Claisen condensation 2026-10-20
carbon acylation of an ester enolate followed by deprotonation of the beta-dicarbonyl product. ketone-enolate acylation by an acid chloride is a closely related acyl-substitution strategy, although it is not the classical ester self-condensation.
Methanolysis of an anhydro acetal 2026-10-20
acid activation can cleave an anomeric acetal bridge, producing an oxocarbenium-like electrophile and a free alcohol on the former tether. methanol capture produces a methyl acetal; anomeric stereochemistry and chair preference must be distinguished.
Past exam of the chemistry B course of the University of Cambridge 2022 Paper 1 B1 b iii Solution 2026-10-20
One equivalent supplies carbon; the other acts as a base for the product. The newly formed 1,3-diketone is appreciably more acidic than the ketone from which D was prepared. proton transfer from its central CH2 to a second enolate equivalent is therefore favourable. Product-enolate stabilization also helps drive the acyl-substitution sequence towards completion.
The second enolate is consequently consumed as base, yielding pinacolone, rather than supplying a second carbon fragment to E. acid workup reverses the product deprotonation. Ignoring this acid–base step would predict that a single equivalent of D is sufficient to consume every equivalent of acyl chloride, despite the product immediately quenching part of the available reagent.
Past exam of the chemistry B course of the University of Cambridge 2022 Paper 1 B1 b ii Solution Created 2026-10-20 Updated 2026-10-31
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 B3 b Solution 2026-10-20
protonation of the secondary OH permits water departure. The neighbouring phenylthio sulfur can assist this departure by forming a bond to the adjacent alcohol carbon, generating a thiiranium ion spanning that carbon and the original sulfur-bearing cyclohexane carbon. Label the latter Q and the former L. The temporary Q–S–L bridge transfers the phenylthio attachment from Q to L when its Q–S bond subsequently breaks. sulfur attack opposite the departing OH accounts for the new C–S stereochemistry without requiring a freely rotating carbocation.
For the protected terminal alcohol, no free tethered OH is available to trap the bridge. Removal of a beta hydrogen from a cyclohexane carbon adjacent to Q forms the ring alkene while Q–S breaks. sulfur remains bonded to L, giving the allylic phenyl thioether. acid is regenerated by the deprotonation, so the catalyst need not be consumed.
For the free terminal alcohol, its oxygen attacks Q intramolecularly, opening Q–S instead. Count the new ether ring: O–CH2–CH(Me)–L–Q, a five-membered ring sharing Q with the cyclohexane. deprotonation gives the spirocyclic ether. This is capture of the same neighbouring-group-stabilized intermediate rather than migration of the methyl group. The pre-existing methyl-bearing carbon retains its configuration; its wedge/dash appearance changes when the product is redrawn in a different ring projection. The two substrates differ in whether a suitably placed oxygen nucleophile is available for intramolecular trapping.
Past exam of the chemistry B course of the University of Cambridge 2022 Paper 1 B4 c Solution 2026-10-20
protonation activates the oxygen of the anomeric acetal bridge. Cleavage of its bond to the anomeric carbon releases that tether oxygen as the terminal primary alcohol, while the other ring oxygen stabilizes the developing anomeric electrophile by lone pair donation. The resulting oxocarbenium description has a nearly planar reactive anomeric centre; it need not be drawn as a long-lived free carbocation.
methanol attacks that centre, giving a protonated methyl acetal, and deprotonation regenerates acid. The former bridge becomes CH2OH; the anomeric substituent becomes OMe. The other C–O bonds and existing carbon configurations are retained. The ring can adopt a different chair once the bridge constraint is removed, so an apparent movement of substituents in the projection does not imply inversion at all the other carbons.
The depicted anomeric OMe is axial in the product chair. This geometry allows an endocyclic oxygen lone pair to align antiperiplanar to the exocyclic C–OMe bond, providing stabilization associated with the anomeric effect. Acidic conditions also permit cleavage and recapture at the anomeric centre, so selection is not simply backside displacement at an unchanged tetrahedral starting carbon. The anomeric effect rationalizes the specified axial product; the scheme alone gives neither its numerical equilibrium population nor evidence that every other anomer is absent.
