Past exam of the chemistry B course of the University of Cambridge / 2023

Number the ring with chlorine at C1 and the methylsulfonyl group at C4. The new nitro group is at C2, equivalently C6 by the starting ring's symmetry: the product is 1-chloro-2-nitro-4-(methylsulfonyl)benzene. Both substituents favour this site. Although chlorine deactivates the ring inductively, its lone pair donation stabilizes ortho/para Wheland intermediates. The sulfone is strongly electron withdrawing and meta directing; attack ortho or para to it produces especially unfavourable cationic contributors to resonance in chemistry. C2 is ortho to chlorine and meta to sulfone, whereas C3 has the opposite, less favourable relationship.

sulfuric acid protonates nitric acid, followed by loss of water to produce the nitronium ion. A ring pi bond attacks nitrogen, giving the Wheland intermediate with the new C–N bond. deprotonation at the attacked carbon restores aromaticity. This is electrophilic aromatic substitution: the aryl chloride and the sulfone both remain attached. The formula check is ${\mathrm{C}}_7{\mathrm{H}}_7\mathrm{Cl}{\mathrm{O}}_2\mathrm{S}+\mathrm{N}{\mathrm{O}}_2-\mathrm{H}={\mathrm{C}}_7{\mathrm{H}}_6\mathrm{ClN}{\mathrm{O}}_4\mathrm{S}$. It confirms replacement of one ring hydrogen rather than chlorine, but directing effects, rather than the formula alone, establish the position.

The malonate carbon receives two allyl groups. With $L=\mathrm{C}{\mathrm{H}}_2\mathrm{CH}=\mathrm{C}{\mathrm{H}}_2$, attached through its first carbon, the product issodium hydride removes a methylene proton from diethyl malonate; molecular hydrogen formation accompanies production of an enolate stabilized by resonance in chemistry. Its carbon attacks the bromomethylene carbon of allyl bromide in an SN2 reaction, expelling bromide. The monoalkylated malonate still has one hydrogen between the carbonyls. Excess base removes that hydrogen, and a second equivalent of allyl bromide produces a second carbon–carbon bond. The final malonate carbon has no remaining hydrogen and cannot undergo a third analogous deprotonation there.

The two alkenes are preserved because the substitution occurs at the primary carbon bearing bromine. Counting the introduced groups gives ${\mathrm{C}}_{13}{\mathrm{H}}_{20}{\mathrm{O}}_4$ for the diallylated diethyl ester. Its two identical allyl groups also mean that the central carbon is not a stereocenter. This stage is the alkylation portion of the malonic ester synthesis.

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 $\mathrm{H}{\mathrm{O}}_2\mathrm{C}-\mathrm{C}(\mathrm{L}{)}_2-\mathrm{C}{\mathrm{O}}_2\mathrm{H}$, with $L=\mathrm{C}{\mathrm{H}}_2\mathrm{CH}=\mathrm{C}{\mathrm{H}}_2$ 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 ${\mathrm{C}}_9{\mathrm{H}}_{12}{\mathrm{O}}_4\to {\mathrm{C}}_8{\mathrm{H}}_{12}{\mathrm{O}}_2+\mathrm{C}{\mathrm{O}}_2$. 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.

The first base equivalent removes O–H; the second removes the alpha C–H. The first equivalent of lithium diisopropylamide converts the carboxylic acid to its lithium carboxylate. A further equivalent removes the remaining hydrogen on the carbon between the two allyl groups, giving a carboxylic acid dianion:The second negative charge has enolate-like delocalization into the carboxyl group; lithium coordination and aggregation are not represented by this simple charge-bookkeeping structure. The carbon nucleophile attacks methyl iodide with C–I bond cleavage, forming the new C–Me bond. acid workup reprotonates the carboxylate, giving $F=\mathrm{H}{\mathrm{O}}_2\mathrm{C}-\mathrm{C}(\mathrm{Me})(\mathrm{L}{)}_2$. The two equivalents are therefore chemically necessary for the stated carbon alkylation rather than simply an arbitrary excess. A single equivalent is consumed predominantly by the much more acidic carboxylic proton. The product has formula ${\mathrm{C}}_9{\mathrm{H}}_{14}{\mathrm{O}}_2$ and still has two identical allyl groups, so this methylation does not create a stereocenter.

