Metal-Carbon σ-bonds Flashcards
What are the 4 ways to make a Metal-Carbon σ-bonds?
1) Salt Elimination
2) Oxidative addition
3) Alkene Insertion
4) Ortho-metallation
Describe oxidative addition to form Metal-Carbon σ-bonds
What groups are added?
How does the geometry change?
- Low-valance complexes (e.g. Ir(I), Ni(0), Pd(0), and Pt(0) ), which are stabilised by phosphines undergo oxidative additions with alkyl, benzyl and aryl halides
- Classically is a d⁸ square planar complex that does oxidative addition to give an octahdedral complex
The alkylating strength of the main group alkyls in salt eliminations decreases with….
Increasing covalent character of the M-C bond (less covalent = less reactive)
E.g. MeLi is more polarised than AlMe₂Cl and hence will methylate 4 times compared to once for AlMe₂Cl
Describe Alkene insertion to form Metal-Carbon σ-bonds
- Hydride or an alcohol chain inserting onto a coordinated alkene to give you an alcohol chain on the metal centre
- (Crucial step in the polymersation of alkenes)
Describe Ortho-methylation to form Metal-Carbon σ-bonds
- Involves the cleavage of the ortho-C-H bonds of an aryl (usually phenyl)
- (it is a common deactivating pathway of noble metal phosphine catalysts)
Metal-Carbon σ-bonds are useful to then undergo other types of chemsitry
Why would you want to cleavage the newly formed M-C bond?
- To form new bonds with electrophiles e.g. halogen
- hydrogenolysis (forming C-H bonds)
Metal-Carbon σ-bonds are useful to undergo other types of chemistry
Why would you want to do a M-C bond insertion
- React with CO to add carbonyl groups between the M-C bond
- React with alkenes, to extend the carbon chain, between the M-C
- React with alkynes to add a C=C between the M-C
True or false:
We can eliminate metal hydrides?
False
Metal hydrides are similar to small alkyls which we cannot elimated
How can we make metal-hydride complexes?
- Beta-hydride elimination
- Hydrogenolysis
- Oxidative addition of H₂
What two types of bonding can occur between metal hydrides?
- The hydrogen has a filled sigma bonding orbital and can donate electron density to the metal centre (σ-donation)
- Also if metal has filled metal d-orbitals and hydrogen has empty σ’ orbital, there is π-back donation
- However the π-back donation will weaken the H-H bond untill is breaks - also known as oxidative addition
What makes a Schrock Carbene?
(metal-carbene complex)
Hint: metal type, carbon properties, ligand types etc
- Schrock Carbenes have early transition metals which are electron poor and high in oxidation state
- their carbons are nucleophilic
- their ligands are π-donors
- They have two unpaired electrons in different orbitals (called a triplet binding)
How does the bonding occur in Schrock Carbenes?
- The carbon is acting as a two electron donor: ne electron through the sigma systen and one electron through the pi system
- In total 4 electrons are involved in the bonding between the metal and carbon
What makes a Fischer Carbene complex?
(metal-carbene complex)
Hint: metal types, carbon properties, ligand types
- Fischer Carbenes tend to be formed with late transition metals that are electron rich and hence a low oxidation state
- Carbon is electrophilic (neutral ligands resulting in a δ⁺ carbon centre as metal is more δ⁻)
- Ligands are π-acceptors
- Ligands will donate their LP from p-orbitals into carbons empty p-orbitals, meaning carbon electrons for the C-M bond are actually paired, called a singlet binding
How does the bonding occur in Fischer Carbenes?
- sp² hybrided carbon
- The LP on the carbon will donte to the metal
- The metal can then back-donate a pair of electron from a d-orbital into an empty p-orbital on carbon
How do you make Schrock Carbenes?
Explain how this takes place with the following Tantalum (V) species
- α-hydrogen abstraction
- a hydride from one of the ligands, transfers and attacks the other ligands
- This breaks one Ta-C bond and allows a new Ta=C bond to form
- Eliminating tBuCH₃
Why can’t this Tantalum (V) penta-alkyl species undergo beta hydride elimination
It has 5 alkyl groups on it which do not have a beta-hydrogen
Instead all have a tert-butyl group in that poisition
How do you make Fischer Carbenes?
E.g. how would it occur with this Chromium (V) species
- Usually from metal hexa-carbonyl species reacting with a nucleophile
- Where the nucleophile attacks one of the C=O ligands
- Then we alkylate (e.g. triethyloxonium)
What is a big advantage of using Schrock Carbenes
- It is good at transferring CR₂ groups onto other things
- E.g. reacting this Tantalum Schrock with a ketone we can form this alkene species
If we cant to form a titanium carbene, how can we overcome the issue of high reactivity?
- By using Tebbe’s reagents
- It is a methylene-transfer reagent (an alternative to a Wittig reagent), and it can be regarded as a protected form of [Cp₂Ti=CH₂] which is unstable when unprotected
All Fischer Carbenes will generally have an oxygen substitued on
How could we remove this?
Using an Amine
How does a cyclopropanation of a Fischer Carbene occur?
- A [2+2] cycloaddition reaction of a carbene with an alkenes give a metallacyclobutane (reversible)
What is a N-Heterocyclic carbene (NHCs)?
- Relatively stable carbene with a lone pair
- Stabilised through the nitrogen atom(s)
In terms of orbitals, how is a N-Heterocylic carbene stabilised?
Nitrogen lone pairs can donate into the p-orbital on carbon, to stabilise it
Making it favourable for the 2E- on carbon to pair up in the sp² lobe (donate to a metal)
singlet diamagnetic ground state
(Note the R groups can also help stabilise due to being bulky and electron donating)
Describe these 4 various binding modes of NHCs which are recognised:
(a) typical model of backbonding in phosphine transition metal complex
(b) Singlet lone pair binds to the transition metal only
(c) Singlet long pair binds to transition metal which back bonds
(d) Electron deficient metal receives lone pair and π-donation
(Depending on the properties of the metal, the NHC can act as either a π-donor ligand or a π-acceptor ligand)
What does a Metal-Carbyne Complex look like?
- The bonding is covalent and results in partial charges: LₙMδ⁺ ← Cδ⁻R
How would you forms a Schrock Carbynes?
- To start off with a Schrock Carbynes and then deprotonate it
- (e.g. this is done there with an alkyl lithium)
How would you form a Fischer Carbyne?
- Starting with a Fischer Carbene, to treat it with a strong Lewis acid BX₃
- It will react with the heteroatom Lone Pair, to remove the heteroatom and a new metal-carbon triple bond is formed
