Synthetic Frontiers of Inorganic Chemistry and Ligand Design

Question 1

Describe an organometallic species.

Answer 1

An organometallic species is anything containing a metal (M) to carbon bond. The nature of the bond doesn’t need to be specified, it can be either covalent or ionic. However, organometallic reagents will be strongly polarised towards C and therefore can be consider to be carbanionic. They are typically nucleophiles and/or strong bases.

Question 2

Describe solid-state structures of s-block organometallics.

Answer 2

• displays a variety of structures, can be related to reactivity
• highly dependent on solvent and other ligands present
• often studied using x-ray diffraction studies
• if crystalline material is available then solid-state structure may not be too challenging to determine
• solid-state structures are often investigated using single-crystal x-ray diffraction

Question 3

How do we understand structure in solution?

Answer 3

• structures in solution are more complex to investigate
• may contain a number of species which may be exchanging with each other relatively quickly
• can use techniques such as NMR, DOSY NMR, cryoscopy

Question 4

Describe how NMR can be used to study structures in solution.

Answer 4

• restricted by the spin/NMR active nuclei available, may need to enrich certain low abundance isotopes
• to prevent fluxionality, spectra are often recorded at low temperatures where a low freezing solvent is required
• need to be aware of equilibria between different aggregation states in the solution phase - often give different splitting patterns to help distinguish

Question 5

Describe the solution structure of Grignard reagents.

Answer 5

• formulated as RMgX but there is a mixture of species in solution which are in equilibrium
• multiple factors influence the position of the equilibrium:
  
  • temperature
  • [RMgX]
  • nature of R and X
  • solvent

Question 6

Describe the Schlenk equilibrium and how to study it.

Answer 6

The Schlenk equilibrium is the dynamic exchange between different species of Grignard reagents in solution.

The primary tool for investigating the species present in Grignard solution phase is NMR - initially ¹H and ²⁵Mg, but can also be ¹³C. Sometimes mass spectrometry studies can support.

Question 7

Describe turbo-charged metallation.

Answer 7

S-block organometallics can be used for many reactions. In many of them, the reactivity, regioselectivity, and functional group tolerance can be enhanced by using more than one s-block element.

Additives like LiCl can be used. However, the mechanism and speciation in solution can be complex.

Question 8

Describe turbo Grignard reagents, such as ⁱPrMgCl.

Answer 8

Question 9

Describe the study of the solution-state structure of (TMP)MgCl.LiCl by NMR in comparison to the x-ray solid-state structure shown below.

Answer 9

NMR shows:

• one type of TMP environment in ¹H
• singlet in ⁷Li NMR (not consistent with LiTMP shift)
• consistent with x-ray structure but it’s also not the only solution consistent with the findings so more studies need to be done

Question 10

Describe the study of the solution-state structure of (TMP)MgCl.LiCl by DOSY NMR in comparison to the x-ray solid-state structure shown below.

Answer 10

DOSY NMR is able to identify different components in solution and indicates their relative sizes. DOSY NMR shown that:

• there’s nothing as big as the x-ray structure in the solution, the dimer must therefore break down to form other species
• 4 other smaller species (3 Mg species, 1 Li) are present
  
  • solution-state structure is most likely a combination of these species
  • rapid equilibration leads to the single TMP environment in ¹H NMR

Question 11

Describe the different kinds of crowns.

Answer 11

Mixture of s-block organometallic bases can form unusual aggregates that are highly reactive and selective reagents for deprotonations.

• crown ethers - Lewis basic donor sites (typically O) in ring centre, they bind cations strongly
• inverse crowns - Lewis acidic materials in ring centre, they bind anions
• pre-inverse crowns - can be used to prepare inverse crowns by deprotonation of substrates

Question 12

Describe the reaction between a pre-inverse crown and napthalene.

Answer 12

Napthalene binds to the Mg atoms in the ring. It causes totally selective deprotonation at the 2-position of napthalene, giving high yields.

Question 13

What happens when the K in the pre-inverse crown is replaced with Na?

Answer 13

A different ring structure is produced, where Na is more rective than the K equivalent. It gives selective dimetallation and different reactivity.

