Past exam of the chemistry B course of the University of Cambridge 2024 Paper 2 B9 b ii Solution 2026-10-20
is a 1:1 nitrogen-bound adduct, with at one ring nitrogen. Its lone pair is diverted from N-to-B pi donation into the new coordinate bond. That nitrogen becomes approximately tetrahedral and the ring can pucker. The two B–N bonds adjacent to it lose pi character and lengthen; the neighboring unbound nitrogen atoms can donate more strongly into those borons, shortening the next pair of B–N bonds. The remaining pair forms a third class. The second structural panel marks these three two-bond sets; it represents connectivity and bond classes, not a measured geometry.
In the slow-exchange structure, two boron atoms flank the bound nitrogen and are equivalent to one another, while the third boron is unique. This gives two B environments in a ratio. At room temperature, rapid exchange of the Lewis acid among the three equivalent nitrogen sites averages every boron over the same environments, giving one boron-11 nuclear magnetic resonance signal. Cooling slows exchange relative to the resonance-frequency difference and resolves the two environments.
This is Chemical exchange in NMR, not restoration of equal instantaneous B–N bonds. A spectrum alone cannot distinguish intramolecular migration from reversible dissociation and reassociation; concentration dependence and crossover experiments would test that distinction. Related N-bound borazine adducts have been characterized by diffraction and variable-temperature NMR, supporting the structural logic while not supplying the exact temperature dependence of this particular hexamethyl compound.[ref]
Aluminium and gallium trihalide adducts of 2,4,6-triorganylborazines and the synthesis of triorganylborazinium tetrabromoaluminates by B. Gemünd and others. 2008, pp. 137–142, N-bound adduct structures and variable-temperature NMR
Past exam of the chemistry B course of the University of Cambridge 2024 Paper 2 B9 d Solution 2026-10-20
In the chromium complex, Cr(0) is and receives substantial donation from four carbonyl ligands. Its occupied orbitals can give pi backbonding into the boron-rich acceptor orbitals of the BN ring, while the nitrogen-rich occupied ring orbitals donate to the metal. This favors binding across the ring face, approximately , with all four ring atoms participating. A nearly planar ring maintains the p-orbital alignment needed for that interaction and accounts for comparable Cr–B and Cr–N distances. A four-electron ring donor also gives the useful count .
Ti(IV) in TiCl is and strongly electron deficient. It is a good acceptor for the two nitrogen lone pairs, but has no filled set for comparable backbonding into the boron acceptors. chromium therefore favors a face-bound pi complex, while titanium favors N,N sigma coordination. Donation makes the nitrogen environments more nearly tetrahedral and permits the 2,4-Diazadiboretidine ring to pucker. The borons are not directly bound in the same way, explaining . The distinction follows from electron availability, rather than simply calling every short metal–ring contact a separate ordinary sigma bond.
Past exam of the chemistry B course of the University of Cambridge 2025 Paper 1 B1 a i Solution 2026-10-20
Number the fused heterocycle as a 2,3-dihydrobenzofuran: oxygen is position 1 and the acetamido substituent occupies aromatic position 6. Position 5 is the preferred site of electrophilic substitution. It is para to the ring oxygen and ortho to the acetamido group, so both groups stabilize its Wheland intermediate by resonance donation. Position 7 also receives donation, but lies between the two substituents and is more crowded. Position 4 lacks the same combination of directing effects.
The aryl oxygen donates a lone pair into the aromatic system. The acetamido nitrogen can also donate into the ring, although its lone pair is shared with its own carbonyl group, so it is less activating than a free amino group. Comparing resonance-stabilized intermediates and steric accessibility gives the position-5 prediction; simply counting heteroatoms does not.
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.
Past exam of the chemistry B course of the University of Cambridge 2025 Paper 1 B4 b iv Solution 2026-10-20
The anomeric effect is the preference, in suitable cyclic acetals, for an electronegative substituent at an anomeric carbon to occupy an axial position even when simple steric reasoning favours equatorial. An important orbital contribution is donation from an endocyclic oxygen lone pair into the antibonding orbital of the adjacent exocyclic C–O bond, . The chair's axial arrangement supplies favourable antiperiplanar overlap. Electrostatic and solvation contributions can also affect the observed conformational preference.
Past exam of the chemistry B course of the University of Cambridge 2025 Paper 1 B5 a Solution 2026-10-20
For hydrogen-bond acceptance, the three strongest groups are the terminal carboxylate, the Cys–CO–N(Pro) tertiary-amide carbonyl, and the Pro–CO–NH(Tyr) secondary-amide carbonyl, in that qualitative order. Negative charge makes carboxylate a particularly strong acceptor. Both amide oxygens accept strongly; greater nitrogen alkyl donation generally makes a tertiary amide oxygen more basic than the comparable secondary-amide oxygen. This last ordering is a functional-group estimate, not a precise calculation of a folded tripeptide's site parameters.
