Past exam of the chemistry A course of the University of Cambridge 2021 Paper 1 A4 b iii Solution 2026-10-20
The nitrogen-centered highest occupied molecular orbital of ammonia supplies a lone pair; planar borane has an empty out-of-plane -derived lowest unoccupied molecular orbital. Approach along this acceptor axis gives constructive orbital overlap and donation of the nitrogen pair into the boron-centered acceptor. The resulting N-to-B coordinate covalent bond forms the ammonia borane adduct.
The occupied molecular orbital formed by this interaction is bonding, while the corresponding antibonding combination remains empty. Borane pyramidalizes as coordination increases from three to four. Ammonia can use its existing outward-pointing lone pair; its inversion to the planar transition geometry is not a prerequisite. The Lewis base and Lewis acid descriptions summarize the donor and acceptor roles, while the frontier molecular orbitals explain their directional interaction.
Past exam of the chemistry A course of the University of Cambridge 2021 Paper 1 A4 b ii Solution 2026-10-20
For planar ammonia, the nitrogen valence atomic orbitals have symmetries , and . The hydrogen symmetry orbitals span . Consequently the in-plane orbitals form a bonding orbital, a bonding pair and their antibonding partners. The out-of-plane has no matching hydrogen combination and remains a nonbonding orbital. With the core omitted, the conventional valence occupation isThe preceding ammonia molecular orbital diagram shows the correlation. On pyramidalization, and both correlate with of the C3v point group, permitting s–p orbital mixing. The pure planar lone pair can acquire bonding character, lowering its energy and helping stabilize the pyramid. Symmetry alone does not provide a numerical inversion barrier.[ref]
Past exam of the chemistry B course of the University of Cambridge 2022 Paper 1 B1 b ii Solution 2026-10-20
amide ion removes a methyl-side alpha proton of pinacolone. The C–H electrons form the enolate C=C bond as the carbonyl pi pair moves onto oxygen; ammonia is formed. The tert-butyl-side alpha carbon has no hydrogen, fixing the ordinary enolate regiochemistry.
The carbon end of D attacks the acyl chloride carbonyl carbon. Move the C=O pi pair onto oxygen, then collapse the tetrahedral intermediate with chloride departure. This nucleophilic acyl substitution gives and constructs the bond between the original enolate methylene and the incoming carbonyl. The observed product requires carbon acylation, rather than an O-acyl enol ester. A further equivalent of D removes the much more acidic central proton of the new 1,3-diketone, generating its extensively delocalized enolate and returning that second D equivalent to pinacolone. Aqueous acid workup protonates the product enolate; keto–enol equilibration gives the conjugated, internally stabilized by a hydrogen bond enol observed by NMR. The acyl chloride contributes its entire pivaloyl group; neither tert-butyl group is lost.
Past exam of the chemistry B course of the University of Cambridge 2022 Paper 2 B7 a ii Solution 2026-10-20
Six neutral ammonia donors leave Cu(II), , soThe donor skeleton has octahedral molecular geometry with a pronounced tetragonal Jahn–Teller distortion, normally four shorter equatorial and two longer axial Cu–N distances. The occupation is orbitally degenerate in the ideal octahedral limit. An electron count above 18 is not a contradiction: ammonia is primarily a sigma donor and the occupied metal-derived antibonding levels make this a weak-field coordination complex rather than a low-valent carbonyl closed shell.
Past exam of the chemistry B course of the University of Cambridge 2023 Paper 2 B6 a i Solution 2026-10-20
ammonia and cyclohexylamine are neutral donors, whereas each coordinated chloride and acetate has charge . Charge balance therefore gives Pt(II) in cisplatin and Pt(IV) in satraplatin:The two acetate groups count once each; their carbonyl and alkoxy oxygen atoms do not make each ligand doubly charged.
Past exam of the chemistry B course of the University of Cambridge 2023 Paper 2 B6 b ii Solution 2026-10-20
cyclohexylamine is expected highest in the spectrochemical series for comparable metal environments. Its alkyl substituent increases the electron-donating ability of nitrogen relative to ammonia, strengthening the sigma interaction and raising the metal–ligand sigma-antibonding levels. The anionic pi donors tend to raise occupied metal levels and give smaller splittings. Steric effects and the metal can modify the exact ordering, so the comparison is a ligand field trend rather than a universal basicity-to-splitting identity.
