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Disconnect the new bond between the two six-carbon units. One portion of bromocyclohexane supplies a grignard reagent; another supplies cyclohexene oxide:A bulky strong base can favour E2 elimination in the alkene-forming step; a peracid such as meta-chloroperoxybenzoic acid provides epoxidation. None of its carbons enters the target.

Treat cyclohexene oxide with cyclohexylmagnesium bromide, then perform acidic aqueous work-up. Backside attack on one epoxide carbon breaks its C–O bond and installs cyclohexyl opposite the retained oxygen. Protonation of the oxygen gives trans-2-cyclohexylcyclohexan-1-ol, the required relative configuration. The two six-carbon reactant portions supply all 12 target carbons. An achiral synthesis produces a racemic mixture trans pair; obtaining only the particular enantiomer represented by the drawing would require resolution or an enantioselective step. The useful stable chair can place the two large trans-1,2 substituents approximately diequatorial.

For the ordinary ester, the carbon nucleophile of a grignard reagent attacks the carbonyl carbon and moves the C=O pi pair onto oxygen. Collapse of the tetrahedral intermediate expels ethoxide, transiently giving ${\mathrm{R}}^1\mathrm{COR}$. This ketone is more electrophilic than the starting ester because it lacks an alkoxy group's stabilizing resonance donation. A second equivalent of the grignard reagent therefore adds to it before isolation:The negatively charged intermediates are associated with magnesium counterions. Merely reducing the amount of reagent to one equivalent does not generally trap the ketone selectively; it can leave unreacted ester together with the double-addition product.

In the first successful preparation, magnesium insertion into the primary C–Br bond gives n-butylmagnesium bromide. The carbon chain has four carbons, so its acetylation produces hexan-2-one, $\mathrm{C}{\mathrm{H}}_3\mathrm{CO}(\mathrm{C}{\mathrm{H}}_2{)}_3\mathrm{C}{\mathrm{H}}_3$, not pentan-2-one. Attack on acetic anhydride forms $\mathrm{C}{\mathrm{H}}_3\mathrm{C}({\mathrm{O}}^{-})(\mathrm{nBu})\mathrm{OCOC}{\mathrm{H}}_3$. Coordination of the anhydride-derived oxygens to magnesium can stabilize this tetrahedral addition complex. Under suitably cold, controlled-addition conditions it survives sufficiently to delay release of the free ketone until aqueous work-up has destroyed the remaining grignard reagent. Protonation followed by collapse releases acetate/acetic acid and the ketone.

This interpretation has direct experimental precedent: Melvin S. Newman and William T. Booth Jr. obtained methyl ketones by cold addition of grignard reagents to acetic anhydride, attributing selectivity to a persistent addition complex. The temperature and reagent-addition regime are genuine conditions on the explanation, rather than proof that an anhydride can never undergo overaddition; the exam scheme omits them.^[ref]

For the benzonitrile route, n-butyl attacks the nitrile carbon and one C≡N pi pair moves to nitrogen, producing the magnesium ketimine salt $\mathrm{PhC}(\mathrm{nBu})={\mathrm{N}}^{-}\mathrm{MgB}{\mathrm{r}}^{+}$. This anionic nitrogen-containing intermediate resists a second nucleophilic addition much more than a neutral carbonyl does. The ketone $\mathrm{PhC}(=\mathrm{O})\mathrm{nBu}$ appears only during acidic aqueous work-up, after organomagnesium has been quenched. Protonation gives the iminium ion; water attacks its carbon, proton transfers give a carbinolamine, and C–N cleavage from its nitrogen-protonated form produces the protonated ketone and ammonia, followed by deprotonation. In the acidic mixture ammonia is present largely as ammonium. This sequence explains why nitrile addition stops at a ketone without requiring the ketone itself to be resistant to grignard reagents.

