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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 57 e Solution by
Codex 0 2026-10-20
Without significant anomalous scattering, the electron density is effectively real and Friedel's law gives , hence .
With anomalous scattering the atomic imaginary term retains its sign when the reflection index reverses. Here , soThus the second expected intensity is approximately . Conjugating the complete anomalous structure factor would incorrectly impose Friedel equality on the second experiment.
Past exam of the chemistry course of the year 3 of the natural sciences course of the University of Cambridge 2022 Paper 4 57 d Solution by
Codex 0 2026-10-20
Write the ordinary structure factor as and use . The intensities implyConsequently , and the two possible ordinary phases areThe twofold ambiguity is expected from a single anomalous-intensity constraint. The calculation assumes that the specified imaginary increment is the only change requiring inclusion; an additional unspecified real dispersive change could not be reconstructed from these data alone.
Past exam of the chemistry course of the year 3 of the natural sciences course of the University of Cambridge 2022 Paper 4 57 c Solution by
Codex 0 2026-10-20
For reflection , the positional phase factor is for each of the four sites just found. Treat the supplied anomalous increment as the imaginary term , with electrons per Se atom. ThenIts magnitude is electrons and its phase is relative to the real axis in this structure-factor convention. Translating the origin rotates both this reference contribution and the ordinary structure factor consistently; the physically relevant phase difference is invariant.
Past exam of the chemistry course of the year 3 of the natural sciences course of the University of Cambridge 2022 Paper 4 57 b Solution by
Codex 0 2026-10-20
Patterson maps give interatomic displacement vectors rather than absolute positions. The three strong half-cell vectors are consistent with a face-centered four-point Se substructure. One permissible coordinate set isEach nonzero vector occurs four times among the twelve ordered Se-Se pairs, explaining equal weights. These positions form a valid four-site orbit of the specified space group in a suitable origin convention.
A common translation of all four sites is invisible to a Patterson map; absolute coordinates are therefore not uniquely fixed by these three peaks. Subsequent phases are stated in the origin convention above. The remaining atoms need not have face-centered translational symmetry merely because the dominant Se substructure does.
Past exam of the chemistry course of the year 3 of the natural sciences course of the University of Cambridge 2022 Paper 4 57 a Solution by
Codex 0 2026-10-20
The useful feature is the selenium atom's wavelength-dependent anomalous X-ray scattering. Close to its absorption edge the atomic scattering factor is complex:Replacing a sulfur-containing side chain by its selenium analogue introduces relatively strong, spatially localized anomalous scatterers. This is the diffraction-physics reason for the label; no biological preparation procedure is needed for the calculation.
The Se substructure can be located from intensity differences or Patterson maps. Measurements at several wavelengths change its known complex contribution while leaving most other atoms' scattering nearly unchanged. Interference between that contribution and the unknown remainder constrains the otherwise missing structure factor phases. Both wavelength-dependent dispersive differences and differences between Friedel pairs contribute to multiwavelength anomalous diffraction. Several wavelength constraints resolve phase ambiguities that remain with one intensity difference. The resulting phased Fourier synthesis yields the electron-density map.
Past exam of the chemistry course of the year 3 of the natural sciences course of the University of Cambridge 2022 Paper 4 56 d Solution by
Codex 0 2026-10-20
First the tertiary amine attacks the primary allylic bromide carbon by SN2 reaction, giving the quaternary ammonium salt. The less hindered nitrogen face is favored by the pre-existing fused-ring stereochemistry.
DBU removes the proton at the carbon alpha to both the ammonium nitrogen and the ester carbonyl. The resulting ammonium ylide undergoes a concerted 2,3-sigmatropic rearrangement: the N–CH(allyl) bond breaks, a bond forms between the ylide carbon and the aryl-bearing terminal allyl carbon, and the double bond moves to give . Nitrogen becomes neutral again. An ordinary alkyl substitution would attach the wrong end of the allyl unit.
An envelope-like five-membered transition structure keeps Ar outside the crowded bicyclic framework. With the pre-existing OMe dash and ring-junction geometry fixed, place the newly installed side chain on the solid-wedge face at the former ylide carbon and choose the benzylic configuration that places Ar outward, as in the displayed intermediate. This is diastereoselective, suprafacial allyl transfer.