For donation, the order is terminal ammonium tyrosine phenolic OH the peptide NH between Pro and Tyr. The thiol is a weaker donor. Protonated ammonium has no available lone pair to accept; the proline amide nitrogen has no N–H bond to donate, and amide nitrogen lone pairs are delocalized into carbonyl groups. The carboxylate has no O–H donor in the drawn protonation state.
Past exam of the chemistry B course of the University of Cambridge 2025 Paper 2 B10 a iii Solution 2026-10-20
At pH 7, a glutamate residue is normally carboxylate. Its most natural role is general-base activation of the attacking water. Draw the carboxylate oxygen lone pair toward a water H, the water O–H pair toward its oxygen, and that oxygen's lone pair toward C3 while the conjugated pi system moves toward the carbonyl oxygen:This can be concerted, avoiding a freely diffusing hydroxide intermediate. The protonated glutamate subsequently returns its proton to the enolate system, directly or through solvent, regenerating carboxylate. A favorable arrangement is needed; ordinary carboxylate does not quantitatively deprotonate bulk water in a prior equilibrium.
The side-chain amide of glutamine is neutral. Its N–H can donate a hydrogen bond to the carbonyl oxygen and stabilize developing enolate charge, while its amide oxygen can accept a hydrogen bond to orient attacking water. Draw , with the dotted interaction from amide N–H to substrate oxygen. This assists polarization and transition state stabilization; the amide nitrogen is not an ordinary amine base because its lone pair is delocalized into its own carbonyl.
The guanidinium group of arginine is positively charged. It can stabilize developing negative charge by hydrogen bonding and electrostatic attraction, drawn . Its delocalized cation is not the preferred neutral general base at pH 7. Productive catalysis requires stronger stabilization of the reacting arrangement than of the unreactive ground-state complex.
Phenylalanine and leucine have hydrocarbon side chains and provide no conventional acid/base or hydrogen-bond donor/acceptor group here. They can position the enone, exclude unproductive solvent and select the attacked face through hydrophobic effect and steric hindrance. Those binding effects can assist catalysis but should not be drawn as an imaginary side-chain proton relay. The original schematic depicts the Glu water relay and the Gln/Arg oxyanion-stabilizing contacts.
Past exam of the chemistry B course of the University of Cambridge 2025 Paper 2 B10 a i Solution 2026-10-20
Use C1 for the carbonyl carbon, C2 for the adjacent carbon, and C3 for the beta carbon of the reactive enone fragment . water adds conjugately at C3, giving a beta-hydroxy ketone. The oxygen lone pair of neutral water attacks C3; move the C2=C3 pi pair toward C1–C2 and the carbonyl pi pair onto oxygen. This produces the charge-separated enolate–oxonium intermediateA second water molecule can relay a proton from the newly attached oxonium oxygen to the enolate system. Using the carbon-centered resonance structure, reform C1=O and protonate C2 while the donating O–H bond returns its pair to the added oxygen. Equivalently, oxygen protonation first gives an enol, followed by water-mediated keto-enol tautomerism. Either path yields a neutral beta-hydroxy ketone and regenerates the relay water. There is no need to introduce externally supplied H or OH; solvent molecules perform the proton transfers. The uncatalyzed charge separation has a large barrier, so this mechanism describes a slow formal pathway, not a claim of rapid hydration in pure water.
Reactive-fragment mechanisms for conjugate hydration, steroid reduction and finasteride trapping
. Details. Past exam of the chemistry B course of the University of Cambridge 2025 Paper 2 B10 b iv Solution 2026-10-20
finasteride is reduced at C1, leaving a C2 enolate that attacks nicotinamide C4 to make the covalent C–C adduct. In the inhibitor's unsaturated lactam, the reactive chain is , with N at steroid position 4. Activate the C3 carbonyl by hydrogen bonding; draw the reduced cofactor's C4–H bond toward inhibitor C1, the C1=C2 pi pair toward C2–C3, and the C3=O pi pair toward oxygen. This gives NADP+ next to the C2-centered resonance form of the inhibitor enolate.
Instead of completing reduction by protonating C2, move the C2 lone pair toward the electrophilic C4 of the oxidized nicotinamide ring. Shift the pyridinium pi system toward its positively charged N, and reform the inhibitor carbonyl if using the oxygen-centered enolate representation. The result is the C2(inhibitor)–C4(nicotinamide) bond and a neutral dihydronicotinamide ring. Numbering the two molecules separately prevents confusing cofactor C4 with the inhibitor's N4. The lactam N–H remains present and can help position the inhibitor through hydrogen bonding; it does not form a covalent bond to the protein.
