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 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 c ii Solution 2026-10-20
The axial penalties depend on the ring position. An axial methyl at C2 encounters the axial hydrogens on C4 and C6. The short C–O links and dioxane geometry make these 1,3-Diaxial interactions particularly unfavorable. In contrast, an axial substituent at C5 points toward ring oxygens O1 and O3 rather than toward axial C–H groups at those positions. Those oxygen sites present much less steric obstruction than axial hydrogens on carbon.
Avoiding the expensive axial C2 methyl outweighs the cost of axial tert-butyl at C5. Transferring cyclohexane tert-butyl preferences unchanged would therefore give the wrong answer. This is primarily a position-dependent steric comparison, not the usual anomeric preference for an axial electronegative substituent at an acetal carbon. Primary NMR work on substituted dioxanes documents this axial C5 tert-butyl conformer.[ref]
Carbon-13 magnetic resonance. Conformation in some tert-butyl-1,3-dioxanes by Alan J. Jones and others. 1971, pp. 4772–4777, discussion of cis-2-methyl-5-tert-butyl-1,3-dioxane
Past exam of the chemistry B course of the University of Cambridge 2024 Paper 1 B3 d Solution 2026-10-20
The cis fusion admits two chair–chair conformers, related by simultaneous ring inversion. The bridgehead hydrogens remain cis, while their axial/equatorial descriptions exchange. In one fold the oxygen next to the fusion occupies the concave, inside position; in the other, the corresponding inside position is occupied by a methylene group. Both chair-only structures are shown in the lower panels of the dioxane diagram.
The O-inside chair–chair conformer is strongly favored. In the unfavorable fold, the inward methylene and its C–H bonds create close nonbonded contacts with the other ring, analogous to costly 1,3-diaxial/gauche contacts in cis-decalin. Placing oxygen in that location removes the corresponding carbon-bound hydrogens and reduces the obstruction. The comparison is between the two conformers of the same cis configuration; making the fusion trans would be a configurational change, not a chair flip. Nor is it appropriate to flatten both six-membered rings into regular hexagons, which would hide the decisive inside contacts.
Past exam of the chemistry B course of the University of Cambridge 2024 Paper 1 B4 a Solution 2026-10-20
The alkene undergoes concerted oxygen transfer from the terminal peroxide oxygen of mCPBA. In the cyclic electron-flow description, the alkene pi bond begins one C–O bond, the peroxide oxygen supplies the second, O–O breaks, and proton transfer/carboxyl rearrangement gives m-chlorobenzoic acid. Both C–O bonds are installed from the same alkene face, so the alkene's relative substituent geometry is retained in the epoxide. There is no freely rotating carbocation intermediate.
The neighboring free OH hydrogen-bonds to the approaching peracid and favors a cyclic, organized transition-state geometry. oxygen is delivered from the OH-directed face, giving the indicated syn relationship between epoxide oxygen and the allylic hydroxyl. In the supplied projection, the epoxide C–O bonds and the original C–OH bond are on the rear/hashed face. Steric interactions still influence the preferred hydrogen-bonded conformer, but “attack from the least hindered face” alone misses the positive directing role of OH. The distant benzyl ether is not the directing free hydroxyl.
The classical experimental comparison of allylic alcohols and protected derivatives established this directing principle; it supports the mechanistic interpretation without itself measuring the selectivity of this acyclic substrate.[ref]
Past exam of the chemistry B course of the University of Cambridge 2024 Paper 1 B4 b Solution 2026-10-20
Acid protonates one methoxy group of 2,2-Dimethoxypropane. methanol departure gives the oxocarbenium species , which is captured by one diol oxygen. After proton transfer, protonation and loss of the remaining methoxy group generate a second oxocarbenium center. Intramolecular attack by the other diol oxygen closes the five-membered ring; deprotonation regenerates acid. The net transacetalization exchanges two methoxy groups for the two diol oxygens, releasing two methanol molecules.
The two starred carbon centers retain their original configurations, and the two carbon side chains on the acetonide are trans. No C–O bond at either original stereocenter is broken; reaction takes place at the reagent-derived CMe acetal carbon. The apparent change in wedge/dash presentation comes from redrawing the backbone into the five-membered ring, not inversion at a starred center. In the planar pentagon used in the diagram, draw the alkynyl-bearing side chain behind and the side chain in front. The unstarred methyl-bearing center also retains its configuration. The acetal carbon itself is not stereogenic because its two methyl substituents are identical.
