Past exam of the chemistry B course of the University of Cambridge 2021 Paper 2 B8 c Solution 2026-10-20
Use neutral electron counting consistently: each terminal Cp supplies five electrons, while the eight pi electrons of neutral pentalene must be allocated across the shared fused framework rather than counted twice as two independent Cp coordination ligands. One set of 18-electron-consistent proposals is:
- Iron: syn - pentalene with an Fe–Fe single bond. The two iron–Cp fragments lie on the same face. Assigning four pentalene electrons to each iron gives before direct bonding; the Fe–Fe bond contributes one electron to each count, giving 18. Its direct bond naturally accounts for the especially short metal–metal separation. An anti arrangement cannot retain this short same-face bond.
- Cobalt: - pentalene without a Co–Co bond. The count is at each metal. Syn and anti arrangements are possible at this formal level because there is no required direct bond. Removal of the bond relative to iron accounts for a larger separation.
- Nickel: slipped - coordination to the two outer allylic portions. The fusion bond remains an uncoordinated alkene in this limiting bonding picture, while each outer three-carbon portion contributes three electrons in neutral counting: . Equivalent ionic counting gives a nickel(II) centre, a six-electron Cp and a four-electron allyl ligand. Moving the metal binding sites toward the outer three-carbon portions increases the separation further. Syn and anti arrangements can again be drawn.
The original schematic shows the donor regions and the required iron–iron bond; dashed metal-to-region lines denote delocalized coordination rather than localized carbon–metal single bonds.
These proposals explain the requested electron counts and distance ordering. electron counting alone does not establish the unique ground-state structure: orbital calculations on related pentalene complexes also identify stable formally electron-deficient arrangements. In particular, the formal requirement of a metal–metal bond in the proposed 18-electron iron model should not be converted into a claim that every known iron pentalene complex adopts that model.[ref]
The Versatility of Pentalene Coordination to Transition Metals: A Density Functional Theory Investigation by S. Bendjaballah and others. Discussion of syn/anti coordination, hapticity and electron-count exceptions
Past exam of the chemistry B course of the University of Cambridge 2021 Paper 2 B8 d ii Solution 2026-10-20
Cleavage of the disulfide supplies one methylthiolate per cobalt. A monomer would have only electrons by ionic counting. Dimerization with two bridging thiolates supplies a second sulfur interaction at each cobalt, and a direct Co–Co bond completes each count. Thus a suitable structure is with a Co–Co single bond, containing an diamond:Each cobalt is formally : Cp and methylthiolate each carry one negative charge. The stoichiometry isThe molecular formula reduces to the required empirical formula. The sulfur–sulfur bond has been cleaved, both sulfur atoms now bridge the metals, and the loss of two CO molecules per cobalt is balanced. Formally cobalt(I) is oxidized to cobalt(II), while the disulfide is reduced to thiolate.
Past exam of the chemistry B course of the University of Cambridge 2021 Paper 2 B9 c Solution 2026-10-20
boron is sufficiently small for effective oxygen-/boron- overlap, and three-coordinate planar boron benefits from the pi donation just described. A six-membered boroxine ring provides this overlap without severe angle strain. The larger aluminium centre has poorer oxygen-/aluminium- matching and remains strongly Lewis acidic. It can lower its electron deficiency by accepting additional oxygen coordination, making a three-dimensional aluminoxane cage favourable. Four-coordinate aluminium does not require a boroxine-like planar pi system.
An appropriate hexamer model is an alternating Al/O hexagonal prism, with six aluminium and six oxygen vertices. Colour the twelve-vertex prism alternately Al and O: each aluminium then has three oxygen neighbours and an outward organic substituent, making it four-coordinate, while each oxygen bridges three aluminium centres. The graph has 18 Al–O edges, hence six Al centres with three Al–O bonds each. Different polyhedral connectivities and sizes can meet similar local coordination requirements, so the hexamer is a proposal rather than a claim that the empirical repeat unit fixes a unique cage.
Past exam of the chemistry B course of the University of Cambridge 2022 Paper 1 B1 a iii Solution 2026-10-20
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 .