This is mechanism-based inhibition: enzyme chemistry generates the species that traps its own oxidized cofactor. Isolation of a cofactor–inhibitor adduct after protein denaturation, rather than persistent attachment to a denatured protein, distinguishes it from covalent modification of an amino acid side chain. The original mechanistic study combined radiolabel recovery, purification and spectroscopy to establish the adduct; its slow breakdown also cautions against treating “irreversible” as an absolute statement for every timescale.[ref]
Past exam of the chemistry B course of the University of Cambridge 2025 Paper 2 B6 a Solution 2026-10-20
The nitrite ion is an ambidentate ligand: its Lewis structures leave donor lone pairs on both nitrogen and oxygen. In the N-bound mode, , the nitrogen lone pair makes a sigma bond to the metal, while a suitable unoccupied nitrite molecular orbital can accept metal electron density by pi backbonding. Thus N-bound nitro is a sigma donor and pi acceptor. The nitrite group remains bent at nitrogen; N coordination does not make it a linear nitrosyl.
In the O-bound mode, , an oxygen lone pair makes the metal–oxygen sigma bond. A second occupied oxygen orbital can donate sideways into an available metal orbital: O-bound nitrito is a sigma and pi donor, rather than the strong pi acceptor of the N-bound description. oxygen's greater electronegativity lowers its donor orbital energies, but the occupied oxygen orbitals still explain its pi-donor classification. Actual mixing depends on metal oxidation state and geometry.
A further possibility is bidentate coordination, , with both oxygen atoms attached to one metal. Each attachment has oxygen-to-metal sigma donation; additional pi donation depends on orbital alignment. nitrite can also be a bridging ligand between metals. Neither alternative should be confused with linkage isomerism, which specifically compares attachment through different atoms at the same binding site.
Past exam of the chemistry B course of the University of Cambridge 2025 Paper 2 B8 b i Solution 2026-10-20
Write for the cyclopentadienyl ligand bearing its –2-pyridyl arm; means that its ring and pyridyl nitrogen both bind the same metal. . The base removes the pyridinium N–H proton in , 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), .
This is an proton transfer followed by chelation, with metal electron count . 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 is already 18-electron. In the preceding acid treatment, protonation of the allyl ligand allows release of propene; the other proton resides on the pendant pyridine, accounting for 's overall charge. One proton is removed here; the cyclopentadienyl tether is not deprotonated or detached.
Past exam of the chemistry B course of the University of Cambridge 2025 Paper 2 B9 a Solution 2026-10-20
The six P–N bonds are not well described as three permanently localized single bonds alternating with three ordinary double bonds. The strongly polar P–N framework has contributions from ionic bonding and nitrogen lone pair donation into neighboring phosphorus-centered acceptor combinations, including P–Cl orbitals. This negative hyperconjugation supplies additional P–N bonding and equivalent environments around the ring. Approximate planarity is compatible with these interactions, but equal lengths and planarity alone do not establish benzene-like aromaticity. Large phosphorus participation is not required.[ref]
Gallium binds to a ring nitrogen, and the two P–N bonds flanking that nitrogen lengthen. Formation of the NGaCl coordinate covalent bond changes the available nitrogen electron density and local geometry, weakening its contribution to the adjacent P–N interactions. The other four P–N bonds retain greater multiple-bond character and are correspondingly shorter. Loss of the threefold equivalence and altered local geometry permit the ring to pucker. This is a Lewis acid-base adduct, not replacement of a P–Cl bond by GaCl. Single-crystal structures and variable-temperature multinuclear nuclear magnetic resonance establish this class of nitrogen-bound group-13 adducts.[ref]
Group 13 Lewis acid adducts of [PCl2N]3 by Zin-Min Tun and others. 2011, pp. 8937–8945, adduct structures and variable-temperature NMR
Past exam of the chemistry B course of the University of Cambridge 2025 Paper 2 B9 c Solution 2026-10-20
, with two cyclohexylamino substituents on each original phosphorus. All six P–Cl bonds have been replaced; the six N–H bonds of those substituents remain. Each equivalent of triethylamine traps one equivalent of HCl.
contains three four-membered spiro rings, one at each original ring phosphorus. Each PCl joins the two geminal NHCy groups: replacement of two of its chlorides forms two new P–N bonds and releases two HCl, consuming both N–H protons. The new phosphorus retains one chloride and a lone pair, making it three-coordinate P(III); the original six-membered ring retains P(V). The product formula is . For each spiro ring, the original phosphorus is bonded to both exocyclic nitrogens, and each of those nitrogens is bonded to the same new PCl center; the braces do not indicate a simple linear substituent.
