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Past exam of the chemistry course of the year 3 of the natural sciences course of the University of Cambridge 2022 Paper 4 52 e Solution by
Codex 0 2026-10-20
In the slowly exchanging cubane A each methyl carbon is bonded to three equivalent Li nuclei with . Its single carbon environment gives a septet, with first-order intensities .
In B the three methyl vertices remain equivalent by the ideal threefold symmetry, but each carbon sees two equivalent Li of one type and one Li of the other. The carbon resonance is therefore a quintet of triplets (15 first-order lines if both couplings are resolved and unequal): the equivalent pair gives , and the third Li gives . Accidental equality of couplings can collapse this to a septet.
D has one methyl-carbon environment coupled to two equivalent Li, giving a quintet . E has no methyl carbon. These patterns assume proton decoupling, resolved , ideal aggregate symmetry and slow exchange; without proton decoupling each methyl-carbon pattern is additionally split by its three attached protons.
Past exam of the chemistry course of the year 3 of the natural sciences course of the University of Cambridge 2022 Paper 4 52 d Solution by
Codex 0 2026-10-20
The zero-ppm signals E are plausibly solvated lithium halide, predominantly LiBr or LiI and their solution ion-pair/aggregate forms. Structural symmetry predicts one Li environment in A, two in B in a ratio, two in C in a ratio, and one in an ideal symmetric D.
The natural lithium assignment is therefore:
- Br mixture: A at approximately (43.4); B at approximately (16.2, three Li) and (5.4, one Li); D at (10.3, broad); E at (24.8).
- I mixture: A at approximately (59.9); B at approximately (8.4, three Li) and (2.8, one Li); C at approximately and (0.6 each); D at (2.3, broad); E at (3.0). The close C/D signals require care about overlap.
The major proton peak near is A. The next peak near to is naturally assigned to B; in the iodide mixture the smaller peaks near and distinguish C and D. Smaller D aggregates can exchange Li or ligands with free halide-containing species more rapidly, giving intermediate-exchange broadening at the two specifically identified broad lithium resonances.
There is a quantitative inconsistency in the printed integrations under these ideal single-species assignments. For example the Br lithium B integrals give molecular amount proportional to , while A gives . Their proton areas should consequently obey , whereas either small proton integral gives . No common proton-versus-lithium calibration removes that ratio discrepancy. Similarly the I B/A ratio predicted from Li is , compared with printed .
Thus the symmetry-based assignment above is useful, but an exact simultaneous quantitative assignment satisfying every printed reliable integral is not established by these data. It would require corrected integrals, an explicitly overlapped signal or an additional speciation model; the discrepancy should not be concealed by treating independently normalized spectra as automatically consistent.
Past exam of the chemistry course of the year 3 of the natural sciences course of the University of Cambridge 2022 Paper 4 52 c Solution by
Codex 0 2026-10-20
The nucleus-count ratios H:Li are A, ; B, ; C, ; D, . Enrichment changes which isotope is counted but not these stoichiometric ratios.
Quantitative lithium integration requires adequate excitation bandwidth, a suitable calibrated pulse, acquisition of the entire broad as well as narrow signal, correct baseline/phasing and no differential saturation. Recycle delays should be long relative to the slowest relevant longitudinal relaxation time, or saturation must be measured and corrected. Exchange and unresolved overlap must be handled in the integrated areas. Proton and lithium spectra also require a stated relative calibration before raw numbers can be compared across nuclei.
Past exam of the chemistry course of the year 3 of the natural sciences course of the University of Cambridge 2022 Paper 4 52 b Solution by
Codex 0 2026-10-20
Useful isotope-selection factors include spin and quadrupole moment, gyromagnetic ratio, natural abundance or enrichment, sensitivity at the available field, line widths and relaxation, chemical-shift dispersion and the couplings needed for structural assignment. Both lithium isotopes are quadrupolar, but Li has a much smaller quadrupole moment and often much narrower lines, allowing inequivalent sites and carbon-lithium couplings to be resolved. Enrichment compensates its low natural abundance and removes isotope-mixture complexity.
Fully enriched Li would generally be more sensitive but broader in these aggregates. Its chemical-shift pattern would remain chemically similar, although isotope shifts and dynamics can make small differences. Resolved H-Li couplings increase in frequency approximately with and have different multiplicities because changes from to : equivalent Li nuclei give lines. If such long-range proton couplings remain unresolved, the main visible difference can instead be broadened lines. One should not invent resolved multiplets absent from the printed spectra.
Past exam of the chemistry course of the year 3 of the natural sciences course of the University of Cambridge 2022 Paper 4 52 a Solution by
Codex 0 2026-10-20
Write . The alternating-vertex cubane A is ; replacing one or two methyl vertices gives and . Balanced redistributions areThe smaller aggregate has composition , since the conventional masses or match the supplied values. Thus is another balanced redistribution.