Past exam of the chemistry B course of the University of Cambridge 2022 Paper 1 B1 a i Solution 2026-10-20
For a monocyclic, approximately planar, continuously conjugated pi system, Hückel's rule predicts aromatic closed-shell filling for pi electrons, with . Every atom on the circuit must have an appropriately oriented orbital; counting double bonds without checking pi conjugation is insufficient.
The two outer benzene rings are clearly aromatic; the central pyrone ring has a weaker, polarized six-pi aromatic contribution. Each outer ring has the familiar six-electron circuit. For the central ring, a useful charge-separated contributor moves the exocyclic carbonyl pi pair onto oxygen and donates one endocyclic oxygen lone pair into the ring. Its four fused carbon p centres contribute four electrons and the ring-oxygen lone pair supplies two: a six-pi, pyrylium-like circuit, with positive charge on the ring oxygen and negative charge on the carbonyl oxygen. This does not make three independent benzene rings of equal aromatic stabilization; localized carbonyl bonding and the fused benzene circuits remain important contributors.
The three oxygens play different roles. The endocyclic ether oxygen donates a p-type lone pair to pi conjugation. The carbonyl oxygen accepts pi density in the charge-separated contributor. The phenolic oxygen donates into the outer ring as a substituent. Its in-plane lone pair is not an extra pair to add to every ring's electron count. In particular, neither the external C=O pair nor both lone pairs on a ring oxygen should be counted as independent extra ring electrons. Structural studies of xanthone support a largely planar conjugated framework, while the division into strongly aromatic outer rings and a more polarized central ring is an electronic-structure interpretation.[ref]
Structure of xanthone by Shigeki Onuma and others. 1990, pp. 1725–1727, molecular geometry
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 1 B2 a Solution 2026-10-20
The five-carbon unit is initially an open chain, not a completed carbocycle. Label it , with phenyl-bearing , carbonyl , amide-substituted , and the two ethylthio groups on . Its double bonds are initially and . Let be the nitro-bearing carbon of nitroethane; its methyl remains a substituent throughout.
DBU removes an alpha proton from nitroethane to give a nitronate. carbon attack at the Michael acceptor forms , with electron redistribution through onto the carbonyl oxygen. proton transfer restores the ketone. Removing the remaining X–H proton gives another carbon-nucleophilic nitronate. It attacks the other conjugated end, , intramolecularly, closing the six-membered circuit . Addition to the ketene dithioacetal and ethylthiolate elimination restore the bond, leaving one SEt group at .
base-assisted loss of the proton at , with C–NO2 cleavage at X, creates and expels nitrite. Enolization at supplies the third aromatic double bond and the phenolic OH. proton transfers to the leaving anions complete the acid–base balance. nitroethane supplies the new ring carbon X and its methyl substituent; its nitro group is eliminated. Aromatization, rather than a separate oxidation of an intact six-membered starting ring, provides the final driving force. This is a proposed stepwise nitroalkane benzannulation; related DBU-promoted tandem Michael addition/cyclization/aromatization has direct primary precedent.[ref]
Original atom maps for the open-chain benzannulation, the two bonds of the norbornanone-forming tandem Michael addition, and the intramolecular Michael closure
. Details. Past exam of the chemistry B course of the University of Cambridge 2022 Paper 1 B2 b Solution 2026-10-20
Number the cyclopentenone carbonyl carbon , its alkene carbons and with OR at , and its saturated carbons and . The ester alkene has alpha carbon , bearing Me and CO2Et, and beta carbon , bearing the protected carbon substituent. lithium diisopropylamide generates the ordinary ketone enolate at the saturated alpha carbon . The acetonide is retained.
The carbon end of this enolate attacks ester beta carbon Y. Its pi system shifts through X onto the ester oxygen, forming the first new bond and an ester enolate at X. That enolate now attacks the original cyclopentenone beta carbon intramolecularly, giving the second new bond . The original enone electrons move onto its ketone oxygen, and protonation of the final ketone enolate completes the product. The OR substituent is retained at the new bridgehead; it need not leave in either Michael addition.
The atom map provides a useful structural check. Between bridgeheads and , the three paths are , , and . Their numbers of internal atoms are 2, 1 and 2, establishing the norbornane skeleton. Both original carbonyls survive and both initial alkenes are consumed. The two successive conjugate additions give this atom map without requiring a concerted cycloaddition.
