Introduction to Palladium Catalysis 1 Flashcards
Why are palladium-catalysed cross-coupling reactions really useful?
These reactions are powerful transformations for C-C bond formation
The reaction shown below is a type of Pd cross-coupling reaction called the Suzuki reaction
What is the driving force for this reaction
Thermodynaics of strong C-C bond
This a more expanded diagram of cross-coupling reactions
Why is Palladium beneficial for these types of reactions?
Because along with other transition metals it can change its oxidation state
i.e. go from Pd(0) to Pd(II) and then back to Pd(0)
What are the names of the core steps within this reaction?
The driving force for most Pd catalysed cross-coupling reactions is thermodynamics, i.e. the formation of stronger bonds from weaker bonds
How strong are C-C/C-H, C-X/C-M and M-X bonds relative to another?
- C-C and C-H bonds are strong
- C-X and C-M bonds are weak
- M-X bonds are strong
Why can we no add metallic palladium into a reaction?
- Ligands are required to stabilise the metal catalyst
- The nature of the ligands affects the oxidation state and number of electrons of a metal complex and therefore its reactivity
What is the formal charge and electrons donated by the following ligands:
- Formal charge: -1
- Electrons donated: 2
What is the formal charge and electrons donated by the following ligands:
- Formal charge: 0
- Electrons donated: 2
What is the formal charge and electrons donated by the following ligands:
- Formal charge: -1
- Electrons donated: 2
What is the formal charge and electrons donated by the following ligands:
Formal charge: 0
Electrons donated: 2
What is the oxidation state and number of electrons in this palladium complex
Pd = 10
PPh₃ = 4x0
Overall charge = 0 = Pd(0) = d¹⁰
Electrons = 10 + (4x2) = 18e⁻
Coordinately saturated
What is the oxidation state and number of electrons in this palladium complex?
Pd = 10
PPh₃ = 2x0
Cl⁻ = 2x-1
Overall charge = -2 = Pd(II) = d⁸
Electrons = 8 + (4x2) = 16e⁻
Coordinately saturated
Will these two palladium complexes react similarly
These palladium complexes will react differently to another
The 18- and 16- electron complexes are coordinately saturated and unreactive -
reactive unsaturated 14e complexes are formed from these complexes by ligand dissociation in solution
What is the oxidation state and number of electrons in this palladium complex?
Pd = 10
PPh₃ = 2x0
MeCN = 2x0
Overall charge = -0 = Pd(0) = d¹⁰
Electrons = 10 + (4x2) = 18e⁻
Coordinately saturated
What is the main chemical difference between 18e- and 16e- complexes
- Pd(0) complexes d¹⁰ are electron-rich and highly nucleophilic
- Pd(II) complexes d⁸ are electron-poor and highly electrophilic
- This is the benefit of palladium as we can switch from an electrophilic to a nucleophilic complex, allowing for several steps in a catalytic cycle
What is the first step in this catalytic cycle
The first step in a catalytic cycle is usually formation of the reactive catalyst from an unreactive pre-catalyst
A reactive Pd(0) catalyst can be generated from Pd(II) pre-castalysts
One way to do this is through the reduction of LₙPdX₂ (X = halide/pseudohalide) with an organometallic
How does this reaction occur
(M-R is often the coupling partner in cross-coupling reactions, hence present in excess)
A reactive Pd(0) catalyst can be generated from Pd(II) pre-castalysts
One way to do this is through reduction of Pd(OAc)₂ with phosphines
What does this involve
Replacing 2xOAc groups with 2xPPh₃
One of the first steps once the reactive catalyst is made is an oxidative addition
How does this reaction occur?
- The reaction requires 2 free coordination sites on Pd
- Selective cis-addition of X and R to the metal centre (concerted reaction)
- The metal is oxidised from Pd(0) to (Pd(II), hence the name oxidative addition
