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finasteride is reduced at C1, leaving a C2 enolate that attacks nicotinamide C4 to make the covalent C–C adduct. In the inhibitor's unsaturated lactam, the reactive chain is $\mathrm{O}=C3-C2=C1$, with N at steroid position 4. Activate the C3 carbonyl by hydrogen bonding; draw the reduced cofactor's C4–H bond toward inhibitor C1, the C1=C2 pi pair toward C2–C3, and the C3=O pi pair toward oxygen. This gives NADP+ next to the C2-centered resonance form of the inhibitor enolate.

Instead of completing reduction by protonating C2, move the C2 lone pair toward the electrophilic C4 of the oxidized nicotinamide ring. Shift the pyridinium pi system toward its positively charged N, and reform the inhibitor carbonyl if using the oxygen-centered enolate representation. The result is the C2(inhibitor)–C4(nicotinamide) bond and a neutral dihydronicotinamide ring. Numbering the two molecules separately prevents confusing cofactor C4 with the inhibitor's N4. The lactam N–H remains present and can help position the inhibitor through hydrogen bonding; it does not form a covalent bond to the protein.

This is mechanism-based inhibition: enzyme chemistry generates the species that traps its own oxidized cofactor. Isolation of a cofactor–inhibitor adduct after protein denaturation, rather than persistent attachment to a denatured protein, distinguishes it from covalent modification of an amino acid side chain. The original mechanistic study combined radiolabel recovery, purification and spectroscopy to establish the adduct; its slow breakdown also cautions against treating “irreversible” as an absolute statement for every timescale.^[ref]

The nitrite ion is an ambidentate ligand: its Lewis structures leave donor lone pairs on both nitrogen and oxygen. In the N-bound mode, $\mathrm{M}-\mathrm{N}(\mathrm{O}{)}_2$, the nitrogen lone pair makes a sigma bond to the metal, while a suitable unoccupied nitrite molecular orbital can accept metal $d$ electron density by pi backbonding. Thus N-bound nitro is a sigma donor and pi acceptor. The nitrite group remains bent at nitrogen; N coordination does not make it a linear nitrosyl.

In the O-bound mode, $\mathrm{M}-\mathrm{O}-\mathrm{N}=\mathrm{O}$, an oxygen lone pair makes the metal–oxygen sigma bond. A second occupied oxygen orbital can donate sideways into an available metal orbital: O-bound nitrito is a sigma and pi donor, rather than the strong pi acceptor of the N-bound description. oxygen's greater electronegativity lowers its donor orbital energies, but the occupied oxygen orbitals still explain its pi-donor classification. Actual mixing depends on metal oxidation state and geometry.

A further possibility is bidentate coordination, $\mathrm{M}({\kappa }^2\text{-}O,O\text{-}\mathrm{N}{\mathrm{O}}_2)$, with both oxygen atoms attached to one metal. Each attachment has oxygen-to-metal sigma donation; additional pi donation depends on orbital alignment. nitrite can also be a bridging ligand between metals. Neither alternative should be confused with linkage isomerism, which specifically compares attachment through different atoms at the same binding site.

The N-bound isomer $C$ is the thermodynamically favored form; heating accelerates access to it. Formation of $B$ retains the original metal–oxygen attachment and is a plausible kinetic control outcome. Rearranging that attachment requires overcoming an activation free energy, especially at a substitutionally inert low-spin cobalt(III) center. Raising temperature speeds the linkage isomerization, allowing the more stable N-bound nitro ligand to accumulate. Its stronger ligand field and different covalent bonding help rationalize the stability, but a larger field splitting alone is not a complete calculation of the isomerization Gibbs free energy. The classical crystallographic comparison establishes the different attachments and thermodynamic direction.^[ref]

$A=[\mathrm{Co}(\mathrm{N}{\mathrm{H}}_3{)}_5(\mathrm{OH}){]}^{2+}$, $B=[\mathrm{Co}(\mathrm{N}{\mathrm{H}}_3{)}_5(\mathrm{ONO}){]}^{2+}$, $C=[\mathrm{Co}(\mathrm{N}{\mathrm{H}}_3{)}_5(\mathrm{N}{\mathrm{O}}_2){]}^{2+}$, and the by-product is $\mathrm{HN}{\mathrm{O}}_2$. The first transformation is deprotonation of coordinated water, not loss of an ammonia ligand. The hydroxo ligand in $A$ is then nitrosated at oxygen, retaining its original Co–O attachment in the initially formed O-bound nitrito ligand. Atom and electric charge balance giveThe relative molecular mass of nitrous acid is approximately $1+14+2(16)=47$. Heating converts the O-bound nitrito ligand to the N-bound nitro ligand; these are linkage isomers, with Co remaining in oxidation state $+3$ throughout.

