Codex @codex  0

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.

The phosphorus centre is approximately tetrahedral, with three P–C sigma bonds and one short terminal P–O bond. Both $\mathrm{M}{\mathrm{e}}_3\mathrm{P}=\mathrm{O}$ and $\mathrm{M}{\mathrm{e}}_3{\mathrm{P}}^{+}-{\mathrm{O}}^{-}$ are useful Lewis contributors; the bond is strongly polar and has additional multiple-bond character. oxygen lone-pair donation into suitable P–C antibonding combinations provides a description of the additional stabilization without requiring a substantial localized phosphorus-3d/oxygen-2p pi bond. A combined diffraction and orbital study illustrates this sigma-bond/negative-hyperconjugation description.^[ref]

Compared with the phosphate triester, this compound's methyl substituents are less electron withdrawing. The comparison of the two terminal P–O stretching force constants and the appropriate experimental control is given with the triester below.

A reasonable product is sodium facial trimetaphosphate–molybdenum tricarbonyl, with the anionOne terminal O from each phosphorus binds Mo; the three O donors occupy one octahedral face, while the three carbonyls occupy the opposite face. The ring bridges remain P–O–P bridges and are not replaced by Mo–P bonds. Mo remains Mo(0), $d^6$, and $6+3(2)+3(2)=18$ electrons. Overall,acetonitrile can transiently replace CO during heating, but it is not required in the final six-coordinate proposal. The same anion has been structurally characterized with a different counterion: that primary study independently supports the facial O${}_3$/CO${}_3$ connectivity, without establishing identical solubility or yield for the sodium salt.^[ref]

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.

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 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.

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.

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.

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.

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.

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.

The relevant chelate rings contain four atoms for dppm, five for dppe, and six for dppp, counting the nickel and both phosphorus atoms. The one-carbon bridge in dppm forces a compact ring and the small bite angle $75.62^{\circ }$. A two-carbon tether accommodates a five-membered chelate closer to the preferred square-planar cis angle, giving $86.99^{\circ }$. The more flexible three-carbon tether can open slightly beyond $90^{\circ }$, giving $91.77^{\circ }$.

For a simplified symmetric chelate with Ni–P distance $r$ and P–P donor separation $d$,Increasing the accessible donor separation therefore increases the bite angle $\beta$. The ligand is a three-dimensional tether, so ring puckering and substituent crowding matter as well as the number of methylene groups. This geometrical explanation does not require different formal oxidation states or electron counts for the three complexes.

The neutral product is cis-$[\mathrm{NiC}{\mathrm{l}}_2(\mathrm{dppe})]$, and the displaced material is six waters per nickel:The two phosphorus donors of dppe chelate the same metal, forcing adjacent donor positions. The diamagnetism identifies a square-planar, paired $d^8$ Ni(II) centre rather than a tetrahedral high-spin $d^8$ centre. Its valence count is $8+4_{\mathrm{P},\mathrm{P}}+4_{\mathrm{Cl},\mathrm{Cl}}=16$ electrons. The reaction is ligand substitution, not reduction of Ni(II).

B is oxidative addition; D is alkyne insertion into Ni–aryl followed by carbonyl-oxygen coordination; E is carbonyl insertion into Ni–alkenyl. The distinction is visible from the bonds made and from consistent electron counting.

For B the depicted Ni(0) starting complex has two phosphine and two MeCN donors, giving $10+4+4=18$ electrons. Addition across the aryl C–Br bond produces two X-type ligands and Ni(II). The drawn product has lost the two solvent donors, giving $8+4+2+2=16$ electrons. Intrinsic oxidative addition to the unsolvated 14-electron $\mathrm{N}{\mathrm{i}}^0(\mathrm{dppe})$ fragment raises its count to 16; the overall drawn $18\to 16$ change also includes loss of four solvent-donor electrons. A falling displayed count does not turn this oxidative addition into a reductive elimination.

Before D, the aryl–alkyne intermediate is Ni(I), $d^9$, with $9+4_{\mathrm{P},\mathrm{P}}+2_{\mathrm{aryl}}+2_{\mathrm{alkyne}}=17$ electrons. Syn migration couples aryl carbon to one alkyne carbon and leaves a Ni–alkenyl bond. Insertion alone removes the separate two-electron alkyne donor and would give a 15-electron alkenyl fragment. The ester carbonyl oxygen then binds, as drawn, restoring the count to $9+4+2_{\mathrm{alkenyl}}+2_{\mathrm{O}}=17$. Ni remains $+1$.

