehyde oxygen is the best acceptor; there is no conventional strong O–H or N–H donor. The N-methylpyrrole nitrogen contributes its lone pair to aromatic bonding and is not an ordinary amine acceptor. A formyl C–H or ring C–H can participate in weak contacts, but is not comparable to the donors in B and C.

B: pyridine nitrogen is the best acceptor and the phenolic OH is the best donor. Its oxygen can accept more weakly, but nitrogen provides the primary basic site in the depicted hydroxypyridine form.

C: lactam carbonyl oxygen is the best acceptor and the amide N–H is the donor. amide nitrogen is not a useful acceptor because its lone pair is delocalized toward the carbonyl.

D: the secondary amine nitrogen is the best acceptor, and its N–H is the only conventional donor, but a weak one here. Its ketone oxygen is also an acceptor, although weaker than the amine site in this comparison. These assignments distinguish functional-group identity and lone-pair availability, rather than counting heteroatoms.

In $A$, the tridentate ligand contains one anionic phenoxide ligand, a neutral secondary amine with its N–H bond intact, and a neutral phosphine ligand. The three chloride ligands contribute another three negative charges. Consequently the neutral complex has Ti(IV), $d^0$, and 12 valence electrons:The N–H label is essential: counting this nitrogen as an anionic amido ligand would give the wrong oxidation state. A six-coordinate early-metal complex need not obey the 18-electron rule.

In $B$, the substituted cyclopentadienyl ligand is ${\eta }^5$ and anionic, while the three-carbon bent ligand is an anionic ${\eta }^3$ allyl ligand, a four-electron donor in ionic counting. The pendant pyridine nitrogen is uncoordinated. Therefore Mo(II), $d^4$, 18 electrons follows fromThe curved delocalization mark across the three-carbon ligand must not be read as a separate alkene or alkyne.

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.

The products are hexanenitrile, $\mathrm{C}{\mathrm{H}}_3(\mathrm{C}{\mathrm{H}}_2{)}_4\mathrm{C}\equiv \mathrm{N}$, and ethyl triphenylmethyl ether, $\mathrm{P}{\mathrm{h}}_3\mathrm{C}-\mathrm{O}-\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_3$. The nitrile carbon comes from cyanide, so replacing chloride in the five-carbon chain gives a six-carbon product.

For the primary chloride, the carbon-end lone pair of $\mathrm{C}{\mathrm{N}}^{-}$ attacks the carbon carrying Cl from the back side while the C–Cl bond pair moves onto Cl. The single concerted nucleophilic substitution step forms the C–C bond and releases $\mathrm{C}{\mathrm{l}}^{-}$. Attack through the carbon end gives the nitrile, not an isocyanide.

For the triphenylmethyl chloride, the C–Cl bond first undergoes heterolysis to $\mathrm{P}{\mathrm{h}}_3{\mathrm{C}}^{+}$ and $\mathrm{C}{\mathrm{l}}^{-}$. An ethanol oxygen lone pair then attacks the carbocation, giving $\mathrm{P}{\mathrm{h}}_3\mathrm{C}-{\mathrm{O}}^{+}(\mathrm{H})\mathrm{Et}$. Finally pyridine accepts that O–H proton, with the O–H bond pair returning to oxygen, giving the neutral ether and pyridinium chloride.

The molecule combines three chemically different binding regions.

These assignments identify the preferred local sites, rather than guaranteeing three independent binding constants: occupancy can alter conformation, electrostatics and the affinity of neighboring sites.

A pyridine ring bearing carboxylic acid substituents at positions 2 and 5. Removing both acidic protons gives a dianion with a neighbouring N,O chelation site and a remote carboxylate site.