Codex @codex  0

Number the nitrogen N1, the phenyl-bearing carbon C2, and then continue through C3–C6 around the piperidine. Call the solid-wedge epoxide face up; the phenyl group is down in both starting structures. The bulky phenyl group favors a pseudoequatorial orientation.

epoxide opening follows the Fürst-Plattner preference for a chair-like, trans-diaxial opening trajectory. Attack at the two epoxide carbons is not equivalent: the favored route evolves toward an ordinary chair, whereas the alternative demands a less favorable twist/boat-like distortion. Such conformational control of epoxy-2-arylpiperidine opening is supported by synthetic studies.^[ref]

For the first epoxide, azide attacks C5 from down, breaking the C5–O bond. The residual C4–O bond is retained, so work-up gives C5–N3 down and C4–OH up, with the original C2–Ph down. For the second, down-face epoxide, azide attacks C4 from up, leaving C4–N3 up and C5–OH down. These assignments are shown in the middle row of the original stereochemical map.

In each mechanism an azide terminal nitrogen attacks the epoxide carbon, the C–O bonding pair goes to oxygen, and protonation gives the alcohol. Each attacked center inverts, and the azide and OH are trans. The phenyl-bearing center is unchanged. Reversing the epoxide face changes which carbon has the favored axial attack trajectory; a steric rule based only on distance from phenyl misses that conformational effect.

Number the cyclohexane consecutively, putting the top methylene group at C1 and the two original epoxides at C2/C3 and C5/C6. Both epoxide oxygens are up.

A is a diazidocyclohexanediol, with each epoxide replaced by one azide and one alcohol. A standard successive chair-favored opening gives N3 down at C2 and C5, OH up at C3 and C6. This is the 1,4-diazido/1,4-diol connectivity drawn in the bottom row of the original map.

For each opening, azide attacks from the face opposite oxygen, breaking one C–O bond with inversion at the attacked carbon. protonation of the two resulting alkoxides gives the two OH groups. After the first opening, the preferred conformation of the opened portion must be used for the second trans-diaxial opening; simply imposing one unchanged chair on both steps can select the wrong regioisomer.

The elemental change is addition of two ${\mathrm{N}}_3^{-}$ groups and two work-up protons to ${\mathrm{C}}_6{\mathrm{H}}_8{\mathrm{O}}_2$, giving ${\mathrm{C}}_6{\mathrm{H}}_{10}{\mathrm{N}}_6{\mathrm{O}}_2$. The formula alone would not distinguish alternative positional isomers; the depicted assignment uses the ordinary successive-opening conformational model.

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.

B is the major product: the newly formed OH is the dashed bond in the printed zigzag structure. Retain the existing solid-wedge OH and the configuration at the carbamate-bearing carbon.

zinc coordinates the ketone oxygen and the adjacent hydroxyl oxygen, forming a five-membered O–Zn–O chelate. This constrains rotation around the carbonyl–alpha-carbon bond. In the favored chelated conformation the bulky $\mathrm{CH}(\mathrm{NHC}{\mathrm{O}}_2{\mathrm{R}}^{\prime })\mathrm{C}{\mathrm{H}}_2\mathrm{OR}$ group shields one face of the ketone, while the alpha hydrogen leaves the other more accessible. hydride attacks from that less crowded face, the carbonyl $\pi$ bond opens onto oxygen, and aqueous work-up gives the second alcohol.

Mapping this Cram chelation model back onto the printed zigzag gives B, not A. Merely placing a new OH “on the same side” of an unrotated page drawing is not a valid chelation argument: the reactive chelate and that drawing use different conformations. Zinc-borohydride reductions of alpha-hydroxy ketones are experimentally known to favor the corresponding anti-1,2-diol family under chelation control.^[ref]

The reagent combination can generate zinc borohydride and related zinc-associated reducing species. The stereochemical justification is coordination-controlled facial selectivity; it does not require identifying a unique solution aggregate from the reagent list. The quoted yield is not itself a measurement of a 95:5 diastereomeric ratio.

