Past exam of the chemistry B course of the University of Cambridge 2023 Paper 1 B4 a ii Solution 2026-10-20
Both permitted alternatives follow the same condensation–activation–migration sequence. sodium acetate liberates a useful concentration of hydroxylamine from its hydrochloride while providing a buffered medium. nitrogen of NH2OH attacks the ketone carbonyl; proton transfer gives a carbinolamine. protonation of its carbon-derived OH and loss of water, with nitrogen lone-pair donation, then gives the oxime after deprotonation. The two C=N geometries can be separated.
For the B-to-D route, oxime O-tosylation begins when oxime oxygen attacks sulfur of tosyl chloride and chloride leaves; triethylamine accepts the released proton. The resulting O-tosyl oxime has an excellent leaving group. The bond, anti to N–OTs, migrates to nitrogen while N–O cleaves, giving a ring-expanded nitrilium ion. Its local connectivity is , with the existing scaffold still joining the two paths. water attacks the nitrilium carbon . proton transfers and imidic-acid/amide tautomerization produce D, with nitrogen inserted on the aromatic side of the old carbonyl.
For the C-to-E route, the same O-tosylation activates the other oxime geometry. Now the anti bond is ; its migration gives , and aqueous trapping produces E, with nitrogen inserted on the spiro-junction side. Each Beckmann rearrangement expands the ketone's six-membered ring to a seven-membered lactam. The original carbonyl carbon remains the amide carbonyl carbon; nitrogen is the inserted atom. In this mechanism the ketone oxygen is lost during oxime condensation, the oxime oxygen leaves with the sulfonate group, and water supplies the new amide oxygen.
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 2023 Paper 1 B4 b ii Solution 2026-10-20
Write the reacting terminus as , where includes the pre-existing stereocentres and the intact substituent. ozone undergoes 1,3-dipolar cycloaddition across the terminal alkene to give a molozonide. In a resonance description of ozone as , the alkene pi pair bonds to one terminal oxygen while the other terminal oxygen forms the second C–O bond and the ozone pi bond shifts onto its central oxygen. Cleavage of this unstable primary ozonide produces a carbonyl compound and a carbonyl oxide, which recombine to form the secondary ozonide. Reductive workup breaks the peroxide bonds and gives the carbonyl fragments and formaldehyde.
Excess sodium borohydride also reduces the newly formed aldehyde: hydride attacks its carbonyl carbon, shifting the pi electrons onto oxygen, and the final acidic workup protonates the alkoxide. Thus the retained chain becomes . The terminal alkene carbon becomes formaldehyde and can be reduced to methanol; it is no longer part of the main product. This accounts for the one-carbon shortening of the retained skeleton.
The original amide carbonyl is present in both H and I and is not reduced under the ordinary borohydride workup. The existing stereocenters and benzyl ether are retained. Stopping the mechanism at an aldehyde, or deleting the amide C=O when reading the smaller printed scheme, would give the wrong product. The new terminal alcohol carbon is CH2 and therefore adds no stereocenter.
Past exam of the chemistry B course of the University of Cambridge 2023 Paper 1 B4 b i Solution 2026-10-20
The lactol is a cyclic hemiacetal and can open to , retaining its pre-existing alkoxy-bearing stereocentre. Its two anomers lose their anomeric configuration on opening to the planar aldehyde, so both can feed the same addition pathway. A first equivalent of decylmagnesium bromide can be consumed by the free OH, producing decane and a magnesium alkoxide; the excess reagent supplies the carbon nucleophile for addition.
magnesium can coordinate both the aldehyde oxygen and the adjacent benzyloxy oxygen. This produces a five-membered chelate, with ring sequence Mg–O(carbonyl)–C(aldehyde)–C(alpha)–O(benzyloxy), that restricts rotation about the aldehyde–alpha-carbon bond. Decyl attack on the less crowded face, away from the alpha-carbon's hydroxyethyl chain in the chelated arrangement, forms the new C–C bond and a magnesium alkoxide at the former aldehyde carbon. Acidic workup protonates both alkoxides to give G. The OBn bond is retained; it is not displaced by the Grignard reagent.