Past exam of the chemistry B course of the University of Cambridge 2024 Paper 1 B4 c Solution 2026-10-20
The reacting alkene pi bond and two Os=O units undergo a concerted cycloaddition to give a five-membered cyclic osmate ester. Both new carbon–oxygen bonds form on the same face. Hydrolysis and proton transfer cleave the Os–O attachments, releasing the vicinal diol and a reduced osmium-containing species.
This is syn dihydroxylation of the terminal alkene. The terminal carbon becomes , and the adjacent carbon becomes ; the remote vinyl chloride remains as drawn. The protected region denoted by is a spectator for this local mechanism. No explanation of alkene chemoselectivity is required here. Since no terminal reoxidant is specified, do not invent a catalytic regeneration step for Os(VIII); osmium reduction and product release suffice for the stated transformation.
Past exam of the chemistry B course of the University of Cambridge 2024 Paper 1 B4 d Solution 2026-10-20
The free propargylic OH first reacts with an aluminium hydride, releasing hydrogen and forming an O–Al bond. Coordination of the nearby alkyne to this aluminium center organizes a five-membered O–C–C–C–Al unit. Hydride addition from the opposite side of the activated alkyne gives a cyclic alkenyl aluminate: aluminium bonds to the distal alkyne carbon, while hydride is delivered to the carbon nearer .
The key step is directed trans hydroalumination. The new C–H and C–Al bonds are anti. Aqueous workup protonolyzes C–Al with retention of its alkene geometry and hydrolyzes the O–Al attachment. Replacing that C–Al bond by C–H therefore produces the E allylic alcohol, with and the rest of the chain on opposite alkene sides. A generic syn hydroalumination drawing followed by simple protonation would predict the wrong geometry.
The free OH provides the directing tether; the silyl-protected oxygen is not equivalently available. Primary electrophile-trapping studies of Red-Al-treated propargylic alcohols support the alkenyl-alanate description.[ref]
trans-Hydroalumination/Alkylation: One-Pot Synthesis of Trisubstituted Allylic Alcohols by Neil F. Langille and Timothy F. Jamison. 2006, pp. 3761–3764, trans-hydroalumination and trapping of alkenyl alanates
Cyclic intermediates in osmium dihydroxylation and directed trans hydroalumination
. Details. Past exam of the chemistry B course of the University of Cambridge 2024 Paper 1 B4 e Solution 2026-10-20
The alcohol oxygen attacks sulfur of methanesulfonic anhydride. Cleavage of the anhydride S–O bond releases methanesulfonate, and the amine base removes the O–H proton, giving the corresponding mesylate. This activation preserves the substrate C–O bond while replacing the poor hydroxide leaving group by .
The base then removes a hydrogen from the adjacent methylene on the side toward the vinyl chloride. In a favorable antiperiplanar arrangement, C–H electrons form the new C=C bond while the C–OMs bond breaks. An E2 elimination creates the alkene between the former alcohol carbon and that methylene. It links the two neighboring unsaturated regions into the extended conjugated triene segment, while the pre-existing carbon skeleton and vinyl chloride are retained.The left-hand polyenoate portion is unchanged. The base is diisopropylethylamine; it accepts acid during mesylation and promotes elimination. The product's E/Z mixture concerns the newly formed alkene, and no single geometry needs to be assigned.
Past exam of the chemistry B course of the University of Cambridge 2024 Paper 1 B5 a iii Solution 2026-10-20
C is the best geometrical match for a strongly bound 1:1 complex with E. Both have an adjacent amide donor–acceptor edge. Arrange them head-to-tail so that the N–H of E donates to the carbonyl oxygen of C, and the N–H of C donates to the carbonyl oxygen of E. This closes the familiar two-hydrogen-bond cyclic amide motif, illustrated below.
C combines a strong acceptor with a useful donor in the correct spacing. B has a stronger individual donor, but its pyridine N and OH are separated around the ring and do not reproduce the compact adjacent amide edge of E. A lacks a matching donor, and D's N–H is weak and has the wrong geometry for the same amide–amide motif. Site strength and spatial complementarity must be considered together; multiplying two isolated probe constants is not a reliable prediction of the complete association equilibrium. Self-association of C or E can compete and must be included in a quantitative binding analysis.
