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Cleavage of a carbon–carbon bond beta to a metal, often from a metal alkoxide, coupled to carbonyl formation and transfer of a carbon fragment to the metal. For a tertiary benzylic alkoxide this can generate a ketone and a metal–aryl bond.

A compound with a halogen and hydroxyl group on adjacent carbons. alkoxide formation can permit intramolecular SN2 reaction to an epoxide.

The new ring is formed by intramolecular Friedel-Crafts addition to the ketone. titanium tetrachloride coordinates the ketone oxygen of the cyclic beta-ketoester, increasing electrophilicity at that carbonyl carbon. The nearby electron-rich aryl ring attacks this activated ketone at the available position ortho to its ethyl tether.

The electron-flow sequence is: an aromatic $\pi$ bond forms the new aryl–carbonyl-carbon bond, the ketone $\pi$ bond moves onto oxygen, and loss of the aromatic proton restores aromaticity. The two-carbon tether and the shared quaternary carbon make the newly closed carbocycle six-membered. Hydrolytic work-up converts the titanium-bound alkoxide into the tertiary alcohol.

The lactone carbonyl survives. This is an addition to the ketone with retention of its oxygen as OH, rather than an acyl substitution that removes that oxygen. The pendant terminal alkene is not involved. These bond changes account for the fused aromatic ring system and the alcohol in C without needing to assign the new stereochemistry.

copper(I) converts the methyl grignard reagent into a softer organocopper nucleophile. Methyl undergoes 1,4 addition to the beta carbon of the cycloheptenone, giving a metal enolate rather than direct 1,2 addition to the ketone carbonyl.

The carbon nucleophilic site of this enolate, at the alpha carbon, attacks the aldehyde carbonyl in the second reagent. Its C=O $\pi$ electrons move to oxygen; protonation of the resulting alkoxide gives the beta-hydroxy ketone side chain. The cyclic acetal is a protected ketone and survives these nucleophilic conditions.

The beta methyl group shields one face of the cyclic enolate. Aldol bond formation from the less crowded opposite face places the new alpha side chain trans to the beta methyl group. Thus the two starred ring centers have an anti relationship; the two mirror-related products are possible from achiral starting materials and reagents. This argument does not assign the separate, unstarred alcohol center, for which the question gives no specified configuration.

sulfide first opens one epoxide by backside attack, giving a carbon-bound thiolate and an alkoxide. Using the same numbering, take this first attack at C2: the C2–S bond is down and C3–O remains up.

The attached sulfur still bears a nucleophilic lone pair and negative charge. It attacks C5 of the second epoxide intramolecularly, breaking C5–O and closing a bridge between C2 and C5. Work-up protonates the oxygen anions at C3 and C6. Thus B is 7-thiabicyclo$[\mathrm{2.2.1}]$heptane-2,5-diol, with both OH groups on the face opposite the sulfur bridge.

Both C–S bonds are formed by inversion at their attacked epoxide carbons. Because the original epoxides are cis, both new C–S bonds point to the same opposite face and one sulfur chemical atom can connect them. The 1,4 bridge gives two five-membered sulfur-containing circuits, unlike the more strained closure between nearby 1,3 positions.

The sulfur therefore replaces two separate external nucleophiles: two epoxides are opened but only one sulfur chemical atom is incorporated. Adding that sulfur and two work-up protons gives ${\mathrm{C}}_6{\mathrm{H}}_{10}{\mathrm{O}}_2\mathrm{S}$, consistent with B.

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 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.

Number the OH-bearing carbon C1 and the adjacent brominated carbon C2. In the trans bromohydrin, base generates an alkoxide. A chair conformation can place O at C1 and Br at C2 trans-diaxial, aligning the oxygen nucleophile for backside attack at C2. Intramolecular SN2 closure expels bromide and inverts C2: the retained C1–O bond and the newly formed C2–O bond are both on the original OH face, giving the epoxide shown. ring inversion can provide the reactive chair even if it is not the dominant ground-state conformer.

The cis bromohydrin cannot place OH and Br trans-diaxial: one is axial when the other is equatorial. In the chair with Br axial, the hydrogen on C1 can instead be antiperiplanar to C2–Br. An alkoxide-assisted semipinacol pathway moves the C1–H bonding pair to C2 while the oxygen lone pair forms C1=O and bromide departs. hydride migration gives the ketone without changing the carbon skeleton. The hydrogen is transferred to C2 rather than expelled as a free hydride. The original phenyl-bearing stereocentre is not involved in either bond-reorganization sequence. Configuration controls which properly aligned displacement or migration can occur; a flat ring drawing alone conceals that distinction.

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.

Base deprotonates the sulfonamide. Its nitrogen attacks one epoxide carbon from the back, breaking the C–O bond at that carbon and producing an adjacent alkoxide. The nitrogen and oxygen attachments on the cyclopentane are therefore trans. The alkoxide then attacks the aromatic carbon bearing the fluorine ortho to the sulfonyl group. Intramolecular nucleophilic aromatic substitution expels fluoride and closes a seven-membered heterocycle with the path $\mathrm{Ar}-\mathrm{S}-\mathrm{N}-{\mathrm{C}}_1-{\mathrm{C}}_2-\mathrm{O}-\mathrm{Ar}$. The para fluorine remains.