The product configuration is the one with OBn and the newly formed OH both projecting forward in the specified zigzag chain, while the opposite configuration at the new alcohol carbon is disfavoured. This is the Cram chelation model for an alpha-alkoxy aldehyde, rather than stereochemical memory of the starting lactol OH. Coordination is a mechanistic explanation consistent with MgBr2 and the product; the scheme does not quantify individual organomagnesium aggregates.[ref]
Felkin-Anh and Cram Chelate by University of Windsor. p. 3, section 2, Chelation and Addition–Cram Chelate Rule
The relevant primary synthetic sequence is documented independently.[ref]Highly Stereoselective Asymmetric Total Synthesis of (−)-Jimenezin via Sequential Intramolecular Amide Enolate Alkylation by Thien Nhan Lu and others. 2022, synthetic sequence leading through the protected oxygen heterocycle
Past exam of the chemistry B course of the University of Cambridge 2023 Paper 2 B7 c iii Solution 2026-10-20
The favoured insertion leaves nickel attached to the methyl-substituted alkyne carbon; the aryl group bonds to the SiMe-substituted carbon. This is the connectivity leading to the illustrated ring-closed product: the carbon carrying methyl subsequently attacks the ester carbonyl.
The bulky trimethylsilyl substituent is kept away from the crowded phosphine-ligated nickel centre in the insertion arrangement. The alternative orientation would place Ni next to the more demanding SiMe end. regioselectivity can also reflect electronic polarization of a silyl-substituted alkyne and stabilization of the resulting alkenyl complex. The ortho ester helps capture the insertion product by oxygen coordination and subsequent ring closure. Steric discrimination at insertion, reinforced by the accessible intramolecular cyclization, accounts for the observed direction; the product drawing alone is not a quantitative separation of steric and electronic contributions.
Past exam of the chemistry B course of the University of Cambridge 2023 Paper 2 B7 c ii Solution 2026-10-20
B is oxidative addition; D is alkyne insertion into Ni–aryl followed by carbonyl-oxygen coordination; E is carbonyl insertion into Ni–alkenyl. The distinction is visible from the bonds made and from consistent electron counting.
For B the depicted Ni(0) starting complex has two phosphine and two MeCN donors, giving electrons. Addition across the aryl C–Br bond produces two X-type ligands and Ni(II). The drawn product has lost the two solvent donors, giving electrons. Intrinsic oxidative addition to the unsolvated 14-electron fragment raises its count to 16; the overall drawn change also includes loss of four solvent-donor electrons. A falling displayed count does not turn this oxidative addition into a reductive elimination.
Before D, the aryl–alkyne intermediate is Ni(I), , with electrons. Syn migration couples aryl carbon to one alkyne carbon and leaves a Ni–alkenyl bond. Insertion alone removes the separate two-electron alkyne donor and would give a 15-electron alkenyl fragment. The ester carbonyl oxygen then binds, as drawn, restoring the count to . Ni remains .
In E the Ni-bound alkenyl carbon attacks the coordinated ester carbonyl carbon, forming the new ring-closing C–C bond. The C=O pi electrons move to oxygen, and an Ni–O alkoxide bond replaces Ni–C. This is carbonyl migratory insertion, equivalently intramolecular addition to a coordinated carbonyl. A neutral O donor plus an alkenyl X ligand become one alkoxide X ligand, so the drawn count falls , while Ni remains . A solvating molecule could fill the resulting vacancy, but it is not shown and should not be silently added to the stated count.
Past exam of the chemistry B course of the University of Cambridge 2023 Paper 2 B8 a i Solution 2026-10-20
The Dewar-Chatt-Duncanson model has two complementary interactions. The filled alkene orbital donates to an empty metal orbital, and an occupied metal orbital back-donates into the empty alkene orbital. The metal acceptor has the symmetry of the sum of the two ligand lobes facing it; the back-donor has opposite phases at the two carbon positions and matches the antisymmetric combination.Both channels strengthen metal–alkene attachment. Donation removes some electron density from the internal bonding orbital, while back-donation populates its antibonding partner, so coordination generally reduces C–C bond order. The schematic shows orbital phase signs, not positive and negative electric charges.
Past exam of the chemistry B course of the University of Cambridge 2023 Paper 2 B8 b ii Solution 2026-10-20
A useful strongly bound limit is Zr(IV), , and Ta(V), , with an aryne dianion in each complex. Charge balance uses two Cp ligands for zirconium, and Cp* plus two methyl ligands for tantalum:Equivalently, an unreduced neutral-aryne description starts from Zr(II) and Ta(III), both , and transfers that pair strongly into the aryne acceptor. These are alternative bookkeeping limits for the same covalent interaction; saying that a formal assignment has no available electrons must not erase the strong metal–carbon bonds in the reduced-ligand description.