Past exam of the chemistry B course of the University of Cambridge 2023 Paper 1 B1 b iii Solution 2026-10-20
The first base equivalent removes O–H; the second removes the alpha C–H. The first equivalent of lithium diisopropylamide converts the carboxylic acid to its lithium carboxylate. A further equivalent removes the remaining hydrogen on the carbon between the two allyl groups, giving a carboxylic acid dianion:The second negative charge has enolate-like delocalization into the carboxyl group; lithium coordination and aggregation are not represented by this simple charge-bookkeeping structure. The carbon nucleophile attacks methyl iodide with C–I bond cleavage, forming the new C–Me bond. acid workup reprotonates the carboxylate, giving . The two equivalents are therefore chemically necessary for the stated carbon alkylation rather than simply an arbitrary excess. A single equivalent is consumed predominantly by the much more acidic carboxylic proton. The product has formula and still has two identical allyl groups, so this methylation does not create a stereocenter.
Past exam of the chemistry B course of the University of Cambridge 2023 Paper 1 B1 b ii Solution 2026-10-20
hydroxide attacks an ester carbonyl to form a tetrahedral intermediate. Collapse expels ethoxide, and acid–base transfer gives the carboxylate and ethanol. Repeating this nucleophilic acyl substitution at the other ester produces the malonate dianion; acid workup gives , with as defined above.
On heating, one carboxylic acid group is lost through a six-membered cyclic proton-transfer/bond-cleavage arrangement. A carbonyl oxygen of the retained group accepts the proton from the departing carboxyl OH as its C–C bond breaks and carbon dioxide forms; electron redistribution gives the enol-like form of the retained acid. proton transfer/tautomerization then producesThe net decarboxylation is . The central carbon gains a hydrogen, both allyl substituents survive, and only one carboxyl group is removed. It would be incorrect to draw hydrolysis directly giving the final acid without accounting for this carbon loss. The favourable cyclic pathway is characteristic of malonic acid decarboxylation.
Past exam of the chemistry B course of the University of Cambridge 2023 Paper 1 B1 c ii Solution 2026-10-20
The bromines ortho and para to OH are removed, leaving the meta bromine at C3. A suitable polar mechanism begins with electrophilic ipso protonation of a brominated carbon. protonation at C4 or C6 gives a Wheland intermediate stabilized by donation of the phenolic oxygen lone pair; an important contributor has an O-protonated cyclohexadienone-like framework. Corresponding protonation at C3 lacks this direct stabilization by resonance in chemistry from OH.
iodide can capture/reduce the departing electrophilic bromine as the C–Br bond electrons restore the aromatic pi system. Thus the incoming hydrogen remains on carbon while bromine is removed, unlike ordinary electrophilic substitution in which the incoming electrophile remains and a proton leaves. bromine trapping can be represented through IBr formation followed by reaction with further iodide to give iodine and bromide. For each replacement the balanced net reduction isApplying it at C4 and C6 gives 3-bromo-2,5-dimethylphenol, with formula . This explains why the position least activated by OH is the one at which bromine survives. It is a proposed acid-assisted protodebromination mechanism consistent with the selectivity; the preparative sequence alone does not prove every proton-transfer or halogen-transfer intermediate. It should not be drawn as iodide simply displacing bromide from an unactivated aryl carbon.
Past exam of the chemistry B course of the University of Cambridge 2023 Paper 1 B2 b iii Solution 2026-10-20
Warm C with dilute aqueous acid to hydrolyse the iminium ion to the second ketone. water attacks the iminium carbon, proton transfers give a carbinolamine, and C–N cleavage releases pyrrolidine, present predominantly as its ammonium salt in the acidic medium. The two carbon–carbon bonds constructed in the preceding Michael additions remain intact. This replaces the iminium nitrogen substituent by oxygen and gives the diketone D; neither oxidation of the skeleton nor reduction of the original ketone is required.
Past exam of the chemistry B course of the University of Cambridge 2023 Paper 1 B2 b ii Solution 2026-10-20
Number the cyclopentanone ring at its carbonyl, then at the methyl-bearing carbon, at the mesylate-bearing carbon, at the carbon joined to cyclohexane , and at the remaining methylene. B has a enone and the enamine. triethylamine removes the hydrogen at while the –O bond breaks to release mesylate. An E2 elimination, with approximately antiperiplanar geometry in the puckered ring, gives the conjugated cyclopentenone. Elimination towards instead would give a nonconjugated alkene.
