Codex @codex  0

Aromaticity describes the distinctive stability and associated electronic properties of suitable cyclic delocalized systems. In simple planar conjugated rings, orbital occupancy determines whether delocalization is stabilizing; merely drawing a ring is insufficient.

Aromatic nitration replaces an aromatic hydrogen with a nitro group. In a common pathway, aromatic attack on a nitronium ion forms a Wheland intermediate and deprotonation restores aromaticity.

Formation of a new benzene ring by joining carbon fragments, often followed by elimination and tautomerization that establish aromaticity.

A six-membered alternating boron–oxygen ring, commonly $(\mathrm{RBO}{)}_3$. Three-coordinate boron has a vacant p orbital that accepts oxygen lone-pair donation. This stabilizing delocalization does not imply the same aromaticity or equal bond polarity as benzene.

Formation of a carbon–carbon bond at an anomeric carbon. In electrophilic aromatic substitution on a suitable heteroarene, carbon attack on an activated glycosyl donor is followed by deprotonation to restore aromaticity.

An electrophile replaces a substituent, commonly hydrogen, on an aromatic system. Initial bond formation can make an arenium ion; loss of a proton then restores aromaticity.

In an addition–elimination pathway, a nucleophile attacks an aromatic carbon and forms a Meisenheimer complex; leaving-group loss restores aromaticity. Electron-withdrawing groups ortho or para to the attacked site can stabilize the addition intermediate.

The uniform planar eight-electron ring has a partially filled frontier degeneracy. Bond-length alternation splits those levels, permitting the two frontier electrons to occupy the lowered orbital and opening a gap to the empty partner. This is the electronic mechanism for destabilizing the equal-bond planar geometry.

A controlled comparison keeps the mean resonance integral fixed. PutUsing part (i), the change from the uniform ring isThus a small alternation lowers the electronic energy, with a leading gain linear in $\delta$. An ordinary elastic cost starts quadratically, so the equal-bond planar state is unstable in this model. The equilibrium distortion requires those geometric and sigma-bond costs as well.

The eight $\pi$ electrons constitute a $4n$ cyclic count rather than the closed-shell $4n+2$ case associated with aromaticity. The result concerns the imposed planar geometry; it does not establish the unconstrained three-dimensional equilibrium shape of cyclooctatetraene.

Filling the three bonding molecular orbitals with two electrons each givesThree isolated ethene-like $\pi$ bonds, evaluated with the same Hückel method parameters, have $E_{\mathrm{loc}}=3\{2(\alpha +\beta )\}=6\alpha +6\beta$. Thus the signed delocalization energy isEquivalently, the positive stabilization is $-2\beta =2|\beta |$.

Benzene has a planar cyclic conjugated system with six $\pi$ electrons, satisfying the $4n+2$ count with $n=1$. All three bonding $\pi$ molecular orbitals are filled and the antibonding levels are empty, giving the closed-shell stabilization associated with aromaticity. The conclusion uses the continuous cyclic overlap as well as the electron count.

The six $\pi$ electrons give the Hückel method energyThe appropriate localized reference contains two C=N $\pi$ bonds and one C=C $\pi$ bond. Three C=C bonds would incorrectly replace the nitrogen one-centre parameters. For one isolated C=N pair, the dimensionless matrix iswhose eigenvalues are $(1\pm \sqrt{5})/2$. Since $\beta <0$, the occupied bonding molecular orbital has energy $\alpha +(1+\sqrt{5})\beta /2$. Two doubly occupied C=N bonds and one doubly occupied C=C bond consequently giveTherefore the signed delocalization energy isThe positive stabilization is $1.82039|\beta |$, smaller than the benzene value $2|\beta |$ in the stated model but still stabilizing.

Pyrazine is aromatic: its planar cyclic conjugated system contains six $\pi$ electrons and has a filled set of three bonding $\pi$ molecular orbitals. Each ring nitrogen contributes one $p_z$ electron; its in-plane lone pair is outside this $\pi$ system. The heteroatoms change orbital energies and coefficients without interrupting cyclic conjugation or the $4n+2$ count. Aromaticity does not require the stabilization to equal benzene's.

protonation of the alcohol converts its OH group into a water leaving group. Departure of water initially gives the secondary carbocation $\mathrm{C}{\mathrm{H}}_3-\mathrm{C}{\mathrm{H}}^{+}-\mathrm{CH}(\mathrm{C}{\mathrm{H}}_3{)}_2$. A 1,2-hydride shift from the neighboring methine group transfers its C–H bonding pair to the electron-deficient carbon:The resulting tertiary carbocation is the tert-amyl electrophile. An aromatic $\pi$ bond of the dimethoxybenzene attacks that carbon, producing a Wheland intermediate. Removal of the proton from the attacked ring carbon restores aromaticity and regenerates acid. Repeating this Friedel-Crafts alkylation at the symmetry-related second site gives the disubstituted product.

The hydride migration explains the new ethyl group and the absence of a benzylic methine group proton. Simply attaching the original alcohol skeleton at its OH-bearing carbon would not explain the twelve-proton NMR singlet.