For the structures, place Li on one tetrahedral set of cube corners and Me/X on the complementary set, joining adjacent unlike vertices. A contains four Me vertices, B three Me and one X, and C two of each. D is a four-membered alternating Li-Me-Li-X ring. The quoted relative masses use conventional elemental masses; a fully Li-enriched D would be approximately two mass units lighter, without changing its composition.
Past exam of the chemistry course of the year 3 of the natural sciences course of the University of Cambridge 2022 Paper 4 51 b Solution by
Codex 0 2026-10-20
All three marked points are at strongly oxidizing potentials, where X is depleted at the electrode: . Assuming only X initially in bulk, and . Oxidation produces X at the interface.
- A: the largest steady limiting current corresponds to the fastest rotation and thinnest diffusion boundary layer. X rises steeply from zero to ; X falls steeply from its surface value to zero.
- B: slower rotation gives a thicker layer and shallower gradients, hence a smaller limiting current.
- C: the nonrotating electrode gives the peaked, subsequently decaying current. Its diffusion layer continues to broaden with time, approximately as , giving the broadest and shallowest profiles at the indicated late-scan point.
For equal diffusion coefficients, the two concentration profiles are complementary, . More generally equality of their interfacial molar fluxes requires , so the product surface concentration need not equal . A and B are approximately steady hydrodynamic profiles; C is a time-dependent diffusion profile.
Past exam of the chemistry course of the year 3 of the natural sciences course of the University of Cambridge 2022 Paper 3 50 f Solution by
Codex 0 2026-10-20
The difficulty is competing chemistry and isolation selectivity. Alkaline treatment can promote dehydrohalogenation at chlorinated backbone units while hydrolysing the acetate side groups. Loss of HCl produces unsaturated segments, potentially followed by further degradation or cross-linking; hydroxyl formation also changes solvent compatibility and intermolecular association. The isolated material can therefore differ from the ideal undegraded C in part (e). Competing dehydrochlorination during copolymer saponification explains why selective conditions are needed.
This is not an absolute prohibition on hydroxyl-containing chloride copolymers. Analytical work on these copolymers used staged alcoholysis/saponification and polymer precipitation to restrain dehydrochlorination. The sensible conclusion is that clean, exhaustive conversion and isolation are difficult under unselective conditions, rather than that no suitable treatment can ever preserve chloride units.[ref]
Determination of vinyl acetate in vinyl chloride-acetate copolymers by Takanori Shimazaki and others. Abstract, pp. 619–623: staged treatment and precipitation used to restrain competing dehydrochlorination
Past exam of the chemistry course of the year 3 of the natural sciences course of the University of Cambridge 2022 Paper 3 50 e Solution by
Codex 0 2026-10-20
Saponification cleaves the acetate ester side groups while the intended carbon backbone remains intact:The ideal target C is therefore a vinyl chloride-vinyl alcohol copolymer, containing and units in the original sequence positions. The carbon–oxygen bond to the backbone remains; hydrolysis removes the acetyl group from oxygen. Incomplete hydrolysis would leave a three-component population of chloride, acetate and alcohol units. New hydroxyl groups can strengthen adhesion and coating interactions through hydrogen bonds.
Past exam of the chemistry course of the year 3 of the natural sciences course of the University of Cambridge 2022 Paper 3 50 d Solution by
Codex 0 2026-10-20
Taking the stated melting point as the reference for a crystallizable A, introducing acetate-bearing units disrupts regular sequences and crystal packing. The expected result is a lower melting point for B, often with broader or less complete crystallization. The ideal copolymer melting-point depression estimate iswhen unlike units are excluded from the crystal. The larger vinyl ester in the drawing is vinyl pentanoate: its acyl substituent has a four-carbon alkyl chain. At comparable incorporation it should generally disrupt regular packing more and favour a still lower melting range or suppressed crystallinity, although the exact trend depends on composition and side-group interactions. The formula concerns melting of crystallites, not the glass transition; a universal numerical melting point cannot be inferred for an unspecified real PVC copolymer.
Past exam of the chemistry course of the year 3 of the natural sciences course of the University of Cambridge 2022 Paper 3 50 c Solution by
Codex 0 2026-10-20
B contains both and monomeric units distributed along the same carbon backbone: a vinyl chloride-vinyl acetate copolymer. One possible local segment isThe ester is attached through oxygen to the backbone carbon. A random copolymer should not be drawn as two long homopolymer blocks. The stated monomer feed is not enough to prove an incorporated unit ratio: monomer reactivity ratios and conversion affect composition and sequence statistics. Under an explicitly ideal equal-incorporation assumption that ratio could be used as a schematic average, not a periodic repeat.
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