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Química Organometálica e Catálise · Organometallic fundamentals

Etapas elementares das reações organometálicas

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O que você vai aprender

  • Track the metal's oxidation state, d-electron count and total electron count through oxidative addition, reductive elimination, migratory insertion, β-hydride elimination, ligand dissociation and transmetalation.
  • Draw curved arrows for each elementary step with the metal as a bookkeeping atom, and distinguish concerted, SN2-type and radical oxidative additions from their evidence.
  • State the geometric and electronic requirements of each step (cis ligands, a vacant site, a β-hydrogen, d electrons to give) and use them to predict products.
  • Assemble elementary steps into catalytic cycles such as alkene hydrogenation, hydroformylation and cross-coupling, and spot impossible steps in a proposed cycle.

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Keep score at the metal

Every organometallic step is followed by three numbers. The oxidation state is the charge left on the metal after every ligand is removed with its bonding pair: hydride, alkyl, aryl, acyl and halide leave as anions (X ligands, −1 each), while phosphines, CO and alkenes leave as neutral molecules (L ligands, 0). The d-electron count is the group number minus the oxidation state: Pd(II) in group 10 is d⁸, Rh(I) in group 9 is d⁸, Ir(III) is d⁶. The total electron count adds 2 for every L or X ligand to the d count. In the mechanisms here the metal is drawn bare, without its spectator phosphines or CO, as a bookkeeping atom: the lone pairs drawn on it are only the pairs that the arrows need, not its d electrons. The numbers above are what tell you what the metal is doing. Bromo(phenyl)palladium, shown, stands for trans-PdBr(Ph)(PPh₃)₂: Pd(II), d⁸, 16 electrons.

Oxidative addition

In oxidative addition the metal breaks an A–B bond and bonds to both A and B. Its oxidation state and coordination number each rise by 2, its d count falls by 2 and its electron count rises by 2, so the metal must have d electrons to give and room for two more ligands: Pd(0) d¹⁰, Rh(I) and Ir(I) d⁸ do it readily, while a d⁰ metal such as Ti(IV) cannot. H–H, C–H and aryl C–X bonds add in a concerted, three-centre step that puts A and B cis, as when Vaska's complex IrCl(CO)(PPh₃)₂ takes up H₂. Polar alkyl halides such as iodomethane add in two steps: the metal's lone pair attacks carbon in an SN2 displacement, and the halide then binds. Like an organic SN2 reaction, this path is second order, fastest for methyl and much slower for secondary halides, and inverts a stereogenic carbon; with a neutral metal complex it is also faster in polar solvents. Some alkyl halides add through one-electron steps and alkyl radicals instead, which reveal themselves by racemisation or by radical-clock rearrangement such as hex-5-enyl to cyclopentylmethyl. For aryl halides with Pd(0) the rate rises as the C–X bond weakens: Ar–I > Ar–Br > Ar–Cl.

Reductive elimination

Reductive elimination is the reverse of oxidative addition: two cis X ligands couple to form an A–B bond that leaves the metal. The oxidation state and coordination number each fall by 2, the d count rises by 2 and the electron count falls by 2. The two groups must be cis; a trans complex has to isomerise first. Elimination is fastest when the metal is electron-poor or crowded, so bulky, weakly donating or wide-bite-angle ligands speed it, and a higher oxidation state such as Pt(IV) eliminates more easily than Pt(II). This is the product-forming step of cross-coupling: cis-Pd(Ph)(CH₃)L₂ releases toluene and returns Pd(0).

Migratory insertion

In a migratory insertion an X ligand moves onto a neighbouring unsaturated L ligand. CO undergoes 1,1-insertion: the alkyl migrates to the carbonyl carbon and gives an acyl, M–C(=O)R. An alkene undergoes 1,2-insertion: the hydride or alkyl moves to one alkene carbon while the metal bonds to the other, through a four-centre transition state that adds M and H to the same face (syn). The oxidation state does not change, because one X ligand (alkyl) becomes another X ligand (acyl or a longer alkyl); the electron count falls by 2 and a vacant site opens, which is why an incoming ligand such as CO often traps the product. Both groups must be cis. When a metal hydride adds to a terminal alkene, the metal usually ends on the terminal carbon, giving the linear alkyl; when an aryl or alkyl group migrates, as in the Heck reaction, that group goes to the terminal carbon and the metal to the internal one.