Question 14

Describe the early proposals for the existence of Mg(I) compounds.

Answer 14

1. H-Mg-Mg-H (matrix isolation)
2. Mg(CN) (circumstellar clouds)
3. Formation of Grignard reagents, possibly by R-Mg-Mg-R

However, 1 and 2 involve bad reaction conditions. It’s not possible to isolate the proposed Mg(I) in 3 as it’s not stable.

Question 15

What is a main problem with Mg(I) compounds?

Answer 15

The Mg(I) species would be thermodynamically stable with respect to disproportionation to form Mg(0) and Mg(II).

Question 16

Describe how the thermodynamic instability of Mg(I) compounds was overcome.

Answer 16

By using very bulky ligands which lend kinetic stability. In particular, ‘priso’ and ‘nacnac’ ligands are useful, which are hard donor chelating ligands that provide steric bulk.

• Mg(II) priso and nacnac complexes containing halides can be reduced to Mg(I)
• solid-state structures of the proposed Mg(I) priso/nacnac complexes demonstrate a bond length shorter than Mg-Mg in elemental Mg, but longer than the sum of Mg covalent radii
  
  • suggests Mg(I) present

Question 17

What was found from DFT calculations about the bonding in Mg(I) L₂Mg-MgL₂ complexes?

Answer 17

• optimised structure close to XRD
• HOMO is an Mg-Mg sigma bond
• LUMO is an Mg-Mg pi bond
• Mg-Mg interaction involves high s-character single bond, not too much sp hybridisation
• Mg-N bond is highly ionic, suggest [Mg₂]²⁺ unit coordinated by priso ligands

Question 18

Describe reactivity studies of L₂Mg-MgL₂ complex.

Answer 18

The Mg-Mg bonded species act as two centre/two electron reducing agents, which has significant potential in synthesis such as in the coupling of isocyanates.

Question 19

Show how ferrocene undergoes one-electron oxidation.

Answer 19

Question 20

What are frustrated Lewis pairs?

Answer 20

Frustrated Lewis pairs form when some chemical feature(s) of the LA and LB prevent or disfavour formation of an adduct. Some of these include:

• bulky substituents on LA or LB preventing close approach and therefore adduct formation
• LA and LB joined by a non-flexible linker so the groups can’t reach each other

Question 21

Why has there been a surge in interested of frustrated Lewis pairs recently?

Answer 21

Because it was shown that a FLP could reversibly activate and eliminate H₂.

Question 22

Why is the H₂ FLP research groundbreaking?

Answer 22

It’s usually a transition metal-catalysed process, activating H₂ by oxidative addition and reductive elimination.

The FLP allows clean activation and elimination without a metal present. This has the potential to deliver new processes utilising hydrogen without the cost/resource issues that come with rare, expensive heavy metals.

Question 23

What is the mechanism of H₂ activation/elimination in the linked FLP system?

Answer 23

Question 24

Give examples of FLP systems that activate H₂ similar to the linked FLP system.

Answer 24

Question 25

What is the mechanism of H₂ activation in free LA and LB FLP systems?

Answer 25

Question 26

What are some limitations of FLP-based catalytic hydrogenation?

Answer 26

• less hindered imines are reduced by the B-H component
• still not as well understood as metal-based systems so can’t be utilised as well
• been shown to react with and activate a range of other substrates
  
  • e.g. ring opening, carbonyls, alkenes

Question 27

Describe general phosphine ligands.

Answer 27

• physical properties vary with R, as do ligand properties
• all phosphines are prone to oxidation due to the high P=O bond strength, although the rate of oxidation varies with R
• often characterised by ³¹P NMR (I = 1/2, 100% abundant)

Question 28

Describe the electronic properties of PR₃.

Answer 28

Question 29

Describe the Tolman electronic parameter.

Answer 29

• quantifies the electronic properties of varying R on the complex [LNi(CO)₃], where L = phosphine
• measures the A₁ stretching mode of C=O attached to Ni by IR
• both sigma and pi effects are combined in this model

Question 30

Describe what happens when PR₃ is a strong sigma donor.

