Organolanthanide Chemistry I Flashcards
What are the main features of organolanthanide chemistry
- Organolanthanide chemistry is not as extensive as organotransition metal chemistry
- Expanded enormously since 1980s, especially in catalysis and C-H Bond Activation studies
- Primarily ionic in their bonding due to the contracted nature of the 4f valence orbitals
- Lanthanides cannot act as π-bases and thus Ln-CO compounds are not stable
- Organolanthanides are extremely air and moisture sensitive reflecting the highly carbanionic character of the organic ligands and the oxophilicity of Ln2+/3+ .
Describe interaction of Lns with carbonyls
- 4f valence orbitals of Lns are too withdrawn to interact with the pi* orbitals of CO - Never been isolated
When can LnCOs be isolated
- Unstable carbonyls Ln(CO)n (n = 1 – 6) may be isolated by co-condensation of lanthanide vapours with CO in an argon matrix at temperatures< – 40oC.
- Makes a system which looks something like Pi-back donation from metal to C-O but very unstable
- None of these complexes have every been definitively characterised as they decompose instantly on raising the temperature
What happens to C-O stretching frequencies of the Ln(CO)n carbonyls as n increases
- The appearance of their IR spectra is surprisingly similar to those of transition metals, suggesting some level of pi-back donation.
- Stretching frequency increases linearly with n
- Suggesting competition for 4f (or 5d?) electron density
What causes stretching frequency to increase
- The amount of energy required to stretch a bond depends on the strength of the bond and the
masses of the bonded atoms. - The stronger the bond, the greater the energy required to stretch it.
What is the bonding from a transition metal to carbonyl
- The first component is a two electron donation of the lone pair on carbon into a vacant metal d-orbital.
- This electron donation makes the metal more electron rich
- in order to compensate for this increased electron density, a filled metal d-orbital may interact with the empty pi* orbital on the carbonyl ligand to relieve itself of the added electron density.
- This second component is called pi-backbonding or pi-backdonation.
Describe the thermodynamics of M-C bond strengths
- In general the larger the metal, the weaker the M-C bond due to less efficient orbital overlap
- Weaker bonds for lanthanides as larger
- More carbanion character - very delta negative
What happens to 4f element alkyls
- Tend to decompose via beta-hydride elimination of alkene
- The f-elements do not form strong alkene complexes (Dewar-Chatt-Duncanson model requires pi-back donation.)
- so the result is a hydride species, which may decompose further
- Intermediate decomposes straight away and is never observed - no orbitals in right symmetry
How is the problem of 4f element alkyls decomposing circumvented
- Use of alkyl ligands with no beta-H substituents
- CH2-Benzyl
- Bis(trimethylsilyl)methyl
- Neopentyl
What is required from ligands of organolanthanide chemistry
- Steric production (bulky)
- Kinetic stability
- Solubilising hydrocarbon substituents
- Tend to rely on carbanions
- Avoid beta-hydrogen atoms
- Big, charged, greasy
What can the use of very bulky ligands result in and give example
- Use of particularly bulky ligands such as bis(trimethylsilyl)methyl can result in low-(three-) coordinate compounds such as the lanthanum trialkyl
Describe why bis(trimethylsilyl)methyl is a good ligand
- 1- Ligands
- Agostic interactions between low coordinate Ln centre and C-H bonds alleviate the electron deficiency of the metal - stabilising interaction from filled orbital to Ln
What happens when smaller alkyl groups are used and give example
- the metal centre may accommodate more alkyl ligands than it needs to satisfy its valency.
- For example, erbium trichloride can react with six equivalents of MeLi to form a (3–) ‘erbiate’ complex.
Why are bulky ligands prefered to smaller alkyl groups
- More control of environment
- Can’t control coordination environment with smaller alkyl groups and are too kinetically stable to do interesting chemistry with
What cyclopentadienyls are available for trivalent lanthanides
- Three types of cyclopentadienyls are available, [LnCpnX3-n].
- [LnCp3]
- [LnCp2X]
- [LnCpX2]
How are the cyclopentadienyls made
- They are synthesised by reaction of anhydrous lanthanide halides (usually chlorides) with Na or K Cp.
- LnCl3 + n NaCp –> [LnCpnCln-3] + nNaCl
How do Cyclopentadienyls show example of Lanthanide contraction
- The larger lanthanides (La – Pr) have polymeric ten coordinate structures (defining Cp as a ‘tridentate’ ligand) where each metal centre has three η5-Cp ligands and one η2-Cp.
- The later (smaller) Tm and Yb form nine coordinate monomeric structures
- while Lu, the smallest lanthanide, forms an eight coordinate polymeric structure.
Describe bonding in Ln-Cp
- Ionic in nature - due to contracted 4f orbitals which result and no accesible d-orbitals either
What happens when [LnCp3] compounds react with FeCl2
- Form ferrocene
- 3FeCl2 + 2 Cp3Ln –> 3 Cp2Fe + 2 LnCl3
- Salt metathesis reaction
What happens to Cp3Ln compounds in solution
- The polarity is also apparent in solution where rapid exchange of Cp ligands occurs between the lanthanide centres.
