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aa The f-elements > Organolanthanide Chemistry I > Flashcards

Organolanthanide Chemistry I Flashcards

1
Q

What are the main features of organolanthanide chemistry

A
  1. Organolanthanide chemistry is not as extensive as organotransition metal chemistry
  2. Expanded enormously since 1980s, especially in catalysis and C-H Bond Activation studies
  3. Primarily ionic in their bonding due to the contracted nature of the 4f valence orbitals
  4. Lanthanides cannot act as π-bases and thus Ln-CO compounds are not stable
  5. Organolanthanides are extremely air and moisture sensitive reflecting the highly carbanionic character of the organic ligands and the oxophilicity of Ln2+/3+ .
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2
Q

Describe interaction of Lns with carbonyls

A
  1. 4f valence orbitals of Lns are too withdrawn to interact with the pi* orbitals of CO - Never been isolated
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3
Q

When can LnCOs be isolated

A
  1. 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.
  2. Makes a system which looks something like Pi-back donation from metal to C-O but very unstable
  3. None of these complexes have every been definitively characterised as they decompose instantly on raising the temperature
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4
Q

What happens to C-O stretching frequencies of the Ln(CO)n carbonyls as n increases

A
  1. The appearance of their IR spectra is surprisingly similar to those of transition metals, suggesting some level of pi-back donation.
  2. Stretching frequency increases linearly with n
  3. Suggesting competition for 4f (or 5d?) electron density
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5
Q

What causes stretching frequency to increase

A
  1. The amount of energy required to stretch a bond depends on the strength of the bond and the
    masses of the bonded atoms.
  2. The stronger the bond, the greater the energy required to stretch it.
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6
Q

What is the bonding from a transition metal to carbonyl

A
  1. The first component is a two electron donation of the lone pair on carbon into a vacant metal d-orbital.
  2. This electron donation makes the metal more electron rich
  3. 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.
  4. This second component is called pi-backbonding or pi-backdonation.
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7
Q

Describe the thermodynamics of M-C bond strengths

A
  1. In general the larger the metal, the weaker the M-C bond due to less efficient orbital overlap
  2. Weaker bonds for lanthanides as larger
  3. More carbanion character - very delta negative
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8
Q

What happens to 4f element alkyls

A
  1. Tend to decompose via beta-hydride elimination of alkene
  2. The f-elements do not form strong alkene complexes (Dewar-Chatt-Duncanson model requires pi-back donation.)
  3. so the result is a hydride species, which may decompose further
  4. Intermediate decomposes straight away and is never observed - no orbitals in right symmetry
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9
Q

How is the problem of 4f element alkyls decomposing circumvented

A
  1. Use of alkyl ligands with no beta-H substituents
  2. CH2-Benzyl
  3. Bis(trimethylsilyl)methyl
  4. Neopentyl
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10
Q

What is required from ligands of organolanthanide chemistry

A
  1. Steric production (bulky)
  2. Kinetic stability
  3. Solubilising hydrocarbon substituents
  4. Tend to rely on carbanions
  5. Avoid beta-hydrogen atoms
  6. Big, charged, greasy
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11
Q

What can the use of very bulky ligands result in and give example

A
  1. Use of particularly bulky ligands such as bis(trimethylsilyl)methyl can result in low-(three-) coordinate compounds such as the lanthanum trialkyl
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12
Q

Describe why bis(trimethylsilyl)methyl is a good ligand

A
  1. 1- Ligands
  2. 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
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13
Q

What happens when smaller alkyl groups are used and give example

A
  1. the metal centre may accommodate more alkyl ligands than it needs to satisfy its valency.
  2. For example, erbium trichloride can react with six equivalents of MeLi to form a (3–) ‘erbiate’ complex.
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14
Q

Why are bulky ligands prefered to smaller alkyl groups

A
  1. More control of environment
  2. Can’t control coordination environment with smaller alkyl groups and are too kinetically stable to do interesting chemistry with
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15
Q

What cyclopentadienyls are available for trivalent lanthanides

A
  1. Three types of cyclopentadienyls are available, [LnCpnX3-n].
  2. [LnCp3]
  3. [LnCp2X]
  4. [LnCpX2]
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16
Q

How are the cyclopentadienyls made

A
  1. They are synthesised by reaction of anhydrous lanthanide halides (usually chlorides) with Na or K Cp.
  2. LnCl3 + n NaCp –> [LnCpnCln-3] + nNaCl
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17
Q

How do Cyclopentadienyls show example of Lanthanide contraction

A
  1. 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.
  2. The later (smaller) Tm and Yb form nine coordinate monomeric structures
  3. while Lu, the smallest lanthanide, forms an eight coordinate polymeric structure.
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18
Q

Describe bonding in Ln-Cp

A
  1. Ionic in nature - due to contracted 4f orbitals which result and no accesible d-orbitals either
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19
Q

What happens when [LnCp3] compounds react with FeCl2

A
  1. Form ferrocene
  2. 3FeCl2 + 2 Cp3Ln –> 3 Cp2Fe + 2 LnCl3
  3. Salt metathesis reaction
20
Q

What happens to Cp3Ln compounds in solution

A
  1. The polarity is also apparent in solution where rapid exchange of Cp ligands occurs between the lanthanide centres.
  2. The bonding is again highly labile because of the absence of ligand field stability.
  3. This behaviour is thus similar to main group organometallics rather than transition metals which obey the 18 electron rule.
  4. Schlenk-type equilibrium:
  5. [Cp3Ln] + [Cp’3Ln] <–> [Cp2LnCp’] + [Cp’2LnCp] <–> etc
21
Q