Past exam of the chemistry B course of the University of Cambridge 2023 Paper 2 B6 b i Solution 2026-10-20
ammonia and cyclohexylamine are sigma-donor ligands; chloride and acetate are sigma- and pi-donor ligands. The nitrogen lone pair of each amine points toward the metal and forms the donor interaction. These saturated amines have no low-lying conjugated acceptor system that would make them conventional pi-acceptor ligands.
chloride uses a lone pair for sigma donation and has additional occupied orbitals perpendicular to the bond that can donate with pi symmetry. An oxygen-bound acetate similarly has occupied oxygen-centred orbitals available for sigma and pi donation, although its carboxylate resonance in chemistry changes their energies and distribution. Neither should be classified as a pi acceptor merely because its bonding can be drawn with multiple bonds or resonance in chemistry contributors. None of these four is a standard sigma-donor/pi-acceptor ligand.
Past exam of the chemistry B course of the University of Cambridge 2023 Paper 2 B6 c Solution 2026-10-20
The average Pt oxidation state is , since six bromides balance the two metal atoms. A natural mixed valence assignment is alternating Pt(II), , and Pt(IV), , rather than two equivalent Pt(III) sites. Use neutral square-planar units and octahedral units. At every site place two ammonia and two bromide ligands in an equatorial plane; the Pt(IV) sites have two additional axial bromides pointing toward the adjacent Pt(II) sites.
Along the chain the connectivity isEvery Pt–Br–Pt bridge is asymmetric: short to Pt(IV), long to Pt(II). The latter contact is a secondary axial interaction, consistent with the preference of Pt(II) for a square-planar primary coordination sphere. The Pt(IV) site is smaller and more strongly attracts the bridging bromides. A drawing with all axial distances identical would obscure the distinct local environments.
There are four equatorial bromides and two axial bridging bromides per two-Pt repeat, giving precisely . A direct Pt–Pt bond is not required. This is a proposed structure consistent with the stated composition and two environments; the formula alone does not determine every equatorial cis/trans arrangement or numerical bond length.
Past exam of the chemistry B course of the University of Cambridge 2024 Paper 2 B10 a Solution 2026-10-20
The reactive fragment is the C6 amidine-like portion of the purine ring; the N9–ribose attachment is retained. Write it locally as , with the rest of the fused ring understood. One productive acid catalysis route protonates N1, making C6 more electrophilic. water then adds at C6 while the C6=N1 pi electrons move onto N1. Subsequent proton transfers generate a carbinolamine and then a protonated amino leaving group:These are local bonding fragments: C6 retains its bond to the adjoining ring carbon at every step. In the addition step, an oxygen lone pair forms C6–O and the C6=N1 pi bond opens. During collapse, an oxygen lone pair reforms C6=O and the C6–N6 bond breaks toward the departing nitrogen. Solvent mediates each proton transfer; no direct unsupported transfer across a long distance is required.
water supplies the new C6 oxygen, and ammonia is expelled after protonation of the leaving group. Final deprotonation gives the lactam form of inosine. In acidic bulk solution the released ammonia is largely trapped as ammonium. The catalytic proton transfers regenerate acid around the substitution cycle; ammonium trapping separately affects the net proton balance. This is hydrolytic deamination, not cleavage of the glycosidic bond.
Past exam of the chemistry B course of the University of Cambridge 2024 Paper 2 B10 e Solution 2026-10-20
A chemically coherent mechanism combines zinc-activated water addition with a proton relay involving His238 and Glu217. In a simple route consistent with the displayed contacts, neutral His238 accepts a proton from zinc-bound water while protonated Glu217 donates to substrate N1. These transfers can be coupled to the bond-forming event rather than requiring accumulation of a free, doubly activated substrate. Zinc-bound hydroxide attacks C6; the C6=N1 pi electrons move onto N1, yielding the tetrahedral C6 intermediate bearing OH and NH while N1 bears H.
