Organic Metallic Chemistry 2: Transverse (Π-Bound) Ligands

Front 1

Describe the bonding between ethylene and metal

Back 1

The ethylene bonds side-on so that both carbons are bonded to the metal. The ethene donates electrons from its occupied π orbital into the empty orbital on the metal, and there is back donation from a filled metal d-orbital into the empty π* on the alkene

Front 2

Draw a diagram showing the bonding between ethene and a transition metal.

Back 2

Front 3

Why, like CO, are there very few main group aleken complexes?

Back 3

Because you cannot get back-donation with main group metals, and it is this that helps stabilise the alkene complexes.

Front 4

Explain how the hydrobortaion of two molecules of isoprene can result in a main group alkene complex.

Back 4

Both double bonds one of isoprene are hydroborated, leaving a third B-H bond to add across the unsubstituted double bond of the second molecule. This leaves one double bond. It is thought there is some donation from the π bond to the electron-deficient boron centre.

Front 5

What evidence is there for the proposed π donation in the isoprene-borohydride complex?

Back 5

Low IR for C=C stretch, as electrons being donated to electron-defficient boron. If dissolved in diethyl ether, this weak donation is replaced by a better Lewis base and C=C IR streching frequency increases,

Front 6

Ethene has two π type orbitals, the symmetrical π and antisymmetrical π*. Draw a diagram to show which metal orbitals that have the correct symmetry and so are able to interact with the ethene π orbital.

Back 6

S, pz, dx^2, d(x^2 - y^2)

Front 7

Ethene has two π type orbitals, the symmetrical π and antisymmetrical π*. Draw a diagram to show which metal orbitals that have the correct symmetry and so are able to interact with the ethene π* orbital.

Back 7

Front 8

The first ever example of a prepared alkene complex was Zeise's salt: K[Pt(C2H4)Cl3]. Draw its structure.

Back 8

Front 9

Why is the coordinated ethene perpendicular to the PtCl3 plane in the solid state in Zeise's salt?

Back 9

To avoid steric clashes

Front 10

Why is the coordinated ethene C=C bond slightly longer in Zeise's salt than in free ethene?

Back 10

The π donation takes electrons away from the bonding orbital, and the π-back donation puts electrons in the π* antibonding orbital. This results in the weakening and lengthening of the bond.

Front 11

Why is the C=C bond longer for [Pt(C2H4)(PPh3)2] compared to Zeise's salt K[Pt(C2H4)Cl3]?

Back 11

Because the Pt is at an oxidation state zero and therefore wants to back donate more electrons.

Front 12

What gives an indication that the carbon atoms are no longe purely sp2 but also have some sp3 character in Zeise's salt?

Back 12

The angle between the hydrogens is no longer coplanar, but 146°. Bond order of the C-C bond reduced towards one.

Front 13

The orbital involved in backbonding must be the dxz as thats the one that lies between the x and z axes. How can we deduce which other orbital will be involved?

Back 13

The alkene can rotate so that the dyz can interact with the metal.

Front 14

Explain the energy profile for alkene rotation in alkene metal complex.

Back 14

Ethene can interact with metal with dxz and dyz orbitals. Lowest energy form is solid form when ethene is in vertical position and dxz can interact with the metal. As it rotates enegry increases as neither dxz od dyz can interact effectively with the metal. A new minimum occurs at 90° where the dyz interacts.

Front 15

Explain why energy would be needed for ethene rotation in Zeise's salt

Back 15

Because the horizontal ethene is at a higher energy than the vertical ethene. Therefore need to overcome activation energy for rotation.

Front 16

Explain how performing NMR at high and low temperatures can be used to show that alkene rotation occurs.

Back 16

The two ends of the ethene should show different spectra as they are adjacent to the NO and CO respectively, and at low temperatures this is the case. However, at high temperatures the activation energy of rotation is overcome and only one single peak appears as the rotation of 90° and plane of symmetry down the model renders the two ends of the ethene equivalent.

Front 17

Explain how many sets of signals you would expect for a 1H NMR spectrum at low temperatures

Back 17

Would expect two sets of signals each integrating to 4 protons, corresponding to the inner and outer protons on the ethene.

Front 18

Explain how many sets of signal you would expect for a 1H NMR spectrum at a high temperature. What is important to note?

Back 18

Eventually one broad singlet as the inner and outer protons become equivalent due to rotation. Note that rotation about the C-C axis would do this, but we still believe it is from the Rh-ethene axis.

Front 19

Explain why in the second compound, the C2H4 ligand still rotates with a virtually unchanged activation energy, but the signals for C2F4 are unchanged wit temperature?

Back 19

Replacing the H's with electronegative atoms makes ligand a poorer electron donor but a better π acceptor. This makes it overall more strongly bonded and activation energy to rotate around to the next orbital is too great.

