Anilinium ion 2026-10-20
An anilinium ion is an aniline protonated at nitrogen. nitrogen protonation removes its lone pair from donation to the aromatic ring and usually gives strong deactivation and meta direction in electrophilic substitution.
Benzylic carbon 2026-10-20
An sp3 carbon directly attached to an aromatic ring.
Ethoxy group 2026-10-20
Friedel-Crafts alkylation 2026-10-20
Nitroalkane benzannulation 2026-10-20
Past exam of the chemistry B course of the University of Cambridge 2021 Paper 1 B1 a iii Solution 2026-10-20
The methoxy groups activate the aromatic ring by electron donation through chemical resonance and direct electrophilic substitution ortho and para. Their para positions are already occupied by one another, so the first alkylation occurs at an available ortho position; the four initially vacant positions are equivalent by symmetry.
After placing the first tert-amyl group at position 2, position 5 is favored for the second substitution. It is para to that bulky group and ortho to the other methoxy group. Position 3 is crowded between methoxy and tert-amyl substituents, while position 6 is near the first tert-amyl group across the intervening methoxy-bearing carbon. The 2,5 arrangement distributes the bulky groups on opposite parts of the ring and preserves the equivalence seen in the spectrum.
Alkyl groups do not electronically deactivate the ring, so the stopping point needs a steric explanation. After two such substitutions the remaining positions are much more congested, each adjacent to a bulky tert-amyl group and a methoxy group. Two large substituents are accommodated much more readily than a third, despite excess alcohol. The formula and NMR identify this sterically favored disubstitution; they do not imply that a Friedel-Crafts reaction always stops after two alkylations.
Past exam of the chemistry B course of the University of Cambridge 2021 Paper 1 B1 b ii Solution 2026-10-20
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 bond forms the new aryl–carbonyl-carbon bond, the ketone 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.
Past exam of the chemistry B course of the University of Cambridge 2022 Paper 2 B9 c Solution 2026-10-20
Join three trigonal BO 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.
Past exam of the chemistry B course of the University of Cambridge 2023 Paper 1 B1 c i Solution 2026-10-20
Number the phenolic carbon C1, with methyl groups at C2 and C5. All three remaining aromatic hydrogens are replaced: is 3,4,6-tribromo-2,5-dimethylphenol. The starting phenol is ; replacing three hydrogens gives , 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]
Total Synthesis of (Nor)illudalane Sesquiterpenes Based on a C(sp3)-H Activation Strategy by Romain Melot and others. 2019, Scheme 6, upper route from 2,5-xylenol
Past exam of the chemistry B course of the University of Cambridge 2023 Paper 1 B4 a i Solution 2026-10-20
B and C are the two geometrical ketoxime isomers, formed by replacing the reactive ring C=O with C=N–OH. The separate spiroacetal remains intact. Define as the original ketone carbon, as the adjacent ring path towards the aromatic ring, and as the adjacent path towards the spiroacetal-bearing junction. These are the two carbon substituents directly bonded to , not detachable independent fragments.
In B, N–OH is anti to the bond across C=N; in C, N–OH is anti to . Those geometries predict exactly the respective nitrogen insertions in D and E. Both products have formula : condensation adds NH2OH to and removes H2O. An alternative ring regioisomer or reaction at the protected ketone would not represent the pair of oxime geometrical isomers.
Past exam of the chemistry B course of the University of Cambridge 2024 Paper 1 B2 b ii Solution 2026-10-20
The same base first makes . This time addition occurs at a carbon bearing hydrogen para to the nitro group, giving a sigma-H adduct. The attacked carbon temporarily bears both its original H and the new carbon substituent. Negative charge is again delocalized through the ring toward the nitro oxygens.
A further base removes the hydrogen on that aromatic addition site. Its C–H electrons form an exocyclic bond toward the incoming carbon, while that carbon expels phenoxide. This is a base-induced elimination across the new ring–side-chain bond. Redistribution of the nitronate/quinoid pi system restores an aromatic ring and gives the side-chain carbanion . protonation during acidic workup yields .
This is vicarious nucleophilic substitution: the ring loses hydrogen while the nucleophile sacrifices its phenoxy leaving group. hydrogen is not expelled as hydride, and the product's side-chain hydrogen can be supplied on workup rather than moving directly from the ring in a compulsory hydride shift. Base and isotope experiments in the primary mechanistic work support reversible addition followed by base-induced elimination, while not uniquely choosing every E2/E1cb limit.[ref]
Reaction of organic anions. Part 108. On the mechanism of the vicarious nucleophilic substitution of hydrogen in nitroarenes by Mieczysław Mąkosza and Tomasz Glinka. 1983, pp. 3860–3861, isotope effects, base dependence and hydrogen/halogen competition
The displayed para product is consistent with access to the less crowded activated position. Past exam of the chemistry B course of the University of Cambridge 2025 Paper 1 B1 b i Solution 2026-10-20
The first stage is a Stork enamine reaction. Pyrrolidine attacks the ketone carbonyl, and proton transfers give a carbinolamine. Loss of water gives an iminium ion; removal of the benzylic alpha proton produces the enamine conjugated with the aromatic ring. Removal of water and prolonged heating favour this condensation. That conjugated enamine is preferred over the alternative with its double bond on the other side of the original carbonyl.
The nucleophilic enamine carbon attacks the methylene carbon of bromoacetonitrile by an SN2 reaction, displacing bromide and making the carbon–carbon bond. It is the carbon beside the aromatic fusion that acquires , as in D. Finally, water adds to the alkylated iminium carbon; proton transfers and cleavage of the C–N bond regenerate the ketone and release pyrrolidine. This explains carbon alkylation, its position, and restoration of the carbonyl rather than treating the amine as a permanent substituent.
Past exam of the chemistry B course of the University of Cambridge 2025 Paper 1 B2 b ii Solution 2026-10-20
The qualitative nucleophilicity order is -OMe -Me H -Cl -NO, matching the product-yield order at the stated stopping time. Electron donation raises the availability of the aniline lone pair for attack. A para methoxy group donates by resonance, while methyl is a weaker donor. Chlorine withdraws inductively and gives an aniline less nucleophilic than the unsubstituted compound. A para nitro group strongly withdraws by resonance: electron flow from nitrogen through the ring into the nitro group stabilizes a charge-separated form and greatly reduces nucleophilicity.
The key arrows for the nitroaniline comparison move the nitrogen lone pair into its own aromatic ring, shift the ring bonds toward the para nitro group, and shift a nitro N=O bond toward oxygen. The methoxy donor produces the opposite electronic bias. These are finite-time yields, so they support a kinetic explanation qualitatively; they are not by themselves measured rate constants or equilibrium constants.
Phenoxide 2026-10-20
Phenoxide is the anion formed by deprotonating phenol. Its oxygen is an effective nucleophilic site, while conjugation distributes some electron density into the aromatic ring.
Pyrazine 2026-10-20
A planar six-membered aromatic ring with nitrogen at opposite ring sites. Its molecular point group is the D2h point group. The six ring atomic orbitals form a six-electron conjugated system; the in-plane nitrogen lone pairs are outside this basis.
Pyrrolyl ligand 2026-10-20
The anion formed by N–H deprotonation of pyrrole. It can bind through nitrogen as a two-electron donor, or through its aromatic ring as a six-electron donor. Specifying hapticity is essential for electron counting.

