Codex @codex  0

A is the predicted major diastereomer. At the carbon adjacent to the ketone, take phenyl as the large group, vinyl as medium, and hydrogen as small. There is no adjacent heteroatom donor that would force a five-membered chelate. The appropriate first model is therefore the non-chelating Felkin-Anh model.

View along the carbonyl-carbon–stereocentre bond. Rotate to place the large alpha phenyl group approximately perpendicular to the carbonyl plane. Of the two resulting approaches opposite that large group, the favourable Bürgi-Dunitz trajectory passes nearer hydrogen rather than vinyl. Propyl transfer to the carbonyl and protonation of its oxygen give the configuration labelled A: in the printed projection the new OH projects forward and the propyl bond backward, while the original vinyl bond remains backward. Rotation used to construct a Newman projection must not be mistaken for inversion of the original stereocentre.

This is a relative-selectivity prediction, not a numerical product ratio. Simple attack on the face that happens to look less crowded in the initial flat drawing is unreliable because the relevant C–C single bond can rotate. B is the competing product from the less favourable approach beside vinyl. The model and its preference for the trajectory beside the small group are described in the institutional treatment.^[ref]

triethylamine removes an alpha proton from nitromethane to generate a nitronate population. Its carbon attacks the cyclic ketone carbonyl of E, forming a new C–CH2NO2 bond while the C=O pi pair moves onto oxygen. protonation by triethylammonium or another proton donor gives the tertiary alcohol F and regenerates base. This is a Henry reaction, also called a nitroaldol reaction: the former ketone carbon now bears both OH and CH2NO2.

The nitro group remains attached to its original nitromethane carbon; attack through oxygen would give a different connectivity. The methyl acetal and the original alcohol substituents need not be cleaved or displaced. This addition produces the isolated connectivity without requiring dehydration to a nitroalkene. The stated complex mixture is compatible with competing additions or stereoisomers, but the scheme supplies no basis for claiming F is the unique or kinetically dominant product. Only its formation is being explained.

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.

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.

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.

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.

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.

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.

Lithium diisopropylamide removes an alpha proton from the ethyl side of the ketone. The C–H electrons form the C=C part of a lithium enolate, while the C=O pi electrons move to oxygen. The phenyl side has no competing alpha C–H site. Diisopropylamine is the conjugate-acid byproduct.The carbon end of this enolate attacks the primary carbon bearing iodine in $\mathrm{IC}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{OMe}$. The C–I bond breaks to iodide in an SN2 reaction, and carbonyl reformation givesThus a three-carbon methoxypropyl chain is attached at the ketone alpha carbon. The remote methyl ether survives. The hindered amide base is chosen to generate the enolate rather than to add to the carbonyl. No absolute stereochemical preference is specified for the new stereocenter under these achiral conditions.

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 $\gg$ 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 $K_a$ corresponds to a binding free-energy advantage $RT\ln 10\simeq 5.7\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$. This does not imply the experiment's temperature was specified, nor that every difference is the energy of one isolated hydrogen bond.

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.

The first stage is a Stork enamine reaction. Pyrrolidine attacks the ketone carbonyl, and proton transfers give a carbinolamine. Loss of water gives an iminium ion; removal of the benzylic alpha proton produces the enamine conjugated with the aromatic ring. Removal of water and prolonged heating favour this condensation. That conjugated enamine is preferred over the alternative with its double bond on the other side of the original carbonyl.

The nucleophilic enamine carbon attacks the methylene carbon of bromoacetonitrile by an SN2 reaction, displacing bromide and making the carbon–carbon bond. It is the carbon beside the aromatic fusion that acquires $\mathrm{C}{\mathrm{H}}_2\mathrm{CN}$, as in D. Finally, water adds to the alkylated iminium carbon; proton transfers and cleavage of the C–N bond regenerate the ketone and release pyrrolidine. This explains carbon alkylation, its position, and restoration of the carbonyl rather than treating the amine as a permanent substituent.

