Adenosine deaminase 2026-10-20
Adenosine deaminase catalyzes conversion of adenosine and water into inosine and ammonia. Its zinc-containing active site activates water and stabilizes the reacting configuration; proton transfers assist addition and leaving-group departure. Mutation kinetics constrain catalytic roles but do not uniquely locate every proton.
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 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 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 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 B2 b i Solution 2026-10-20
The cyclohexanone enamine attacks the beta carbon of the cyclopentenone. Its nitrogen lone pair forms an iminium C=N bond as the enamine pi bond forms the new C–C bond; simultaneously the enone pi system shifts onto its oxygen. proton transfer converts the resulting cyclopentanone enolate back to the ketone.
A final iminium deprotonation regenerates an enamine, but the position of its double bond matters. Label the nitrogen-bearing cyclohexane carbon , its initially nucleophilic neighbour , and the other neighbour . After addition, bears the cyclopentanone substituent. deprotonation at gives , the less substituted, less hindered enamine drawn in A. Merely restoring the original bond would give a different regioisomer. Interconversion through the iminium ion permits this enamine regioisomerization under the reaction conditions. The methyl and mesylate substituents are spectators during this first Michael addition.
Past exam of the chemistry B course of the University of Cambridge 2023 Paper 1 B2 c iii Solution 2026-10-20
base removes an alpha-amide proton from G, giving a carbon nucleophile. Its intramolecular attack on the nitroaryl ipso carbon, which currently bears the ether oxygen, forms a five-membered cyclic Meisenheimer complex. Count its atoms along the existing tether: alpha carbon, the original phenyl carbon bearing the side chain, its adjacent phenoxy carbon, ether oxygen, and attacked nitroaryl ipso carbon. The ortho nitro group again stabilizes the temporarily dearomatized anion.
Rearomatization now breaks the nitroaryl–O bond, leaving phenoxide on the original ring. proton transfer gives . This is a Truce-Smiles rearrangement: the nitroaryl group migrates from oxygen to carbon, while the original phenoxy oxygen stays on its original ring and becomes OH. No nitro-group migration or amide cleavage is needed. G and H have identical formulas; only the bond connections and a proton position change. carbonate need not produce a high equilibrium enolate concentration: a small population can be consumed by favourable intramolecular cyclization and rearomatization. The specified transformation alone does not establish its detailed proton-transfer sequence.
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.
Past exam of the chemistry B course of the University of Cambridge 2024 Paper 2 B10 a Solution 2026-10-20
The reactive fragment is the C6 amidine-like portion of the purine ring; the N9–ribose attachment is retained. Write it locally as , with the rest of the fused ring understood. One productive acid catalysis route protonates N1, making C6 more electrophilic. water then adds at C6 while the C6=N1 pi electrons move onto N1. Subsequent proton transfers generate a carbinolamine and then a protonated amino leaving group:These are local bonding fragments: C6 retains its bond to the adjoining ring carbon at every step. In the addition step, an oxygen lone pair forms C6–O and the C6=N1 pi bond opens. During collapse, an oxygen lone pair reforms C6=O and the C6–N6 bond breaks toward the departing nitrogen. Solvent mediates each proton transfer; no direct unsupported transfer across a long distance is required.
water supplies the new C6 oxygen, and ammonia is expelled after protonation of the leaving group. Final deprotonation gives the lactam form of inosine. In acidic bulk solution the released ammonia is largely trapped as ammonium. The catalytic proton transfers regenerate acid around the substitution cycle; ammonium trapping separately affects the net proton balance. This is hydrolytic deamination, not cleavage of the glycosidic bond.
Past exam of the chemistry B course of the University of Cambridge 2024 Paper 2 B10 d iii Solution 2026-10-20
Glu217 is important for catalysis, most naturally through proton transfer at the purine ring. Changing a glutamic acid residue to glutamine preserves much of the side-chain size and hydrogen-bonding geometry, while removing an ordinary carboxylic acid/base pair. Its placement near N1 makes acid donation there a plausible way to activate C6 for water addition and stabilize the resulting nonaromatic intermediate. The large turnover defect with little change in the Michaelis constant supports that catalytic interpretation, subject to the same and protein-integrity qualifications as above.
The protonated carboxyl drawn at pH 7 need not behave like free glutamate in water: local electrostatics and burial can shift its p, and protonation may change during the cycle. The mutation alone cannot choose between an initially protonated Glu217 acid and a Glu217 residue that first accepts and then relays a proton. Primary mutational work supports catalytic roles for active-site histidine and glutamate groups while explicitly comparing purified enzymes and zinc content.[ref]
Mutational analysis of active site residues of human adenosine deaminase by D. Bhaumik and others. 1993, pp. 5464–5470, kinetic and zinc-content comparisons
Past exam of the chemistry B course of the University of Cambridge 2024 Paper 2 B10 d iv Solution 2026-10-20
The oxygen installed at C6 is derived from solvent water. An attacking zinc-bound water or hydroxide therefore provides the straightforward explanation of the oxygen-18 label. The result supports hydrolytic addition rather than direct oxygen transfer from a pre-existing protein carboxyl group as the sole oxygen source.
It does not by itself establish the identity of the base, the timing of proton transfers, or exclusion of a transient covalent enzyme intermediate: water could introduce the labeled oxygen during breakdown of such an intermediate. Product oxygen exchange must also be excluded. Incubate preformed unlabeled inosine in labeled water with and without enzyme, and use early-time quenched reactions to compare incorporation with product formation. These controls test whether labeling reports the formation step rather than later exchange.