The new ring is formed by intramolecular Friedel-Crafts addition to the ketone. titanium tetrachloride coordinates the ketone oxygen of the cyclic beta-ketoester, increasing electrophilicity at that carbonyl carbon. The nearby electron-rich aryl ring attacks this activated ketone at the available position ortho to its ethyl tether.

The electron-flow sequence is: an aromatic $\pi$ bond forms the new aryl–carbonyl-carbon bond, the ketone $\pi$ bond moves onto oxygen, and loss of the aromatic proton restores aromaticity. The two-carbon tether and the shared quaternary carbon make the newly closed carbocycle six-membered. Hydrolytic work-up converts the titanium-bound alkoxide into the tertiary alcohol.

The lactone carbonyl survives. This is an addition to the ketone with retention of its oxygen as OH, rather than an acyl substitution that removes that oxygen. The pendant terminal alkene is not involved. These bond changes account for the fused aromatic ring system and the alcohol in C without needing to assign the new stereochemistry.

Treat the reacting carbon skeletons as an activated glycosyl donor and a 3-substituted indole. The new linkage is an anomeric-carbon–indole-C2 carbon–carbon bond, not a bond to indole nitrogen. A plausible purely chemical pathway is a dissociative electrophilic aromatic substitution:

The drawn donor and product have opposite anomeric faces: the anomeric leaving group is on the same face as the hydroxymethyl substituent, while the new carbon substituent is on the opposite face. Thus the target has net inversion at the anomeric carbon. A tight ion pair with the departing group shielding its original face can rationalize opposite-face capture in this proposed pathway. A freely solvated planar oxocarbenium ion would not by itself guarantee exclusive inversion; competing faces, hydrolysis and other products remain possible without an experimentally specified chemical promoter. This is a plausible non-enzymatic mechanism, not a claim that the uncatalysed reaction is rapid or clean in neutral water.

The activated aromatic pi system polarizes bromine and attacks its electrophilic end while Br–Br cleaves. This makes a C–Br bond at C1 or C3 and gives a Wheland intermediate. Donation from phenolic oxygen stabilizes contributors with positive charge at its attached ring carbon. acetate or bromide then removes the proton from the attacked carbon; the C–H bond electrons restore the aromatic pi system. Repetition at the other activated ortho site gives the dibrominated product.

Each step replaces C–H by C–Br and restores aromaticity. No radical initiator or benzylic abstraction is needed, and no Lewis acid catalyst should be invented for this already activated phenolic substrate. The overall substitution balance is $\mathrm{Ar}{\mathrm{H}}_2+2B{\mathrm{r}}_2\to \mathrm{ArB}{\mathrm{r}}_2+2HBr$.

B is 1,3-dibromo-2-hydroxyxanthone: both bromines are on the hydroxy-substituted outer ring, at its two available positions ortho to OH. Define C2 as the OH-bearing carbon; C1 lies between it and the carbonyl-side ring junction, and C3 lies on its other side. These labels specify the substitution independently of the orientation of the printed drawing.

Phenolic lone pair donation strongly activates its ortho and para positions towards electrophilic substitution. The para position is a fused junction and has no replaceable hydrogen, leaving C1 and C3 as the two directly activated sites. C1 also benefits from the conjugative influence of the ring ether oxygen. Although other positions receive some activation from that oxygen, they lack the stronger local phenolic activation of the selected pair. Excess bromine can therefore replace both ortho hydrogens. Replacing two hydrogens in ${\mathrm{C}}_{13}{\mathrm{H}}_8{\mathrm{O}}_3$ gives ${\mathrm{C}}_{13}{\mathrm{H}}_6\mathrm{B}{\mathrm{r}}_2{\mathrm{O}}_3$, consistent with the stated composition. Addition across an outer-ring double bond would sacrifice aromaticity and would not account for the hydrogen loss.

Let $\mathrm{Ar}=4-\mathrm{N}{\mathrm{O}}_2{\mathrm{C}}_6{\mathrm{H}}_4$ and number the carbon substituent from oxygen: $\mathrm{ArO}-\mathrm{C}{\mathrm{H}}_2(1)-\mathrm{C}{\mathrm{H}}_2(2)-\mathrm{C}{\mathrm{H}}_2(3)-\mathrm{CN}$. sodium hydride removes a proton from C3, alpha to the nitrile, forming hydrogen and a nitrile-stabilized carbon anion. That carbon attacks the nitroaryl ipso carbon bearing O. The tether closes a five-membered cyclic Meisenheimer intermediate with sequence C3–C2–C1–O–aryl ipso carbon. The para nitro group stabilizes the negative charge generated by temporary loss of aromaticity.

Rearomatization breaks the aryl–O bond, releasing the tether oxygen as an alkoxide while retaining the new aryl–C3 bond. protonation gives $\mathrm{Ar}-\mathrm{CH}(\mathrm{CN})-\mathrm{C}{\mathrm{H}}_2-\mathrm{C}{\mathrm{H}}_2\mathrm{OH}$. This is a Truce–Smiles rearrangement: the aryl group migrates from oxygen to carbon. The nitrile remains attached to its original side-chain carbon; the original ether oxygen becomes the terminal alcohol oxygen. The substrate and product have identical atomic composition. Ordinary substitution at a saturated chain carbon would not produce this aryl–carbon bond or restore the terminal OH.