Answer 30

Question 31

What are the limitations of how Tolman determined the electronic properties of PR₃?

Answer 31

• Ni(CO)₃L are made from Ni(CO)₄ which is highly toxic
• v(CO) is affected by ligand sterics
• may have complications from vibrational coupling due to multiple CO ligands
• doesn’t allow a range of ligand classes to be compared (just phosphines)

Question 32

How was Tolman’s electronic parameter improved upon?

Answer 32

Gusev proposed a new system for assessing the donor properties of a range of different ligands (not just phosphines). DFT was used to model the structures and carbonyl stretching frequencies of (η⁵-C₅H₅)Ir(CO)L complexes.

• computational so no toxicity
• less sterically congested
• only one CO so no vibrational coupling problems
• allows a range of ligands to be compared

Question 33

Describe Tolman’s steric parameter.

Answer 33

• the angle of the cone, whose apex is 2.28 Å from the centre of P, that touches the van der Waals radii of the outermost atoms of PR₃
• making the R groups more bulky has the effect of increasing overall steric demand of the PR₃ ligand, which can be quantified by measuring equilibrium constants
• also affects the maximum number of PR₃ ligands which can coordinate to the metal

Question 34

Why is selecting the correct phosphine important?

Answer 34

It’s possible to tune the metal centre reactivity by selecting a ligand with a particular combination of sterics and electronics.

Choosing the right phosphine allows us to:

• exert control over the amount of electron density at the metal
• exert control over the steric crowding at the metal

Question 35

Give an example of a catalyst used for catalytic hydrogenation.

What mechanism is thought to occur?

Answer 35

Wilkonson’s catalyst: [RhCl(PPh₃)₃]

Many pathways are possible, but the ‘hydride mechanism’ it thought to operate, where hydrogen binds before the alkene.

Question 36

Describe the possible electronic effects in catalytic hydrogenation.

Answer 36

More electron donating phosphines give more active catalysts. The step most affected by electronics is the formation of the metal hydride (addition of H₂).

A more electron rich metal gives greater back donation to the H-H sigma orbital and pushes the equilibrium towards dihydride (RHS). More electron donating phosphines will increase the electron density at the metal and therefore favour dihydride formation.

More electron donating ligands will also stabilise the higher M ox. state on the RHS.

Question 37

Describe the possible steric effects in catalytic hydrogenation.

Answer 37

The step most affected by sterics is the formation of the 5-coordinate intermediate, which is formed prior to alkene binding (after dihydride formation). Generation of this intermediate is crucial for catalysis. Loss of phosphine is driven by steric crowding, so bulkier phosphines give more active catalysts up to a point.

Question 38

Describe N-heterocyclic carbenes (NHCs).

Answer 38

Share some similarities with phosphines, but are generally better sigma donors (and worse pi acceptors) and their sterics can be different (more asymmetric).

NHCs can be isolated as thermodynamically stable products. The N lone pairs donate electron density into the ‘empty’ p-orbital on the carbon, stabilising it.

Question 39

Describe the general synthesis to an NHC.

Answer 39

Question 40

How can you avoid unwanted side reactions in the synthesis of NHCs?

Answer 40

Question 41

How are silver salts useful when it comes to NHCs?

Answer 41

Use of silver salts as bases allows the ‘trapping’ of NHCs prior to their transfer to another more active transition metal.

Some additional benefits are:

• solubility of the base - poor in organic solvents
• coordination of silver - ‘traps’ NHC and prevents dimerisation
• allows the NHC to be easily stored until transfer to another metal - air, mosture and light stable

Question 42

Describe the relative activities of Grubbs-type catalysts.

Answer 42

Ligand loss is essential for generating an active catalyst species.

• replacing one PCy₃ with an NHC increases the reactivity - NHC promotes PCy₃​ loss by pushing electron density onto Rh
• replacing both PCy₃​ with NHC decreases the activity - NHC loss is much less favourable than PCy₃​ loss, because NHC binds more strongly to Rh

Question 43

Describe isolobal analogues of NHCs.

Answer 43

Many have different properties to the NHC complexes.

Question 44

Give an example of a boryl NHC analogue as a ligand.

Answer 44