Number the phenolic carbon C1, with methyl groups at C2 and C5. All three remaining aromatic hydrogens are replaced: $H$ is 3,4,6-tribromo-2,5-dimethylphenol. The starting phenol is ${\mathrm{C}}_8{\mathrm{H}}_{10}\mathrm{O}$; replacing three hydrogens gives ${\mathrm{C}}_8{\mathrm{H}}_7\mathrm{B}{\mathrm{r}}_3\mathrm{O}$, agreeing with the intermediate's composition. Bromination of methyl side chains would not follow the stated Lewis acid-promoted aromatic pathway.

aluminium tribromide polarizes Br–Br. The aromatic ring attacks the electrophilic bromine end, forming a Wheland intermediate; removal of the proton on the attacked carbon restores aromaticity. Repetition fills C4 and C6, the available para and ortho positions activated by OH, and C3, which is activated by the methyl substituents when the other positions are occupied. The formula fixes the extent of substitution without requiring an experimentally unreported order for the individual brominations. The preparative perbromination/selective-debromination sequence is documented in the primary synthetic work.^[ref]

The bromines ortho and para to OH are removed, leaving the meta bromine at C3. A suitable polar mechanism begins with electrophilic ipso protonation of a brominated carbon. protonation at C4 or C6 gives a Wheland intermediate stabilized by donation of the phenolic oxygen lone pair; an important contributor has an O-protonated cyclohexadienone-like framework. Corresponding protonation at C3 lacks this direct stabilization by resonance in chemistry from OH.

iodide can capture/reduce the departing electrophilic bromine as the C–Br bond electrons restore the aromatic pi system. Thus the incoming hydrogen remains on carbon while bromine is removed, unlike ordinary electrophilic substitution in which the incoming electrophile remains and a proton leaves. bromine trapping can be represented through IBr formation followed by reaction with further iodide to give iodine and bromide. For each replacement the balanced net reduction isApplying it at C4 and C6 gives 3-bromo-2,5-dimethylphenol, with formula ${\mathrm{C}}_8{\mathrm{H}}_9\mathrm{BrO}$. This explains why the position least activated by OH is the one at which bromine survives. It is a proposed acid-assisted protodebromination mechanism consistent with the selectivity; the preparative sequence alone does not prove every proton-transfer or halogen-transfer intermediate. It should not be drawn as iodide simply displacing bromide from an unactivated aryl carbon.

The first reagent supplies a vinyl nucleophile, $\mathrm{C}{\mathrm{H}}_2=\mathrm{CH}-$, rather than an allyl group. Two successive conjugate additions install vinyl at C2 and a 2-formylethyl chain at C1. Number the original nitro-bearing alkene carbon C1. Vinylmagnesium bromide attacks the beta carbon C2 of the nitroalkene, shifting the original C1=C2 electrons into a magnesium nitronate. The new C2–vinyl bond remains intact during the second stage; its terminal double bond is not the electrophilic site in that stage.

The carbon end of the nitronate at C1 then attacks the terminal beta carbon of acrolein. Move the acrolein C=C electrons towards its carbonyl and the C=O electrons onto oxygen. This forms the C1–CH2CH2CHO connectivity through an aldehyde enolate; protonation by methanol and tautomerization restore the aldehyde. The nitronate returns to a neutral nitro group as it makes the second C–C bond. Thus the chain is attached through the carbon formerly beta to the acrolein carbonyl, rather than by a direct addition to its aldehyde carbon.

After the first addition, C2 bears the vinyl group while the nitronate carbon C1 remains approximately planar. Approach of the second electrophile to the less crowded face, opposite the adjacent vinyl substituent, gives the new formylethyl chain trans to vinyl. Consequently nitro and vinyl are cis, as depicted: at C1 the chain and nitro group occupy opposite faces. This is a substrate-controlled diastereoselective tandem Michael addition. The reagents and starting material contain no chiral source, so the depicted product represents a racemic pair rather than an enantioselective synthesis. Steric facial selection explains the relative configuration without requiring a uniquely established magnesium aggregate or transition-state conformation.