- The bonding is again highly labile because of the absence of ligand field stability.
- This behaviour is thus similar to main group organometallics rather than transition metals which obey the 18 electron rule.
- Schlenk-type equilibrium:
- [Cp3Ln] + [Cp’3Ln] <–> [Cp2LnCp’] + [Cp’2LnCp] <–> etc
How are [Cp2LnX] compounds generally found
- [Cp2LnX] compounds are generally dimers or oligomers
What happens if you add THF to [Cp2LnX]
- Can also form base adducts such as the THF complex
- Adduction
- O bonds to Ln instead of a chloride
Describe properties of [Cp2LnX]
- Fairly insoluble in hydrocarbons
- Not well defined as not easy to work with
- Lots of redistributions
What is Cp
- C5H5-
What happens when C5H5- is replaced with bulkier analogues
- Results in complexes with much more convenient properties
- Less prone to ligand redistribution reactions
- More soluble in organic solvents
What is Cp*
- Cp with 5 methyl groups attached
- Bulky analogue
- C5Me5-
What is another bulky analogue of Cp
- Cp with 2 SiMe3 groups attached
- C5H3(SiMe3)2-
What are 2 Bigger Cp ligand complexes with Ln
- Tris Cp*3Ln
- Cp*2LnCl
Describe Tris Cp*3Ln
- Derivatives are not accessible by simple metathesis of KCp* and LnX3 due to the extreme crowding induced by the bulk of the ligand
- Too big- only get 2 on
Describe Cp*2LnCl
- Synthesised readily from LnCl3 and NaCp* or KCp*
- Generally exist as dimers
- Soluble in solvents - ideal starting materials
How do you form Mixed Cp/alkyls, Cp*2LnR
- By reaction of Cp*2LnX and an appropriate alkylating agent.
- Salt metathesis
What are 3 things that can be added to Cp2LnX to form Cp2LnR
- LiCH(SiMe3)2 - loses LiCl
- MeLi replaces Cl in bridged dimer- loses LiCl
- Al2Me6 adducts replacing LnCp*2- loses AlMe2Cl
What are the 2 distinct types of activity Ln-C compounds show
- Insertion of C=C unsaturation into polar Ln-R bonds
- Sigma-bond metathesis
How does insertion work
- Ln is so delta +ve and Me is so delta -ve, can take unpolarised alkene and is forcibly polarised
- Allows insertion of unpolarised things
What does sigma-bond metathesis involve
- Ln-R sigma bond reaction with a polarised other bond
- Allows new Ln-element bond from P-block with loss of R group
- Mostly protonolysis where RH is lost
- But if H is attached to a delta positive element like Si, B or Sn- transfer Hydride onto Ln e.g. R3SiH
What can [Cp*2LuMe] do
- [Cp*2LuMe] is one of the few systems capable of mediating the exchange of C-H bonds in unactivated hydrocarbons - sigma-bond metathesis.
- 13C-enriched methane reacts via a four-membered transition state to give the 13C enriched Lu methyl.
- the methyl group can be thought of as acting as a very strong base in deprotonating the methane.
Why can [Cp*2LuMe] deprotonate methane
- Due to the lanthanide contraction
- Lu is last in series
- High charge density- allows it to deprotonate methane
What will charge of p-block elements bonded to Ln
- Always delta negative
- Ln is always delta +
What will charge of p-block elements bonded to Ln
- Always delta negative
- Ln is always delta +
Which Lns are involved in low oxidation state organometallics
- Sm2+
- Eu2+
- Yb2+
How can you isolate dialkyls of the divalent Sm2+, Eu2+ and Yb2+
- Even bulkier methyl ligands are required to isolate dialkyls of the divalent Sm2+, Eu2+ and Yb2+
- 2(Me3Si)3CK + EuI2 –> Eu(C(SiMe3)3)2 +2KI
How can you form Ln ‘Grignards’
- Using Eu/Yb not Sm as to reactive so would over oxidise itself
- (Me3Si)3CI + Eu + Et2O–> A bridged Grignard like compound
Describe how to form bisCp*Ln compound
- Reaction of KCp* with LnI2 (Ln = Sm, Eu, Yb) in THF gives the THF solvated compounds Cp*2Ln(THF)2.
- The THF molecules may be removed by vacuum sublimation to give the unsolvated compounds [Cp*2Ln].
Describe the structure of [Cp*2Ln]
- The structures of all three compounds are bent (cf the Cp* derivatives of Ca, Sr and Ba).
- Rationalised as a polarisation effect
Describe reactivity of (eta5-C5Me5)2Sm
- The divalent Cp* derivatives of Sm, Eu and Yb display a wide reductive reactivity
- (eta5-C5Me5)2Sm react with a wide range of reducible substrates as unhappy in +2 OS
- Goes to +3 oxidation state
What happens when (eta5-C5Me5)2Sm is exposed to N2
- the dinuclear side-on bridged complex [(eta5-C5Me5)2Sm]2(μ- eta2-eta2-N2) can be crystallised.
- The N2 is bound very weakly and the coordinated N2 can be removed in the solid state under vacuum.
- Adduct
4.