A undergoes a Robinson annulation. For an unambiguous connectivity description, call the original ring carbonyl carbon C1 and the side-chain-bearing alpha carbon C6. Label the appended chain $C6-\mathrm{a}-\mathrm{b}-\mathrm{c}(=\mathrm{O})-\mathrm{d}$, where d is initially methyl. Base makes the enolate at d, which attacks C1 intramolecularly to close the six-membered ring $C1-C6-\mathrm{a}-\mathrm{b}-\mathrm{c}-\mathrm{d}$. The carbonyl $\pi$ electrons move to oxygen, and protonation gives the beta-hydroxy ketone.

An E1cb elimination then removes the proton on d and the hydroxy group on C1, giving the conjugated enone with C1=d and a carbonyl at c. The original ring methyl remains on the carbon next to C1 on the other side. The two rings share the C1–C6 bond. The molecular formula is consistent with loss of one water molecule from the diketone.

Pyrrolidine and 2-methylcyclohexanone first form an enamine, through carbonyl addition, carbinolamine formation, dehydration to an iminium ion, and alpha deprotonation. The less hindered enamine carbon is on the side opposite the methyl-bearing alpha carbon. It attacks the terminal beta carbon of methyl vinyl ketone, while the conjugated $\pi$ system shifts to oxygen. Protonation gives an alkylated iminium species; aqueous acid then hydrolyses it back to the ketone.

The result is a Michael addition product: the previously unsubstituted alpha carbon bears $-\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{COC}{\mathrm{H}}_3$. In drawing the arrows, start the new C–C bond at the enamine carbon, not at nitrogen. The iminium bond is restored during addition and removed by hydrolysis. The sequence is an enamine-mediated conjugate addition, rather than direct addition to the methyl vinyl ketone carbonyl.

Apply the Felkin-Anh model to the stereocentre alpha to the ketone. Its large group is the chain continuing through the phenyl-bearing carbon, its medium group is methyl, and its small group is hydrogen. In the preferred transition-state arrangement, put the large group approximately perpendicular to C=O and approach from the opposite face, passing the small hydrogen along a Bürgi-Dunitz trajectory. The remote ether cannot form the usual small alpha-alkoxy chelate, and the coordinating solvent further supports using an open-addition prediction.^[ref]

The predicted new stereocentre is $S$. In the question's zigzag projection, this corresponds to the new OH bond directed behind the page and the newly introduced methyl toward the reader, with the original stereocentres unchanged. The new-centre priority order is OH $>$ phenyl $>$ the substituted alpha-carbon chain $>$ methyl. This is a model-based major-product prediction, rather than a quantitative diastereomeric ratio.

The strong $1724\mathrm{c}{\mathrm{m}}^{-1}$ absorption is the carbonyl stretching vibration of a saturated ketone. Its exact wavenumber depends on substitution and measurement conditions; the important discrimination is that it is far below the high-frequency acyl chloride band in B and is accompanied by a carbonyl carbon-13 NMR resonance near 200 ppm.

A is 3-Chlorobutan-2-one, $\mathrm{C}{\mathrm{H}}_3\mathrm{C}(=\mathrm{O})\mathrm{CH}(\mathrm{Cl})\mathrm{C}{\mathrm{H}}_3$. Counting chlorine as one hydrogen in the index of index of hydrogen deficiency gives $(2\times 4+2-7-1)/2=1$. A single carbonyl accounts for that unsaturation; the resonance at $203.2$ ppm identifies a ketone, while the $59.0$ ppm methine bears chlorine.

The carbon-13 NMR assignments are ${\delta }_C=203.2$ for $\mathrm{C}=\mathrm{O}$, $25.7$ for $\mathrm{C}{\mathrm{H}}_3\mathrm{CO}$, $59.0$ for $\mathrm{CHCl}$ and $20.1$ ppm for $\mathrm{CH}(\mathrm{Cl})\mathrm{C}{\mathrm{H}}_3$. The four distinct carbon environments agree with the unsymmetrical structure. The small signal near 77 ppm belongs to the usual deuterated chloroform solvent, rather than to a fifth carbon of A.