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 $h_1$, its initially nucleophilic neighbour $h_2$, and the other neighbour $h_6$. After addition, $h_2$ bears the cyclopentanone substituent. deprotonation at $h_6$ gives $h_1=h_6$, the less substituted, less hindered enamine drawn in A. Merely restoring the original $h_1=h_2$ 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.

Number the cyclopentanone ring $p_1$ at its carbonyl, then $p_2$ at the methyl-bearing carbon, $p_3$ at the mesylate-bearing carbon, $p_4$ at the carbon joined to cyclohexane $h_2$, and $p_5$ at the remaining methylene. B has a $p_2=p_3$ enone and the $h_1=h_6$ enamine. triethylamine removes the hydrogen at $p_2$ while the $p_3$–O bond breaks to release mesylate. An E2 elimination, with approximately antiperiplanar geometry in the puckered ring, gives the conjugated cyclopentenone. Elimination towards $p_4$ instead would give a nonconjugated alkene.

The enamine carbon $h_6$ now attacks enone beta carbon $p_3$ intramolecularly. nitrogen donates into $h_1$ 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 $h_6-p_3$; the first was $h_2-p_4$. These two connections account for the bridged skeleton without a concerted cycloaddition or an aldol step. The formula of B is ${\mathrm{C}}_{16}{\mathrm{H}}_{23}\mathrm{NO}$: relative to A, elimination removes $\mathrm{C}{\mathrm{H}}_4{\mathrm{O}}_3\mathrm{S}$ overall, namely methanesulfonic acid after proton transfers.

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.

The phenolic O–H is most acidic; the benzylic CH2 adjacent to the amide carbonyl is second. Phenolic deprotonation gives a phenoxide stabilized by resonance in chemistry. Removing an alpha-amide C–H gives an enolate/carbanion stabilized by the carbonyl and the adjoining aryl group, but requires substantially stronger basicity for a comparable equilibrium concentration. The morpholine nitrogen is an acylated tertiary amide nitrogen and has no N–H proton. The other morpholine and aromatic C–H positions do not have the same direct carbonyl stabilization.

G is the diaryl ether $(2-\mathrm{N}{\mathrm{O}}_2{\mathrm{C}}_6{\mathrm{H}}_4)-\mathrm{O}-(2-\mathrm{C}{\mathrm{H}}_2\mathrm{C}(=\mathrm{O}){\mathrm{N}}_{\mathrm{morph}}{\mathrm{C}}_6{\mathrm{H}}_4)$. Here $N_{morph}$ denotes the nitrogen of the intact morpholine ring; the phenoxy oxygen and phenylacetamide chain are ortho on their original ring. potassium carbonate first generates phenoxide from F. Its oxygen attacks the fluorine-bearing carbon of 1-fluoro-2-nitrobenzene, forming a Meisenheimer complex. Delocalization of its negative charge into the ortho nitro group stabilizes this addition intermediate. Fluoride elimination restores aromaticity and completes nucleophilic aromatic substitution.

This first step forms an O–aryl bond, rather than immediately arylating the less acidic alpha carbon. The neutral product has formula ${\mathrm{C}}_{18}{\mathrm{H}}_{18}{\mathrm{N}}_2{\mathrm{O}}_5$, obtained from the two substrates by net loss of HF. The connectivity is shown beside B in the independently drawn diagram above. The amide-substrate O-to-C migration strategy has direct primary precedent.^[ref]

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 $(2-\mathrm{HO}{\mathrm{C}}_6{\mathrm{H}}_4)-\mathrm{CH}(2-\mathrm{N}{\mathrm{O}}_2{\mathrm{C}}_6{\mathrm{H}}_4)-\mathrm{C}(=\mathrm{O}){\mathrm{N}}_{\mathrm{morph}}$. 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.