Past exam of the chemistry B course of the University of Cambridge 2022 Paper 1 B2 c Solution 2026-10-20
The new bridge is made by an intramolecular Michael addition. On the cyclohexanone side, the methyl-bearing fused alpha carbon is quaternary and cannot be deprotonated there. triethylamine can instead generate a small equilibrium population of the enolate at the other, methylene alpha carbon. Its carbon attacks the beta carbon of the enone in the seven-membered ring. The enone pi electrons move towards its carbonyl and then onto oxygen, creating the new bridge and a second enolate.
protonation of that enolate at the former methyl-substituted enone alpha carbon restores the left-hand ketone. The initial enone C=C has therefore become a saturated C–C bond while one new transannular C–C bond forms. Both ketones are retained, and the initial two-ring framework becomes a three-ring framework without changing the total index of hydrogen deficiency. The methyl originally at the fused quaternary centre does not migrate; the other methyl ends up on a protonated, formerly unsaturated carbon. A strong-base concentration large enough for quantitative enolate generation is unnecessary for a favourable intramolecular trapping reaction, and neither an aldol dehydration nor a methyl shift is needed.
Past exam of the chemistry B course of the University of Cambridge 2022 Paper 1 B5 a i Solution 2026-10-20
water achieves substantial binding through a localized O–H···O hydrogen bond; benzene achieves a comparable total through a larger set of weaker contacts. In the water dimer, one water is the hydrogen-bond donor and the other presents an oxygen lone pair region as acceptor. The interaction includes electrostatics, induction interaction, dispersion and short-range exchange repulsion; a hydrogen bond is not a separate fundamental force that can simply be added to all those terms again.
For benzene, draw a T-shaped edge-to-face pair or a slipped parallel pair. Its larger, more polarizable electron clouds provide appreciable London dispersion. Its quadrupolar charge distribution also makes the energy orientation dependent, despite each molecule having no permanent dipole by symmetry. An edge C–H approaching the other ring's pi region is therefore preferable to a model of two structureless, noninteracting dipole-free objects. Dispersion attracts while Pauli repulsion limits close overlap. Several distributed contacts can sum to the scale of one water–water hydrogen bond; counting only the number of conventional hydrogen bonds would miss them.
The comparison concerns the given gas-phase pair interaction energies, not an equality of every solvent-dependent association free energy. Correlated primary calculations explicitly identify both T-shaped and parallel-displaced benzene-dimer minima, rather than a uniquely preferred perfectly eclipsed sandwich.[ref]
Estimates of the Ab Initio Limit for π–π Interactions: The Benzene Dimer by Mutasem Omar Sinnokrot and others. 2002, abstract and pp. 10887–10893, comparison of the three dimer geometries
Original schematic geometries of a hydrogen-bonded water dimer and T-shaped and parallel-displaced benzene dimers
. Details. Past exam of the chemistry B course of the University of Cambridge 2022 Paper 2 B6 b ii Solution 2026-10-20
A Co(III) centre has , and charge balance requires one closed-shell catecholate ligand and one semiquinonate ligand. For low-spin Co(III), , the metal has , giving an independent-spin magnetic moment ofThus contains low-spin Co(III), , with its observed moment residing on the free radical ligand. High-spin Co(III), , would have and give under the stated independent-spin approximation, so it is inconsistent with this measurement.
For Co(II), , charge balance instead requires two semiquinonate ligands. High-spin Co(II) has and , soTherefore contains high-spin Co(II), . Low-spin Co(II), , would give when the two ligand spins are included. These four checks distinguish both oxidation state and metal spin.
The larger octahedral splitting at Co(III) exceeds the effective pairing energy and favours a low-spin complex; the smaller octahedral splitting at Co(II) favours a high-spin complex with occupation of both orbitals before further pairing. Interconversion also transfers an electron between cobalt and a dioxolene ligand, so this is valence tautomerism, rather than a metal-only spin change. Squared moments add here because the supplied model treats the centres as independent; adding the three spins first would be a different, exchange-coupled model.