Front 20

What is the η notation used to denote?

Back 20

How many carbons are bonded to the metal

Front 21

What is the μ notation used to denote?

Back 21

Ligands that are bridging 2 or more metals such as μ-CO

Front 22

Explain why alkynes can act as 4 electron donors as opposed to alkenes which act as 2 electron donors.

Back 22

Alkynes have two π bonds. One is π parallel and lies in the same plane as the alkyne substituents. There is possible overlap from s, p and dz2 orbitals in the same way an alkene can. In addition one is π perpendicular because it lies in the same plane as the alkene substituent and can overlap with its dyz orbitals. More electron density donated.

Front 23

Discuss back donation in alkyne complexes and compare to alkenes.

Back 23

Unsymmetrical π* orbital can interact with dxz and px orbitals. However, orbital overlap between dxy and π* is poor as the two orbitals side by side. So second backbonding only small contribution to bonding pictures. Still better π acceptors than alkenes.

Front 24

Explain the following bonding picture.

Back 24

As back donation increases, transition from sp hybridised carbon to sp2 like character as electron density from π orbitals is pushed into the π*. This lengthens the C≡C bond and bends the substituents away from the metal. As this carries on get extreme sp3 carbon, in reality bonding somehwere between the first two, but depends on e density of emktal, substituents, alkyne etc.

Front 25

Compare rotation in alkenes and alkynes.

Back 25

Rotation for alkynes is possible but less common than alkenes and with higher activation energies due to strogner bonding.

Front 26

Compare the strength of the alkyne-metal bonds and C≡C bond lengths in each complex.

Back 26

In both cases the metal is Mo(II) and d4. We are aiming to get an 18 electron complex. In compound 1; each Cp- donates 6 electrons, therefore there is a total of 16e and the alkyne only needs to donate 2 electrons to make 18. In compound 2 each N donates 2e and each N- donates 3e giving a total of 14e. Therefore 4 e needed from alkyne. Compound 2 therefore has stronger metal-alkyene bond and a longer C≡C bond length due to more e density being removed from π orbitals.

Front 27

Show how the dithiocarbamate ion bonds to a metal

Back 27

Front 28

How can educated guesses be made about the amount of electrons donated by the alkyene from 13C NMR spectra?

Back 28

Chemical shifts are much higher for 4 electron donating alkynes compared to 2 electron donating alkynes.

Front 29

Explain why and how each alkyne donates 3 electrons

Back 29

The Mo(dtc)2 fragment has 12 electrons so the two alkynes have to supply a total of 6e. The alkynes are cis lying parallel to each other. Both alkenes can donate from their π parallel orbitals to metal orbitals, however, their perpendicular π orbitals are both attempting to overlap with the same metal d orbital. 3-centre 4e interaction occurs which with C2v smmetry gives 2 possible orbital combinations. The top one can overlap effectively but the second is non-bonding. Overall the alkyne donates 4e per π-parallel and 2e per π-perpendicular = 6. 3 each.

Front 30

Explain the alkene contribution in the following complex

Back 30

Mo is d6 + 2e from CO gives 8e. therefore need 10e from alkynes. Each alkynes must donate 3.3e. Structure is basically tetrahedral with parallel alkynes. Each alkyne can donate 2 from π parallel, but there is only a total of 2 d orbitals available for them to donate to from their π perpendicular. Overall 6 from π parallel and 4 from π perpendicular.

Front 31

Explain the dramatic increase in 13C shift values. What can be ntoed about the 1H shift value change?

Back 31

As Cl- ligand is lost, metal centre requires an extra 2e from alkyne making it a 4e donor increasing 13C shift. Because it is a terminal alkyne, 1H chemical shift of CH proton increases.

Front 32

Cyclooctyne is the smallest stable cyclic alkyne as it is the smallest stable cyclic alkyne that can accomadate a 180°. How can smaller cyclic alkynes be made such as cyclohexyne?

Back 32

When a complexed alkyne is made the angle is typically reduced to 140-160°.

Front 33

Give two methods for synthesising benzyne complexes

Back 33

Loss of methyl group, or dehalogenation

Front 34

The phospine is linear and so shoiuld not be able to act as a chelating ligand. Explain how it can.

Back 34

By attaching it to a metal through the C≡C bond, the phosphines are bent back and can act as chelating ligands to a second metal fragment.

Front 35

Show how the alkyne can act as a bridge to form an 18 electron complex. Name the alkyne using appropriate nomenclature. In which case would the alkyne only be able to donate 2e?

Back 35

4 e donor so can interact with both metals by donating 2 to each. Co is d9 and CO 2e donor, donation of 1e from Co-CO. 9+6+2 +1 = 18.

Alkyne is a μ,η2,η2-alkyne.