A and C are chiral; B is achiral. In A, a quaternary ammonium stereocenter occurs because nitrogen has four different substituents: phenyl, methyl, ethyl, and the sulfinylethyl chain. It has no nitrogen lone pair available for ordinary pyramidal inversion of an amine. The sulfoxide sulfur is also stereogenic: its pyramidal arrangement distinguishes oxygen, methyl, the nitrogen-containing chain, and the lone-pair direction. There are two independent stereocenters and no internal symmetry, giving four configurations in two pairs of enantiomers. This is chirality of the cation; a bromide counterion does not remove it.

B can adopt a trans-1,4-dimethylcyclohexane chair with both methyl groups equatorial. An inversion centre at the chair centre exchanges its opposite carbon atoms and methyl groups. Its mirror image is therefore superposable: trans substituents in a flat ring drawing do not by themselves prove chirality. ring inversion gives the less favourable diaxial conformer without changing the trans configuration.

In C, tetrahedral phosphorus has four different substituent directions: oxo, methoxy, ethoxy, and phenoxy. C has a phosphorus stereocenter. Delocalization in the P–O bonds does not make these four ligand environments identical. A tetrahedral drawing and its mirror cannot be superposed without exchanging two ligands.

The bridged, sterically hindered biphenyl framework is twisted and does not rapidly interconvert the two faces of each methylene group on the NMR timescale. Label one hydrogen on each CH2 $H_A$ and the other $H_B$. Replacing $H_A$ versus $H_B$ by a test substituent gives diastereomeric environments, so these hydrogens are diastereotopic and need not have equal chemical shifts even in an achiral solvent. Molecular symmetry relates the two $H_A$ sites to each other and the two $H_B$ sites to each other.

Each methylene hydrogen undergoes geminal coupling to its partner, giving the two doublets, each accounting for two hydrogens overall. Their separation is $0.21$ ppm; at operating frequency ${\nu }_0$ in MHz it corresponds to $0.21{\nu }_0$ Hz. This separation is not the geminal coupling constant, which must be measured from the line spacing within a doublet. If the shift separation is not large compared with that coupling, an AB pattern with second-order distortion is expected rather than ideal first-order doublets. A racemic pair of twisted conformations still has these signals: enantiomers have the same spectrum in an achiral medium, while the two hydrogen environments within each molecule remain different.

In both cases the alkene undergoes concerted oxygen transfer from metachloroperbenzoic acid. The alkene pi bond forms one C–O bond to the terminal peroxide oxygen; electron redistribution forms the second C–O bond, breaks O–O, and transfers the peracid proton to its carbonyl oxygen as the carboxylic acid product forms. This cyclic epoxidation pathway delivers both C–O bonds from one face, without a freely rotating carbon intermediate.

For 1-methylcyclohexene, the two alkene faces are enantiotopic. Attack from opposite faces gives the two mirror-related epoxides, and an achiral peracid in an achiral environment cannot distinguish their Gibbs energies of activation. The product is racemic, 1:1.

For the cis-fused bicyclic alkene, the junction configuration and its puckered ring conformation make approach from the two faces sterically different. The folded saturated ring shields the concave face; oxygen transfer preferentially approaches the more exposed convex face, directed away from that ring. The alternative attack places the incoming peracid closer to the folded skeleton. The 95:5 ratio represents facial diastereoselectivity, while concerted syn oxygen delivery is retained for either product. It should not be explained by free rotation or by one epoxide subsequently changing configuration. For scale only, if this ratio were measured under irreversible kinetic conditions at 298 K, it would correspond to $RT\ln (95/5)\simeq 7.30\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$ difference between the competing Gibbs energies of activation; the exam does not specify that temperature.

E, with the axial anomeric acetate, hydrolyses faster. The tert-butyl substituent strongly favours the chair in which it is equatorial, so the two substrates cannot freely interchange axial and equatorial acetate simply by an inexpensive ring inversion. protonation activates the anomeric acetoxy leaving group. As its C–O bond cleaves to release acetic acid, a lone pair on the endocyclic oxygen donates into the developing anomeric orbital, producing an oxocarbenium ion. water attacks that carbon, and deprotonation gives the cyclic hemiacetal. Attack from either face accounts for the unspecified anomeric configuration of OH in the product.

