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The opposite insertion orientation gives a benzylic metal–carbon bond, $\mathrm{Co}-\mathrm{CH}(\mathrm{Ph})\mathrm{C}{\mathrm{H}}_3$. C–Si reductive elimination then gives the branched productA plausible explanation is stabilization of the developing and final benzylic Co–C interaction by the adjacent aromatic pi system, which can outweigh the steric preference for a primary alkyl. That makes an electronically different choice from an ordinary alkyl substituent. This is a mechanistic rationale for the proposed cycle, not proof of its rate-determining step. In the absence of a chiral catalyst or another source of asymmetric induction, the new stereogenic carbon would give a racemic mixture.

oxidative addition of an intact Si–H bond initially places H and SiR${}_3$ cis in the metal coordination sphere; this is the local arrangement generated by addition of a sigma bond. hydride and alkene must also attain adjacent sites for intramolecular insertion, and the eventual alkyl and silyl groups must be cis for direct reductive elimination. Subsequent rearrangements should not be confused with the stereochemistry of the bond-addition event itself.

For an alkyl-substituted terminal alkene, insertion puts Co on the terminal carbon and H on the substituted carbon, producing $\mathrm{Co}-\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{R}$. It avoids the more crowded secondary metal–alkyl arrangement. Replacement of that Co–C bond by C–Si therefore gives the linear product $\mathrm{RC}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{Si}{\mathrm{R}}_3$, an Anti-Markovnikov addition outcome. Hydrometallation is syn at the coordinated alkene, although this simple terminal substrate generally does not retain two substituents whose relative stereochemistry could diagnose the complete sequence.

Step 1 is oxidative addition of Si–H to Co(I), giving Co(III)–H and Co(III)–SiR${}_3$. It cleaves the Si–H bond and forms two metal–ligand bonds, increasing the formal metal oxidation state by two. The phosphine dissociation shown before this numbered step makes room for the reaction; alkene binding and phosphine exchange are ligand-substitution events rather than oxidative additions.

Step 2 is migratory insertion of the coordinated alkene into Co–H, equivalently intramolecular hydride migration to the alkene. One alkene carbon gains H and the other gains the Co–C bond. Co remains formally Co(III); a two-electron alkene ligand plus a hydride is replaced by one alkyl ligand, reducing the electron count by two before phosphine reassociation.

Step 3 is C–Si reductive elimination between the alkyl and silyl ligands. It releases the hydrosilylated alkane and regenerates Co(I). The electron count for the drawn sequence is consistent: the one-PMe${}_3$ alkene adduct is 16-electron; oxidative addition gives 18; insertion gives 16; PMe${}_3$ association gives 18; reductive elimination returns the 16-electron catalyst. Because insertion occurs into Co–H, this is the conventional Chalk-Harrod mechanism.

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 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.

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 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.

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.

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 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.

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.

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.

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]

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.

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]

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.