How are [Cp2LnX] compounds generally found

A
  1. [Cp2LnX] compounds are generally dimers or oligomers
22
Q

What happens if you add THF to [Cp2LnX]

A
  1. Can also form base adducts such as the THF complex
  2. Adduction
  3. O bonds to Ln instead of a chloride
23
Q

Describe properties of [Cp2LnX]

A
  1. Fairly insoluble in hydrocarbons
  2. Not well defined as not easy to work with
  3. Lots of redistributions
24
Q

What is Cp

A
  1. C5H5-
25
Q

What happens when C5H5- is replaced with bulkier analogues

A
  1. Results in complexes with much more convenient properties
  2. Less prone to ligand redistribution reactions
  3. More soluble in organic solvents
26
Q

What is Cp*

A
  1. Cp with 5 methyl groups attached
  2. Bulky analogue
  3. C5Me5-
27
Q

What is another bulky analogue of Cp

A
  1. Cp with 2 SiMe3 groups attached
  2. C5H3(SiMe3)2-
28
Q

What are 2 Bigger Cp ligand complexes with Ln

A
  1. Tris Cp*3Ln
  2. Cp*2LnCl
29
Q

Describe Tris Cp*3Ln

A
  1. Derivatives are not accessible by simple metathesis of KCp* and LnX3 due to the extreme crowding induced by the bulk of the ligand
  2. Too big- only get 2 on
30
Q

Describe Cp*2LnCl

A
  1. Synthesised readily from LnCl3 and NaCp* or KCp*
  2. Generally exist as dimers
  3. Soluble in solvents - ideal starting materials
31
Q

How do you form Mixed Cp/alkyls, Cp*2LnR

A
  1. By reaction of Cp*2LnX and an appropriate alkylating agent.
  2. Salt metathesis
32
Q

What are 3 things that can be added to Cp2LnX to form Cp2LnR

A
  1. LiCH(SiMe3)2 - loses LiCl
  2. MeLi replaces Cl in bridged dimer- loses LiCl
  3. Al2Me6 adducts replacing LnCp*2- loses AlMe2Cl
33
Q

What are the 2 distinct types of activity Ln-C compounds show

A
  1. Insertion of C=C unsaturation into polar Ln-R bonds
  2. Sigma-bond metathesis
34
Q

How does insertion work

A
  1. Ln is so delta +ve and Me is so delta -ve, can take unpolarised alkene and is forcibly polarised
  2. Allows insertion of unpolarised things
35
Q

What does sigma-bond metathesis involve

A
  1. Ln-R sigma bond reaction with a polarised other bond
  2. Allows new Ln-element bond from P-block with loss of R group
  3. Mostly protonolysis where RH is lost
  4. But if H is attached to a delta positive element like Si, B or Sn- transfer Hydride onto Ln e.g. R3SiH
36
Q

What can [Cp*2LuMe] do

A
  1. [Cp*2LuMe] is one of the few systems capable of mediating the exchange of C-H bonds in unactivated hydrocarbons - sigma-bond metathesis.
  2. 13C-enriched methane reacts via a four-membered transition state to give the 13C enriched Lu methyl.
  3. the methyl group can be thought of as acting as a very strong base in deprotonating the methane.
37
Q

Why can [Cp*2LuMe] deprotonate methane

A
  1. Due to the lanthanide contraction
  2. Lu is last in series
  3. High charge density- allows it to deprotonate methane
38
Q

What will charge of p-block elements bonded to Ln

A
  1. Always delta negative
  2. Ln is always delta +
39
Q

What will charge of p-block elements bonded to Ln

A
  1. Always delta negative
  2. Ln is always delta +
40
Q

Which Lns are involved in low oxidation state organometallics

A
  1. Sm2+
  2. Eu2+
  3. Yb2+
41
Q

How can you isolate dialkyls of the divalent Sm2+, Eu2+ and Yb2+

A
  1. Even bulkier methyl ligands are required to isolate dialkyls of the divalent Sm2+, Eu2+ and Yb2+
  2. 2(Me3Si)3CK + EuI2 –> Eu(C(SiMe3)3)2 +2KI
42
Q

How can you form Ln ‘Grignards’

A
  1. Using Eu/Yb not Sm as to reactive so would over oxidise itself
  2. (Me3Si)3CI + Eu + Et2O–> A bridged Grignard like compound
43
Q

Describe how to form bisCp*Ln compound

A
  1. Reaction of KCp* with LnI2 (Ln = Sm, Eu, Yb) in THF gives the THF solvated compounds Cp*2Ln(THF)2.
  2. The THF molecules may be removed by vacuum sublimation to give the unsolvated compounds [Cp*2Ln].
44
Q

Describe the structure of [Cp*2Ln]

A
  1. The structures of all three compounds are bent (cf the Cp* derivatives of Ca, Sr and Ba).
  2. Rationalised as a polarisation effect
45
Q

Describe reactivity of (eta5-C5Me5)2Sm

A
  1. The divalent Cp* derivatives of Sm, Eu and Yb display a wide reductive reactivity
  2. (eta5-C5Me5)2Sm react with a wide range of reducible substrates as unhappy in +2 OS
  3. Goes to +3 oxidation state
46
Q

What happens when (eta5-C5Me5)2Sm is exposed to N2

A
  1. the dinuclear side-on bridged complex [(eta5-C5Me5)2Sm]2(μ- eta2-eta2-N2) can be crystallised.
  2. The N2 is bound very weakly and the coordinated N2 can be removed in the solid state under vacuum.
  3. Adduct
    4.

aa The f-elements (11 decks)

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