The axial C–OAc bond in E can adopt antiperiplanar geometry relative to an endocyclic oxygen lone pair. This favourable $n_O\to {\sigma }_{C-O}^{\ast }$ interaction assists departure in the ionization transition state. The corresponding bond in F is poorly aligned in its preferred chair; achieving comparable overlap requires an unfavourable chair change, placing tert-butyl axial, or substantial ring distortion. The distinction is stereoelectronic control of acetal hydrolysis, rather than merely counting steric crowding in the products.

An axial acetal can itself receive ground-state stabilization from the anomeric effect, so “axial means unstable” is not an adequate rate argument. What matters is the relative Gibbs energy of activation for the departing bond and the geometry of lone-pair assistance. Conformationally locked model acetals experimentally demonstrate faster acid-catalysed cleavage of the axially oriented isomer, although their numerical rate ratios should not be transferred to these different substrates.^[ref]

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]

B and C are the two geometrical ketoxime isomers, formed by replacing the reactive ring C=O with C=N–OH. The separate spiroacetal remains intact. Define $K$ as the original ketone carbon, $R_a$ as the adjacent ring path towards the aromatic ring, and $R_s$ as the adjacent path towards the spiroacetal-bearing junction. These are the two carbon substituents directly bonded to $K$, not detachable independent fragments.

In B, N–OH is anti to the $K-R_a$ bond across C=N; in C, N–OH is anti to $K-R_s$. Those geometries predict exactly the respective nitrogen insertions in D and E. Both products have formula ${\mathrm{C}}_{18}{\mathrm{H}}_{21}\mathrm{N}{\mathrm{O}}_5$: condensation adds NH2OH to ${\mathrm{C}}_{18}{\mathrm{H}}_{20}{\mathrm{O}}_5$ and removes H2O. An alternative ring regioisomer or reaction at the protected ketone would not represent the pair of oxime geometrical isomers.

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 $K-R_a$ bond, anti to N–OTs, migrates to nitrogen while N–O cleaves, giving a ring-expanded nitrilium ion. Its local connectivity is $R_s-K\equiv N^{+}-R_a$, with the existing scaffold still joining the two paths. water attacks the nitrilium carbon $K$. 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 $K-R_s$; its migration gives $R_a-K\equiv N^{+}-R_s$, 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.

The C–C bond anti to the activated N–O bond migrates. This is antiperiplanar migration in the Beckmann rearrangement. Donation from that sigma bond into the developing electron deficiency at nitrogen requires the appropriate antiperiplanar geometry. Migration and leaving-group departure are coupled, giving a nitrilium ion rather than an equilibrating free carbon carbocation. B places the aromatic-side bond in the migrating orientation and therefore gives D; C places the spiro-side bond in that orientation and gives E. The two products consequently record the oxime geometries. A rule based only on which group seems intrinsically more electron rich would miss this geometric constraint; retention of configuration in a migrating group is also consistent with concerted bond migration.

The lactol is a cyclic hemiacetal and can open to $\mathrm{HOC}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2-\mathrm{CH}(\mathrm{OBn})-\mathrm{CHO}$, retaining its pre-existing alkoxy-bearing stereocentre. Its two anomers lose their anomeric configuration on opening to the planar aldehyde, so both can feed the same addition pathway. A first equivalent of decylmagnesium bromide can be consumed by the free OH, producing decane and a magnesium alkoxide; the excess reagent supplies the carbon nucleophile for addition.

magnesium can coordinate both the aldehyde oxygen and the adjacent benzyloxy oxygen. This produces a five-membered chelate, with ring sequence Mg–O(carbonyl)–C(aldehyde)–C(alpha)–O(benzyloxy), that restricts rotation about the aldehyde–alpha-carbon bond. Decyl attack on the less crowded face, away from the alpha-carbon's hydroxyethyl chain in the chelated arrangement, forms the new C–C bond and a magnesium alkoxide at the former aldehyde carbon. Acidic workup protonates both alkoxides to give G. The OBn bond is retained; it is not displaced by the Grignard reagent.