Both ligands have therefore lost much of the extra in-plane C–C bonding of free benzyne and approach the same reduced, approximately double-bond-length C1–C2 framework. Their identical C–C distances are consistent with similar aryne reduction in both complexes. The distance is intermediate in a delocalized ring, so it should not be converted into an exact integer bond order.
The M–C distances depend additionally on metal size, effective nuclear attraction and orbital overlap. The high formal charge of Ta(V), the size effects associated with the lanthanide contraction, and effective covalent bonding favour shorter Ta–C contacts than Zr–C contacts. This is consistent with versus . Different coordination envelopes and ancillary ligands also contribute, so this pair is not a controlled measurement of a radius or a metal–carbon bond energy. Similar internal ligand bond order can coexist with different metal–ligand distances.
Past exam of the chemistry B course of the University of Cambridge 2023 Paper 2 B8 b i Solution 2026-10-20
The metal interacts with the extra in-plane bonding system of benzyne, not by replacing the ordinary out-of-plane aromatic pi system. Take the ring and metal to lie in the plane, with the metal-to-C–C-midpoint direction along and C1–C2 along . The carbon orbitals directed toward the metal form an occupied symmetric combination and an empty antisymmetric combination.
The occupied in-plane combination donates into an empty metal orbital of matching symmetry. An occupied metal -type orbital overlaps with the in-plane combination, providing back-donation. These are the same symmetry principles as alkene binding, rotated into the plane of the ring. The carbon orbitals perpendicular to the ring continue to participate in its aromatic system. The strained formal triple bond therefore does not supply two independent, ordinary out-of-plane alkene interactions.
Strong population of the in-plane acceptor can be described as reduction of the aryne to a dianionic, metallacyclopropene-like ligand. That description emphasizes two M–C sigma bonds and loss of the extra C–C pi bond. It is particularly useful for early-metal aryne complexes. The lower panels of the orbital schematic distinguish the interacting in-plane orbitals from the perpendicular aromatic set.
Past exam of the chemistry B course of the University of Cambridge 2024 Paper 1 B1 a iii Solution 2026-10-20
The initial nucleophile attacks silicon, not the nitrile or an alkene carbon. Each trimethylsilyl group has an electrophilic Si center. A methyl group from methyllithium attacks Si, passing through a pentacoordinate silicon arrangement; Si–O cleavage releases and the lithium salt of the oxygen-containing fragment. Cleaving the silyl enol ether generates a lithium enolate; cleaving the aryl silyl ether generates a phenoxide. Neither operation needs a new deprotonation at a carbon stereocenter.Here represents either the phenoxide or the oxygen-centered enolate contributor. Excess reagent permits both cleavages. The pre-existing methyl/nitrile stereocenter retains its configuration: it has no hydrogen to remove, and neither bond attached to that center is broken.
After HMPA is added, each anionic oxygen attacks the carbonyl carbon of allyl chloroformate. The C=O pi electrons move to the carbonyl oxygen, giving a tetrahedral intermediate; collapse expels chloride. The enolate becomes an allyl enol carbonate, and the phenoxide becomes an aryl allyl carbonate, each containing . oxygen acylation accounts for preservation of the enol double bond and for the two newly installed carbonate groups. carbon attack would give a different carbon skeleton. The two oxygens are trapped because they were selectively unmasked by desilylation, rather than because every ether oxygen in the molecule is equally reactive.
Past exam of the chemistry B course of the University of Cambridge 2024 Paper 1 B1 a ii Solution 2026-10-20
Lithium diisopropylamide removes an alpha proton from the ethyl side of the ketone. The C–H electrons form the C=C part of a lithium enolate, while the C=O pi electrons move to oxygen. The phenyl side has no competing alpha C–H site. Diisopropylamine is the conjugate-acid byproduct.The carbon end of this enolate attacks the primary carbon bearing iodine in . The C–I bond breaks to iodide in an SN2 reaction, and carbonyl reformation givesThus a three-carbon methoxypropyl chain is attached at the ketone alpha carbon. The remote methyl ether survives. The hindered amide base is chosen to generate the enolate rather than to add to the carbonyl. No absolute stereochemical preference is specified for the new stereocenter under these achiral conditions.