Past exam of the chemistry B course of the University of Cambridge 2024 Paper 1 B5 a ii Solution 2026-10-20
4-Fluorophenol probes acceptor strength because it donates its OH hydrogen. The qualitative order is C D B A: the electron-rich lactam oxygen is the strongest acceptor here, the secondary amine is next, pyridine nitrogen is intermediate, and aldehyde oxygen is weakest. Resonance donation from amide nitrogen increases the lactam oxygen's basicity relative to an ordinary aldehyde or ketone oxygen. The N-methylpyrrole site in A does not add an independent amine-like acceptor.
N-Methylpyrrolidin-2-one is a carbonyl acceptor without N–H, so it instead probes the solute's donor ability. B C A and D: B's phenolic OH is a strong donor, the lactam N–H of C is appreciable, A lacks a conventional donor, and D's neutral secondary-amine N–H is weak. The amine can be a good acceptor and a poor donor at the same time; the two properties are not opposites on one scale.
These are association equilibria in cyclohexane, where solvent competition for hydrogen bonds is weak. The quoted constants should be compared within each probe series, not treated as solvent-independent measures. At 298 K, as an illustrative common temperature, a tenfold increase in corresponds to a binding free-energy advantage . This does not imply the experiment's temperature was specified, nor that every difference is the energy of one isolated hydrogen bond.
Past exam of the chemistry B course of the University of Cambridge 2024 Paper 1 B5 a i Solution 2026-10-20
A: aldehyde oxygen is the best acceptor; there is no conventional strong O–H or N–H donor. The N-methylpyrrole nitrogen contributes its lone pair to aromatic bonding and is not an ordinary amine acceptor. A formyl C–H or ring C–H can participate in weak contacts, but is not comparable to the donors in B and C.
B: pyridine nitrogen is the best acceptor and the phenolic OH is the best donor. Its oxygen can accept more weakly, but nitrogen provides the primary basic site in the depicted hydroxypyridine form.
C: lactam carbonyl oxygen is the best acceptor and the amide N–H is the donor. amide nitrogen is not a useful acceptor because its lone pair is delocalized toward the carbonyl.
D: the secondary amine nitrogen is the best acceptor, and its N–H is the only conventional donor, but a weak one here. Its ketone oxygen is also an acceptor, although weaker than the amine site in this comparison. These assignments distinguish functional-group identity and lone-pair availability, rather than counting heteroatoms.
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 b Solution 2026-10-20
A glutamic acid residue supplies a carboxyl group whose protonation state can change with the active site environment. The following five schematic contributions are possible in a hypothetical enzyme; they are alternatives and overlapping physical effects, rather than five independently established reactions of this enzyme. In the diagrams is the protein tether, and is the reacting purine carbon.
general acid catalysis: . Donation of the carboxyl proton activates the amidine-like C6 center for water attack. The same acid could instead assist the amino leaving group: . These are two locations for general acid catalysis, rather than two different categories.
general base catalysis: , coupled to oxygen attack on C6. Accepting a proton makes water a more effective nucleophile. This does not require an appreciable pool of free hydroxide in bulk solution.
covalent catalysis: a glutamate oxygen can in principle attack activated C6 to form . After addition/elimination replaces the amino group, water must hydrolyze that transient enzyme–substrate bond to release product and regenerate . Such a pathway is a chemically distinct possibility, not established by the given structure.
electrostatic catalysis: the carboxylate can stabilize a developing positive charge, for example in a protonated purine or amino leaving group; a protonated carboxyl can also donate a hydrogen bond to an oxygen developing negative charge. This contributes only if the reacting configuration is stabilized relative to the bound starting state. A carboxylate placed next to an oxyanion without compensating geometry would instead be unfavorable.
proximity and orientation catalysis: donor and acceptor hydrogen bonds can hold the purine and attacking water in a productive arrangement, schematically . Binding reduces unproductive orientations and can distort the substrate toward a reactive geometry. This last contribution is sometimes described through substrate binding or strain; a binding interaction is not automatically catalytic unless it improves access to the transition state.
