Methyl vinyl ketone Created 2026-10-20 Updated 2026-11-05
Methyl vinyl ketone is , an alpha,beta-unsaturated ketone. Conjugation connects its carbon-carbon double bond with the carbonyl group.
A nucleophile can therefore react at the carbonyl carbon or through conjugate addition at the beta carbon. The product depends on the nucleophile and conditions, illustrating how conjugation creates more than one electrophilic site.
Monosaccharide Created 2026-10-14 Updated 2026-10-18
A monosaccharide is a single-unit carbohydrate that cannot be split into smaller sugars by hydrolysis. Ribose, deoxyribose, and glucose are examples. The unit is a whole molecule containing many chemical atoms; a disaccharide instead contains two linked monosaccharide units.[ref]
25.1 Classification of Carbohydrates. McMurry, Organic Chemistry, §25.1: simple and complex carbohydrates
Classification combines carbon count with open-chain connectivity. An aldose has an aldehyde group; a ketose has a ketone group. Thus glucose is an aldohexose, while fructose is a six-carbon ketose. These names describe the open-chain reference forms even when cyclic forms predominate in water.[ref]
25.1 Classification of Carbohydrates. McMurry, Organic Chemistry, §25.1: aldoses, ketoses and carbon-count classification
Open-chain D-glucose and D-fructose: the same molecular formula, different carbonyl group connectivity
. Original numbered Fischer projections. At tetrahedral stereocenters, horizontal chemical bonds point toward the viewer and vertical chemical bonds away. The carbonyl group carbon is not a tetrahedral stereocenter. The C5 hydroxy group on the right specifies D here; D does not specify the sign of optical rotation. These reference structures do not show relative abundances in water. Details.Check the distinction by counting the groups in the drawings. Open-chain glucose contains one , four and one group. Open-chain fructose contains two , three and one group. The respective hydrogen totals are and ; each structure also has six carbon and six oxygen chemical atoms. Their shared molecular formula therefore cannot identify the carbonyl group position or the stereochemistry.
The beauty is that a familiar sugar name can embrace an interconverting family of forms. Monosaccharide ring-chain interconversion connects those drawings to mutarotation, a change visible with polarimetry. Spatial arrangement also matters to biology: an enzyme's active site encounters a particular arrangement of hydroxy groups, rather than merely an inventory of chemical atoms. The existing enzyme and chirality articles explain that recognition principle.
Nef reaction 2026-10-20
Nitrile 2026-10-20
Nitroalkane benzannulation 2026-10-20
Nitroethylene 2026-10-20
Oxaloacetate Created 2026-10-18 Updated 2026-11-05
Oxaloacetate is a four-carbon dicarboxylate with a ketone group adjacent to one carboxylate. It is regenerated in the citric acid cycle rather than consumed as a net reactant on every turn.
Condensation with acetyl-CoA forms citrate; later oxidation of malate regenerates oxaloacetate. Removing oxaloacetate for biosynthesis therefore requires replenishing the cycle's intermediates to sustain flux.[ref]
Past exam of the chemistry B course of the University of Cambridge 2021 Paper 1 B1 b ii Solution 2026-10-20
The new ring is formed by intramolecular Friedel-Crafts addition to the ketone. titanium tetrachloride coordinates the ketone oxygen of the cyclic beta-ketoester, increasing electrophilicity at that carbonyl carbon. The nearby electron-rich aryl ring attacks this activated ketone at the available position ortho to its ethyl tether.
The electron-flow sequence is: an aromatic bond forms the new aryl–carbonyl-carbon bond, the ketone bond moves onto oxygen, and loss of the aromatic proton restores aromaticity. The two-carbon tether and the shared quaternary carbon make the newly closed carbocycle six-membered. Hydrolytic work-up converts the titanium-bound alkoxide into the tertiary alcohol.
The lactone carbonyl survives. This is an addition to the ketone with retention of its oxygen as OH, rather than an acyl substitution that removes that oxygen. The pendant terminal alkene is not involved. These bond changes account for the fused aromatic ring system and the alcohol in C without needing to assign the new stereochemistry.
