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Química Organometálica e Catálise · Catalytic cycles

Acoplamento cruzado com paládio e reação de Heck

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O que você vai aprender

  • Draw the oxidative addition, transmetalation and reductive elimination steps of Suzuki, Stille, Negishi, Kumada and Sonogashira couplings with exact oxidation states and electron counts.
  • Explain the role of the base in Suzuki couplings, of copper(I) in the Sonogashira reaction and of bulky phosphines in couplings of aryl chlorides.
  • Predict Heck products from syn migratory insertion and syn β-hydride elimination, including regiochemistry and alkene geometry.
  • Draw the Buchwald–Hartwig C–N coupling cycle and choose the coupling reaction and partners that make a given bond.

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One cycle, three steps

Almost every palladium cross-coupling runs the same cycle. The active catalyst is a Pd(0) complex, usually PdL₂ (L = a phosphine): palladium is d¹⁰ and the complex has 10 + 2 × 2 = 14 valence electrons. Oxidative addition of an aryl halide Ar–X gives ArPd(X)L₂: palladium gains two anionic ligands, so its oxidation state rises from 0 to +2 (d⁸), and the count rises to 16 electrons. Transmetalation then swaps X for the carbon group R of an organometallic reagent R–M, leaving Pd(II) and 16 electrons. Finally reductive elimination joins Ar and R in a new C–C bond and returns palladium to Pd(0), ready for the next aryl halide. In the drawings here palladium is shown bare, without its phosphines, as a bookkeeping atom; phenylpalladium(II) bromide is shown.

Oxidative addition decides which halide reacts

Oxidative addition is easiest for the weakest carbon–halogen bond. Aryl C–X bond energies are about 280 kJ/mol for C–I, 340 kJ/mol for C–Br, 400 kJ/mol for C–Cl and 530 kJ/mol for C–F, so aryl iodides react fastest, bromides next and chlorides slowest; aryl fluorides are essentially inert. In a molecule carrying two different halogens, such as 1-bromo-4-chlorobenzene (shown), palladium inserts selectively into the C–Br bond. For aryl chlorides oxidative addition is usually the slow step, and bulky, electron-rich phosphines such as P(t-Bu)₃ or SPhos are used: their electron donation makes palladium a better nucleophile toward the C–Cl bond, and their bulk favours the very reactive monoligated PdL species.

Transmetalation partners name the reaction

The organometallic partner names the coupling: an organoboron compound in the Suzuki–Miyaura reaction, an organostannane in the Stille reaction, an organozinc in the Negishi reaction and a Grignard reagent in the Kumada reaction (with nickel or palladium). Boronic acids are neutral and trigonal; a base (hydroxide, carbonate, phosphate) adds to boron and gives an anionic tetracoordinate boronate (shown) whose aryl group transfers to palladium. Stannanes transfer without base but leave toxic tin by-products. Grignard and organozinc reagents are the most reactive partners, but Grignard reagents attack esters, ketones and nitriles, so Kumada couplings tolerate the fewest functional groups. In the Sonogashira reaction a terminal alkyne, an amine base and catalytic copper(I) iodide form a copper acetylide, which transfers the alkynyl group to palladium.

Reductive elimination needs cis groups

The two carbon groups must be cis on palladium to bond to each other. Transmetalation often gives a trans complex, which isomerises to cis before elimination. Reductive elimination lowers the oxidation state by two (Pd(II) → Pd(0)), the coordination number by two and the electron count by two. Bulky ligands crowd the cis groups together and speed this step. A Pd(II) salt such as Pd(OAc)₂ is only a precatalyst: it must first be reduced to Pd(0) before any oxidative addition can happen.

The Heck reaction: insertion and β-hydride elimination

In the Heck reaction the arylpalladium(II) halide meets an alkene instead of an organometallic reagent. The alkene coordinates and inserts into the Pd–Ar bond in a syn fashion: the aryl group goes to the less hindered alkene carbon, which is the terminal CH₂ of acrylates and styrenes, and palladium to the other carbon. After rotation about the new C–C single bond, a β-hydrogen syn to palladium is removed in a syn β-hydride elimination, releasing the substituted alkene, mostly as the E isomer, and H–Pd(II)–X. The E isomer wins because the Pd–C and C–H bonds must be eclipsed for this elimination, and the conformer that gives the Z isomer would eclipse the two large groups with each other. A base such as Et₃N removes HX and returns palladium to Pd(0). Methyl acrylate and iodobenzene give methyl (E)-cinnamate (shown). In rings, where the C–C bond cannot rotate, the hydrogen removed must be syn to palladium, so the new double bond forms away from the aryl-bearing carbon. With electron-rich alkenes such as enol ethers the regiochemistry can reverse: under cationic conditions (aryl triflates with a chelating diphosphine) the aryl group goes to the oxygen-bearing carbon.

