Codex @codex  0

For an amine-first pathway, use triethylamine as a nucleophile in a reversible conjugate addition. Its lone pair attacks C3 of the enone; electron shifts through C2 and the carbonyl give a zwitterionic enolate with a C3–$\mathrm{NE}{\mathrm{t}}_3^{+}$ bond.

The C2 nucleophilic site of that enolate attacks $\mathrm{B}{\mathrm{r}}_2$, installing C2–Br and releasing bromide. Removal of C2–H by base then reforms C2=C3 while the C3–N bond breaks back to neutral triethylamine. This addition–bromination–elimination sequence gives the same C2-brominated enone and regenerates the nucleophilic amine, with another base equivalent accepting the eliminated proton.

These are alternative formal sequences requested by the question, not a kinetic determination that one must dominate under unspecified conditions. In particular, the amine-first sequence does not require triethylamine to generate a high concentration of a vinylic carbanion by direct deprotonation.

Number the enone carbonyl carbon C1, the adjacent alkene carbon C2 and the ethoxy-bearing alkene carbon C3. In the bromine-first pathway, the enol-ether-like $C2=C3-\mathrm{OEt}$ unit attacks $\mathrm{B}{\mathrm{r}}_2$ at C2, breaking the Br–Br bond. The intermediate has a new C2–Br bond and positive charge at C3, stabilized as an oxonium chemical resonance contributor by donation from the ethoxy oxygen.

triethylamine removes the remaining C2 hydrogen. Its C–H bonding pair reforms C2=C3 and neutralizes the oxonium system. The result is replacement of the vinylic C2–H by C2–Br, with the original enone and ethoxy group retained. If bromide first captures the cationic intermediate, the equivalent vicinal dibromide can eliminate HBr to reach the same product.

The ethoxy group makes this alkene sufficiently electron-rich for the electrophilic-addition description. It is important to place bromine at C2 and restore the double bond, rather than stop at a saturated addition product.

The alcohol oxygen attacks sulfur of methanesulfonyl chloride, chloride leaves, and triethylamine removes the O–H proton. This gives an O-mesylate with retention of the carbon configuration: the original carbon–oxygen bond has not been displaced.

An accessible chair places the mesylate axial. triethylamine then removes the anti axial hydrogen on the carbon between the ketone and mesylate. Its C–H bond forms the new C=C bond while the C–O bond expels mesylate. The result is the conjugated cyclohexenone.

The neighboring carbon on the acetonide side has its relevant axial position occupied by its oxygen substituent, rather than an anti axial hydrogen. The elimination toward the ketone is therefore favored both by the trans-diaxial geometry and by conjugation of the resulting alkene with C=O. The acetonide remains intact. No inversion of the original alcohol carbon is required during mesylation, and it becomes planar during elimination.

The new bridge is made by an intramolecular Michael addition. On the cyclohexanone side, the methyl-bearing fused alpha carbon is quaternary and cannot be deprotonated there. triethylamine can instead generate a small equilibrium population of the enolate at the other, methylene alpha carbon. Its carbon attacks the beta carbon of the enone in the seven-membered ring. The enone pi electrons move towards its carbonyl and then onto oxygen, creating the new bridge and a second enolate.

protonation of that enolate at the former methyl-substituted enone alpha carbon restores the left-hand ketone. The initial enone C=C has therefore become a saturated C–C bond while one new transannular C–C bond forms. Both ketones are retained, and the initial two-ring framework becomes a three-ring framework without changing the total index of hydrogen deficiency. The methyl originally at the fused quaternary centre does not migrate; the other methyl ends up on a protonated, formerly unsaturated carbon. A strong-base concentration large enough for quantitative enolate generation is unnecessary for a favourable intramolecular trapping reaction, and neither an aldol dehydration nor a methyl shift is needed.

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.

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.

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]

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.

$B={\mathrm{N}}_3{\mathrm{P}}_3(\mathrm{NHCy}{)}_6$, with two cyclohexylamino substituents on each original phosphorus. All six P–Cl bonds have been replaced; the six N–H bonds of those substituents remain. Each equivalent of triethylamine traps one equivalent of HCl.

$C$ contains three four-membered ${\mathrm{P}}^{\mathrm{V}}-\mathrm{N}(\mathrm{Cy})-{\mathrm{P}}^{\mathrm{III}}(\mathrm{Cl})-\mathrm{N}(\mathrm{Cy})$ spiro rings, one at each original ring phosphorus. Each PCl${}_3$ joins the two geminal NHCy groups: replacement of two of its chlorides forms two new P–N bonds and releases two HCl, consuming both N–H protons. The new phosphorus retains one chloride and a lone pair, making it three-coordinate P(III); the original six-membered ${\mathrm{P}}_3{\mathrm{N}}_3$ ring retains P(V). The product formula is ${\mathrm{N}}_3{\mathrm{P}}_3\{(\mathrm{NCy}{)}_2\mathrm{PCl}{\}}_3$. For each spiro ring, the original phosphorus is bonded to both exocyclic nitrogens, and each of those nitrogens is bonded to the same new PCl center; the braces do not indicate a simple linear substituent.

The three new P–Cl directions permit an all-cis isomer and a cis,trans isomer. Relative to the central ring, write these as $(+,+,+)$ and $(+,+,-)$, with cyclic relabeling giving equivalent choices. Reversing all signs does not create a third relative configuration. This describes the configuration pattern; puckering and substituent conformations are additional structural details.

The oxygen substituent changes position through a halohydrin–epoxide sequence. First, hydroxide attacks the acetate carbonyl of $\mathrm{BrC}{\mathrm{H}}_2\mathrm{C}(\mathrm{Me}{)}_2\mathrm{OAc}$, giving a tetrahedral intermediate. Collapse expels the substrate alkoxide, and proton transfer gives $\mathrm{BrC}{\mathrm{H}}_2\mathrm{C}(\mathrm{Me}{)}_2\mathrm{OH}$ plus acetate. This is saponification, breaking the acyl–oxygen bond rather than replacing bromide directly with hydroxide.

A small equilibrium population of $\mathrm{BrC}{\mathrm{H}}_2\mathrm{C}(\mathrm{Me}{)}_2{\mathrm{O}}^{-}$ undergoes intramolecular SN2 reaction at the adjacent primary carbon: the O lone pair attacks $\mathrm{C}{\mathrm{H}}_2\mathrm{Br}$ and its C–Br pair leaves with bromide. The result is 2,2-Dimethyloxirane. Addition of HBr protonates the ring oxygen. bromide then attacks preferentially at the more substituted carbon, where the protonated ring has greater carbocation character, breaking that C–O bond and leaving oxygen attached to the former primary carbon:The middle formula denotes a three-membered epoxide, not an open-chain ether. Acidic opening of this highly unsymmetrical epoxide has the opposite regioselectivity to a simple basic nucleophilic opening at its less substituted carbon.

Finally, the primary alcohol oxygen attacks acetyl chloride. The tetrahedral intermediate collapses with chloride departure, and triethylamine removes the O–H proton and captures HCl. The product is $\mathrm{AcOC}{\mathrm{H}}_2\mathrm{C}(\mathrm{Br})(\mathrm{Me}{)}_2$. oxygen is carried through the epoxide to the other carbon; it is not necessary to invent a methyl migration or a carbon-skeleton rearrangement.

Triethylamine is a tertiary amine, $\mathrm{NE}{\mathrm{t}}_3$, often used to capture acid generated during substitution. Protonation gives triethylammonium.