Past exam of the chemistry B course of the University of Cambridge 2021 Paper 1 B1 b i Solution 2026-10-20
Let denote the 2,5-dimethoxyphenyl group. A is the doubly C-alkylated beta-ketoesterThe central carbon has no hydrogen after the second step. The alkene, ester and tert-butyl ether remain intact.
carbonate reversibly removes a proton from the methylene group between the two carbonyl groups. The resulting resonance-stabilized enolate attacks the primary iodoethyl carbon of the aryl-containing reagent by an displacement, releasing iodide. The monoalkylated intermediate still has one proton at the same activated carbon. A second deprotonation followed by attack on the primary bromide introduces the but-3-en-1-yl group.
The beta-dicarbonyl position is much more acidic than the methylene group next to the ketone and ether at the other end, explaining the alkylation site. Counting the retained carbon skeleton gives , including the two carbonyl groups and one terminal alkene. The second panel of the original product drawing specifies this connectivity.
Past exam of the chemistry B course of the University of Cambridge 2021 Paper 1 B2 b Solution 2026-10-20
A is a fused five-/six-membered bicyclic diketone containing sulfur. To specify its structure unambiguously, retain the cyclohexanone numbering C1=O, C2 and C3 for the original alkene. Add the five-membered ringThe original C1 ketone remains, and the former methyl-ester carbonyl becomes the second ketone. The product is drawn in the first panel below.
methoxide first deprotonates the thiol. The soft thiolate attacks the beta carbon C3 in a Michael addition, shifting the enone electrons onto oxygen. The resulting enolate is nucleophilic at C2. It then attacks its tethered ester carbonyl intramolecularly, forming a five-membered ring through a tetrahedral intermediate. Collapse expels methoxide. proton transfers and work-up give the neutral bicyclic beta-diketone.
The initial 1,4 addition followed by intramolecular Claisen acylation explains the regiochemistry: it connects C3 to sulfur and C2 to the ester carbonyl, giving a favorable five-membered closure. Addition to the original ketone would not give this same enolate-and-ester arrangement. The net loss of methanol from the combined starting materials gives the observed formula.
Past exam of the chemistry B course of the University of Cambridge 2021 Paper 1 B2 c Solution 2026-10-20
Label the enamine carbon bonded to nitrogen C1 and its alkene partner C2; C6 is the other carbon next to C1. B is the C-alkylated iminium ion bromide. Its new substituent at C2 is , while C1 is double-bonded to the pyrrolidine nitrogen, which bears positive charge. bromide is the counterion, accounting for Br in the empirical formula without requiring a retained C–Br bond.
The C2 end of the enamine attacks the bromomethyl carbon by displacement. Simultaneously, donation of the nitrogen lone pair creates the C1=N bond. This is ordinary enamine C-alkylation, not N-alkylation.
proton transfer removes an alpha proton at C6, converting the iminium ion into the enamine regioisomer C1=C6. That enamine attacks the terminal of the tethered acrylate in an intramolecular Michael addition. nitrogen donation restores the iminium ion, and protonation of the ester enolate completes the sequence.
The new bridge is . Together with the original ring and its C1 bridge this produces the bicyclononane iminium ion skeleton in C. It retains the pyrrolidine group and the ester; hydrolysis to a ketone is not part of the displayed reaction. The second panel of the original annulation figure shows the chemical atom connectivity and distinguishes ionic bromide in B from a covalent bromine substituent.
Past exam of the chemistry B course of the University of Cambridge 2021 Paper 1 B2 d Solution 2026-10-20
copper(I) converts the methyl grignard reagent into a softer organocopper nucleophile. Methyl undergoes 1,4 addition to the beta carbon of the cycloheptenone, giving a metal enolate rather than direct 1,2 addition to the ketone carbonyl.
The carbon nucleophilic site of this enolate, at the alpha carbon, attacks the aldehyde carbonyl in the second reagent. Its C=O electrons move to oxygen; protonation of the resulting alkoxide gives the beta-hydroxy ketone side chain. The cyclic acetal is a protected ketone and survives these nucleophilic conditions.
The beta methyl group shields one face of the cyclic enolate. Aldol bond formation from the less crowded opposite face places the new alpha side chain trans to the beta methyl group. Thus the two starred ring centers have an anti relationship; the two mirror-related products are possible from achiral starting materials and reagents. This argument does not assign the separate, unstarred alcohol center, for which the question gives no specified configuration.
Past exam of the chemistry B course of the University of Cambridge 2021 Paper 1 B4 a Solution 2026-10-20
The alcohol oxygen attacks sulfur of methanesulfonyl chloride, chloride leaves, and triethylamine removes the O–H proton. This gives an O-mesylate with retention of the carbon configuration: the original carbon–oxygen bond has not been displaced.