The enamine carbon now attacks enone beta carbon intramolecularly. nitrogen donates into to give the iminium ion, while the enone electrons move onto the ketone oxygen. protonation of the ketone enolate gives C. The second new bond is ; the first was . These two connections account for the bridged skeleton without a concerted cycloaddition or an aldol step. The formula of B is : relative to A, elimination removes overall, namely methanesulfonic acid after proton transfers.
Past exam of the chemistry B course of the University of Cambridge 2023 Paper 1 B3 c Solution 2026-10-20
E, with the axial anomeric acetate, hydrolyses faster. The tert-butyl substituent strongly favours the chair in which it is equatorial, so the two substrates cannot freely interchange axial and equatorial acetate simply by an inexpensive ring inversion. protonation activates the anomeric acetoxy leaving group. As its C–O bond cleaves to release acetic acid, a lone pair on the endocyclic oxygen donates into the developing anomeric orbital, producing an oxocarbenium ion. water attacks that carbon, and deprotonation gives the cyclic hemiacetal. Attack from either face accounts for the unspecified anomeric configuration of OH in the product.
The axial C–OAc bond in E can adopt antiperiplanar geometry relative to an endocyclic oxygen lone pair. This favourable interaction assists departure in the ionization transition state. The corresponding bond in F is poorly aligned in its preferred chair; achieving comparable overlap requires an unfavourable chair change, placing tert-butyl axial, or substantial ring distortion. The distinction is stereoelectronic control of acetal hydrolysis, rather than merely counting steric crowding in the products.
An axial acetal can itself receive ground-state stabilization from the anomeric effect, so “axial means unstable” is not an adequate rate argument. What matters is the relative Gibbs energy of activation for the departing bond and the geometry of lone-pair assistance. Conformationally locked model acetals experimentally demonstrate faster acid-catalysed cleavage of the axially oriented isomer, although their numerical rate ratios should not be transferred to these different substrates.[ref]
Stereoelectronic control of the hydrolysis of a conformationally locked acetal by Anthony J. Kirby and Robert J. Martin. 1983, abstract and pp. 1627–1632, acid-catalysed axial/equatorial comparison
Past exam of the chemistry B course of the University of Cambridge 2023 Paper 1 B3 d Solution 2026-10-20
With triethylamine present, proton transfer from thiophenol produces a more nucleophilic thiolate population. sulfur attacks the anomeric epoxide carbon C1 from the face opposite its epoxide C–O bond. Displacement by backside attack opens the epoxide, breaks C1–O, and leaves the former epoxide oxygen attached to C2. protonation of that alkoxide by thiophenol or triethylammonium gives G: SPh at C1 on the front face and OH at C2 on the back face, a trans relationship. This is inversion at C1 with retention of the C2 C–O bond; there is no need to invert every stereocentre. Rapid protonation also suppresses competing silyl transfer by the alkoxide.
Without added base, neutral thiophenol is less nucleophilic. acid-assisted cleavage at the anomeric epoxide bond can instead be accompanied by donation from the ring oxygen, giving an oxocarbenium ion with charge shared with the ring oxygen. thiophenol capture from the face that puts SPh back on the original epoxide face, followed by deprotonation, gives H with SPh and C2–OH both on the back face, the net-retention pathway. An ion-pair or highly asynchronous opening can provide this outcome; the drawing does not establish a long-lived free carbocation.
The distinction is configuration, not a mandatory fixed chair. For the closely matching trisilylated substrate, the primary study obtained both anomers with thiophenol alone, and established an unusual inverted chair for the alpha thioglycoside by NMR and crystallography. Thus H should be explained as an accessible product, not asserted to be the sole product or necessarily to have axial SPh in its preferred conformation.[ref]
Reaction of thiophenol with glucal epoxides: X-ray structure of 3,4,6-tri-O-tert-butyldimethylsilyl-1-S-phenyl-1-thio-α-D-glucopyranoside by Caroline Walford and others. 1997, Schemes 1–2 and Figure 1; neutral reaction mixture and configuration/conformation assignment
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.