β-Hydride elimination

β-Hydride elimination is the reverse of 1,2-insertion: a hydrogen on the carbon β to the metal moves to the metal, and the metal–carbon bond becomes a C=C bond. It needs a β-hydrogen, a vacant site cis to the alkyl, and a syn-coplanar M–Cα–Cβ–H arrangement. The oxidation state is unchanged and the electron count rises by 2. Alkyls without β-hydrogens (methyl, neopentyl, benzyl, CH₂SiMe₃) are therefore robust, and metallacycles that cannot rotate the β-hydrogen into the syn-coplanar position decompose far more slowly than open-chain alkyls. In cyclic systems the syn requirement decides which alkene forms.

Ligand dissociation and transmetalation

An 18-electron complex cannot add another ligand without becoming a 20-electron species, so it substitutes by first losing a ligand: the rate is first order in complex and independent of the incoming ligand, as for Ni(CO)₄ with PPh₃. When a pre-equilibrium dissociation precedes a slow step, added free ligand slows the reaction and the rate falls as 1/[L]. Wilkinson's catalyst, RhCl(PPh₃)₃, enters its cycle by losing PPh₃ to form 14-electron RhCl(PPh₃)₂. Transmetalation moves an organic group from an electropositive main-group element (Mg, Zn, Sn, B) to the transition metal in exchange for a halide; the oxidation state is unchanged. The more polar the C–M bond of the donor, the more carbanion-like and nucleophilic its carbon: C–Mg > C–Zn > C–Sn > C–B. Grignard reagents transfer readily but also attack carbonyl groups, while the nearly covalent C–B bond of a boronic acid transfers only with base activation, either as a boronate or through a hydroxo- or alkoxopalladium complex.

What to carry forward

Name each step by what happens at the metal: oxidative addition raises the oxidation state by 2, reductive elimination lowers it by 2, and migratory insertion, β-hydride elimination, ligand dissociation and transmetalation leave it unchanged. Check the requirements (d electrons, a vacant site, cis partners, a β-hydrogen) before drawing any arrow, and a catalytic cycle must return the metal to the oxidation state it started with.

Problemas de prática (40)

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Desenhe o mecanismo 12

Ligand dissociation and transmetalation

  • Kumada coupling of chlorobenzene with methylmagnesium bromideDraw oxidative addition, transmetalation from magnesium and reductive elimination at nickel, following nickel's oxidation state around the cycle.

Migratory insertion and β-hydride elimination

  • β-Hydride elimination from (2-phenylethyl)palladium(II)Draw the syn β-hydride elimination that turns a (2-phenylethyl)palladium complex into styrene and a palladium hydride, and state its requirements.
  • 1,1-Migratory insertion of CO into a Pd–CH₃ bondDraw CO coordination and the 1,1-migratory insertion that turns a methylpalladium complex into an acetylpalladium complex, tracking the oxidation state and electron count.
  • 1,2-Insertion of ethylene into a Pt–H bondDraw the four-centre 1,2-migratory insertion of ethylene into a platinum–hydride bond and show that the oxidation state does not change.

Oxidative addition

  • Oxidative addition of bromobenzene to palladium(0)Draw the concerted oxidative addition of bromobenzene to Pd(0) and follow the oxidation state, d count and electron count of palladium.
  • SN2-type oxidative addition of iodomethane to rhodium(I)Draw the two-step SN2-type oxidative addition of iodomethane to the Monsanto rhodium(I) anion and track the oxidation state and electron count.
  • Oxidative addition of H₂ to Vaska's complexDraw the concerted oxidative addition of dihydrogen to Ir(I) and follow the iridium oxidation state, d count and electron count.

Reductive elimination

  • Reductive elimination of toluene from palladium(II)Draw the concerted C–C reductive elimination from a cis methyl(phenyl)palladium(II) complex and show that palladium(0) is regenerated.
  • Reductive elimination of ethane from platinum(IV)Draw C–C reductive elimination from a trimethylplatinum(IV) complex and explain why ethane, not iodomethane, is eliminated.

Steps in sequence

  • Decomposition of dibutylbis(triphenylphosphine)platinum(II)Draw the phosphine dissociation, β-hydride elimination and C–H reductive elimination that convert a dibutylplatinum(II) complex into but-1-ene and butane, and explain why added phosphine slows it.
  • Rhodium-catalysed hydroformylation of propeneDraw the five elementary steps of rhodium hydroformylation (alkene insertion, CO binding, CO insertion, H₂ oxidative addition, reductive elimination) that convert propene into butanal.
  • Hydrogenation of cyclohexene with Wilkinson's catalystDraw H₂ oxidative addition, 1,2-insertion of the alkene into Rh–H and C–H reductive elimination, and track rhodium between Rh(I) and Rh(III).

Preveja o produto 4

Reductive elimination

  • Product of an aryl–aryl reductive eliminationPredict the organic product released when two different cis aryl groups reductively eliminate from palladium(II).