The first step is a vinylogous aldol condensation at C5 of the unsaturated lactam. Number nitrogen N1, the lactam carbonyl C2, and then the methoxy-substituted alkene C3=C4 and the methylene group C5.

hydroxide reversibly removes a C5 proton. The resulting vinylogous enolate is delocalized through C5–C4–C3–C2–O: it is not necessary to create an unstabilized isolated carbanion beside nitrogen. Its C5 nucleophilic contributor attacks acetone. protonation gives a C5-linked tertiary alcohol, $C5-\mathrm{C}(\mathrm{OH})(\mathrm{Me}{)}_2$. Further removal of the remaining C5 hydrogen and elimination through an aldol-dehydration sequence forms C5=C(Me)2, leaving the original methoxy enol ether in place.

In the second step, aqueous acid hydrolyses that enol ether to a ketone at C4. protonation at C3 gives an oxygen-stabilized cation at C4. water attacks C4, proton transfers form a hemiacetal-like intermediate, and loss of methanol followed by deprotonation produces C4=O and the C3 methylene group.

The lactam carbonyl and N–H group remain, as does the exocyclic isopropylidene group. This accounts for replacement of the C3=C4–OMe unit by C3H2–C4=O without inserting a new oxygen into the ring skeleton.

The hydroxyl group first binds aluminium, normally by deprotonation to an aluminium alkoxide with release of hydrogen. Coordination also brings aluminium close to the alkyne. directed trans-hydroalumination then puts hydride on the alkyne carbon proximal to CH2O and aluminium on the distal carbon.

One useful mechanistic description has external hydride attacking the coordinated alkyne opposite the aluminium-containing face. The resulting vinylaluminate is stabilized by a five-membered O–Al–C–C–CH2 chelate. This is not the ordinary undirected syn addition of a neutral dialkylaluminium hydride. Propargylic-alcohol-directed trans-hydroalumination with Red-Al gives vinylaluminate intermediates that can be intercepted by electrophiles.^[ref]

iodine replaces the distal C–Al bond with C–I without scrambling the alkene geometry; aqueous work-up releases the alcohol. hydride and iodine are consequently trans across the alkene, while iodine and the CH2OH group are on the same side.

By Cahn-Ingold-Prelog priority rules priority the product is therefore $Z$: CH2OH outranks H at one alkene carbon and I outranks CH2R' at the other. Describing the two carbon chains as trans would not make this an $E$ alkene, because the iodine changes which substituent has higher priority.

Let $\mathrm{Ar}=4-\mathrm{N}{\mathrm{O}}_2{\mathrm{C}}_6{\mathrm{H}}_4$ and number the carbon substituent from oxygen: $\mathrm{ArO}-\mathrm{C}{\mathrm{H}}_2(1)-\mathrm{C}{\mathrm{H}}_2(2)-\mathrm{C}{\mathrm{H}}_2(3)-\mathrm{CN}$. sodium hydride removes a proton from C3, alpha to the nitrile, forming hydrogen and a nitrile-stabilized carbon anion. That carbon attacks the nitroaryl ipso carbon bearing O. The tether closes a five-membered cyclic Meisenheimer intermediate with sequence C3–C2–C1–O–aryl ipso carbon. The para nitro group stabilizes the negative charge generated by temporary loss of aromaticity.

Rearomatization breaks the aryl–O bond, releasing the tether oxygen as an alkoxide while retaining the new aryl–C3 bond. protonation gives $\mathrm{Ar}-\mathrm{CH}(\mathrm{CN})-\mathrm{C}{\mathrm{H}}_2-\mathrm{C}{\mathrm{H}}_2\mathrm{OH}$. This is a Truce–Smiles rearrangement: the aryl group migrates from oxygen to carbon. The nitrile remains attached to its original side-chain carbon; the original ether oxygen becomes the terminal alcohol oxygen. The substrate and product have identical atomic composition. Ordinary substitution at a saturated chain carbon would not produce this aryl–carbon bond or restore the terminal OH.

protonation of the secondary OH permits water departure. The neighbouring phenylthio sulfur can assist this departure by forming a bond to the adjacent alcohol carbon, generating a thiiranium ion spanning that carbon and the original sulfur-bearing cyclohexane carbon. Label the latter Q and the former L. The temporary Q–S–L bridge transfers the phenylthio attachment from Q to L when its Q–S bond subsequently breaks. sulfur attack opposite the departing OH accounts for the new C–S stereochemistry without requiring a freely rotating carbocation.