The product configuration is the one with OBn and the newly formed OH both projecting forward in the specified zigzag chain, while the opposite configuration at the new alcohol carbon is disfavoured. This is the Cram chelation model for an alpha-alkoxy aldehyde, rather than stereochemical memory of the starting lactol OH. Coordination is a mechanistic explanation consistent with MgBr2 and the product; the scheme does not quantify individual organomagnesium aggregates.^[ref] The relevant primary synthetic sequence is documented independently.^[ref]

Write the reacting terminus as $R-\mathrm{C}{\mathrm{H}}_2-\mathrm{CH}=\mathrm{C}{\mathrm{H}}_2$, where $R$ includes the pre-existing stereocentres and the intact $\mathrm{OC}{\mathrm{H}}_2\mathrm{C}(=\mathrm{O})\mathrm{NM}{\mathrm{e}}_2$ substituent. ozone undergoes 1,3-dipolar cycloaddition across the terminal alkene to give a molozonide. In a resonance description of ozone as ${\mathrm{O}}^{-}-{\mathrm{O}}^{+}=\mathrm{O}$, the alkene pi pair bonds to one terminal oxygen while the other terminal oxygen forms the second C–O bond and the ozone pi bond shifts onto its central oxygen. Cleavage of this unstable primary ozonide produces a carbonyl compound and a carbonyl oxide, which recombine to form the secondary ozonide. Reductive workup breaks the peroxide bonds and gives the carbonyl fragments $R-\mathrm{C}{\mathrm{H}}_2\mathrm{CHO}$ and formaldehyde.

Excess sodium borohydride also reduces the newly formed aldehyde: hydride attacks its carbonyl carbon, shifting the pi electrons onto oxygen, and the final acidic workup protonates the alkoxide. Thus the retained chain becomes $R-\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{OH}$. The terminal alkene carbon becomes formaldehyde and can be reduced to methanol; it is no longer part of the main product. This accounts for the one-carbon shortening of the retained skeleton.

The original amide carbonyl is present in both H and I and is not reduced under the ordinary borohydride workup. The existing stereocenters and benzyl ether are retained. Stopping the mechanism at an aldehyde, or deleting the amide C=O when reading the smaller printed scheme, would give the wrong product. The new terminal alcohol carbon is CH2 and therefore adds no stereocenter.

The acceptor-strength trend is amide oxygen and amine nitrogen strongest, ketone oxygen weaker, hydrocarbon essentially nonaccepting. trimethylamine and isopropylamine each provide a relatively available nitrogen lone pair; adding N–H bonds changes donor ability much more conspicuously than it changes the acceptor ranking in this comparison. Alkyl substitution and solvation modify the precise values, so the matching amine parameters should be read as the stated empirical strengths rather than a universal equality of all amines.

In acetamide, nitrogen lone-pair donation contributes an ${\mathrm{O}}^{-}-\mathrm{C}={\mathrm{N}}^{+}$ form contributing to resonance in chemistry, increasing electron density at oxygen and making it a strong hydrogen bond acceptor. The amide acceptor is oxygen, not nitrogen: the nitrogen lone pair is delocalized into the carbonyl. acetone also accepts at oxygen, but lacks this nitrogen-to-carbonyl donation and has the smaller hydrogen-bond acceptor parameter. isobutane has neither a heteroatom lone pair nor a comparable acceptor site; its zero value is consistent with the absence of an ordinary directional hydrogen bond acceptor.

The clearest comparison is between the two amines, which have the same molecular formula and molar mass. isopropylamine can donate N–H hydrogen bonds to neighbouring nitrogen lone pairs; trimethylamine has no N–H donor. The resulting intermolecular association raises the primary amine's boiling point, allowing it to remain liquid at room temperature, whereas trimethylamine is gaseous. Independent phase-change data place their normal boiling points on opposite sides of room temperature.^[ref]

acetone has a substantial carbonyl dipole and London dispersion forces sufficient for a room-temperature liquid, even though it cannot build a strong donor–acceptor hydrogen bond network by itself. isobutane has mainly London dispersion forces and a compact shape and remains gaseous. acetamide combines carbonyl acceptors with two N–H donors, allowing an extended, directionally bonded crystal lattice; favourable packing as well as intermolecular association stabilizes its solid state.