Oxidative addition

  • Stereochemistry of an SN2-type oxidative additionPredict the configuration at carbon after an SN2-type oxidative addition of a stereogenic benzylic chloride to palladium(0).

Migratory insertion and β-hydride elimination

  • Syn β-hydride elimination in a ringPredict the alkene released by syn β-hydride elimination from a cis-2-phenylcyclopentylpalladium intermediate.

Ligand dissociation and transmetalation

  • What does transmetalation make?Predict the palladium complex formed when an organozinc reagent transmetalates to an arylpalladium(II) halide.

Classificar 4

Migratory insertion and β-hydride elimination

  • Which product comes from a 1,2-insertion?Classify migratory insertions as 1,1 or 1,2 from the product formed.

Oxidative addition

  • Which metal cannot do an oxidative addition?Identify the complex that cannot undergo oxidative addition because it has no d electrons to give.
  • What oxidative addition does to the metalClassify the changes in oxidation state, coordination number, d count and electron count caused by an oxidative addition.

Ligand dissociation and transmetalation

  • Which step leaves the oxidation state unchanged?Classify elementary steps by whether they change the metal's oxidation state.

Comparar 3

Oxidative addition

  • Rank aryl halides for oxidative addition to Pd(0)Order aryl halides by their rate of oxidative addition to palladium(0) using carbon–halogen bond strengths.

Migratory insertion and β-hydride elimination

  • Which alkyl survives β-hydride elimination?Compare alkyl ligands and identify the one that cannot undergo β-hydride elimination.

Ligand dissociation and transmetalation

  • Rank C–M bond polarity of transmetalating agentsOrder organometallic transmetalating agents by the polarity of their carbon–metal bond from electronegativity differences.

Calcular 7

Migratory insertion and β-hydride elimination

  • Oxidation state after CO insertionCalculate the manganese oxidation state in an acylmanganese complex and see that migratory insertion leaves it unchanged.
  • Electron count after an alkene insertionCalculate the electron count of a metal complex after 1,2-insertion of a coordinated alkene into a metal–hydride bond.

Steps in sequence

  • d count in the hydroformylation acyl dihydrideCalculate the d-electron count of rhodium in the acyl dihydride intermediate of hydroformylation.

Ligand dissociation and transmetalation

  • Rate inhibition by added phosphineCalculate how an observed rate constant falls with free-ligand concentration when a pre-equilibrium dissociation precedes the slow step.
  • Count the species after phosphine lossCalculate the electron count of the three-coordinate rhodium species formed when Wilkinson's catalyst loses a phosphine.

Oxidative addition

  • d count of rhodium after iodomethane additionCalculate the d-electron count of rhodium in the Monsanto oxidative-addition product from its ligands and overall charge.

Reductive elimination

  • Electron-count change in reductive eliminationCalculate the change in total valence electron count when ethane reductively eliminates from a platinum(IV) complex.

Encontre o erro 2

Steps in sequence

  • Find the flaw in a proposed Heck cycleDiagnose a catalytic cycle that fails to return palladium to its starting oxidation state.

Reductive elimination

  • Find the bookkeeping errorDiagnose a mislabelled oxidation-state and electron-count change in a proposed C–H reductive elimination.

Interprete as evidências 3

Migratory insertion and β-hydride elimination

  • Where does the labelled CO go?Interpret a ¹³CO labelling result to locate the origin of the acyl carbonyl in a CO insertion.

Ligand dissociation and transmetalation

  • Reading a substitution rate lawInterpret a rate law that is independent of the incoming ligand as evidence for dissociative substitution.

Oxidative addition

  • A radical clock for oxidative additionInterpret a radical-clock product to decide whether an alkyl halide added to a metal through a free radical.

Planeje uma síntese 2

Reductive elimination

  • Speed up a slow reductive eliminationChoose a ligand change that accelerates C–C reductive elimination from a palladium(II) diaryl complex.

Steps in sequence

  • Order the steps of Wilkinson hydrogenationSequence the elementary steps of alkene hydrogenation by Wilkinson's catalyst along the hydride route.

Explique por quê 3

Ligand dissociation and transmetalation

  • Why 18-electron complexes lose a ligand firstExplain from the electron count why substitution at an 18-electron complex begins with ligand dissociation.

Migratory insertion and β-hydride elimination

  • Why is the platinacycle so stable?Explain why a metallacyclopentane resists β-hydride elimination compared with its open-chain dialkyl analogue.

Reductive elimination

  • A trans-locked complex that will not eliminateExplain why two groups held trans on a square-planar metal cannot reductively eliminate.

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