Finally methanesulfonic acid protonates the methyl acetal. Methanol leaves to give an oxocarbenium/iminium-stabilized electrophilic centre, which is attacked intramolecularly by the electron-rich arene at its available ortho position. Deprotonation restores aromaticity and closes the bridge. The terminal isopropenyl group survives; it is not the bond consumed in this final Intramolecular Friedel-Crafts reaction. The constrained tether selects the displayed bridged diastereomer.
Past exam of the chemistry course of the year 3 of the natural sciences course of the University of Cambridge 2022 Paper 4 56 c Solution by
Codex 0 2026-10-20
This is a thermal Claisen rearrangement followed by intramolecular acetal exchange. Identify the six-atom array as the cyclopentenone enol-ether C=C–O–CH–C=C unit. In its concerted rearrangement:
- The O–CH bond breaks.
- A new C-C bond joins the other end of the cyclopentenone enol double bond to the cyclohexene carbon bearing CH(OEt).
- The former ether-bearing cyclopentene carbon becomes a ketone, and the cyclohexene double bond becomes the exocyclic methylene.
The intermediate is thus a substituted cyclopentane-1,3-dione, with the acetal tether now correctly placed for six-membered-ring closure. Its enol oxygen attacks the acetal carbon, replacing OEt and releasing ethanol. Enolization gives the conjugated cyclic enol ether drawn in the product; the second ethoxy group remains.
A chair-like Claisen rearrangement transition structure with the phenyl-bearing part of the cyclohexane directed outward avoids severe axial crowding. Suprafacial bond formation and the pre-existing configuration give the two indicated ring-junction hydrogens on opposite faces: the left H is a wedge and the right H a dash in the supplied orientation. Closure favors the acetal epimer with outward OEt, drawn as a wedge, while the original Ph remains dashed. This is substrate-controlled diastereoselection followed by conformationally selective acetal closure, rather than independent arbitrary assignment of three centres.
Past exam of the chemistry course of the year 3 of the natural sciences course of the University of Cambridge 2022 Paper 4 56 b Solution by
Codex 0 2026-10-20
The hydroxylactam is a cyclic hemiaminal. Its alkoxide can open to the corresponding aldehyde/amidate, in its protonated representation. This exposes an aldehyde without reducing or replacing the lactam carbonyl.
The ketophosphonate anion attacks that aldehyde. Formation and collapse of the phosphorus-containing intermediate give the predominantly E enone and phosphate by Horner-Wadsworth-Emmons reaction. The amide nitrogen then closes by intramolecular aza-Michael addition at the enone beta carbon. Protonation of the ketone enolate gives the observed five-membered lactam with a substituent.
A folded, relatively unencumbered closure places OBn and the developing side chain in outward-facing positions; the expected major product has OBn and the C5 acetonyl substituent cis. With the OBn wedge retained, show the new C5–CHCOCH bond as a solid wedge. The closure is diastereoselective, while the original wavy hemiaminal centre is erased by aldehyde formation. This facial prediction rests on the preferred folded conformer rather than retention of either starting hemiaminal epimer.
Past exam of the chemistry course of the year 3 of the natural sciences course of the University of Cambridge 2022 Paper 4 56 a Solution by
Codex 0 2026-10-20
Osmium tetroxide adds to the alkene through a concerted cyclic addition, giving a five-membered osmate ester with both C-O bonds on the same face. Hydrolysis releases the vicinal diol; NMO reoxidizes reduced osmium so that OsO can be catalytic.
The two new hydroxyl groups are therefore syn. For the drawn substrate, approach opposite the two existing solid-wedge substituents is less crowded: draw both new C-OH bonds dashed, while retaining the solid-wedge COBn and NHBoc bonds. The major diol is cis at its two new centres and anti to the pre-existing substituted face.
Syn delivery is the stereospecificity of cyclic osmylation; preference for one of the two diastereotopic alkene faces is substrate-controlled diastereoselectivity. Attack from the other face would still be syn but would give the minor diastereomer.
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