Past exam of the chemistry B course of the University of Cambridge 2022 Paper 2 B6 b v Solution 2026-10-20
The Co(III) centre in has : every occupied d subset is orbitally balanced. There is no first-order metal-centred Jahn–Teller driving force in . This does not require all six experimental bond lengths to be equal, since pyridine and dioxolene are different donors.
The Co(II) centre in has . Its two orbitals each contain one electron, so that subset gives no first-order preference. The hole does give a weaker Jahn–Teller instability. To compare tetragonal distortions, take a positive small splitting parameter and retain each subset's orbital energy barycentre. On axial elongation let shift by and by . Five electrons then giveFor equal-magnitude compression the shifts reverse, and filling the lower orbital first givesThe two singly occupied orbitals contribute zero net shift in either case. Thus the elementary tetragonal splitting model favours elongation for high-spin , rather than equal electronic stabilization by the two signs. In a strictly sigma-only octahedral molecular orbital model is nonbonding, so this weak effect is absent; it arises when the electrostatic and non-sigma interactions neglected in that diagram are restored. spin-orbit coupling, unequal donors and chelate constraints can modify the actual distortion, so the argument does not establish which experimental Co–donor bonds must lengthen.
Past exam of the chemistry B course of the University of Cambridge 2022 Paper 2 B7 a iii Solution 2026-10-20
Neutral CO ligands require formal Cr(), hence . With five two-electron carbonyl ligands,A trigonal-bipyramidal molecular geometry is expected, analogous in electron count to . Strong pi-acceptor ligands stabilize a low oxidation state through backbonding, and this five-coordinate structure accommodates the closed-shell 18-electron configuration. The negative oxidation state is a formal assignment; it does not mean that a free chromium dianion is trapped inside the molecule.
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 2022 Paper 2 B7 a iv Solution 2026-10-20
With oxide charge and four thiocyanates each , gives V(IV), . Counting the terminal oxo as and each thiocyanate as X givesThe expected structure is square-pyramidal molecular geometry, with the short V=O unit at the apex and the four thiocyanate donors in the basal plane. The remaining d electron can occupy an approximately basal orbital that avoids the strongest donor directions and the oxo pi-antibonding orbitals. This is a high-oxidation-state oxovanadium complex; reaching 18 electrons is not its governing stability criterion. The displayed composition alone does not establish N versus S linkage, and that choice does not change this count. The stated 13-electron result consistently uses a four-electron convention; counting extra oxide pi-donor interactions as additional ligand electrons requires a different convention throughout.
Past exam of the chemistry B course of the University of Cambridge 2022 Paper 2 B7 b Solution 2026-10-20
In ionic counting, two cyclopentadienyl anions donate electrons. Therefore ferrocene, cobaltocene and nickelocene contain 18, 19 and 20 valence electrons, respectively: Fe(II) is , Co(II) and Ni(II) . ferrocene fills the bonding and approximately nonbonding levels while leaving the next metal–ring antibonding pair empty. cobaltocene and nickelocene add one and two electrons to that antibonding set.
Increasing antibonding occupation lowers the net metal–ring bond order. It therefore explains the simultaneous increase in M–C distance and decrease in dissociation enthalpy from Fe to Co to Ni. The trend does not follow the shrinking size of the isolated metal ions; the orbital occupation is decisive here. The specified enthalpy separates the entire sandwich into a gaseous metal dication and two gaseous ring anions. It measures that complete ionic dissociation process, not ten independent localized M–C bond energies, so dividing it by ten is not a justified bond-energy determination.
Past exam of the chemistry B course of the University of Cambridge 2022 Paper 2 B7 c Solution 2026-10-20
Oxidizing cobaltocene removes its antibonding nineteenth electron and gives the particularly favourable 18-electron cobaltocenium cation. iodine can accept the electrons, giving cobaltocenium iodide at the simple stoichiometryEach cobalt changes from Co(II), , to Co(III), ; each iodine atom changes from 0 to . Excess iodine can change the isolated counterion to a polyiodide such as . In contrast, oxidation of 18-electron ferrocene would remove an electron from its occupied closed-shell manifold to form 17-electron ferrocenium. iodine is insufficient to drive that reaction under the conditions implicit here. ferrocene can still be oxidized by suitable stronger oxidants; an electron count is a useful rationale for the comparative redox behaviour, not a universal proof of a numerical electrode potential.