Buchwald–Hartwig amination

The same cycle forms C–N bonds. After oxidative addition the amine binds palladium in place of the halide, a strong base such as NaOt-Bu removes its N–H proton to give a palladium amido complex, and C–N reductive elimination releases the aryl amine and Pd(0). Morpholine and bromobenzene give 4-phenylmorpholine (shown). An amido group with a hydrogen on the carbon next to nitrogen can instead undergo β-hydride elimination, giving an imine and, after reductive elimination of Ar–H, the reduced arene; bulky biaryl phosphines speed the C–N reductive elimination so that it wins.

What to carry forward

Track palladium through every step: Pd(0) and 14 electrons for PdL₂, Pd(II) and 16 electrons after oxidative addition and transmetalation, Pd(0) again after reductive elimination. The weakest C–X bond adds first, the organometallic partner names the reaction, and in the Heck reaction syn insertion followed by syn β-hydride elimination sets the regiochemistry and the E geometry.

Problemas de prática (40)

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Desenhe o mecanismo 15

Buchwald–Hartwig and planning

  • Buchwald–Hartwig amination of 4-chlorotoluene with anilineDraw the Buchwald–Hartwig coupling of an aryl chloride with aniline and explain why a bulky, electron-rich biaryl phosphine is needed.
  • Buchwald–Hartwig amination of bromobenzene with morpholineDraw the Buchwald–Hartwig coupling of bromobenzene and morpholine: oxidative addition, amine binding, deprotonation by tert-butoxide and C–N reductive elimination.

The Heck reaction

  • Heck reaction of iodobenzene with allyl alcoholDraw the Heck reaction of iodobenzene with allyl alcohol in which β-hydride elimination toward the CH₂OH carbon gives an enol that tautomerises to 3-phenylpropanal.
  • Heck reaction of 4-bromobenzonitrile with styreneDraw the Heck reaction of 4-bromobenzonitrile with styrene and explain why the E-stilbene forms.
  • Heck reaction of iodobenzene with methyl acrylateDraw the Heck reaction of iodobenzene with methyl acrylate: oxidative addition, syn migratory insertion with aryl delivery to the terminal carbon, syn β-hydride elimination to the E-alkene, and base-promoted regeneration of Pd(0).

Kumada and Sonogashira

  • Nickel-catalysed Kumada coupling of 2-bromothiophene with 2-thienylmagnesium bromideDraw the nickel-catalysed Kumada coupling that joins two thiophene rings into 2,2′-bithiophene through Ni(0) and Ni(II).
  • Kumada coupling of (Z)-1-bromoprop-1-ene with phenylmagnesium bromideDraw the palladium-catalysed Kumada coupling of a (Z)-alkenyl bromide and show that the alkene geometry is retained.
  • Selective Sonogashira coupling of 1-bromo-4-iodobenzene with trimethylsilylacetyleneDraw the Sonogashira coupling that substitutes only the iodine of 1-bromo-4-iodobenzene, and explain the halide selectivity.
  • Sonogashira coupling of iodobenzene with phenylacetyleneDraw the Sonogashira coupling of iodobenzene and phenylacetylene, including the formation of copper(I) phenylacetylide and its transfer to palladium.

Suzuki, Stille and Negishi

  • Negishi coupling of methyl 4-iodobenzoate with ethylzinc bromideDraw the Negishi coupling of an aryl iodide with a primary alkylzinc reagent and explain how reductive elimination outruns β-hydride elimination.
  • Negishi coupling of 4-iodoacetophenone with benzylzinc bromideDraw the Negishi coupling of 4-iodoacetophenone with benzylzinc bromide, showing why a ketone survives an organozinc partner.
  • Stille coupling of iodobenzene with tributyl(vinyl)tinDraw the Stille cycle for iodobenzene and tributyl(vinyl)tin: oxidative addition, transfer of the vinyl group from tin to palladium, and reductive elimination of styrene.
  • Stille ketone synthesis from benzoyl chloride and tetramethyltinDraw the Stille coupling of benzoyl chloride with tetramethyltin: oxidative addition into the acyl C–Cl bond, methyl transfer from tin, and reductive elimination of acetophenone.
  • Suzuki coupling of bromobenzene with 4-methylphenylboronic acidDraw one turn of the Suzuki–Miyaura cycle: oxidative addition of bromobenzene to Pd(0), activation of the boronic acid by hydroxide, transmetalation, and reductive elimination of 4-methylbiphenyl.
  • Suzuki coupling through a palladium hydroxideDraw the Suzuki–Miyaura cycle in which hydroxide first replaces bromide on palladium and the Pd–OH oxygen then delivers the neutral boronic acid to palladium for transmetalation.