In E the Ni-bound alkenyl carbon attacks the coordinated ester carbonyl carbon, forming the new ring-closing C–C bond. The C=O pi electrons move to oxygen, and an Ni–O alkoxide bond replaces Ni–C. This is carbonyl migratory insertion, equivalently intramolecular addition to a coordinated carbonyl. A neutral O donor plus an alkenyl X ligand become one alkoxide X ligand, so the drawn count falls $17\to 15$, while Ni remains $+1$. A solvating molecule could fill the resulting vacancy, but it is not shown and should not be silently added to the stated count.

zinc reduction is required in A, C and F. Reading the drawn ligands with consistent ionic bookkeeping givesIn A the two chlorides of the precursor are removed and neutral solvent donors occupy the reduced centre. In C the bromide is removed while the aryl ligand remains and the neutral alkyne coordinates; the drawn neutral intermediate has Ni(I). In F an anionic alkoxide-derived product is released and the neutral solvated Ni(0) complex is regenerated. zinc supplies electrons by oxidation to Zn(II); halide and alkoxide ions can be taken up in zinc-containing products.

A is precatalyst activation. The two one-electron reductions C and F together require two electrons per catalytic turnover, corresponding to one Zn atom if its oxidation ends at $+2$. The diagram does not establish whether each formal reduction is a single elementary electron-transfer event or a sequence of surface reactions and ligand transfers.

A useful strongly bound limit is Zr(IV), $d^0$, and Ta(V), $d^0$, with an aryne dianion in each complex. Charge balance uses two Cp${}^{-}$ ligands for zirconium, and Cp*${}^{-}$ plus two methyl ligands for tantalum:Equivalently, an unreduced neutral-aryne description starts from Zr(II) and Ta(III), both $d^2$, and transfers that pair strongly into the aryne acceptor. These are alternative bookkeeping limits for the same covalent interaction; saying that a formal $d^0$ assignment has no available electrons must not erase the strong metal–carbon bonds in the reduced-ligand description.

Both ligands have therefore lost much of the extra in-plane C–C bonding of free benzyne and approach the same reduced, approximately double-bond-length C1–C2 framework. Their identical $1.364\text{A˚}$ C–C distances are consistent with similar aryne reduction in both complexes. The distance is intermediate in a delocalized ring, so it should not be converted into an exact integer bond order.

The M–C distances depend additionally on metal size, effective nuclear attraction and orbital overlap. The high formal charge of Ta(V), the size effects associated with the lanthanide contraction, and effective $5d$ covalent bonding favour shorter Ta–C contacts than Zr–C contacts. This is consistent with $2.075$ versus $2.250\text{A˚}$. Different coordination envelopes and ancillary ligands also contribute, so this pair is not a controlled measurement of a radius or a metal–carbon bond energy. Similar internal ligand bond order can coexist with different metal–ligand distances.

The terminal nitrosyls are linear in both proposed electron-precise structures; the formal M–M bond orders are V–V double and Fe–Fe single. Use three-electron terminal linear NO donors, three-electron ${\mu }_2$-NO bridges, and five-electron neutral Cp donors in covalent counting.

For the vanadium dimer the two metal fragments together containA metal–metal bond of order $b$ adds $b$ to each local count, so $32+2b=36$ gives $b=2$ and 18 electrons per V. For an ionic cross-check, Cp is $-1$, each linear terminal NO is conventionally NO${}^{+}$, and each bridge is represented as NO${}^{-}$. ThusEach V then has $4+6_{\mathrm{Cp}}+2_{\mathrm{terminal}\mathrm{NO}}+4_{\mathrm{two}\mathrm{bridges}}+2_{\mathrm{V}=\mathrm{V}}=18$ electrons.

The iron species is monoanionic, and ${\mathrm{L}}^{-}$ is a pyrazolate bridge binding through its two nitrogens. This is a four-electron donor overall in ionic counting, or three overall in neutral counting. The carbonyl bridge supplies two electrons overall. The covalent fragment total is consequentlyTherefore $34+2b=36$ gives $b=1$. In the formal NO${}^{+}$ description,Each Fe receives $10+4_{\mathrm{two}\mathrm{NO}}+2_{\mathrm{pyrazolate}\mathrm{N}}+1_{\mu \mathrm{CO}}+1_{\mathrm{Fe}-\mathrm{Fe}}=18$ electrons. The negative formal oxidation state follows from the chosen nitrosyl assignment and the overall charge; omitting that charge would change the count.

Bent terminal NO ligands would supply fewer neutral-count electrons and demand much higher formal M–M bond orders to attain 18-electron counts. The linear assignments give the ordinary double and single bond solutions. Since NO is a non-innocent ligand, these oxidation states are formal bookkeeping limits, not measured metal charges; actual nitrosyl bending and M–M bond strengths require structural and spectroscopic evidence.^[ref]

A neutral, simple ${\mu }_2$-NO contributes three electrons to the two-metal unit overall. In neutral bookkeeping it is an LX-type bridge: a two-electron donor component plus the odd electron of neutral NO. For a symmetric bridge, this can be apportioned as $1.5$ electrons to each metal in a covalent count; it is not three electrons independently to each metal.

An equivalent ionic description represents the bridge as NO${}^{-}$ donating two electrons to each metal, four overall. The extra negative ligand charge must then be included when assigning metal oxidation states. Mixing the four-electron ionic donation with a neutral-NO charge assignment would double-count an electron.