An accessible chair places the mesylate axial. triethylamine then removes the anti axial hydrogen on the carbon between the ketone and mesylate. Its C–H bond forms the new C=C bond while the C–O bond expels mesylate. The result is the conjugated cyclohexenone.
The neighboring carbon on the acetonide side has its relevant axial position occupied by its oxygen substituent, rather than an anti axial hydrogen. The elimination toward the ketone is therefore favored both by the trans-diaxial geometry and by conjugation of the resulting alkene with C=O. The acetonide remains intact. No inversion of the original alcohol carbon is required during mesylation, and it becomes planar during elimination.
Past exam of the chemistry B course of the University of Cambridge 2021 Paper 1 B4 b Solution 2026-10-20
B is the major product: the newly formed OH is the dashed bond in the printed zigzag structure. Retain the existing solid-wedge OH and the configuration at the carbamate-bearing carbon.
zinc coordinates the ketone oxygen and the adjacent hydroxyl oxygen, forming a five-membered O–Zn–O chelate. This constrains rotation around the carbonyl–alpha-carbon bond. In the favored chelated conformation the bulky group shields one face of the ketone, while the alpha hydrogen leaves the other more accessible. hydride attacks from that less crowded face, the carbonyl bond opens onto oxygen, and aqueous work-up gives the second alcohol.
Mapping this Cram chelation model back onto the printed zigzag gives B, not A. Merely placing a new OH “on the same side” of an unrotated page drawing is not a valid chelation argument: the reactive chelate and that drawing use different conformations. Zinc-borohydride reductions of alpha-hydroxy ketones are experimentally known to favor the corresponding anti-1,2-diol family under chelation control.[ref]
The reagent combination can generate zinc borohydride and related zinc-associated reducing species. The stereochemical justification is coordination-controlled facial selectivity; it does not require identifying a unique solution aggregate from the reagent list. The quoted yield is not itself a measurement of a 95:5 diastereomeric ratio.
Past exam of the chemistry B course of the University of Cambridge 2021 Paper 1 B4 c Solution 2026-10-20
The first step is a vinylogous aldol condensation at C5 of the unsaturated lactam. Number nitrogen N1, the lactam carbonyl C2, and then the methoxy-substituted alkene C3=C4 and the methylene group C5.
hydroxide reversibly removes a C5 proton. The resulting vinylogous enolate is delocalized through C5–C4–C3–C2–O: it is not necessary to create an unstabilized isolated carbanion beside nitrogen. Its C5 nucleophilic contributor attacks acetone. protonation gives a C5-linked tertiary alcohol, . Further removal of the remaining C5 hydrogen and elimination through an aldol-dehydration sequence forms C5=C(Me)2, leaving the original methoxy enol ether in place.
In the second step, aqueous acid hydrolyses that enol ether to a ketone at C4. protonation at C3 gives an oxygen-stabilized cation at C4. water attacks C4, proton transfers form a hemiacetal-like intermediate, and loss of methanol followed by deprotonation produces C4=O and the C3 methylene group.
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 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 B3 a Solution 2026-10-20
Number the OH-bearing carbon C1 and the adjacent brominated carbon C2. In the trans bromohydrin, base generates an alkoxide. A chair conformation can place O at C1 and Br at C2 trans-diaxial, aligning the oxygen nucleophile for backside attack at C2. Intramolecular SN2 closure expels bromide and inverts C2: the retained C1–O bond and the newly formed C2–O bond are both on the original OH face, giving the epoxide shown. ring inversion can provide the reactive chair even if it is not the dominant ground-state conformer.
The cis bromohydrin cannot place OH and Br trans-diaxial: one is axial when the other is equatorial. In the chair with Br axial, the hydrogen on C1 can instead be antiperiplanar to C2–Br. An alkoxide-assisted semipinacol pathway moves the C1–H bonding pair to C2 while the oxygen lone pair forms C1=O and bromide departs. hydride migration gives the ketone without changing the carbon skeleton. The hydrogen is transferred to C2 rather than expelled as a free hydride. The original phenyl-bearing stereocentre is not involved in either bond-reorganization sequence. Configuration controls which properly aligned displacement or migration can occur; a flat ring drawing alone conceals that distinction.