These compounds have similar molar masses, so mass alone cannot explain the differences. The hydrogen-bond acceptor parameter also cannot predict the phase on its own: a pure substance needs complementary donors to exploit strong acceptor sites, and freezing additionally depends on crystal packing and entropy. A stronger acceptor does not automatically imply a higher melting point for arbitrary molecules.

Transfer is favourable when new water–solute interactions compensate the cavity and reorganization costs of accommodating the solute. The Gibbs energy also includes the loss of solute–alkane interactions; it is not simply a count of hydrogen bonds. isobutane cannot compensate those costs with strong directional interactions, explaining its positive transfer Gibbs energy. acetone and both amines can accept hydrogen bonds from water and have favourable electrostatic solvation, making their transfer moderately favourable. isopropylamine can additionally donate to water, consistent with a somewhat greater preference for water than trimethylamine.

acetamide combines a strong carbonyl acceptor with N–H donors and has the most favourable hydration in the group, accounting for its much more negative transfer Gibbs energy. Its pure solid lattice is not an extra sublimation penalty to insert here: the stated process starts from a dilute alkane solution and ends in dilute water, rather than dissolving the pure crystal. The nonmonotonic relationship between hydrogen-bond acceptor parameter and transfer Gibbs energy is expected because donor strength, nonpolar surface, solvent cavity formation, and all other solvation contributions also enter.

For matched standard states, $\Delta G_{\mathrm{tr}}^{\circ }=-RT\ln K_{\mathrm{tr}}$, with $K_{\mathrm{tr}}$ the water-to-alkane activity ratio. As an illustrative 298 K calculation, the hydrocarbon value implies $K_{\mathrm{tr}}\simeq 7.0\times 10^{-4}$, whereas the amide value implies approximately $3.6\times 10^4$. These are standard-state partition preferences, not absolute aqueous solubilities. amine protonation could introduce an additional pH-dependent contribution to an experimentally measured distribution coefficient, but no such speciation information is supplied and it is not needed for the molecular comparison.

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ammonia and cyclohexylamine are neutral donors, whereas each coordinated chloride and acetate has charge $-1$. Charge balance therefore gives Pt(II) in cisplatin and Pt(IV) in satraplatin:The two acetate groups count once each; their carbonyl and alkoxy oxygen atoms do not make each ligand doubly charged.

cisplatin has square planar molecular geometry and a $5d^8$ metal centre. Choose $x$ and $y$ along the four Pt–donor directions. The strongly antibonding $d_{x^2-y^2}$ orbital is highest and remains empty; the other four d orbitals are doubly occupied. The square-planar splitting is represented by the conventional qualitative orderThe ordering among the lower occupied levels can depend on the ligand set, and the cis arrangement removes exact fourfold symmetry. The robust result is the filled lower four levels and empty highest level.

satraplatin has approximately octahedral molecular geometry around Pt(IV), $5d^6$. Its large octahedral splitting favours a low-spin complex:All three $t_{2g}$ orbitals are paired. Mixed ligands remove exact degeneracies, but do not change this qualitative occupancy.

platinum is in group 10, so subtracting its oxidation state gives $d^8$ for Pt(II) and $d^6$ for Pt(IV). Adding two electrons from each coordinated donor gives the local valence electron counting resultsEach acetate in the depicted structure is monodentate, binding through one oxygen. The 16-electron count of the square-planar complex is normal for a low-spin late-metal $d^8$ centre; it is not a missing ligand inferred from the 18-electron rule.