In this case the alkynes lie perpendicular to the Co-Co bond, if they were parallel they only use 2e.

Front 36

How would you synthesise this product?

Back 36

Addition of alkene to vaska's complex.

Front 37

How would you synthesise this product?

Back 37

Substitution of Co 2e donor with alkyne.

Front 38

The allyl (propenyl ligand) -CH2CH≡CH2 is an η3 ligand with all 3 carbons bonded to the metal. Draw the possible 3 possible bonding combinations.

Back 38

Front 39

Show which allyl orbitals are capable of of overlapping with the metal orbitals for each bonding combination.

Back 39

Ψ1 can overlap with s, pz and dz2 like with alkenes and alyknes. Ψ2 can overlap with py and dyz orbitals. Ψ3 can overlap with px and dxz.

Front 40

Discuss the π bonding for metals and allyl anions (C3H5)- in terms of their bonding combinations.

Back 40

Ψ1 and Ψ2 will both be full, meaning they can be used asπe donors to the metal whilst Ψ3 is empty and can be used as a π acceptor for back donation.

Front 41

Discuss the π bonding for metals and allyl cations (C3H5)+ in terms of their bonding combinations.

Back 41

Ψ1 will be full, meaning it can be used as πe donor to the metal whilst both Ψ2 and Ψ3 are empty and can be used as π acceptors for back donation.

Front 42

Discuss the π bonding for metals and allyl radicals (C3H5)' in terms of their bonding combinations.

Back 42

Ψ1 will be full, meaning it can be used as πe donor to the metal, Ψ3 is empty and can be used as a π acceptor for back donation. Ψ2 contains one electron and can do either.

Front 43

Summarise the bonding of allyl ligands with metals

Back 43

Ψ1 will always be full, meaning it can be used as πe donor to the metal, Ψ3 is always empty and can be used as a π acceptor for back donation. Ψ2 can be either depending on the metal centre it is attached to.

Front 44

In contrast to even numbered ligands (η2,η4,η6 etc.), odd numbered ligands are often very unstable and need to be introduced as derivatives. Explain how you would synthesise an Mn(CO)4(allyl) starting from [Mn(CO)5]-

Back 44

React [Mn(CO)5]- with an allyl chloride to obtain [Mn(CO)5(allyl)]. This is anη1 alkyl ligand attached by the alkyl group with an uncoordinated double bond. By heating or UV a CO group is lost and the double bond coordinates to the Mn resulting in aη3 ligand (π-allyl ligand).

Front 45

Explain why only UV can be used to lose the CO and the appearance of 2 isomers

Back 45

Complex decomposes with heating. 2 isomers formed depending on whether the allyl ligand is facing away or to the Cp ligand.

Front 46

Discuss the bonding in a halide bridged dimer

Back 46

Each halide joined to one metal by a normal σ-bond and also donates a pair of electrons to the other.

Front 47

What type of addition occurs in the following reaction to synthesise an allyl complex?

Back 47

Oxidative addition. Ni(0) becomes Ni(II) assumnig that the allyl ligands are uninegative

Front 48

Give an alternate method to oxidative addition so synthesise the folowing allyl complex from reactant

Back 48

Addition of allyl Grignard with metal salt.

Front 49

How can an allyl complex be synthesised from this starting material?

Back 49

Deprotonation of the methyl group adjacent to the double bond. Cannot occur for free alkenes, but can occur if they are coordinated.

Front 50

What method is typically used to synthesise a methyl substituted allyl complex?

Back 50

Insertion of M-H into a diene.

Front 51

Describe what you would be looking for on a characteristic NMR for an allyl complex

Back 51

H3 will be coupled to 2 equivalent H2 and H1 giving a triplet of triplets (multiplet). H2 and H1 will be coupled to H3 respectively giving two doublets. Unusually there is no or little coupling between H1 and H2 giving a spectra containing 2 doublets and a multiplet.

Front 52

Explain why the NMR of an allyl complex simplifies from 2 doublets and a mulitplet to one doublet and a quintet if a ligand is added to the solution, or if it is done in a coordinating solvent such as DMSO.

Back 52

From the NMR can see that the syn and anti protons are now equivalent. An equilibrium is set up in which the dimer is cleaved by the solvent ligand to give a mononuclear complex in which the allyl is η1 or σ-allyl. This species allowd free rotation about M-C and C-C bonds. When we revert to η3 the syn and anti protons have changed place. See an averaged picture.

Front 53

Outline how coordinating an alkene to a metal changes its reactivity?

Back 53

Alkenes are generally susceptible to electrophilic attack (e.g. from Br+) due to high π e density. By coordinating to metal they lose some e density and can be subjected to nucleophilic attack.

Front 54

Explain why nucleophiles will attack terminal carbons of allyl ligands if attached to an electron-poor centre?