The strong $1810\mathrm{c}{\mathrm{m}}^{-1}$ band is an acyl chloride carbonyl stretching vibration. Compared with an ester, donation from a chlorine lone pair into the carbonyl antibonding molecular orbital is relatively ineffective, while chlorine's inductive effect withdraws electron density. The resulting stronger effective carbonyl force constant places the absorption above an ordinary saturated ketone band. Bond polarity alone is not a universal quantitative frequency rule.

The $1621\mathrm{c}{\mathrm{m}}^{-1}$ absorption is predominantly the carbon–carbon double-bond stretching vibration of the electron-rich vinyl ether. resonance donation from an oxygen lone pair changes both its frequency and its electric dipole moment variation during stretching. There is no carbonyl in the assigned structure, consistent with the absence of a carbon-13 NMR signal in the usual ketone or acyl chloride region.

For the ordinary ester, the carbon nucleophile of a grignard reagent attacks the carbonyl carbon and moves the C=O pi pair onto oxygen. Collapse of the tetrahedral intermediate expels ethoxide, transiently giving ${\mathrm{R}}^1\mathrm{COR}$. This ketone is more electrophilic than the starting ester because it lacks an alkoxy group's stabilizing resonance donation. A second equivalent of the grignard reagent therefore adds to it before isolation:The negatively charged intermediates are associated with magnesium counterions. Merely reducing the amount of reagent to one equivalent does not generally trap the ketone selectively; it can leave unreacted ester together with the double-addition product.

In the first successful preparation, magnesium insertion into the primary C–Br bond gives n-butylmagnesium bromide. The carbon chain has four carbons, so its acetylation produces hexan-2-one, $\mathrm{C}{\mathrm{H}}_3\mathrm{CO}(\mathrm{C}{\mathrm{H}}_2{)}_3\mathrm{C}{\mathrm{H}}_3$, not pentan-2-one. Attack on acetic anhydride forms $\mathrm{C}{\mathrm{H}}_3\mathrm{C}({\mathrm{O}}^{-})(\mathrm{nBu})\mathrm{OCOC}{\mathrm{H}}_3$. Coordination of the anhydride-derived oxygens to magnesium can stabilize this tetrahedral addition complex. Under suitably cold, controlled-addition conditions it survives sufficiently to delay release of the free ketone until aqueous work-up has destroyed the remaining grignard reagent. Protonation followed by collapse releases acetate/acetic acid and the ketone.

This interpretation has direct experimental precedent: Melvin S. Newman and William T. Booth Jr. obtained methyl ketones by cold addition of grignard reagents to acetic anhydride, attributing selectivity to a persistent addition complex. The temperature and reagent-addition regime are genuine conditions on the explanation, rather than proof that an anhydride can never undergo overaddition; the exam scheme omits them.^[ref]

For the benzonitrile route, n-butyl attacks the nitrile carbon and one C≡N pi pair moves to nitrogen, producing the magnesium ketimine salt $\mathrm{PhC}(\mathrm{nBu})={\mathrm{N}}^{-}\mathrm{MgB}{\mathrm{r}}^{+}$. This anionic nitrogen-containing intermediate resists a second nucleophilic addition much more than a neutral carbonyl does. The ketone $\mathrm{PhC}(=\mathrm{O})\mathrm{nBu}$ appears only during acidic aqueous work-up, after organomagnesium has been quenched. Protonation gives the iminium ion; water attacks its carbon, proton transfers give a carbinolamine, and C–N cleavage from its nitrogen-protonated form produces the protonated ketone and Ammonia, followed by deprotonation. In the acidic mixture Ammonia is present largely as ammonium. This sequence explains why nitrile addition stops at a ketone without requiring the ketone itself to be resistant to grignard reagents.

The atom-labelled hydrolysis sequence isWater's lone pair attacks the iminium carbon while its C=N pi pair moves to nitrogen; the carbinolamine oxygen then reforms C=O while the C–N pair moves to the departing protonated amino group. Final deprotonation gives the neutral ketone. The arrows involving protonation and hydrolysis also exchange protons with the acidic aqueous reservoir; the condensed sequence tracks the carbon–nitrogen and carbon–oxygen transformations rather than displaying every solvent molecule.