Next, the exocyclic amino group receives a proton, through His238, Glu217 and/or intervening water as permitted by the changing geometry. The oxygen is deprotonated as it forms the C6 carbonyl, and the C6–N6 bond cleaves to release ammonia. These transfers can be coupled in the collapse step, avoiding expulsion of . Zinc stabilizes the reacting oxygen and the organized intermediate; the ring regains conjugation in the inosine product. Product dissociation, fresh water binding and proton exchange with solvent reset the catalytic groups.
The core result is zinc-assisted addition–elimination, with acid/base proton transfers and water supplying C6 oxygen. The contributions are metal ion catalysis, general base catalysis, general acid catalysis, electrostatic catalysis and proximity and orientation catalysis. The proposed route does not require covalent catalysis by a protein side chain: a substrate–water C–O bond is not an enzyme–substrate covalent linkage. Asp296 can retain a structural hydrogen-bonding role without acting as an indispensable acid/base.
The data establish the importance of His238 and Glu217 but do not uniquely assign the entire proton relay. In particular, an alternative has Glu217 first deprotonate water and then deliver that proton to N1, with His238 assisting hydroxide stabilization and later amino-group protonation. The primary observation that His238 substitution need not abolish analogue hydration cautions against calling its initial-base role proven.[ref]
Site-directed mutagenesis of histidine 238 in mouse adenosine deaminase: substitution of histidine 238 does not impede hydroxylate formation by V. Sideraki and others. 1996, pp. 15019–15028, retained hydroxylate formation and proposed electrostatic contribution
pH-rate profiles, solvent isotope effects and time-resolved intermediate experiments would discriminate these refinements. The diagram is a proposed catalytic route, not a structure directly observed by the crystal experiment. Past exam of the chemistry B course of the University of Cambridge 2025 Paper 2 B6 b iii Solution 2026-10-20
The relevant spectrochemical series order is O-bound nitrito ammonia N-bound nitro. A pi-donor ligand tends to raise the metal orbitals derived from , reducing the octahedral splitting; a pi-acceptor ligand stabilizes those orbitals, increasing the splitting. Since photon energy is , the stronger-field complex absorbs at the shorter wavelength. ThusThe corresponding spectroscopic wavenumbers are approximately , and . These are the observed band energies, not three exact independent determinations of : low-spin transition energies also contain electron repulsion contributions, and replacing one ligand lowers the ideal octahedral symmetry. The relative assignment is nevertheless robust.
Past exam of the chemistry B course of the University of Cambridge 2025 Paper 2 B6 b i Solution 2026-10-20
, , , and the by-product is . The first transformation is deprotonation of coordinated water, not loss of an ammonia ligand. The hydroxo ligand in is then nitrosated at oxygen, retaining its original Co–O attachment in the initially formed O-bound nitrito ligand. Atom and electric charge balance giveThe relative molecular mass of nitrous acid is approximately . Heating converts the O-bound nitrito ligand to the N-bound nitro ligand; these are linkage isomers, with Co remaining in oxidation state throughout.
Past exam of the chemistry B course of the University of Cambridge 2025 Paper 2 B6 b v Solution 2026-10-20
Both complexes contain cobalt(III), hence a metal ion. The approximately octahedral molecular geometry, relatively high metal oxidation state and five ammonia donors make the octahedral splitting large enough that pairing is favorable. Their low-spin configuration isAll six metal electrons are paired, and the ligands in coordination chemistry have closed shells, so neither complex has a permanent spin magnetic moment: both exhibit diamagnetism. The difference between N and O attachment is insufficient to change this low-spin assignment.
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 . 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, , not pentan-2-one. Attack on acetic anhydride forms . 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]
The Preparation of Ketones from Grignard Reagents by Melvin S. Newman and William T. Booth Jr.. 1945, printed p. 154: proposed intermediate I and low-temperature dependence
For the benzonitrile route, n-butyl attacks the nitrile carbon and one C≡N pi pair moves to nitrogen, producing the magnesium ketimine salt . This anionic nitrogen-containing intermediate resists a second nucleophilic addition much more than a neutral carbonyl does. The ketone 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.
