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The two starting ester groups are adjacent on the benzene ring. Select one ester carbon as the site receiving both incoming methyl groups. A first methyl addition gives a tetrahedral intermediate; its collapse expels methoxide to give the ortho-substituted ketone $\mathrm{o}-!\mathrm{Me}{\mathrm{O}}_2\mathrm{C}-{\mathrm{C}}_6{\mathrm{H}}_4-\mathrm{C}(=\mathrm{O})\mathrm{Me}$. A second methyl addition produces $\mathrm{o}-!\mathrm{Me}{\mathrm{O}}_2\mathrm{C}-{\mathrm{C}}_6{\mathrm{H}}_4-\mathrm{C}({\mathrm{O}}^{-})\mathrm{M}{\mathrm{e}}_2$, with the alkoxide bound to magnesium.

The newly generated oxygen is now positioned to attack the remaining ester intramolecularly. Draw an arrow from its lone pair to that ester carbonyl and another from the C=O pi bond to its oxygen. Collapse of the resulting tetrahedral intermediate reforms the carbonyl and expels methoxide, closing a five-membered lactone. Protonation of magnesium-bound species in the work-up gives 3,3-Dimethylphthalide. Equivalently, protonation can first give the ortho hydroxy ester, followed by acid-assisted lactonization with loss of methanol; both descriptions show the same required bond formation.

The oxygen in the new ring is the oxygen of the first ester's original carbonyl; the surviving lactone carbonyl comes from the second ester. The product retains the six ring carbons and both original acyl carbons, loses the two ester methoxy carbons, and gains two carbons from methylmagnesium bromide. The favourable intramolecular five-membered closure accounts for the displayed product. It does not make the isolated lactone universally inert to a large excess of grignard reagent; stoichiometry and trapping/work-up conditions matter.

In molecular boron trichloride, boron uses three approximately $s{p}^2$ hybrids to form three equivalent B–Cl sigma bonds in a trigonal plane. Its perpendicular empty $2p$ orbital can accept electron density from the compatible combination of filled chlorine $3p$ lone pairs. The resulting pi donation is delocalized over the three B–Cl connections.

In the elementary model, one net pi bond is distributed over three equivalent sigma bonds, giving formal average B–Cl bond order $1+1/3=4/3$. The corresponding resonance contributors each put a B=Cl connection in a different position, with formal negative boron and positive donating chlorine. They do not mean one experimentally distinct double bond and two single bonds. The $4/3$ value is the idealized sigma-plus-delocalized-pi bookkeeping value; a computed bond index need not equal that exact rational number.

The product-forming pairing is $n(\mathrm{Et}{\mathrm{O}}^{-})\to {\pi }^{\ast }(\mathrm{C}=\mathrm{O})$. Orient an occupied ethoxide oxygen lobe toward the large carbon lobe of the carbonyl acceptor with matching phase. An oblique approach permits constructive overlap while the pi bond weakens; this is the usual Bürgi-Dunitz trajectory to acyl carbon. The high-energy occupied orbital on the anion makes ethoxide the stronger electron-pair donor than the neutral ester oxygen.

The reverse frontier pairing, $n(\mathrm{ester}\mathrm{O})\to \mathrm{empty}\mathrm{Li}$, gives oxygen-to-lithium coordination. It is a plausible stabilizing contact and can coexist with the first interaction, rather than constituting an alternative way to replace the ester alkoxy group. In the elementary frontier picture, the anionic donor-to-carbonyl interaction is the important pairing for the new covalent bond and is favoured by its donor strength and appropriate accepting coefficient at carbon. The Li contact can polarize the carbonyl and organize the reacting pair. Precise energy-gap rankings require a specified solvated lithium species; the sketches alone do not justify pretending that lithium coordination cannot occur.

There are $\boxed{4+2(1)+2(5)=16}$ valence electrons, all paired, so diazomethane exhibits diamagnetism. A localized orbital occupancy consistent with the double-bond construction in 2(c)(i) isHere $c,t$ mean central and terminal. Each listed occupied spatial orbital contains two electrons of opposite spin. The central N's two p atomic orbitals are perpendicular, so they can support the two perpendicular pi bonds. No central-N lone pair is left over: its positive formal charge accompanies four bonding pairs in this Lewis structure.

Allowing resonance mixes a terminal-N lone pair into the C–N–N $\pi$ orbitals. In the alternative terminal-$sp$ basis, the same electron count is four occupied $\sigma$ orbitals ($8$ electrons), one terminal-$sp$ lone pair ($2$), one perpendicular N–N $\pi$ orbital ($2$), and two occupied orbitals of a three-centre C–N–N $\pi$ system ($4$). The $16$-electron count and the paired ground state survive this change of description; the localized picture does not assert two perfectly isolated double bonds.

One consistent localized orbital hybridization description uses the Lewis structuresFor methyldiazonium, take C as $s{p}^3$ and both N atoms as $sp$. For diazomethane, the double-bond resonance contributor suggests C as $s{p}^2$, central N as $sp$, and terminal N as $s{p}^2$. The central N uses two opposite $sp$ hybrids for the linear C–N–N sigma bond framework. In the localized double-bond picture, terminal N uses one $s{p}^2$ hybrid for its sigma bond, two for lone pairs, and its remaining p atomic orbital for a pi bond.