This tandem epoxide-opening/ring-closing pathway is required by the formula of F: simple epoxide addition would retain both fluorines, whereas F contains only one. In the second stage, pyrrolidine replaces the remaining para fluorine by another nucleophilic aromatic substitution, producing G. The mass balance is F plus pyrrolidine minus HF, as required by the given formula of G.

The meso epoxide has enantiotopic ends. With achiral reagents, opening at either end gives an enantiomeric product, so F is a racemic pair with the same trans relative configuration; G retains that configuration.

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.

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.

Build the phenethyl grignard reagent, use one portion to make a phenethyl-substituted ester, and add two further equivalents to that ester. Write $R=\mathrm{PhC}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2$; the target is ${\mathrm{R}}_3\mathrm{COH}$, so all three carbon substituents must be phenethyl, not benzyl.

hydroboration followed by oxidation of styrene gives $\mathrm{PhC}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{OH}$: use $\mathrm{B}{\mathrm{H}}_3$ in a suitable ether, followed by alkaline hydrogen peroxide. The Anti-Markovnikov addition regiochemistry is essential; acid-catalysed Markovnikov addition of water would give the wrong alcohol. Convert the primary alcohol to $\mathrm{PhC}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{Br}$ with phosphorus tribromide, then form $\mathrm{RMgBr}$ with magnesium in dry ether. The organomagnesium reagent must be prepared and used without a protic solvent.

carboxylate one portion with carbon dioxide and acidify to obtain $\mathrm{RC}{\mathrm{O}}_2\mathrm{H}$, then esterify with ethanol under acid catalysis to give $\mathrm{RC}{\mathrm{O}}_2\mathrm{Et}$. Treat the isolated ester with at least two equivalents of another portion of $\mathrm{RMgBr}$ and then perform an aqueous acidic work-up:The first addition to the ester gives a ketone after ethoxide elimination; the second addition gives the tertiary alkoxide. Work-up supplies its OH proton. The extra central carbon comes from carbon dioxide, and each original styrene contributes one intact eight-carbon phenethyl unit. The resulting carbon count is $1+3\times 8=25$, matching the target skeleton.

B is 4-hydroxy-1-methylpyrrolidin-2-one; C is 1-methylpyrrolidin-3-ol. Their ring connectivities are shown below. Each ring remains five membered; the attached N-methyl group is retained.

For B, a B–H bond of sodium borohydride supplies a hydride equivalent to the ketone carbonyl carbon, and the carbonyl $\pi$ pair moves to oxygen. This gives an alkoxide, whose protonation during the reaction or workup gives the secondary alcohol. The lactam carbonyl remains. Converting one $\mathrm{C}=\mathrm{O}$ to $\mathrm{CH}(\mathrm{OH})$ adds two H atoms and gives ${\mathrm{C}}_5{\mathrm{H}}_9\mathrm{N}{\mathrm{O}}_2$.

For C, lithium aluminium hydride reduces that ketone in the same net way and also reduces the lactam. Coordination of the amide oxygen to aluminium assists hydride attack at its carbonyl carbon, giving a tetrahedral intermediate. The N lone pair then forms a C=N bond while the C–O bond breaks to an aluminium-bound oxygen group. A resulting iminium ion accepts a second hydride at carbon, yielding the cyclic tertiary amine. Aqueous workup protonates the ketone-derived alkoxide.

The lactam conversion replaces $\mathrm{N}-\mathrm{C}(=\mathrm{O})$ by $\mathrm{N}-\mathrm{C}{\mathrm{H}}_2$; its oxygen is removed rather than retained as another alcohol. Together with ketone reduction, the net change is $+4\mathrm{H}-\mathrm{O}$, giving ${\mathrm{C}}_5{\mathrm{H}}_{11}\mathrm{NO}$. No stereochemistry is specified: the new alcohol stereocentre is formed as a racemic mixture in an achiral environment.

Disconnect the tertiary alcohol at its bond to the alkenyl group. The required carbonyl partner is $\mathrm{MeCOCH}(\mathrm{Me})\mathrm{Et}$; the other partner is an E-configured $\mathrm{PhC}{\mathrm{H}}_2\mathrm{CH}=\mathrm{CH}$ organometallic reagent. This disconnection fixes the alkene geometry before the alcohol is made.

First form the ethyl-side enolate of butan-2-one under equilibrating conditions, then trap it with ethyl iodide and isolate the required monoalkylated ketone:The incoming group is Et: methylation would give a carbon skeleton one carbon too short. Because an unsymmetrical ketone can form two enolates and undergo repeated alkylation, regioisomer separation and controlled conversion are part of this step.

