1,3-diketone 2026-10-20
Two ketone carbonyls separated by one carbon. The intervening C–H bonds are relatively acidic, and the enol can be stabilized by pi conjugation and an intramolecular hydrogen bond.
Aldol addition 2026-10-20
Addition of an enolate or enol to an aldehyde or ketone, giving a beta-hydroxy carbonyl compound. Subsequent dehydration is a distinct step and gives an aldol condensation.
Chelated enol 2026-10-20
Enol 2026-10-20
An enol has an OH group bonded to a carbon of a C=C bond. Protonation and pi-bond rearrangement can interconvert it with a carbonyl compound through keto-enol tautomerism.
Enolization 2026-10-20
Interconversion of a carbonyl compound with its enol or enolate-derived forms. Reprotonation of a planar intermediate can epimerize an alpha stereocentre.
Intramolecular aldol reaction 2026-10-20
An aldol reaction between two reactive sites of the same molecule. A tethered enol or enolate attacks a carbonyl group to close a ring; the tether, conformation and competing chemical reactions influence the product.
Keto-enol tautomerism 2026-10-20
Manganese(III)-mediated radical cyclization 2026-10-20
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 B1 b i Solution 2026-10-20
Write . D is the sodium enolate . Its two resonances in the alkene-proton region belong to the two hydrogens of the terminal methylene group. One is cis to oxygen and the other trans to oxygen, so they are not related by a symmetry operation even though both belong to the same carbon. Replacing one or the other gives different alkene geometries, so they are diastereotopic. Their weak or unresolved geminal coupling is consistent with the reported singlet descriptions. Without an additional spatial assignment, it is not justified to decide uniquely which numerical shift belongs to which of those two orientations. The nine-proton singlet is the tert-butyl group. The formula follows from removing one alpha hydrogen from pinacolone.
E is the enol-dominated 1,3-diketone , observed as . The very downfield broad signal belongs to the enol OH in a six-membered intramolecular O–H···O chelate. The one-proton signal in the vinyl region is its central enol CH; a predominantly diketo structure would instead require a two-proton central CH2 signal and no enol OH. The eighteen-proton singlet accounts for both tert-butyl groups. Rapid exchange between the two degenerate enol orientations can average their different local environments; accidental unresolved equality cannot be excluded by the supplied spectrum alone. Thus one should not infer a permanently symmetric single enol Lewis structure from that singlet. The total integration is , agreeing with .
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 2025 Paper 2 B10 a i Solution 2026-10-20
Use C1 for the carbonyl carbon, C2 for the adjacent carbon, and C3 for the beta carbon of the reactive enone fragment . water adds conjugately at C3, giving a beta-hydroxy ketone. The oxygen lone pair of neutral water attacks C3; move the C2=C3 pi pair toward C1–C2 and the carbonyl pi pair onto oxygen. This produces the charge-separated enolate–oxonium intermediateA second water molecule can relay a proton from the newly attached oxonium oxygen to the enolate system. Using the carbon-centered resonance structure, reform C1=O and protonate C2 while the donating O–H bond returns its pair to the added oxygen. Equivalently, oxygen protonation first gives an enol, followed by water-mediated keto-enol tautomerism. Either path yields a neutral beta-hydroxy ketone and regenerates the relay water. There is no need to introduce externally supplied H or OH; solvent molecules perform the proton transfers. The uncatalyzed charge separation has a large barrier, so this mechanism describes a slow formal pathway, not a claim of rapid hydration in pure water.
Reactive-fragment mechanisms for conjugate hydration, steroid reduction and finasteride trapping
. Details. Past exam of the chemistry B course of the University of Cambridge 2025 Paper 2 B10 b iii Solution 2026-10-20
Hydride is delivered to steroid C5 and a proton to C4; the C3 ketone is retained. Use the conjugated fragment . Carbonyl hydrogen bonds from Tyr91–OH and protonated Glu57 lower the barrier to conjugate reduction. Draw a curved arrow from the nicotinamide C4–H bond of NADPH to steroid C5, shift the C4=C5 pi pair toward C3–C4, and shift C3=O pi electrons onto oxygen. The nicotinamide ring rearomatizes to NADP+; the steroid becomes a bound enolate, or its protonated enol equivalent. electron shifts across the nicotinamide ring must accompany C–H bond cleavage rather than leaving an unexplained neutral carbon radical.
Next, reform C3=O and protonate steroid C4 from an appropriately positioned acid, directly or through an enol/proton-relay route. This regenerates the catalytic acid after solvent reprotonation. The enzyme fixes the hydride-accepting face so that the new C5–H has the 5α configuration, giving dihydrotestosterone; delivery to the opposite face would give a different ring-junction configuration. This is not simple 1,2 reduction of the C3 carbonyl to an alcohol. The net local bookkeeping is one hydride plus one proton across C4=C5, with two electrons supplied by NADPH. The original reactive-fragment diagram shows the enolate-forming arrows and the separate protonation step.