Both complexes have all their $d$ electrons paired, so $S=0$. Their spin-only magnetic moment is thereforeThey exhibit diamagnetism. The formula describes the spin contribution; it does not say that a closed-shell sample has no diamagnetic response to an applied field.

ammonia and cyclohexylamine are sigma-donor ligands; chloride and acetate are sigma- and pi-donor ligands. The nitrogen lone pair of each amine points toward the metal and forms the donor interaction. These saturated amines have no low-lying conjugated acceptor system that would make them conventional pi-acceptor ligands.

chloride uses a lone pair for sigma donation and has additional occupied $p$ orbitals perpendicular to the bond that can donate with pi symmetry. An oxygen-bound acetate similarly has occupied oxygen-centred orbitals available for sigma and pi donation, although its carboxylate resonance in chemistry changes their energies and distribution. Neither should be classified as a pi acceptor merely because its bonding can be drawn with multiple bonds or resonance in chemistry contributors. None of these four is a standard sigma-donor/pi-acceptor ligand.

cyclohexylamine is expected highest in the spectrochemical series for comparable metal environments. Its alkyl substituent increases the electron-donating ability of nitrogen relative to ammonia, strengthening the sigma interaction and raising the metal–ligand sigma-antibonding levels. The anionic pi donors tend to raise occupied metal $d_{\pi }$ levels and give smaller splittings. Steric effects and the metal can modify the exact ordering, so the comparison is a ligand field trend rather than a universal basicity-to-splitting identity.

The average Pt oxidation state is $+3$, since six bromides balance the two metal atoms. A natural mixed valence assignment is alternating Pt(II), $d^8$, and Pt(IV), $d^6$, rather than two equivalent Pt(III) sites. Use neutral square-planar $\mathrm{P}{\mathrm{t}}^{\mathrm{II}}(\mathrm{N}{\mathrm{H}}_3{)}_2\mathrm{B}{\mathrm{r}}_2$ units and octahedral $\mathrm{P}{\mathrm{t}}^{\mathrm{IV}}(\mathrm{N}{\mathrm{H}}_3{)}_2\mathrm{B}{\mathrm{r}}_4$ units. At every site place two ammonia and two bromide ligands in an equatorial plane; the Pt(IV) sites have two additional axial bromides pointing toward the adjacent Pt(II) sites.

Along the chain the connectivity isEvery Pt–Br–Pt bridge is asymmetric: short to Pt(IV), long to Pt(II). The latter contact is a secondary axial interaction, consistent with the preference of Pt(II) for a square-planar primary coordination sphere. The Pt(IV) site is smaller and more strongly attracts the bridging bromides. A drawing with all axial distances identical would obscure the distinct local environments.

There are four equatorial bromides and two axial bridging bromides per two-Pt repeat, giving precisely $\mathrm{P}{\mathrm{t}}_2(\mathrm{N}{\mathrm{H}}_3{)}_4\mathrm{B}{\mathrm{r}}_6$. A direct Pt–Pt bond is not required. This is a proposed structure consistent with the stated composition and two environments; the formula alone does not determine every equatorial cis/trans arrangement or numerical bond length.

The more radially extended Pt $5d$ orbitals give strong metal–ligand overlap and covalent bonding. This helps compensate the energetic cost of reaching Pt(IV), and the large octahedral ligand field splitting strongly stabilizes its low-spin $d^6$ configuration, with a crystal field stabilization energy contribution $-2.4{\Delta }_o$ before separately counting pairing and the full bond energies. Ni $3d$ bonding is less effective in this respect; Pd $4d$ compounds also generally favour the very stable $d^8$ M(II) arrangement more strongly than Pt does. The stronger bonding and large ligand-field stabilization available to Pt(IV) make oxidation to $+4$ more favourable relative to its congeners. This is a comparative trend, not a claim that Ni(IV) or Pd(IV) complexes cannot exist.

For the usual octahedral Ni(II) hydration-isomer series, distinguish coordinated water from water of crystallization:The first two coordination spheres have no cis/trans choice because they contain zero or one chloride. The last hydration class has two geometrical isomers: the chlorides occupy adjacent or opposite octahedral positions. Thus there are three hydration classes and four structures when cis/trans isomerism is included. Every formula retains two chlorides and six waters overall; the coordination number remains six. The schematic in B6(c) displays the distinct octahedral connectivities. This enumeration states the usual six-coordinate interpretation, rather than treating unspecified changes in coordination number as further members of the same elementary series.