The three new P–Cl directions permit an all-cis isomer and a cis,trans isomer. Relative to the central ring, write these as and , with cyclic relabeling giving equivalent choices. Reversing all signs does not create a third relative configuration. This describes the configuration pattern; puckering and substituent conformations are additional structural details.
Past exam of the chemistry B course of the University of Cambridge 2025 Paper 2 B9 d Solution 2026-10-20
Replace the two chlorides on one ring phosphorus with the two ends of the diiron unit. has a spiro triangle, with an Fe–Fe bond, a –Fe bond to each metal and four terminal CO ligands on each Fe. The other two phosphorus atoms retain their groups. Its composition is . In neutral electron counting, each metal has .
has a triangular Fe cluster, one bridging CO, and an adjacent ring NFe bond. The two original metals remain bonded to and to one another, and each bonds to the added third Fe. Each Fe bears three terminal CO ligands; one additional bridging carbonyl ligand spans the original Fe–Fe edge. A nitrogen directly adjacent to donates its lone pair to the third Fe. The formula is therefore , satisfying retained CO ligands. The original diagram above gives all metal–metal and metal–ring connections.
The counts are an especially useful structural check. Each original Fe has . The added Fe has . Omitting nitrogen coordination leaves the third Fe two electrons short; omitting the bridging CO leaves both original metals one electron short.
In , the two untouched phosphorus sites are equivalent: an phosphorus-31 nuclear magnetic resonance system gives their doublet with integral 2, and the unique triplet with integral 1. The two equivalent nuclei do not split each other. In , coordination of one adjacent nitrogen distinguishes the two sides of the ring, producing three inequivalent phosphorus sites. Each couples to the other two and gives a doublet of doublets in the first-order limit; equal couplings or strong coupling could distort this simple pattern.
The high-frequency infrared spectroscopy bands belong to terminal carbonyl stretching modes. The extra low-frequency band is diagnostic of a strongly backbonded bridging carbonyl ligand. The number of IR bands counts allowed, possibly overlapping normal modes, not individual CO ligands. Formula, 18-electron rule, NMR equivalence and IR together select this model; the historical experimental work independently established diiron spiro rings and triiron products containing both P–Fe and N–Fe bonds.[ref]
Iron-Containing Metallophosphazenes and Their Clusters Derived from Chlorophosphazenes and Organometallic Dianions by Harry R. Allcock and others. 1984, pp. 4966–4977, diiron compound 2, triiron compound 3, and conversion with iron carbonyls
Acetonitrile is suitable: it dissolves the organic reagents, is polar enough to stabilize the developing ionic products and is an aprotic, relatively unreactive medium. It does not strongly hydrogen-bond-solvate the amine lone pair or compete efficiently as the methylating nucleophile. Sufficient solubility and chemical compatibility matter as well as polarity.
Bromonium formation and intramolecular opening give the six-membered cyclic bromoether
. Details. The alkene pi bond attacks Br as the Br-Br bond breaks to bromide, forming a bridged, unsymmetrical bromonium ion. The carbon bearing the two methyl groups has greater carbocation-like character.
The tethered alcohol oxygen attacks that carbon from the side opposite the bromine bridge. Its lone pair forms the new O-C bond while the attacked C-Br bridge bond opens, giving a six-membered cyclic oxonium ion. Bromide or solvent removes the O-H proton to give the neutral bromoether.
This explains why O becomes attached at the more substituted alkene carbon while Br remains at the other carbon. Polar solvent assists charge separation, and intramolecular capture competes effectively with external bromide attack. Opening is anti; attack on the two faces of the achiral starting material gives a racemic product at the newly formed CHBr stereocentre.
The ring-forming step is bromoetherification.
B-F has a large favorable polar/ionic contribution because electron-deficient boron and fluorine have a large electronegativity difference. Fluorine lone-pair donation into boron's empty 2p orbital can additionally strengthen bonding in planar BF. C-F is a strong, polar second-row sigma bond, but tetrahedral CF has no equivalent empty carbon 2p acceptor for that pi donation.
N-F has less favorable bond polarity and substantial repulsion between nitrogen and fluorine lone pairs, making it weaker. Invoking occupied boron d orbitals is unnecessary and inappropriate for this second-row comparison.
C-C is strongest; N-N is much weaker despite nitrogen's smaller size. Carbon's effective nuclear attraction and good overlap of compact second-row valence orbitals favor a strong C-C sigma bond. The larger, more electron-deficient boron centres give less effective B-B bonding in the stated molecule, despite its relatively high-s-character trigonal bonding.
In hydrazine, the two neighboring nitrogen lone pairs strongly repel one another, destabilizing the N-N bond. Bond strength therefore cannot be predicted from atomic size alone; lone-pair repulsion explains the especially low N-N value.
This illustrates bonding trends in second-period single bonds.