A consistent proposed sequence is desilylation, five-membered hemiketal formation, retro hetero-Diels–Alder opening, intramolecular aldol addition and hemiacetal closure. The essential carbon rearrangement is loss of the original ketone–side-chain bond and formation of a bond between its two neighbouring carbons; it is not a simple Michael addition.
For an explicit atom map, label the starting six-membered ring , with the ketone at , methyl substituents at and , the hydroxy group at , and the side chain . Here is the methylene and is the carbon bearing the protected alcohol. Let be the original ketone oxygen and the protected alcohol oxygen. These labels identify atoms, not new substituents.
First, Fluoride removes the silyl protecting group from . The resulting alcohol attacks the protonated ketone at , forming the five-membered hemiketal . The original oxygen is now a hydroxy group; the original pyran ring remains intact at this stage.
Next, The original pyran ring can undergo a retro hetero-Diels-Alder reaction. The three electron-pair movements are , and : break and , and form the three indicated double bonds. The result is the tethered aldehyde/dienol esterThe temporary bond keeps both reacting ends within the same molecule. At , both oxygen substituents remain attached, and the new tether prevents separation into independent fragments.
Then The enol attacks the tethered aldehyde at . Formation of , restoration of , and proton transfer give the six-membered lactone ring , with . This is an intramolecular aldol reaction, and it supplies the crucial new carbon–carbon bond without losing any carbon atom.
Finally, Keto-enol tautomerism at gives the methyl ketone at and introduces the hydrogen at . The alcohol then attacks that ketone, closing the five-membered hemiacetal ring . The resulting fused rings have exactly the product's connectivity: is the lactone carbonyl, is its ring oxygen, and is the five-membered-ring oxygen.
The stereocentre at survives throughout this sequence. The configuration initially at is erased at the aldehyde stage, and those at , , and arise during the later cyclization and proton-transfer steps. A folded intramolecular aldol reaction followed by reversible hemiacetal formation is consistent with the drawn cis ring junction, but connectivity and orbital bookkeeping alone do not prove that it is the unique selected diastereomer. In particular, a claim of simple retention at would contradict this atom map.
A related pyranone-to-fused-lactone rearrangement under aqueous hydrogen fluoride is experimentally reported; that account also explicitly qualifies the relative stereochemistry as assigned on mechanistic grounds. The sequence above is a proposed explanation of the examination structure, rather than a mechanism claimed to have been experimentally established in that paper.[ref]
Total synthesis of (±)-tirandamycin A by Philip DeShong and others. Pages 5220–5221, Schemes II–III and the discussion of pyranones 10 and 13 giving lactone 12
A suitable sequence is selective alpha-bromination, then enol O-methylation. Use one equivalent of Br under controlled acidic enolization conditions to introduce bromine at the enolizable alpha carbon that becomes the brominated vinyl carbon. The other alpha site is a constrained bridgehead and cannot form the required ordinary enol double bond. Then O-alkylate the ketone with trimethyloxonium tetrafluoroborate and remove the brominated alpha proton with a nonnucleophilic base such as triethylamine. Collapse of the O-methylated carbonyl oxonium gives the alpha-bromo methyl enol ether E.
The methyl group belongs on oxygen, not on the alpha carbon. Simply writing methyl iodide with an unspecified enolate would not guarantee this outcome because competitive C-alkylation is possible. Use conditions designed for carbonyl O-alkylation followed by deprotonation, and limit bromine to avoid overbromination or damage to the protecting group.
Manganese(III) removes one electron from the enol/enolate of H, and loss of its hydroxyl proton gives a carbon-centred radical at the tether-bearing alpha carbon of the resulting 1,3-dicarbonyl system. The enolic OH becomes the second carbonyl oxygen. This radical attacks the terminal CH of the tethered isopropenyl group, forming a new C–C bond and a six-membered ring. The residual radical is on the substituted carbon bearing methyl, which is more stable than the primary radical from the alternative regiochemical closure.
Dioxygen traps that radical as a peroxyl intermediate. reduction/proton transfer and O–O cleavage in the ensuing oxygenation/workup sequence give the new tertiary alcohol on the methyl-bearing carbon. The drawn net transformation specifies that oxygenated product, not a persistent hydroperoxide; individual electron/proton transfers require the reaction medium and workup and cannot be fixed uniquely from the abbreviated reagent line alone. The rigid cage and folded tether select the closure face and therefore the relative configuration of the new ring junction.
Finally the new hydroxyl oxygen attacks silicon of TBSCl, with imidazole assisting deprotonation and chloride departure, giving the tertiary silyl ether I. The hindered pre-existing hydroxyls remain as shown. Track the two key atom changes: the terminal alkene carbon becomes a methylene in the new ring, and the former substituted alkene carbon acquires oxygen and retains its methyl group. Calling the first step a two-electron aldol reaction would miss both the single-electron oxidant and the role of dioxygen.
Single-electron transfer 2026-10-20
Tautomerization 2026-10-20
Interconversion of constitutional forms by proton transfer accompanied by redistribution of pi bonds. Keto–enol interconversion is a familiar example.