Back 54

The allyl ligand will be in the form Ψ2 and so have a node at the middle carbon. Therefore terminal carbons are most affected by changes in electron density and if attached to electron poor centre will be donating more electrons and so be more susceptible to nucleophilic attack.

Front 55

Explain why electrophiles will attack terminal carbons of allyl ligands if attached to an electron-rich centre?

Back 55

The allyl ligand will be in the form Ψ2 and so have a node at the middle carbon. Therefore terminal carbons are most affected by changes in electron density and if attached to electron rich centre will be accepting more electrons and so be more susceptible to electrophilic attack.

Front 56

Why are cyclopropenyls only known for first row transition metals?

Back 56

Because poor overlap of π system with larger 4d and 5d orbitals.

Front 57

What are the three basic types of diene complexes?

Back 57

Those containing non-conjugated dienes, conjugated dienes or cyclobutadiene.

Front 58

What is a non-conjugated diene?

Back 58

One or more saturated groups between the double bonds.

Front 59

How do non-conjugated dienes act? Why does this make them good ligands for metals?

Back 59

Essentially double bon ds act as two independent alkenes. They can act as chelating ligands and are also often easily displaced meaning diene complexes can sometimes act as stabilised synthons for metal fragments. E.g. M(cod)2 can act as a useful source of M(0).

Front 60

Why does Mo(CO)6 + 2PPh3 result in a mix of products Mo(CO)6,Mo(CO)5(PPh3)cis + trans, Mo(CO)4(PPh3)2 andMo(CO)3(PPh3). Whilst Mo(CO)6 + nbd --> Mo(CO)4(nbd) --> cis-Mo(CO)4(PPh3)2(just cis)

Back 60

Because acts as chelating ligand and selectively chooses where the ligands attach before being displaced.

Front 61

What is a conjugated diene?

Back 61

Where the double bonds are adjacent to each other, e.g. 1,3-butadiene

Front 62

Show which metal orbitals can interact with the MO of butadiene

Back 62

Orbitals of the right symmetry to overlap with 4 π type MO's of butadiene.

Front 63

Discuss the effect of overlap of metal orbitals with Ψ1 in a butadiene complex

Back 63

Ψ1 is full so will donate electron density to the metal. It is a bonding unit all along the C4 unit, so by removing electron density all of the bond lengths will increase

Front 64

Discuss the effect of overlap of metal orbitals with Ψ2 in a butadiene complex

Back 64

Bonding type orbitals at the terminal bonds but antibonding in the centre. If electron density is removed from this orbital terminal bonds will lengthen but the central bond will shorten.

Front 65

Discuss the effect of overlap of metal orbitals with Ψ3 in a butadiene complex

Back 65

Ψ3 is empty so will be accepting electron density from the metal in a backbonding interaction. It is antibonding over the two terminal bonds but bonding at the centre. But we are now pushing electron density back, so terminal bonds will lengthen and central bond will shorten.

Front 66

Discuss the effect of overlap of metal orbitals with Ψ4 in a butadiene complex

Back 66

Antibonding the whole length of the C4 unit, addition of electron density will increase all the bond lengths

Front 67

Overall discuss the net effect of metal alkene coordination on C-C bond length in butadiene

Back 67

The terminal C=C bonds will lengthen whilst the central C-C bond will shorten. In other words the single bond will act more like a double bond and the double bond more like a single.

Front 68

Other tyhan coordinating it to a transition metal, what can we do to butadiene to lengthen terminal C=C bonds and shorten middle C-C? If this was carried on to extreme what could we end up with in theory?

Back 68

Excite an electron from Ψ2 to Ψ3. If taken to extreme the two terminal bonds would become single bonds and the central one a double which would give us a metallocylcopentene structure.

Front 69

Why is the metallocyclopentene structure favoured by early transition metals?

Back 69

Because they favour higher oxidation states

Front 70

What structure is favoured by middle and late metals with conjugated dienes?

Back 70

A diene π-complex structure.

Front 71

Explain why in a free state cyclobutadiene, C4H4, is an unstable species an all attempts to prepare it results in its dimer in which it has undergone Diers-Alder on itself.

Back 71

Because it is highly strained. It would also be a diradical with an electron in each of Ψ2 and Ψ3. Both Ψ2 and 3 are now degenerate and non-bonding.

Front 72

Give an example of how this cyclobutadiene complex can be made

Back 72

Some of the iron is lost as FeCl2 as this is a dechlorination reaction. Treat with Ce4+ at low temperature to oxidise off the iron, and in the presence of another reagent, the soilution acts as cyclobutadiene.

Front 73

Another method of making cyclobutadiene complexes is the combination of two alkynes. How can they be converted to give just the cyclobutadiene species?