These labels specify an approximate bonding construction, not uniquely measurable atomic properties. An alternative representation takes terminal N as $sp$, with one $sp$ lone pair and one p lone pair; the latter can participate in the delocalized C–N–N $\pi$ system. It describes the same resonance more explicitly. Neither construction requires an unpaired electron or a bent central N.

An N-bound thiocyanate need not make an exactly linear Eu–N–C linkage. Eu(III) bonding is predominantly electrostatic, with weak directional covalent contributions from the shielded $4f$ orbitals. There is consequently less energetic reward for enforcing a precisely collinear metal–donor arrangement than in a strongly directional transition-metal pi bond. In the crowded eight-coordinate sphere, bending the external NCS groups can reduce ligand–ligand repulsion while retaining favourable Eu–N distances. Angles of $150^{\circ }$–$160^{\circ }$ are compatible with bent N coordination; they do not require changing the linkage to sulfur. This concerns the Eu–N–C angle, not bending the internal N–C–S skeleton.

The Hückel orbital charge density and Hückel pi bond order giveEach site receives $1/2$ electron from the lowest orbital and one from the occupied nonbonding subspace, so $n_i=3/2$ at every carbon, summing to six. Relative to one neutral pi electron per carbon this is a pi charge of $-e/2$ at each site.

For adjacent sites, only the lowest orbital contributes, giving $p_{12}=p_{23}=p_{34}=p_{41}=1/2$. For opposite sites the nonbonding contribution reverses the sign: $p_{13}=p_{24}=-1/2$. These non-neighbour negative density-matrix entries are not extra sigma bonds or extra resonance integrals in the Hamiltonian. If only skeletal nearest-neighbour pi bond orders are requested, the four values $1/2$ are the relevant ones.

Write $H=\alpha I+K$ with the same carbon diagonal parameter throughout, and define $P$ to have entries $+1$ on one sublattice and $-1$ on the other. Bipartite connectivity impliesThus $Kc=ϵc$ implies $K(Pc)=-ϵ(Pc)$. This proves energy pairing about $\alpha$, $E=\alpha \pm ϵ$, and equal site-probability magnitudes in each paired pair, since $|(Pc{)}_i{|}^2=|c_i{|}^2$. This is the Coulson-Rushbrooke pairing property.^[ref]

For the usual neutral half-filled closed-shell occupation, the lower member of each pair is doubly occupied. Combining paired coefficients with completeness of the orthonormal basis gives one pi electron per carbon, so the model predicts no net pi charge separation. The same argument makes off-diagonal pi bond orders between sites on the same sublattice zero. These statements assume a neutral half-filled density and do not apply unchanged to the charged six-electron four-site ion considered separately.

An imbalance of sublattice sizes also guarantees at least $|N_A-N_B|$ nonbonding orbitals at energy $\alpha$, because the rectangular coupling block has insufficient rank to remove all its null space. None of the spectral pairing, equal paired densities, neutral charge uniformity or sublattice bond-order restrictions is required for a non-alternant graph. Long-range same-sublattice couplings or unequal diagonal parameters can likewise destroy the simple proof.

Silver(I) promotes iodide departure, with precipitation of AgI. The oxygen of dimethyl sulfoxide substitutes at the iodine-bearing carbon to form an alkoxydimethylsulfonium intermediate. Triethylamine removes the hydrogen on that carbon; collapse forms C=O and breaks O–S, releasing dimethyl sulfide. The result is ketone D by a Kornblum oxidation. In arrow terms, the C–H electrons form the carbonyl pi bond as the O–S bond breaks. The carbon bearing iodine becomes the carbonyl carbon; the existing bridged ether and acetonide are spectators. The iodine centre loses its tetrahedral stereochemistry because it becomes trigonal planar.

This is an acid-catalysed alpha-ketol rearrangement. Label the original ketone carbon $C_A$ and the adjacent tertiary alcohol carbon $C_B$. Protonate the oxygen on $C_A$ to activate that carbonyl. The appropriately aligned cage C–C sigma bond attached to $C_B$ migrates onto $C_A$, while the hydroxyl oxygen at $C_B$ supplies its lone pair to form the new $C_B=O$ bond. At the same time the old $C_A=O$ pi bond moves onto oxygen, which becomes the new $C_A$–OH. deprotonation regenerates the acid.

The migration reroutes the bridge rather than removing water: the original alcohol carbon becomes the ketone, and the original ketone carbon becomes the tertiary alcohol. A productive conformation aligns the migrating sigma orbital with the accepting carbonyl pi-star orbital; the oxygen lone pair simultaneously overlaps the developing p orbital at the donor carbon. The cage's geometry controls which adjacent sigma bond has the necessary alignment and which less strained skeletal isomer results. A freely rotating isolated carbocation or a simple proton shift cannot account for the changed cage connectivity.

A formal fourfold bond. In the familiar d-metal example, one sigma bond, two pi bonds and one delta bond are occupied, giving the configuration ${\sigma }^2{\pi }^4{\delta }^2$ and formal bond order four.