Past exam of the chemistry B course of the University of Cambridge 2024 Paper 2 B10 c Solution 2026-10-20
The zinc ion is coordinated by protein donors and a water oxygen placed near the reacting C6 center. Zinc acts as a Lewis acid that activates and positions water. Its positive charge withdraws electron density from oxygen, lowers the effective p of coordinated water and facilitates formation of the attacking hydroxide, potentially with assistance from a nearby base. The same metal–oxygen interaction can stabilize developing negative charge as the addition intermediate forms. Zinc stays Zn(II); this is metal ion catalysis, not a zinc redox cycle.
The structure identifies a plausible reaction geometry, but a static X-ray model does not by itself establish the water's protonation state, proton-transfer order or the rate-limiting step. Structural work with bound transition-state analogues independently supports a zinc-containing catalytic center.[ref]
Atomic structure of adenosine deaminase complexed with a transition-state analog: understanding catalysis and immunodeficiency mutations by David K. Wilson and others. 1991, pp. 1278–1284, catalytic zinc and the bound transition-state analogue
An appropriate experimental test would compare native enzyme, a metal-depleted preparation and stoichiometrically zinc-reconstituted enzyme, checking retained fold and active enzyme concentration. Loss and recovery of activity with those controls distinguishes a catalytic metal requirement from nonspecific denaturation. Past exam of the chemistry B course of the University of Cambridge 2024 Paper 2 B10 d i Solution 2026-10-20
His238 contributes strongly to turnover, with a plausible acid/base or electrostatic role near the attacking water. Replacing its imidazole by the methyl side chain of alanine removes both proton-transfer capacity and a positioned polar group, so the large fall in turnover number fits impaired chemical catalysis. A candidate role is acceptance of a proton from zinc-bound water, or stabilization and subsequent proton shuttling around the attacking oxygen.
Little change in the Michaelis constant suggests that the major kinetic defect is not a gross loss of substrate recognition, but is not generally a binding dissociation constant: for the elementary scheme. Nor does reduced uniquely prove that His238 is the initial general base. Purified-protein controls should check fold, zinc occupancy and active concentration. Importantly, primary mouse-enzyme experiments found that His238 mutants could still hydrate a substrate analogue, directly illustrating this mechanistic limitation.[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, abstract and pp. 15019–15028, purine-riboside hydration by mutant enzyme
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 2024 Paper 2 B6 a Solution 2026-10-20
The carbon end of a carbonyl ligand has a filled molecular orbital with an outward-facing lobe. Its overlap with a vacant metal orbital of the same axial symmetry gives CO-to-metal sigma donation. Occupied metal orbitals can simultaneously overlap with the empty CO antibonding molecular orbitals, giving metal-to-CO pi backbonding. Two perpendicular pi channels are possible; the schematic represents one. Orbital signs indicate phase, not electric charge.
The two interactions reinforce one another: sigma donation transfers density toward the metal, and pi backbonding distributes it into the ligand acceptor orbitals. Both strengthen M–C attachment, while backbonding weakens C–O. Occupation of CO decreases the internal C–O bond order, generally increasing its bond length and lowering its stretching spectroscopic wavenumber. This is why a strongly backbonded carbonyl ligand can have a strong metal–carbon bond and a weakened internal carbon–oxygen bond at the same time.
Past exam of the chemistry B course of the University of Cambridge 2024 Paper 2 B7 a i Solution 2026-10-20
In the ionic assignment, both tetrahedral anions would require Mn(VII), . Such a high oxidation state is strongly oxidizing and is best supported by strong bonding to oxygen. Short Mn–O bonds combine good orbital overlap with oxygen's resistance to further oxidation. This stabilizes permanganate relative to a hypothetical all-sulfide analogue.
sulfide is larger, more polarizable and much easier to oxidize than oxide. An attempted Mn(VII) sulfide would therefore combine an exceptionally oxidizing metal center with reducing ligands: internal redox can reduce manganese while oxidizing sulfur, for example toward S–S-bonded or elemental sulfur products. The larger sulfur donors also provide less favorable short multiple bonding to the small, highly charged manganese center. The oxygen compound supports Mn(VII); the sulfide combination is redox-incompatible. This predicts instability without requiring a specific unprovided decomposition product.