Past exam of the mathematics course of the University of Cambridge 2016 iii Paper 315 4 j Solution Created 2026-10-03 Updated 2026-10-06
Three solar-system effects illustrate chemical disequilibrium. First, vertical mixing carries carbon monoxide from hot deep regions into cooler observable layers of giant planets, where its local-equilibrium abundance would be small. Second, phosphine survives in the observable atmospheres of Jupiter and Saturn because deep supply competes with its destruction, making it a tracer of transport. Third, methane photolysis creates hydrocarbons such as ethane and acetylene, and can help form upper-atmospheric atmospheric hazes in giant planets and Titan.
Three exoplanet effects are complementary. First, carbon monoxide–methane quenching can keep carbon monoxide abundant and methane depleted in cool observable gas, including hot-Neptune/giant atmospheres that would locally favor methane. Second, atmospheric photochemistry dissociates methane/ammonia and produces radicals, hydrogen cyanide, hydrocarbons and possibly atmospheric hazes, changing spectral bands from their equilibrium strengths. Third, horizontal chemical quenching can preserve hot-dayside carbon chemistry on the cooler nightside/terminator and reduce the abundance contrast expected from separate local-equilibrium columns.
Transport, photon-driven reactions and finite chemical adjustment times change the observable composition. The effects are conditional on the species' chemical timescale relative to mixing, advection and irradiation; they do not imply that every molecule is out of equilibrium everywhere.
Past exam of the mathematics course of the University of Cambridge 2018 iii Paper 315 3 a Solution Created 2026-10-03 Updated 2026-10-05
Four mechanisms that can produce disequilibrium chemistry in an exoplanet atmosphere are:
- Atmospheric photochemistry. Stellar ultraviolet photons initiate reactions whose products need not follow local thermochemical equilibrium. The ozone layer on Earth is a solar-system example. Calculations for HD 189733 b predict enhanced hydrogen cyanide and acetylene from the processing of methane and ammonia; these are model examples rather than assertions of an unambiguous detection.
- Vertical transport and chemical quenching. When the eddy mixing time is shorter than the chemical relaxation time, gas retains a deeper abundance above its chemical quench level. The excess carbon monoxide in Jupiter's cool atmosphere exemplifies carbon monoxide–methane quenching. Models of HD 189733 b predict quenched methane and ammonia abundances differing from their local chemical equilibrium values. The enhancement or depletion depends on the underlying atmospheric pressure-temperature profile.
- Horizontal chemical quenching. If the atmospheric advection time is short, winds move chemically processed gas into regions with different irradiation or temperature faster than it can re-equilibrate. Transport of gas within Earth's ozone layer moves material away from its local photochemical production regions. Models of HD 209458 b show that dayside carbon monoxide-rich composition can persist into the cooler nightside instead of forming the local chemical equilibrium amount of methane.
- Condensation with sedimentation or rainout. Finite-rate cloud formation can depart from phase equilibrium, while atmospheric condensate rainout removes elements from a layer and changes its gas composition. Earth's water atmospheric cold trap limits the supply of water to the stratosphere. In HD 209458 b models, titanium-bearing condensates can settle and suppress upper-atmospheric titanium monoxide. The remaining gas can still be in local chemical equilibrium with its depleted inventory: rainout is an open-column effect, not necessarily a failure of equilibrium among all gas reactions.
The exoplanet transport and photochemical examples follow kinetic atmosphere calculations and models including horizontal transport; the condensate example is examined in cold-trap calculations.
Past exam of the mathematics course of the University of Cambridge 2018 iii Paper 315 3 b iii Solution Created 2026-10-03 Updated 2026-10-05
Four possible molecular signatures of disequilibrium chemistry in an exoplanet atmosphere, measured through an exoplanet transmission spectrum or exoplanet emission spectrum, are:
- Excess carbon monoxide in the cool upper layers, where local chemical equilibrium would place most carbon in methane. This can indicate carbon monoxide–methane quenching from deeper hot gas.