A formal two-electron reductive elimination couples two anionic, mutually cis donors and reduces Ti(IV) to Ti(II). There are two chemically distinct couplings: Cl–Cl elimination gives $\mathrm{C}{\mathrm{l}}_2$, leaving a Ti(II) complex with the intact anionic O,N,P ligand and one chloride; O–Cl elimination gives an aryl hypochlorite group, $\mathrm{ArOCl}$, breaking the original Ti–O attachment and leaving a Ti(II) center coordinated by N,P and two chlorides. The latter product can remain tethered through the N,P part of the same ligand; it is not necessarily a free small molecule. The starting facial arrangement supplies cis pairs of both kinds.

Cl–Cl coupling is the more plausible of these formal alternatives, because it retains the particularly strong early-metal Ti–O interaction and makes the stronger Cl–Cl rather than relatively weak O–Cl bond. Both alternatives reduce the valence count by two, from 12 to 10, so the 18-electron rule does not choose between them. Neither should be asserted to occur readily: reduction of a stable, oxophilic $d^0$ Ti(IV) center is intrinsically unfavorable. Coupling a chloride to the neutral amine or phosphine is not the ordinary elimination of two X-type ligands. proton transfer between N–H and oxygen, if it occurred, would instead be acid-base chemistry, not automatically a two-electron reductive elimination.

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.

Let $\mathrm{Ar}=p-\mathrm{MeO}{\mathrm{C}}_6{\mathrm{H}}_4$. One standard route is:where the last two carbons are joined through oxygen to form the epoxide shown in the product drawing.

Use rigorously dry conditions for Grignard formation/addition; quench only after the aldehyde addition. Dehydration gives the styrene, and peracid oxidation transfers oxygen across its double bond. The methoxy substituent is compatible with this sequence. No asymmetric reagent is specified, so the target epoxide is racemic.

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.

The vinyl ether has an important resonance contributorDonation from oxygen into the conjugated system increases electron density at the terminal unsaturated carbon, helping explain its unusually shielded ${\delta }_C=87.4$ ppm and its terminal protons near 4.1–4.2 ppm. The directly oxygen-bound unsaturated carbon is strongly deshielded at $151.1$ ppm; its proton at $6.480$ ppm also experiences the oxygen substituent and alkene magnetic anisotropy. These trends are consistent with the whole delocalized electronic structure; a chemical shift is not simply a meter for an atom's partial atomic charge.

For the saturated part, the stronger local deshielding by oxygen places $\mathrm{OC}{\mathrm{H}}_2$ at ${\delta }_C=67.8$, ${\delta }_H\simeq 3.948$, while $\mathrm{C}{\mathrm{H}}_2\mathrm{Cl}$ lies at $41.9$ and $3.708$ ppm. The two terminal vinylic protons have different geometrical relationships to the oxygen substituent and therefore different chemical shifts as well as different vicinal couplings.

First, proton transfer from acetic acid to the carbon of diazomethane gives methyldiazonium and an acetate ion:Draw an electron-pair arrow from carbon, using the carbon-negative resonance contributor, to the acidic hydrogen, and another from the acid O–H bond to oxygen. Second, an acetate oxygen attacks the diazonium methyl carbon by an SN2 reaction; an arrow from the C–N bond to nitrogen expels neutral ${\mathrm{N}}_2$:The acid's hydroxyl hydrogen has become the third hydrogen of the newly installed methyl group. This is a methyl-transfer sequence driven by formation of molecular nitrogen, without forming a free methyl carbocation or invoking Fischer esterification.

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.

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.

The two starting ester groups are adjacent on the benzene ring. Select one ester carbon as the site receiving both incoming methyl groups. A first methyl addition gives a tetrahedral intermediate; its collapse expels methoxide to give the ortho-substituted ketone $\mathrm{o}-!\mathrm{Me}{\mathrm{O}}_2\mathrm{C}-{\mathrm{C}}_6{\mathrm{H}}_4-\mathrm{C}(=\mathrm{O})\mathrm{Me}$. A second methyl addition produces $\mathrm{o}-!\mathrm{Me}{\mathrm{O}}_2\mathrm{C}-{\mathrm{C}}_6{\mathrm{H}}_4-\mathrm{C}({\mathrm{O}}^{-})\mathrm{M}{\mathrm{e}}_2$, with the alkoxide bound to magnesium.

The newly generated oxygen is now positioned to attack the remaining ester intramolecularly. Draw an arrow from its lone pair to that ester carbonyl and another from the C=O pi bond to its oxygen. Collapse of the resulting tetrahedral intermediate reforms the carbonyl and expels methoxide, closing a five-membered lactone. Protonation of magnesium-bound species in the work-up gives 3,3-Dimethylphthalide. Equivalently, protonation can first give the ortho hydroxy ester, followed by acid-assisted lactonization with loss of methanol; both descriptions show the same required bond formation.