Preveja o produto 6

The Heck reaction

  • Heck arylation of an electron-rich alkenePredict the regiochemistry of a cationic Heck reaction with an enol ether and the product after hydrolysis.
  • Heck reaction on a ring alkenePredict the alkene formed by syn β-hydride elimination after syn insertion into a cyclic alkene.
  • Predict a Heck productPredict the regiochemistry and alkene geometry of a Heck reaction with an acrylate ester.

Suzuki, Stille and Negishi

  • Which halogen couples?Predict which carbon–halogen bond of a dihaloarene reacts in a Suzuki coupling.
  • Predict a Suzuki productPredict the cross-coupled biaryl from an aryl bromide and an arylboronic acid.

The catalytic cycle

  • Put the Suzuki cycle in orderOrder the events of one Suzuki–Miyaura turnover starting from Pd(PPh₃)₄.

Classificar 2

Suzuki, Stille and Negishi

  • Name the couplingClassify a cross-coupling by the metal of its organometallic partner.

The catalytic cycle

  • What reductive elimination changesClassify the changes in oxidation state, coordination number and electron count during reductive elimination.

Comparar 1

  • Rank halobenzenes by oxidative addition rateRank aryl halides by their rate of oxidative addition to Pd(0) using C–X bond energies.

Calcular 5

The catalytic cycle

  • How much Pd(PPh₃)₄ to weigh outConvert a catalyst loading in mol% into the mass of precatalyst to weigh.
  • Electron count of the active Pd(0) catalystCount the valence electrons of the coordinatively unsaturated Pd(0) species formed by phosphine dissociation.
  • Electron count of the oxidative-addition productCount the valence electrons of a square-planar arylpalladium(II) halide bis(phosphine) complex.
  • Oxidation state after oxidative additionAssign the oxidation state of palladium in the arylpalladium halide formed by oxidative addition.
  • Turnover number of a Suzuki couplingCalculate a catalyst turnover number from the catalyst loading and the product yield.

Encontre o erro 2

The catalytic cycle

  • Find the error in a Suzuki cycleIdentify an oxidation-state error in a proposed Suzuki catalytic cycle.

Kumada and Sonogashira

  • A by-product from a leaky flaskIdentify the alkyne homocoupling by-product of a Sonogashira reaction run in air and its cause.

Trabalhe de trás para frente 2

Buchwald–Hartwig and planning

  • Choose the Buchwald–Hartwig partnersWork backwards from an N-aryl morpholine to the aryl halide and amine for a Buchwald–Hartwig coupling.
  • Choose the Heck partnersWork backwards from a cinnamate ester to the aryl halide and alkene for a Heck reaction.

Planeje uma síntese 2

Kumada and Sonogashira

  • Which coupling would fail?Choose the cross-coupling partner that is incompatible with an ester in the substrate.

Buchwald–Hartwig and planning

  • Choose the reaction for an aryl alkyneSelect the cross-coupling that forms an aryl–alkyne C(sp²)–C(sp) bond.

Explique por quê 5

Buchwald–Hartwig and planning

  • Why the arene was reducedExplain the arene reduction and imine by-products of a Buchwald–Hartwig amination with an acyclic secondary amine.

The catalytic cycle

  • Why trans complexes do not eliminateExplain why two carbon ligands must be cis on palladium for reductive elimination.
  • How Pd(OAc)₂ enters the cycleExplain how a Pd(II) precatalyst becomes the active Pd(0) catalyst in a Heck reaction.

The Heck reaction

  • Why the Heck reaction needs a baseExplain why the Heck reaction needs a stoichiometric base to be catalytic in palladium.

Suzuki, Stille and Negishi

  • Why process chemists avoid Stille couplingsExplain the main practical drawback of the Stille coupling compared with the Suzuki coupling.

Travou em acoplamento cruzado com paládio e reação de Heck? Escaneie o problema.

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