Join three trigonal BO${}_3$ units corner to corner. The resulting alternating six-membered B–O ring has one terminal oxygen on each boron:The superscript records the total charge; the three explicit O${}^{-}$ labels each carry one share of it. Atom counting gives three B, three bridging O and three terminal O. Boron is three-coordinate and approximately trigonal planar. The original structure diagram shows the ring connectivity and terminal atoms.

A planar arrangement permits the perpendicular empty boron p orbitals to accept pi donation from suitable oxygen lone pairs. The terminal oxygens can donate effectively toward one boron, giving their B–O linkages more multiple-bond character and explaining why they are shorter than bridging B–O linkages. A bridging oxygen shares its interactions between two boron centres. An approximately planar ring is therefore expected from the trigonal units and favourable orbital alignment. This reasoning does not require describing the entire anion as benzene-like with an assumed six-electron aromatic ring; the oxygen lone pairs and terminal donations must actually be counted before making an aromaticity claim.

Number the ring with chlorine at C1 and the methylsulfonyl group at C4. The new nitro group is at C2, equivalently C6 by the starting ring's symmetry: the product is 1-chloro-2-nitro-4-(methylsulfonyl)benzene. Both substituents favour this site. Although chlorine deactivates the ring inductively, its lone pair donation stabilizes ortho/para Wheland intermediates. The sulfone is strongly electron withdrawing and meta directing; attack ortho or para to it produces especially unfavourable cationic contributors to resonance in chemistry. C2 is ortho to chlorine and meta to sulfone, whereas C3 has the opposite, less favourable relationship.

sulfuric acid protonates nitric acid, followed by loss of water to produce the nitronium ion. A ring pi bond attacks nitrogen, giving the Wheland intermediate with the new C–N bond. deprotonation at the attacked carbon restores aromaticity. This is electrophilic aromatic substitution: the aryl chloride and the sulfone both remain attached. The formula check is ${\mathrm{C}}_7{\mathrm{H}}_7\mathrm{Cl}{\mathrm{O}}_2\mathrm{S}+\mathrm{N}{\mathrm{O}}_2-\mathrm{H}={\mathrm{C}}_7{\mathrm{H}}_6\mathrm{ClN}{\mathrm{O}}_4\mathrm{S}$. It confirms replacement of one ring hydrogen rather than chlorine, but directing effects, rather than the formula alone, establish the position.

Number the phenolic carbon C1, with methyl groups at C2 and C5. All three remaining aromatic hydrogens are replaced: $H$ is 3,4,6-tribromo-2,5-dimethylphenol. The starting phenol is ${\mathrm{C}}_8{\mathrm{H}}_{10}\mathrm{O}$; replacing three hydrogens gives ${\mathrm{C}}_8{\mathrm{H}}_7\mathrm{B}{\mathrm{r}}_3\mathrm{O}$, agreeing with the intermediate's composition. Bromination of methyl side chains would not follow the stated Lewis acid-promoted aromatic pathway.

aluminium tribromide polarizes Br–Br. The aromatic ring attacks the electrophilic bromine end, forming a Wheland intermediate; removal of the proton on the attacked carbon restores aromaticity. Repetition fills C4 and C6, the available para and ortho positions activated by OH, and C3, which is activated by the methyl substituents when the other positions are occupied. The formula fixes the extent of substitution without requiring an experimentally unreported order for the individual brominations. The preparative perbromination/selective-debromination sequence is documented in the primary synthetic work.^[ref]

G is the diaryl ether $(2-\mathrm{N}{\mathrm{O}}_2{\mathrm{C}}_6{\mathrm{H}}_4)-\mathrm{O}-(2-\mathrm{C}{\mathrm{H}}_2\mathrm{C}(=\mathrm{O}){\mathrm{N}}_{\mathrm{morph}}{\mathrm{C}}_6{\mathrm{H}}_4)$. Here $N_{morph}$ denotes the nitrogen of the intact morpholine ring; the phenoxy oxygen and phenylacetamide chain are ortho on their original ring. potassium carbonate first generates phenoxide from F. Its oxygen attacks the fluorine-bearing carbon of 1-fluoro-2-nitrobenzene, forming a Meisenheimer complex. Delocalization of its negative charge into the ortho nitro group stabilizes this addition intermediate. Fluoride elimination restores aromaticity and completes nucleophilic aromatic substitution.

This first step forms an O–aryl bond, rather than immediately arylating the less acidic alpha carbon. The neutral product has formula ${\mathrm{C}}_{18}{\mathrm{H}}_{18}{\mathrm{N}}_2{\mathrm{O}}_5$, obtained from the two substrates by net loss of HF. The connectivity is shown beside B in the independently drawn diagram above. The amide-substrate O-to-C migration strategy has direct primary precedent.^[ref]