The atom-labelled hydrolysis sequence isWater's lone pair attacks the iminium carbon while its C=N pi pair moves to nitrogen; the carbinolamine oxygen then reforms C=O while the C–N pair moves to the departing protonated amino group. Final deprotonation gives the neutral ketone. The arrows involving protonation and hydrolysis also exchange protons with the acidic aqueous reservoir; the condensed sequence tracks the carbon–nitrogen and carbon–oxygen transformations rather than displaying every solvent molecule.

First, proton transfer from acetic acid to the carbon of diazomethane gives methyldiazonium and an acetate ion:Draw an electron-pair arrow from carbon, using the carbon-negative resonance contributor, to the acidic hydrogen, and another from the acid O–H bond to oxygen. Second, an acetate oxygen attacks the diazonium methyl carbon by an SN2 reaction; an arrow from the C–N bond to nitrogen expels neutral ${\mathrm{N}}_2$:The acid's hydroxyl hydrogen has become the third hydrogen of the newly installed methyl group. This is a methyl-transfer sequence driven by formation of molecular nitrogen, without forming a free methyl carbocation or invoking Fischer esterification.

An intense band near $2088\mathrm{c}{\mathrm{m}}^{-1}$ is consistent with the high-frequency coupled C–N–N stretching motion of a diazo group, with substantial N–N multiple-bond character. It lies above a typical isolated C=N double-bond stretch, and supports contributions with a strong N–N connection. It does not establish that the molecule is a literal localized C–N single bond plus an unperturbed N≡N triple bond: a normal mode moves the coupled framework, and the resonance contributors describe one electronic structure.

Protonation at the nucleophilic carbon gives methyldiazonium, $\mathrm{C}{\mathrm{H}}_3-{\mathrm{N}}^{+}\equiv \mathrm{N}$. The carbon-rich highest occupied molecular orbital and the carbon-negative resonance contributor rationalize this site. Subsequent displacement of molecular nitrogen can be strongly favourable; the initial protonation product still contains both nitrogens.

The four sigma bonds are the two C–H connections, C–N and N–N, containing eight electrons. A terminal-nitrogen axial lone pair contains two more. Of the remaining six electrons, two occupy an N–N pi bond lying in the molecular plane, while four occupy a perpendicular three-centre four-electron pi bond extending over C–N–N. There are therefore 16 valence electrons in total.

For orientation, a three-site model with equal on-site energy $\alpha$ and adjacent interaction $\beta <0$ gives perpendicular pi levels at $\alpha +\sqrt{2}\beta$, $\alpha$, and $\alpha -\sqrt{2}\beta$. The lower two are doubly occupied and the upper is empty. The middle level has a terminal-end combination with little central-atom amplitude; unequal C and N orbital energies change its coefficients and make it only approximately nonbonding.

The highest occupied molecular orbital is the upper occupied perpendicular pi orbital, with substantial carbon amplitude, broadly above the strongly bonding sigma framework and the lower occupied pi orbitals. It is not the terminal axial nitrogen lone pair. Real diazomethane has C–N bonding and N–N antibonding character in this frontier orbital, rather than being an exactly nonbonding equal-site model. Photoelectron and Rydberg-state experiments support this assignment.^[ref]

An internally consistent elementary orbital hybridisation model is $s{p}^2$ at carbon and $sp$ at both nitrogen atoms; the hydrogens use $1s$ orbitals. The three carbon hybrids form two C–H sigma bonds and one C–N sigma bond, leaving one perpendicular p orbital. The central nitrogen uses opposite $sp$ hybrids for its two collinear sigma connections, leaving two perpendicular p orbitals for the two pi systems.