Past exam of the chemistry B course of the University of Cambridge 2024 Paper 2 B10 e Solution 2026-10-20
A chemically coherent mechanism combines zinc-activated water addition with a proton relay involving His238 and Glu217. In a simple route consistent with the displayed contacts, neutral His238 accepts a proton from zinc-bound water while protonated Glu217 donates to substrate N1. These transfers can be coupled to the bond-forming event rather than requiring accumulation of a free, doubly activated substrate. Zinc-bound hydroxide attacks C6; the C6=N1 pi electrons move onto N1, yielding the tetrahedral C6 intermediate bearing OH and NH while N1 bears H.
Next, the exocyclic amino group receives a proton, through His238, Glu217 and/or intervening water as permitted by the changing geometry. The oxygen is deprotonated as it forms the C6 carbonyl, and the C6–N6 bond cleaves to release ammonia. These transfers can be coupled in the collapse step, avoiding expulsion of . Zinc stabilizes the reacting oxygen and the organized intermediate; the ring regains conjugation in the inosine product. Product dissociation, fresh water binding and proton exchange with solvent reset the catalytic groups.
The core result is zinc-assisted addition–elimination, with acid/base proton transfers and water supplying C6 oxygen. The contributions are metal ion catalysis, general base catalysis, general acid catalysis, electrostatic catalysis and proximity and orientation catalysis. The proposed route does not require covalent catalysis by a protein side chain: a substrate–water C–O bond is not an enzyme–substrate covalent linkage. Asp296 can retain a structural hydrogen-bonding role without acting as an indispensable acid/base.
The data establish the importance of His238 and Glu217 but do not uniquely assign the entire proton relay. In particular, an alternative has Glu217 first deprotonate water and then deliver that proton to N1, with His238 assisting hydroxide stabilization and later amino-group protonation. The primary observation that His238 substitution need not abolish analogue hydration cautions against calling its initial-base role proven.[ref]
Site-directed mutagenesis of histidine 238 in mouse adenosine deaminase: substitution of histidine 238 does not impede hydroxylate formation by V. Sideraki and others. 1996, pp. 15019–15028, retained hydroxylate formation and proposed electrostatic contribution
pH-rate profiles, solvent isotope effects and time-resolved intermediate experiments would discriminate these refinements. The diagram is a proposed catalytic route, not a structure directly observed by the crystal experiment. Past exam of the chemistry B course of the University of Cambridge 2025 Paper 2 B8 b i Solution 2026-10-20
Write for the cyclopentadienyl ligand bearing its –2-pyridyl arm; means that its ring and pyridyl nitrogen both bind the same metal. . The base removes the pyridinium N–H proton in , restoring a pyridine lone pair. Intramolecular nitrogen coordination then replaces one acetonitrile ligand and closes a chelate ring. Both CO ligands are retained; the molybdenum remains Mo(II), .
This is an proton transfer followed by chelation, with metal electron count . The chelate effect, favorable formation of the Mo–N bond, and release of a solvent molecule explain the driving force. Merely invoking a 16-to-18-electron change would be misleading because is already 18-electron. In the preceding acid treatment, protonation of the allyl ligand allows release of propene; the other proton resides on the pendant pyridine, accounting for 's overall charge. One proton is removed here; the cyclopentadienyl tether is not deprotonated or detached.
First, proton transfer from acetic acid to the carbon of diazomethane gives methyldiazonium and an acetate ion:Draw an electron-pair arrow from carbon, using the carbon-negative resonance contributor, to the acidic hydrogen, and another from the acid O–H bond to oxygen. Second, an acetate oxygen attacks the diazonium methyl carbon by an SN2 reaction; an arrow from the C–N bond to nitrogen expels neutral :The acid's hydroxyl hydrogen has become the third hydrogen of the newly installed methyl group. This is a methyl-transfer sequence driven by formation of molecular nitrogen, without forming a free methyl carbocation or invoking Fischer esterification.
The oxygen substituent changes position through a halohydrin–epoxide sequence. First, hydroxide attacks the acetate carbonyl of , giving a tetrahedral intermediate. Collapse expels the substrate alkoxide, and proton transfer gives plus acetate. This is saponification, breaking the acyl–oxygen bond rather than replacing bromide directly with hydroxide.
A small equilibrium population of undergoes intramolecular SN2 reaction at the adjacent primary carbon: the O lone pair attacks and its C–Br pair leaves with bromide. The result is 2,2-Dimethyloxirane. Addition of HBr protonates the ring oxygen. bromide then attacks preferentially at the more substituted carbon, where the protonated ring has greater carbocation character, breaking that C–O bond and leaving oxygen attached to the former primary carbon:The middle formula denotes a three-membered epoxide, not an open-chain ether. Acidic opening of this highly unsymmetrical epoxide has the opposite regioselectivity to a simple basic nucleophilic opening at its less substituted carbon.
Finally, the primary alcohol oxygen attacks acetyl chloride. The tetrahedral intermediate collapses with chloride departure, and triethylamine removes the O–H proton and captures HCl. The product is . oxygen is carried through the epoxide to the other carbon; it is not necessary to invent a methyl migration or a carbon-skeleton rearrangement.