Past exam of the chemistry B course of the University of Cambridge 2022 Paper 2 B7 d Solution 2026-10-20
The anionic nitrogen donor replaces iodide to giveHere Cp denotes cyclopentadienyl and specifies attachment through the ring nitrogen of the pyrrolyl anion, . Fe(II), , receives six electrons from Cp, four from the two CO ligands and two from an N-bound pyrrolyl ligand: .
Two coupled CO stretching modes, approximately symmetric and antisymmetric combinations, account for the two strong carbonyl-region absorptions. The bent two-CO arrangement of this piano-stool complex permits both combinations to be infrared active. They are not evidence for two different chemical products.
Heating expels two CO molecules and changes the binding of the pyrrolyl ligand from to , giving the sandwich azaferrocene,The four electrons lost with CO are replaced by the four additional electrons donated on changing the hapticity of the pyrrolyl ligand. Thus is also Fe(II), , and 18-electron: . Each complex has one time-averaged Cp proton environment, two pairs of protons from the pyrrolyl ligand, one adjacent to N and one farther from N. Hence three proton environments, with integrals , fit both structures. For proton nuclear magnetic resonance, three signals means three chemical shift environments; the individual resonances can still show coupling.
Past exam of the chemistry B course of the University of Cambridge 2022 Paper 2 B8 a i Solution 2026-10-20
Choose along M–N–N. The occupied dinitrogen orbital, with amplitude on the terminal N atoms, overlaps an empty metal orbital of sigma symmetry to give N-to-metal sigma donation. Filled metal and orbitals separately overlap the two empty N orbitals, giving metal-to-N pi backbonding. The lower-energy combination of each interacting pair is bonding and the higher is antibonding; each donation channel starts with two electrons in the occupied fragment orbital. The lower panel of the original orbital diagram shows these three channels.
As for a terminal carbonyl ligand, the donor and acceptor interactions reinforce one another: sigma donation helps make electron density available at the metal, while pi backbonding strengthens M–N and weakens N–N. For conventional electron counting an intact end-on neutral N ligand is an L-type two-electron donor; counting the backbonded electrons a second time as extra donated ligand electrons would double count them.
Past exam of the chemistry B course of the University of Cambridge 2022 Paper 2 B8 b Solution 2026-10-20
Use ionic electron counting, counting each ordinary donor atom once and keeping ligand charge distinct from denticity. A is 16-electron, B is 14-electron and C is 16-electron, so none is an 18-electron complex under this convention.
A contains a modified, monoanionic PNP pincer: the exocyclic double bond and altered pyridine-ring bond pattern in the drawing matter. Its two phosphines and its central N donor provide six electrons altogether, while neutral N provides two. Neutrality then gives Co(I), , and . Treating the modified pincer as a neutral intact pyridine ligand would incorrectly produce Co(0) and a 17-electron count. A four-coordinate approximately square planar molecular geometry in a complex can leave the strongly sigma-antibonding orbital empty and be stable at 16 electrons. The underlying modified-pincer chemistry has been examined using structure, spectroscopy and electronic-structure calculations.[ref]
Four-Coordinate Cobalt Pincer Complexes: Electronic Structure Studies and Ligand Modification by Homolytic and Heterolytic Pathways by Scott P. Semproni and others. JACS 136, 9211–9224; modified PNP framework and dinitrogen derivative
B has a neutral tertiary amine donor at the centre of its tripod and three anionic amido arms. Their combined charge is , so Re is Re(III), . The four N donors give eight electrons and N gives two: . Such an electron count is not inherently unexpected in an amido-supported complex: strong amido pi donation and the imposed five-coordinate geometry alter the orbital energies, and the empty orbitals are not nine equivalent slots that must all be filled. Steric protection and chelation also help stabilize this coordinatively restricted structure. The 14-electron figure uses the usual two-electron ionic count for each amido donor; pi bonding should not be appended to it inconsistently.
C has an anionic acyl carbon donor and a neutral phosphine in the chelate, together with two neutral PMe donors. Thus Co(I), , receives ligand electrons, giving . This is again a reasonable four-coordinate structure. The acyl C=O is part of a carbon-bound acyl ligand, not an additional free CO donor.