Past exam of the chemistry B course of the University of Cambridge 2024 Paper 1 B1 a i Solution 2026-10-20
An enolate is an ambident nucleophile: the negative charge is concentrated toward oxygen, while its occupied conjugated orbital also has substantial reactive amplitude at the alpha carbon. An alkyl iodide presents an accessible, polarizable carbon electrophile. carbon attack through an SN2 reaction is generally kinetically favorable, while the more strongly charged oxygen is often coordinated to the counterion or solvated. The carbonyl reforms as the new C–C bond is made.carbon alkylation restores the carbonyl and makes the new C–C bond. oxygen alkylation instead gives an enol ether. The two outcomes are not resonance contributors of one product, and their ratio is not determined just by where the largest formal negative charge is drawn. electrophile, counterion, solvent and aggregation can change selectivity; hard silicon electrophiles and suitable acylating reagents often favor oxygen trapping. Product stability alone would not explain an irreversible kinetic choice.
Past exam of the chemistry B course of the University of Cambridge 2024 Paper 1 B1 b i Solution 2026-10-20
The carbamate carbonyl oxygen attacks electrophilic sulfur of triflic anhydride, with cleavage of the anhydride S–O bond to release triflate. Donation of the nitrogen lone pair toward the carbonyl carbon then gives the indicated imidoyl-type cation: the former carbonyl oxygen now bears Tf, and the N–H nitrogen participates in a C=N bond. This converts a relatively unreactive carbamate carbonyl into a strongly electrophilic carbon bearing an excellent leaving group.
The aromatic pi bond para to ethoxy attacks that carbon. The C=N pi electrons move to nitrogen as the new aryl–carbon bond forms. Loss of the aromatic proton restores aromaticity; nitrogen donation and departure of the O-triflylated group restore the imine bond. proton transfers during the sequence and carbonate workup remove the remaining N–H proton. The surviving five-membered ring is an oxazoline: the product is 2-(4-ethoxyphenyl)-2-Oxazoline. An equivalent arrow order can eliminate the triflyloxy leaving group before arene capture; these are limiting descriptions of the activated electrophile, not a need to postulate an unactivated carbamate undergoing ordinary Friedel–Crafts acylation.
Ethoxy donates a lone pair into the ring, stabilizing the ortho and para Wheland intermediates. The para pathway avoids the crowding of an incoming heterocycle beside ethoxy, giving the observed position. Mechanistic experiments on this transformation include in situ carbon NMR of the activated reagent, competition experiments and isotope-effect measurements supporting electrophilic aromatic substitution; they do not directly image every intermediate.[ref]
Simple and Efficient Aromatic C–H Oxazolination by Qiu Shi and others. 2023, Figure 7 and supporting information, product 3a and electrophilic-substitution evidence
Past exam of the chemistry B course of the University of Cambridge 2024 Paper 1 B2 a Solution 2026-10-20
In the strongly acidic nitrating mixture, the dimethylamino nitrogen is predominantly protonated. The substituent is therefore during aromatic substitution, rather than the neutral electron-donating group. Its lone pair is unavailable for conjugation. Ortho or para attack would generate a sigma-complex contributor with positive ring charge at the carbon attached to an already positive nitrogen; meta attack avoids that especially unfavorable contributor.
Sulfuric acid protonates nitric acid, and loss of water produces the nitronium ion. Aromatic attack at a meta position forms a Wheland intermediate; deprotonation restores aromaticity. The bicarbonate workup then returns the amino group to its neutral form. is 3-nitro-N,N-dimethylaniline.
There are six inequivalent aromatic carbons in this unsymmetrical meta-disubstituted ring and one carbon environment for the two equivalent N-methyl groups, giving seven carbon NMR signals. The para isomer would have only four aromatic environments plus the methyl environment. An ortho isomer could also give seven signals, so signal counting supports but does not alone establish the meta assignment; protonation-dependent directing effects are essential.
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 2024 Paper 1 B2 b i Solution 2026-10-20
tert-Butoxide removes the acidic hydrogen from . The resulting carbon anion is stabilized by the nitrile group and the phenyl substituent. It attacks the aromatic carbon bearing fluorine, giving a Meisenheimer addition intermediate. Ring pi bonds shift toward the para nitro group, where the negative charge can be delocalized onto oxygen.
Loss of fluoride restores aromaticity and gives . This is an addition–elimination nucleophilic aromatic substitution. The phenoxy group remains attached to the incoming carbon. The ipso aromatic carbon already has a leaving group, so rearomatization can occur by fluoride departure without removing phenoxide. Acid workup neutralizes residual anions; it is not a separate step that replaces the aryl fluoride.