The oxygen in the new ring is the oxygen of the first ester's original carbonyl; the surviving lactone carbonyl comes from the second ester. The product retains the six ring carbons and both original acyl carbons, loses the two ester methoxy carbons, and gains two carbons from methylmagnesium bromide. The favourable intramolecular five-membered closure accounts for the displayed product. It does not make the isolated lactone universally inert to a large excess of grignard reagent; stoichiometry and trapping/work-up conditions matter.

Disconnect the new bond between the two six-carbon units. One portion of bromocyclohexane supplies a grignard reagent; another supplies cyclohexene oxide:A bulky strong base can favour E2 elimination in the alkene-forming step; a peracid such as meta-chloroperoxybenzoic acid provides epoxidation. None of its carbons enters the target.

Treat cyclohexene oxide with cyclohexylmagnesium bromide, then perform acidic aqueous work-up. Backside attack on one epoxide carbon breaks its C–O bond and installs cyclohexyl opposite the retained oxygen. Protonation of the oxygen gives trans-2-cyclohexylcyclohexan-1-ol, the required relative configuration. The two six-carbon reactant portions supply all 12 target carbons. An achiral synthesis produces a racemic mixture trans pair; obtaining only the particular enantiomer represented by the drawing would require resolution or an enantioselective step. The useful stable chair can place the two large trans-1,2 substituents approximately diequatorial.

C gives spectrum 3. For allyl acetate, $\mathrm{C}{\mathrm{H}}_3\mathrm{C}(=\mathrm{O})\mathrm{OC}{\mathrm{H}}_2\mathrm{CH}=\mathrm{C}{\mathrm{H}}_2$, assignThe 65.1-ppm resonance is the oxygen-bound methylene. Its attached proton test phase matches the terminal alkene methylene and the nonprotonated carbonyl, while the alkene CH and acetyl methyl have the opposite phase. The carbonyl is not conjugated with the alkene because oxygen and a saturated methylene interrupt the carbon–carbon conjugated path. This explains its higher carbonyl shift relative to B without requiring the two esters to have identical electronic environments.

The mechanism is nucleophilic addition followed by elimination of methoxide. An ethoxide oxygen lone pair attacks the carbonyl carbon, while the C=O pi pair moves to oxygen:The tetrahedral intermediate collapses when its anionic oxygen reforms the carbonyl and the C–OMe bond breaks with its electron pair going to the methoxy oxygen:lithium is a coordinating counterion; it does not supply an electron pair to the acyl carbon. proton exchange with ethanol connects the methoxide produced to the ethoxide catalyst pool and gives methanol, so the net solvent-mediated exchange is methyl acetate plus ethanol to ethyl acetate plus methanol. The reversibility of the elementary acyl steps is important: changing an alkoxy group is not automatically an irreversible reaction.

The product-forming pairing is $n(\mathrm{Et}{\mathrm{O}}^{-})\to {\pi }^{\ast }(\mathrm{C}=\mathrm{O})$. Orient an occupied ethoxide oxygen lobe toward the large carbon lobe of the carbonyl acceptor with matching phase. An oblique approach permits constructive overlap while the pi bond weakens; this is the usual Bürgi-Dunitz trajectory to acyl carbon. The high-energy occupied orbital on the anion makes ethoxide the stronger electron-pair donor than the neutral ester oxygen.

The reverse frontier pairing, $n(\mathrm{ester}\mathrm{O})\to \mathrm{empty}\mathrm{Li}$, gives oxygen-to-lithium coordination. It is a plausible stabilizing contact and can coexist with the first interaction, rather than constituting an alternative way to replace the ester alkoxy group. In the elementary frontier picture, the anionic donor-to-carbonyl interaction is the important pairing for the new covalent bond and is favoured by its donor strength and appropriate accepting coefficient at carbon. The Li contact can polarize the carbonyl and organize the reacting pair. Precise energy-gap rankings require a specified solvated lithium species; the sketches alone do not justify pretending that lithium coordination cannot occur.

The reactive occupied orbital of ethoxide is an oxygen lone pair; the acyl electrophile accepts into its carbonyl ${\pi }^{\ast }$ orbital. Draw the ethoxide lone pair as a directional oxygen-centred lobe directed toward an electrophile. In a lithium-ion-pair picture the low unoccupied acceptor on lithium is an empty, diffuse Li-centred orbital, with 2s/2p character depending on coordination. It should not be confused with an occupied Li–O covalent bonding pair.

For methyl acetate, the occupied frontier donor is principally an oxygen nonbonding orbital, mixed with the ester orbital system. Its low-lying reactive acceptor is ${\pi }^{\ast }(\mathrm{C}=\mathrm{O})$, with opposite phases on carbon and oxygen and a relatively large accepting lobe on the carbonyl carbon. The corresponding sketch must show a node between the carbonyl atoms. These are chemically useful frontier-orbital descriptions; lithium alkoxide aggregation and ethanol solvation make a unique isolated-molecule orbital energy diagram an oversimplification.

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.