- Suppressed methane bands relative to the same cool equilibrium model, consistent with transport preventing complete carbon monoxide conversion or with photochemical loss.
- An ammonia abundance inconsistent with the local equilibrium nitrogen partition, potentially recording nitrogen–ammonia quenching. Its direction must be evaluated for the actual atmospheric pressure-temperature profile; a depleted value is possible when hotter, molecular nitrogen-rich gas is transported upward.
- Enhanced hydrogen cyanide or acetylene bands, consistent with atmospheric photochemistry acting on the transported carbon and nitrogen reservoirs.
The first two signatures can be consequences of the same process and are not independent evidence for two mechanisms. Changes in atmospheric carbon-to-oxygen ratio, atmospheric metallicity of a giant planet, exoplanet cloud coverage, and the atmospheric pressure-temperature profile can mimic abundance changes. A convincing inference compares several molecular bands with a chemically consistent equilibrium model rather than identifying one unusual band alone.
Past exam of the mathematics course of the University of Cambridge 2018 iii Paper 315 3 b ii Solution Created 2026-10-03 Updated 2026-10-05
For solar elemental abundances, take a molecular hydrogen- and helium-dominated atmosphere with atmospheric carbon-to-oxygen ratio near . The principal oxygen-bearing molecule is generally water; the carbon and nitrogen carriers depend on both temperature and pressure.
At the cool observable upper layers, methane is the expected main carbon reservoir under chemical equilibrium, with abundant water and much less carbon monoxide. At the hotter layers approaching , carbon monoxide becomes the main carbon carrier and water contains much of the oxygen not bound in it. The relevant law of mass action follows fromLower temperature favors the exothermic methane-forming direction, while increasing pressure favors the side with fewer molecules. This explains why the change of dominant carrier cannot be specified by temperature alone.
Nitrogen is distributed between molecular nitrogen and ammonia. Cooling favors ammonia, but low pressure favors molecular nitrogen; it is therefore unsafe to call ammonia dominant throughout the low-pressure region. The exact partition requires the equilibrium constant for .
Carbon dioxide is usually a minor constituent at solar composition; hydrogen cyanide and acetylene are much less abundant than the principal carbon carriers in this oxygen-rich equilibrium case. Condensation can remove refractory species where a condensation curve is crossed. The robust cool-atmosphere expectation is a molecular hydrogen–helium background with methane and water, changing toward carbon monoxide in hotter layers. Precise mixing ratios require thermodynamic data and an explicitly specified elemental inventory after rainout.
Sodium amide 2026-10-20
The ionic compound , commonly used as a strong base. ketone deprotonation produces an enolate and ammonia.
Trigonal pyramidal molecular geometry 2026-10-13
In trigonal pyramidal molecular geometry, three neighbouring chemical atoms form a triangular base and the central chemical atom lies outside its plane. The pyramid has four atomic nuclei; its central chemical atom is the apex, not the centre of the base. Ammonia, , has this arrangement. The [ref]
https://cccbdb.nist.gov/exp2x.asp?casno=7664417&charge=0. NIST CCCBDB experimental ammonia entry, geometric data, citing Herzberg (1966)
lists an H–N–H bond angle of , commonly rounded to or about .The Lewis structure of ammonia allocates eight valence electrons as three bond pairs and one lone pair. Four electron domains therefore suggest tetrahedral electron pair geometry; removing the lone pair marker leaves three hydrogen chemical atoms below nitrogen. The VSEPR model treats the lone pair as more space-demanding and predicts compression relative to , but it does not calculate the precise value. See Chemistry 2e, §7.6: lone-pair effects.
Three equal bond-dipole contributions in this pyramid cancel sideways, but retain a component along its axis. By contrast, three equal contributions lying in one plane at would cancel completely. This shows how a small change in molecular geometry can permit a permanent electric dipole moment. Ammonia is a polar molecule; NIST lists a measured magnitude of about debye in the same [ref]
https://cccbdb.nist.gov/exp2x.asp?casno=7664417&charge=0. experimental entry, electric dipole moment, citing Shimizu (1970)
. The vector argument predicts allowed direction and cancellation, not the full measured magnitude.