For the protected terminal alcohol, no free tethered OH is available to trap the bridge. Removal of a beta hydrogen from a cyclohexane carbon adjacent to Q forms the ring alkene while Q–S breaks. sulfur remains bonded to L, giving the allylic phenyl thioether. acid is regenerated by the deprotonation, so the catalyst need not be consumed.

For the free terminal alcohol, its oxygen attacks Q intramolecularly, opening Q–S instead. Count the new ether ring: O–CH2–CH(Me)–L–Q, a five-membered ring sharing Q with the cyclohexane. deprotonation gives the spirocyclic ether. This is capture of the same neighbouring-group-stabilized intermediate rather than migration of the methyl group. The pre-existing methyl-bearing carbon retains its configuration; its wedge/dash appearance changes when the product is redrawn in a different ring projection. The two substrates differ in whether a suitably placed oxygen nucleophile is available for intramolecular trapping.

The alkene attacks electrophilic mercury(II), with acetate departure as appropriate, to give a bridged mercurinium ion between the ring carbon and the exocyclic methylene. That bridge is polarized: the more substituted ring carbon can better stabilize developing positive charge than the terminal CH2 carbon. water attacks the ring carbon, opening the mercury bridge there, and deprotonation gives the tertiary alcohol with an exocyclic CH2–HgOAc substituent.

Subsequent sodium-borohydride demercuration replaces the terminal C–Hg bond by C–H. Thus the original exocyclic CH2 becomes Me, and the new OH remains on the substituted ring carbon. The result is Markovnikov hydration of the exocyclic alkene. There is no need for a freely equilibrating tertiary carbocation, and no rearrangement of the carbohydrate carbon skeleton or cleavage of the methyl acetal is required. The question does not require the mercury-stage facial selection or a detailed demercuration mechanism.

protonation activates the oxygen of the anomeric acetal bridge. Cleavage of its bond to the anomeric carbon releases that tether oxygen as the terminal primary alcohol, while the other ring oxygen stabilizes the developing anomeric electrophile by lone pair donation. The resulting oxocarbenium description has a nearly planar reactive anomeric centre; it need not be drawn as a long-lived free carbocation.

methanol attacks that centre, giving a protonated methyl acetal, and deprotonation regenerates acid. The former bridge becomes CH2OH; the anomeric substituent becomes OMe. The other C–O bonds and existing carbon configurations are retained. The ring can adopt a different chair once the bridge constraint is removed, so an apparent movement of substituents in the projection does not imply inversion at all the other carbons.

The depicted anomeric OMe is axial in the product chair. This geometry allows an endocyclic oxygen lone pair to align antiperiplanar to the exocyclic C–OMe bond, providing $n_O\to {\sigma }_{C-O}^{\ast }$ stabilization associated with the anomeric effect. Acidic conditions also permit cleavage and recapture at the anomeric centre, so selection is not simply backside displacement at an unchanged tetrahedral starting carbon. The anomeric effect rationalizes the specified axial product; the scheme alone gives neither its numerical equilibrium population nor evidence that every other anomer is absent.

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.

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 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]

Draw a curly arrow from the nitrogen lone pair to methyl carbon, and a second from the C-I bond to I:This is concerted SN2 quaternization. The product is choline iodide; nitrogen has four carbon substituents and positive charge, while the alcohol remains OH.

First hydrolyze the benzylidene acetal. Protonate one acetal oxygen and break its bond to the Ph-CH centre, giving one free alcohol and a benzylic oxocarbenium ion. Water attacks that centre; proton transfers give a hemiacetal. Protonate the remaining acetal-derived OR group and cleave it to release the second alcohol and benzaldehyde.

The resulting unsaturated hydroxy ester isThe OH on the allyl-substituted methine is gamma to the ester carbonyl. Protonation of that carbonyl enables intramolecular OH attack to give a five-membered tetrahedral intermediate. Proton transfer to methoxy and collapse expel methanol, followed by deprotonation to give the alpha,beta-unsaturated gamma-lactone.