Past exam of the chemistry B course of the University of Cambridge 2024 Paper 1 B2 c ii Solution 2026-10-20
Base again generates catalytically. In , attack at the central allene carbon moves the adjacent C=C pi electrons onto the phosphorus-bearing carbon. This makes the stabilized alpha-phosphoryl carbanionprotonation at that carbon gives , the second product drawn above. The remaining double bond is terminal, so no E/Z pair is possible. The new sulfur bond is at the central carbon of the original allene, not at the terminal methylene.
This regiochemistry places the developing negative charge beside the electron-withdrawing phosphine-oxide group, instead of at a less stabilized vinylic position. It is an ionic addition/proton-transfer cycle, rather than radical thiol–ene chemistry. proton donation by another thiophenol molecule regenerates thiolate. One thiophenol is incorporated without loss of atoms, giving .
Past exam of the chemistry B course of the University of Cambridge 2024 Paper 1 B2 c i Solution 2026-10-20
Hydroxide initially deprotonates a little thiophenol, generating the reactive thiolate . Number the extended conjugated system , where C1–C2–C3 is the ring enone and C4=C5 is the attached vinyl group. sulfur attacks the terminal delta carbon C5. The pi bonds relay toward oxygen, generating an extended enolate.
proton transfer at C4, either directly from thiophenol or through solvent-mediated enolization, gives the product retaining the conjugated ring enone. The product is 3-[2-(phenylsulfanyl)ethyl]cyclohex-2-en-1-one, drawn below. Its side chain is ; the ring C2=C3 double bond remains. This is net sulfur/hydrogen addition to the terminal part of the extended system, proceeding through 1,6-Conjugate addition and protonation of the extended enolate.
The terminal site is less crowded than the ring beta carbon, and the isolated product retains favorable carbonyl–alkene conjugation. Reversible proton transfers and conjugate addition under thiolate catalysis can favor that product over an unconjugated enone-derived alternative. Transfer of a proton from another PhSH molecule regenerates PhS, explaining why a catalytic quantity of hydroxide suffices. The net balance is one thiophenol added, ; carbonyl 1,2 addition would instead produce a different functional group and is not favored for this soft sulfur nucleophile.
Past exam of the chemistry B course of the University of Cambridge 2024 Paper 1 B3 b Solution 2026-10-20
For both substrates, the alkene pi bond attacks bromine and polarizes/breaks Br–Br. A bromine lone pair completes a three-membered bromonium ion, and bromide subsequently attacks a carbon from the opposite face, cleaving that C–Br bridge bond. The two new C–Br bonds are anti. A free carbocation that rotates before capture is not needed.
In trans-fused , the fixed conformation makes the alternative anti-opening pathways diastereomeric. A chair-like, trans-diaxial opening is favored over a geometrically strained alternative that requires a twist/boat-like arrangement of the six-membered ring. The Fürst-Plattner rule describes this conformational preference. One depicted enantiomer of therefore gives essentially one dibromide diastereomer, with the two bromines occupying the trans-diaxial arrangement in its favored chair. If a racemic sample of were used, its other enantiomer would of course give the mirror product as well.
For cis-fused , the accessible mirror-related folds and the equivalent anti pathways produce mirror-related dibromides with equal rates in an achiral environment. Adding the two bromines destroys the symmetry that made the starting configuration achiral. gives a racemic dibromide pair. The racemate follows from equivalent enantiomer-forming pathways, not from a loss of anti stereospecificity.
Past exam of the chemistry B course of the University of Cambridge 2024 Paper 1 B3 c iii Solution 2026-10-20
P is H2, the acetal methine between two oxygens; its only resolved coupling is to the attached methyl, giving the quartet. S is that C2 methyl, whose doublet has the same 6.9 Hz coupling. T is tert-butyl, integrating to nine equivalent hydrogens with no adjacent carbon-bound methine proton across its quaternary attachment.
U is equatorial H5, coupled to the four hydrogens at C4 and C6. The two equivalent equatorial methylene hydrogens are Q, each a doublet of doublets with a 12.3 Hz geminal coupling to its axial partner and a 1.3 Hz coupling to H5. The two equivalent axial methylene hydrogens are R, each with the same 12.3 Hz geminal coupling and a 3.7 Hz vicinal coupling to H5. Equivalence between C4 and C6 makes two pairs rather than four separately shifted methylene hydrogens.Coupling U to two equivalent R hydrogens and two equivalent Q hydrogens produces its triplet of triplets. The different small vicinal couplings are consistent with oblique H–C–C–H torsions for equatorial H5; the 12.3 Hz value is geminal and must not be mistaken for a large vicinal trans-diaxial coupling.