Back 73

Ligand transfer reactions with metal carbonyls

Front 74

Explain the effect of each possbility for R on the η4 alkene complex

Back 74

The simple cyclic ketone undergoes Diers Alder with itself to form a dimer. If you stick bulky substituents on the ring it becomes more stabilised. Some exist as dimers than dissociate to monomers on heating, some dont dimerise at all and just exist as monomers.

Front 75

The Fe complex can act as a stabiliser to produce what product that would be unstable in the free state?

Back 75

Front 76

Distinguish between Cp, Cp' and Cp*

Back 76

Cp is η-C5H5. Cp' is η-C5H4Me. Cp* is η-C5Me5

Front 77

Why is the cyclopentadienyl ligand so ubiquitous?

Back 77

The M-Cp bond dissociation energy is high so fairly inert and it blocks up coordination sites.

Front 78

The cyclopentadienyl ligand is a useful NMR handle, what signals can be used to identify it?

Back 78

A strong singlet in both 1H NMR and 13C NMR

Front 79

How can Cp change the steric and electronic properties of a complex?

Back 79

By changing into Cp' or Cp*. Cp* is more electron rich and pushes electron density onto the metal. If CO ligands attached can tell from IR as CO stretching frequency drops due to increased backbonding.

Front 80

How can the cyclopentadienyl ion be made?

Back 80

Cyclopentadiene is an unstable molecule that exists as a dimer (Diels-Alders). By heating it to 170°C dimer reverts to a monomer and can be easily deprotonated due to aromatic stability of the cyclopentadienyl ion product.

Front 81

What is occuring in the reaction pictured?

Back 81

The η4 complex of the diene has been isolated and can undergo ready oxidative addition of the C-H bond to give Cp complexes containing hydride ligands.

Front 82

What is the easiest method for making ferrocene?

Back 82

Deprotonate cracked cyclopentadiene with KOH in dme and then add FeCl2

Front 83

What further experiments can be done to ferrocene?

Back 83

Ferrocene can be easily (and reversibly) oxidised to the ferricinium ion, [FeCp2]+. It can also undergo electrophilic substitution to give acetyl ferrocene

Front 84

What evidence is there that the ferrocenyl unit acts as more of an electron donor than benzene?

Back 84

FcCO2H is a weaker acid than PhCO2H and FcNH2 is a stronger base than PhNH2

Front 85

Explain the simplified ferrocene bonding diagram.

Back 85

If we use crystal field theory and say that the metal rings above and below the Fe atom lie on the z axis; then the orbitals in the xy plane will be furthest away and feel less repulsion from the π system and hence be lowest energy. The dz^2 will point towards the centre of the rings and be the next lowest, whilst the highest energy will be the dxz and dyz that will point directly at the rings. Fe is group 8 and in a +2 state leaving us with 6 electrons to fill the diagram.

Front 86

What is the main basic method for making other first row metallocenes?

Back 86

MCl2 + NaCp

Front 87

Explain why cobaltacene (CoCp2) is a good recuding agent?

Back 87

Because it is d9 and so will have 7 electrons. The extra electron compared to ferrocene will go into the top e1 orbitals and is very easily lost gicing the [CoCp2]+ ion which contains Co(III) and is much more stable.

Front 88

What is different about MnCp2?

Back 88

In the solid state it is not a discrete sandwich complex, instead it has a chain structure.

Front 89

Why does MnCp2 show an unusual dependence of its magnetic moment on temperature or molecular environmen?

Back 89

It has both high spin and low spin conformations which only differ marginally in energy, so 5 electrons can therefore occupy either of these two arrangements. Ignore the fact that higher energy orbital doubly occupied, exact orbitals vary from compound (can tell from photoelectron spectroscpy)

Front 90

Metal-carbon bond length increases from ferrocene to cobaltacene and nickelocene, despite the fact that the meta atom is getting smaller. Why?

Back 90

Because from looking at MO diagram can see that electrons are being added to anti-bonding orbital.

Front 91

Why does oxidising ferrocene result in an increased Fe-C bond length, whulst oxidising cobaltocene results in a decreased Co-C bond length?

Back 91

Because electron is being lost from bonding orbital of ferrocene, whilst it is being lost from an antibonding cobaltacene orbital.

Front 92

Explain why TiCp2 isn't a true metallocene

Back 92

Ti does not like being in a (+2) oxidation state and so performs oxidative addition to one of the CH bonds of the ring which is then followed by reductive elminiation with a C-C bond being formed giving a 'fulvalene' complex.

Front 93

Most metallocenes in the 2nd and 3rd row of the transitio series follow a similar process to TiCp2 e.g. Nb, explain why RU and Os are capable of forming proper MCp2 compounds.

Back 93

Because they're 18 electron complexes and so msot stable.

Front 94

Towards the left hand side of the periodic table we get metallocene dihalides such as Cp2TiCl2. Explain why Cp2MoH2 is a good metal base?