At terminal nitrogen, one $sp$ hybrid points toward the central nitrogen and the other houses an axial lone pair. One perpendicular p orbital forms the second, N–N-only pi system; the other participates in the three-atom C–N–N pi system. This description remains the same across both resonance contributors; counting lone-pair domains in just one drawing and assigning a different terminal hybridisation to each contributor would obscure the delocalization. orbital hybridisation is a useful choice of localized basis, not a uniquely measurable label: advanced treatments of diazomethane require a superposition of bonding patterns.^[ref]

The framework has a planar ${\mathrm{H}}_2\mathrm{C}$ group attached to a straight C–N–N axis. Two useful octet resonance contributors areIn the left contributor, the terminal nitrogen has two lone pairs; in the right, the terminal nitrogen and carbon each have one. Each formula has 16 valence electrons and zero net electric charge. The double-headed resonance arrow changes the electronic allocation while retaining the same nuclear framework; these are not two different equilibrium species.

Take the bond axis as $z$. The heavy-atom $2s$ and $2p_z$ atomic orbitals can mix with hydrogen $1s$ to make three sigma-symmetry molecular orbitals. In the qualitative valence ordering, these are a low predominantly heavy-atom $2s$ level, a bonding $\sigma$ level and a higher ${\sigma }^{\ast }$ level. The perpendicular $2p_x,2p_y$ orbitals form a degenerate approximately nonbonding $\pi$ pair because hydrogen has no valence orbitals of the matching sideways symmetry. The filled heavy-atom $1s$ core is omitted from the diagram.

Counting valence electrons givesThus both exhibit paramagnetism: CH has one unpaired electron and NH has two in its ground state. For NH, Hund's rules favour separate parallel-spin occupation of the two degenerate $\pi$ orbitals. The corresponding observed ground terms are $X{}^2\Pi$ and $X{}^3{\Sigma }^{-}$; the total bond order is approximately one in this simple sigma-bond/nonbonding-pi description.^[ref]

The nitrogen-derived orbital energies lie lower than the corresponding carbon-derived ones on an approximate common energy scale, while the isolated hydrogen $1s$ reference is unchanged. The bonding orbitals consequently have different polarizations and mixing coefficients; drawing identical energies for C and N would miss the screening trend. The figure specifies ordering and occupancy, not measured gaps or a numerical prediction from (a).

The $2s$ atomic orbital is spherically symmetric and has one radial node. A real $2p$ atomic orbital has two lobes of opposite quantum phase, one atomic-orbital nodal plane through the nucleus and no radial node. The three spatial orientations $2p_x,2p_y,2p_z$ remain degenerate in a spherically symmetric atom when external fields and fine structure are neglected.

In a many-electron atom, $2s$ penetrates the screened core more effectively than $2p$, experiences a stronger attraction and is lower in energy: ${\epsilon }_{2s}<{\epsilon }_{2p}$. The hydrogen-like ion degeneracy therefore does not survive screening. Both valence orbital types are generally more tightly bound and spatially contracted in nitrogen than in carbon, because of the larger effective nuclear charge. These are central-field orbital statements; the detailed atomic multiplet structure also involves electron repulsion and Hund's rules.

The configurations are $\mathrm{C}:1s^{2}2s^{2}2p^2$ and $\mathrm{N}:1s^{2}2s^{2}2p^3$. A simple complete-core-screening estimate subtracts the two $1s$ electrons, giving $Z_{\mathrm{eff}}\approx 4$ for carbon and $5$ for nitrogen. This neglects screening by the other valence electrons, and is an explicitly rough estimate.

A more systematic estimate using Slater's rules includes $0.85$ for each $1s$ electron and $0.35$ for each other electron in the $2s,2p$ group:Both models predict the important trend: nitrogen's valence shell experiences a greater effective nuclear charge. There is no unique numerical effective nuclear charge without specifying an orbital and a screening prescription; the one-electron formula in (a) does not by itself define neutral-atom orbital energies.^[ref]

The one-electron species are ${\mathrm{C}}^{5+}$ and ${\mathrm{N}}^{6+}$, so there is no electron–electron screening and the nuclear charges remain $Z=6$ and $7$.Within the stated hydrogen-like ion model, $2s$ and all three $2p$ atomic orbitals of a given species have the same energy. The energy zero is a separated electron and the residual nucleus.