Independently perform a halo-Wittig reaction on phenylacetaldehyde using $\mathrm{P}{\mathrm{h}}_3\mathrm{P}=\mathrm{CHBr}$ and isolate E-$\mathrm{PhC}{\mathrm{H}}_2\mathrm{CH}=\mathrm{CHBr}$ from any geometrical mixture. magnesium insertion in dry ether gives the corresponding vinylmagnesium halide; add the isolated ketone at low temperature and protonate the resulting alkoxide. Addition to C=O preserves the pre-existing C=C geometry:The E alkene and the complete fifteen-carbon skeleton are obtained. No absolute configuration is specified at the newly formed alcohol centre, so an unselected stereoisomeric mixture there is acceptable. An alternative carbon–carbon disconnection is acetylide addition to the same ketone followed by a controlled trans-selective alkyne reduction; the vinyl-reagent route avoids relying on reduction of a tertiary propargylic alcohol.

The first equivalent of titanocene methylidene methylenates the ester: the carbonyl oxygen and titanium form an oxatitanacyclobutane, whose fragmentation gives $\mathrm{PhC}(\mathrm{OMe})=\mathrm{C}{\mathrm{H}}_2$ and a titanium–oxygen product. The second equivalent adds across this electron-rich double bond. The initial adduct is a titanacyclobutane with cyclic connectivity $\mathrm{C}{\mathrm{p}}_2\mathrm{Ti}-\mathrm{C}{\mathrm{H}}_2-\mathrm{C}(\mathrm{Ph})(\mathrm{OMe})-\mathrm{C}{\mathrm{H}}_2-$, closing from the last methylene to titanium. The ester's phenyl-substituted carbon lies opposite Ti; the two new methylenes originate from the two methylidene equivalents.

Beta-methoxy elimination opens that metallacycle to B, $\mathrm{C}{\mathrm{p}}_2\mathrm{Ti}(\mathrm{OMe})[\mathrm{C}{\mathrm{H}}_2\mathrm{C}(\mathrm{Ph})=\mathrm{C}{\mathrm{H}}_2]$, the allyltitanium reagent. The C–OMe bond breaks, OMe transfers to Ti, one Ti–C bond becomes the new C=C bond, and the other Ti–C bond remains as the allylic nucleophile. The cyclic adduct is therefore a precursor to the opened reagent B; treating B as merely the one-equivalent enol ether loses the second reagent carbon.

The aldehyde oxygen coordinates to titanium. Allylic carbon transfer to its carbonyl carbon, accompanied by movement of the C=O pi electrons to oxygen, gives the titanium alkoxide $\mathrm{PhC}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{CH}(\mathrm{O}-\mathrm{Ti})\mathrm{C}{\mathrm{H}}_2\mathrm{C}(\mathrm{Ph})=\mathrm{C}{\mathrm{H}}_2$. Acidic aqueous workup cleaves O–Ti and protonates oxygen, giving $\mathrm{PhC}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{CH}(\mathrm{OH})\mathrm{C}{\mathrm{H}}_2\mathrm{C}(\mathrm{Ph})=\mathrm{C}{\mathrm{H}}_2$. A six-membered allyl-transfer arrangement is a convenient mechanistic model. No enantioselectivity is imposed by these achiral reagents.

Let $\mathrm{Ar}=2,6$-dimethylphenyl and $\mathrm{A}{\mathrm{r}}^{\prime }=2$-methoxyphenyl. After conversion of the Pd(OAc)${}_2$ precatalyst into an active phosphine–Pd(0) species, oxidative addition of $\mathrm{A}{\mathrm{r}}^{\prime }\mathrm{Cl}$ gives ${\mathrm{L}}_{\mathrm{n}}\mathrm{P}{\mathrm{d}}^{\mathrm{II}}(\mathrm{A}{\mathrm{r}}^{\prime })\mathrm{Cl}$. Carbonate-assisted replacement of chloride by the alcohol's alkoxide produces ${\mathrm{L}}_{\mathrm{n}}\mathrm{P}{\mathrm{d}}^{\mathrm{II}}(\mathrm{A}{\mathrm{r}}^{\prime })[\mathrm{OC}(\mathrm{Me}{)}_2\mathrm{Ar}]$.

In the key beta-carbon elimination, the alkoxide oxygen forms a C=O bond while the bond between its carbinol carbon and $\mathrm{Ar}$ cleaves, transferring $\mathrm{Ar}$ to palladium. This gives acetone and a diarylpalladium(II) complex, ${\mathrm{L}}_{\mathrm{n}}\mathrm{P}{\mathrm{d}}^{\mathrm{II}}(\mathrm{A}{\mathrm{r}}^{\prime })(\mathrm{Ar})$. reductive elimination creates the new biaryl bond and regenerates Pd(0). Thus the tertiary alcohol functions as an aryl donor by carbon–carbon bond cleavage, rather than being incorporated as a carbinol substituent. The overall organic products are the cross-coupled biaryl and acetone. The carbinol carbon has no hydrogen, so ordinary beta-hydride elimination from this alkoxide cannot explain the observed aryl transfer. The bulky electron-rich phosphine helps access the aryl-chloride oxidative addition and the sterically congested biaryl-forming elimination.