Back 94

Because it has a lone pair in the d2 orbital.

Front 95

the reation of CpFe(CO)2Cl with NaCp gives the product Fe(CO)2(C5H5)2. Both rings cannot be coordinated as the compound would have more than 18 electrons (22), what is the product?

Back 95

One ring is η5 coordinated one is η1 coordinated, so overall 18 electrons.

Front 96

What would you expect to see for the H NMR and C NMR of the η5 Cp ring? Why is the compound [CpRu(PPh3)2(η1-CH2=CHPh)]+ an exception?

Back 96

Rotational barrier very low so HNMR gives singlet for 5 protons of the ring and a singlet for five carbons. [CpRu(PPh3)2(η1-CH2=CHPh)]+ can be slowed down efficiently at temperatures around -100 degrees so the 5 protons are no longer equivalent.

Front 97

When cooled this is seen along with the η5 Cp ligand. However at room temperature a fluxional process occurs, what is NMR spectra is seen?

Back 97

Just two singlets of equal intensity.

Front 98

What evidence is there that the iron moves around the ring and spends time attached to each of the carbons to explain the fluxionality of CpFe(CO)2(η1-C5H5), as opposed to protons moving around the ring?

Back 98

If we treat CpFe(CO)2Cl with LiC5Me5, we get a similar compound CpFe(CO)2(η1-C5Me5). If we look at the NMR of this one, we again see just two singlets, this time in a ratio of 1:3 (i.e. 5:15). While it's possible that in our original compound the H's were moving around the ring (after all, H shifts are pretty common in organic chemistry), it's a lot less likely that we could break carbon-carbon bonds so easily and move the methyls round the ring.

Front 99

What evidence is there from the CpFe(CO2)(η1-idenyl) NMR spectrum that the Fe moves around the ring via a 1,2-shift for CpFe(CO)2(η1-C5H5)?

Back 99

A 1,3 shift could occurr feasibly as moves from a place next to a ring junction to a place next to a ring junction and those 2 positions would become equivalent. A 2,1 shift would not be a move to next to a ring junction and would result in the redistribution of double bonds that would destroy aromaticity of benzene ring. The compound is static on the NMR timescale, therefore indicates that 1,2 shift resonsible for other compounds.

Front 100

Ferrocene has 18 electrons, to the right in the periodic table e.g. Co, we've seen that we get 19-electron CoCp2. However: by changing one Cp ring to a C4H4 cyclobutadiene ring get 18e. For Ni, we now need to replace our C5 ring by a 3-electron donor to get an 18-electron compound. In this case it's CpNi(allyl). Predict 18e Mn sandwich compound.

Back 100

Front 101

How can 'half-sandwiches' be formed across the transition metals?

Back 101

By keeping a metal tricarbonyl group as a permanent feature and adjusting ring size accordingly to achieve an 18 electron compound, or alter number of CO ligands.

Front 102

If we had CpFe(CO)2 we would have 17 electrons and this is not a stable species. How is this normally stabilised?

Back 102

Adding or losing an electron is possible to give an ion, or dimerising to make Cp2Fe2(CO)4 with a metal metal bond.

Front 103

If we look at the following substitution reaction can see the rate is dependent on PR3 so is an associative process. How does this occur?

Back 103

Ring slippage: The coordination mode of the Cp ligand is changed from η5 to η3 enabling the coordination of ligand L. In this case an attached CO will dissociate allowing the Cp ligand to resume η5 bonding mode.

Front 104

Explain the relative rates of these reactions.

Back 104

For the Cp complex the rate will be faster with phosphine over phosphite complexes as it is a better electron donor. Cp* worse electron acceptor as more electron rich so slower than Cp, also rate with phosphite faster than phosphine due to less steric interference.

Front 105

We would expect the Cp idenyl analogue to be slower because it is a bigger ligand, why isn't it?

Back 105

When we ring slip the idenyl ligand an aromatic ring is created, this is therefore more favourable than for Cp. The transition state is stabilised and reaction is faster.

Front 106

Summarise the idenyl effect

Back 106

The fact that associative substitution reactions are much faster with idenyl complexes than with the analogous Cp ones.

Front 107

This intermediate can be isolated, explain why it is evidence for ring slippage.

Back 107

Because if we did the electron count we would find it to be 20 electrons so both rings can't be η5. One must be η3 and one must be η5

Front 108

Pictures is a simple associative substitution reaction, what can occur in excess PMe3?

Back 108

The ring can slip further to an η1 bonding mode to accomodate more PMe3.

Front 109

Compelxes where benzene is acting as a ligand can also be synthesised as in the pictured reaction. This can be applied to all metals in groups 6,8 and 9 as well as V, Re and Ni. What are its limitations?