The other liberated OH remains CH${}_2$OH. Both C=C units survive: one is the allyl substituent and the other becomes the lactone's conjugated double bond. Hydrolysis to a hydroxy acid followed by lactonization is an alternative sequence under the aqueous acid; direct acid-catalysed intramolecular ester exchange gives the same atom mapping.

The alkene pi bond attacks Br${}_2$ as the Br-Br bond breaks to bromide, forming a bridged, unsymmetrical bromonium ion. The carbon bearing the two methyl groups has greater carbocation-like character.

The tethered alcohol oxygen attacks that carbon from the side opposite the bromine bridge. Its lone pair forms the new O-C bond while the attacked C-Br bridge bond opens, giving a six-membered cyclic oxonium ion. Bromide or solvent removes the O-H proton to give the neutral bromoether.

This explains why O becomes attached at the more substituted alkene carbon while Br remains at the other carbon. Polar solvent assists charge separation, and intramolecular capture competes effectively with external bromide attack. Opening is anti; attack on the two faces of the achiral starting material gives a racemic product at the newly formed CHBr stereocentre.

The ring-forming step is bromoetherification.

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.

ethanol undergoes proton transfer with the methoxide produced. The oxygen lone pair of methoxide attacks the hydroxyl proton of ethanol, while the ethanol O–H bond pair returns to its oxygen:The alcohol acidities are similar, so this is an equilibrium, not an overwhelmingly one-way deprotonation determined solely by intrinsic acidity. The large ethanol reservoir and initially small methanol concentration nevertheless matter. This exchange means that lithium ethoxide need not be consumed stoichiometrically in the net transesterification. Neutral ethanol can also participate in proton shuttling; it is not acting as a reducing reagent or as a substrate for elimination under these conditions.

Excess ethanol drives a reversible ester exchange mainly through the actual mixture composition, not through a large new bond-energy advantage. The net reaction exchanges one ester and one alcohol for a closely related ester and alcohol. It preserves the main bond types and has no obvious large exothermicity. Solvation and ion pairing can contribute to its enthalpy, but there is no basis here for asserting a large negative standard reaction enthalpy.

There are two molecules on each side, so an increase in particle count is not the entropy explanation. Instead useWith ethanol as the solvent and little methanol initially, the reaction quotient is small; product formation and mixing lower the Gibbs free energy until the exchange equilibrium is reached. ethanol can be treated approximately as having fixed activity near its pure-solvent value, rather than assigned an arbitrarily enormous activity, while methanol remains dilute. This composition-dependent chemical-potential contribution is the appropriate entropic/mixing argument. It explains forward conversion under these conditions without claiming that ${\Delta }_r{S}^{\circ }$ must be positive, or that complete conversion follows without an equilibrium constant or removal of methanol.

The first product is $\mathrm{BrCH}=\mathrm{C}(\mathrm{C}{\mathrm{H}}_3)\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{OC}{\mathrm{H}}_3$, with the original alkene geometry retained. methoxide attacks the primary saturated carbon bearing bromine from the backside. Its oxygen lone pair forms the new C–O bond as the C–Br bonding pair leaves with bromide. The other bromine is vinylic, directly bonded to an sp${}^2$ carbon: ordinary SN2 reaction is unfavourable there, and ionization would require an unstable vinyl cation. Thus the primary alkyl bromide is replaced selectively, rather than replacing the vinylic halide or moving the double bond. There is no observable inversion label at the attacked carbon because it is a methylene.

The second product is sodium 4-propoxybutane-1-sulfonate, $\mathrm{N}{\mathrm{a}}^{+}{}^{-}{\mathrm{O}}_3\mathrm{S}(\mathrm{C}{\mathrm{H}}_2{)}_4\mathrm{OC}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_3$. sodium hydride first removes the alcohol proton, giving propoxide and hydrogen. The propoxide oxygen then attacks the primary ring carbon bonded to the ring oxygen; simultaneous cleavage of that carbon–oxygen bond opens the sultone. The departing ring oxygen remains attached to sulfur as part of the resonance-stabilized sulfonate anion. The carbon attached directly to sulfur is not the substitution site, because cleavage there would not expel the same stable sulfonate leaving group. This is carbon-centred ring opening of a cyclic sulfonate ester, not addition across an S=O bond.