Back 109

The harsh reaction conditions mean that alkylbenzenes are isomerised by the Lewis acid AlCl3, and arenes have that have a lone pair form an a acid-base adduct with it. Things like phenols, chlorobenzene, anilines, benzaldehydes etc. impossible to use successfully.

Front 110

Give an alternate synthesis of dibenzene chromium

Back 110

Chromium evaporated from an ingot by firing electron gun and condensed in a cooled trap. When trap is warmed up the meal atoms react with first thing they encounter, in this case benzene. At the end of the reaction benzene drained off and remainder filtered to remove Cr metal giving dibenzene chromium.

Front 111

What are the advantages and disadvantages of preparing metal dibenzene complexes via the method of using evaporated metal in a cooled trap?

Back 111

Advantages: Can be done for lots of arenes and also lots of metals, even high boiling ones.

Disadvantages: yields tend to be low, but only-by-product is usually metal, and also expensive apparatus.

Front 112

Compare the acid/base behaviour of ferrocene, dibenzene chromium and phenyl groups

Back 112

Can tell in ferrocene Fc is acting as an electron donating group compared to Ph as fcCOOH is a weaker acid than PhCOOH, whilst dibenzene chromium is an electron withdrawing unit compared to Ph as C6H6Cr(PhCOOH) is a stronger acid than PhCOOH

Front 113

Why (unlike ferrocene) does dibenzene chromium not undergo electrophilic substitution?

Back 113

The electrophile preferentially oxidises Cr(0)to Cr(I)

Front 114

Benzene chromium tricarbonyl is highly air sensitive (although less so than bis(arene)complexes. It's synthesis can be applied to lots of different arenes. How can it be synthesised?

Back 114

Front 115

Why can you not go from Cr(CO)3(arene) to dibenzene chromium?

Back 115

Because as you gradually replace CO groups with other ligands, those ligands will be poorer backbonders so the remaining CO ligands will take on more backbonding and become tighter bound to the Cr

Front 116

What useful benefits occur from attaching a Cr(CO)3 unit to an arene?

Back 116

Makes it more reactive and introduces planar chirality

Front 117

Explain why the deprotonation at low temperatures of the pictured Cr tricarbonyl complex of indene followed by the addition of an electrophile RX (e.g. methyl iodide) at room temperature results in the attached R group being stereospecifically on the same side as the Cr?

Back 117

The 5-membered ring will be deprotonated and the Cr group will migrate to it changing its 'hapticity' (η number) giving comething similar to the [CpCr(CO)3] anion. When this is treated with RX the R group attacks the Cr. When warmed up the methyl group migrates to the 5-membered ring and the Cr migrates back.

Front 118

Cycloheptatriene can be synthesised as pictured. It is coordinated as a triene though the three double bonds. However this is an η-6 complex, how can the η7 be made?

Back 118

By removing H- as this will maintain the 6 pi electrons but also make it planar so the ring is aromatic. Gives [Mo(CO)3(η7C7H7)]

Front 119

Draw the stereochemistry of these two species to explain why the anion is aromatic and the neutral species is not

Back 119

Must be planar to be aromatic.

Front 120

Explain why this reaction is possible

Back 120

This is a 17-electron paramagnetic species, the hydrogen from the ligand is presumably lost as H2.

Front 121

Why do we find η7-C7H7 most often on metals on the left of the transition metal series but η3 and η5 with metals further to the right?

Back 121

Because if we regard C7H7 as a 7 electron donor, then need some other ligands to make up coordination number resulting in too many electrons for a stable complex.

Front 122

Cyclooctatraene can act as a diene ligand by coordinating through two non-adjacent double bonds. However it has 8π electrons and it isn't aromatic. How can C8H8 be introduced as a ligand?

Back 122

Can reduce using potassium to make it 10π electron and aromatic.

Front 123

Reaction of iron carbonyls with cyclooctatetraene gives η4 Fe(CO)3(C8H8). the COT ligand is not planar but more so than the free state. Why is there only 1 H NMR peak?

Back 123

There must be a fluxional process that renders all hydrogens on the ring equivalent. This is 'ring whizzing' where the metal gradually moves around the ring, as one carbon comes into coordination another one leaves.

Front 124

Very few examples of completely coordinate η8-C8H8 complexes exist, some examples include Ti(η8-C8H8)(η4-C8H8) (this is a 16e species) which when synthesised in reducing conditions as shown produces a dinuclear compound where each Ti has a C8H8 ring, as well as mixed ring sandwiches like CpTi(ηC8H8). what other complexes can be made?

Back 124

One way to get around the restriction of only having 18 electrons and only having orbitals t put them in is to use f-block metals.

Front 125

Explain why it may be possible to replace a CH unit with an N in a cyclopentadienyl ligand

Back 125

The CH group has 3 remaning sp3 orbitals, and so does the N atom, although it has a line pair in the fourth orbital as opposed to an H. Would give us NC4H4- which will be isoelectronic with the Cp ligand.

Front 126

Pictured is a method for making ferrocene. What can be altered to produce azaferrocene?

Back 126

Instead of adding NaCp, add potassium salt of pyrrole K[NC4H4]. Effectively replacing a CH unit with an N.

Front 127

How can phosphaferrocene be synthesised?

Back 127

At high temperatures the Ph group will fall off as benzene.

Front 128

How can di-phosphaferrocene be synthesised?

Back 128

The Ph group can be cleaved by Li metal which reacts with FeCl2.

Front 129

How can penta-phosphaferrocene be synthesised? What is this another good example of?

Back 129

This is an example where a species doesn't exist in the free state but is stabilised when coordinated to a transition metal

Front 130

What can be noted about these heterocyclic compounds?

Back 130

Both N and P retain their lone pairs which can be used to coordinate additional metal fragments

Front 131

If we replace one of the CH groups in benzene with N we get pyridine. Why is pyridine only a good ligand to start with when trying to produce an η6 complex e.g. with Cr(CO)6?

Back 131

Because the lone pair on the N will coordinate to the metal instead of the ring. In this example Cr(CO)6 + pyridine --> fac-Cr(CO)3(py)3

Front 132

How can the the lone pair on N in pyridine be prevented from coordinating to the metal allowing an η6-complex to be formed upon reaction with Cr(CO)6?

Back 132

If we put bulky substituents on the ring (e.g. SiMe3 groups) they can act as a steric block to the N lone pair. If heated with Cr(CO)6 will give us an η6 coordination. The silyl groups can then be removed with fluoride ions.

Front 133

How can the bis(pyridine) chromium complex Cr(η6-pyridine)2 be synthesised?

Back 133

Front 134

How can B be included into a heterocycle?

Back 134

A BH- unit is isoelectronic with CH. So can replace CH's by BH-, e.g. instead of Cp could have C4H5B 2-

Front 135

Can make η6-complexes of borazines, why are these not completely like Cr(CO)3 arenes?

Back 135

The ring is not planar it is puckered.

Front 136

Why are we actually interested in coordinating these ligands to transition metals?

Back 136

Because it changes their reactivity and allows us to carry out organic transformations that wouldn't be possible otherwise. Coordination removes electron density from the ligand making it more succeptible to nucleophilic attack as opposed to electrophilic (although can still sometimesoccur e.g. acetylation of ferrocene).

Front 137

What is the Wacker process?

Back 137

An example where an aldehyde can be produced by the attack of water or hydroxide on a coordinated ethene.

C2H4 + PdCl2 + H2O --> CH3CHO + Pd + 2HCl

Front 138

Why do these two reactons occurr at very different temperatures?

Back 138

The transition metal complex takes some electron density from the ring and so makes it more succeptible to nucleophillic attack and thrrefore can occurr at a much lower temperature.

Front 139

There are often multiple possible reaction pathways, Davies-Green-Mingos rules can be used to determine which is the most likely. However these rules have certain restrictions, what are they and why do they apply?

Back 139

Davies-Green-Mingos rules allow us to predict the site of kinetically controlled nucleophilic attack on 18 electron cationic complexes. If the reaction is thermodynamically controlled the nucleophile may attack one site and migrate to another, and if the metal centre is not 18 electrons then the metal will be the most likely point of attack.

Front 140

What are Daies-Green-Mingos rules?

Back 140

1. Nucleophilic attack occurs preferentially at EVEN coordinated polyenes (closed means fully conjugated ring like Cp)

2. Nucleophilic attack at OPEN coordinated polyenes is preferred to addition to CLOSED ones

3. For even open polyenes, nucleophilic attack always occurs at the terminal C atom; for odd open polyenyls, attack only occurs at the terminal carbon if LnM+ is a strongly electron withdrawing fragment

Rules to be applied in the order 1 2 3

Front 141

Using D.G.M. rules deduce where will the nucleophile attack.

Back 141

1) Both odd numbered coordination ligands

2) Preferred attack at open polyene therefore the allyl ligand

3) Cationic and so not particularly electron rich, will attack terminal C to give alkene complex.

Front 142

Use D.G.M. rules to deduce where the nucleophiule will attack

Back 142

1) All odd numbered coordination

2) Will attack open allyl

3) WCp2 not particularly electron withdrawing so will attack central carbon

Front 143

Use D.G.M. rules to deduce where the nucleophiule will attack

Back 143

1) Will attack even

2) Will attack open before closed

3) Will attack terminal C

Front 144

Explain why the following complex is an exception to D.G.M. rules

Back 144

Because if you folowed the rules and attacked the C6Me6 ligand from the exo direction, you would push the Me towards the metal which is unfavourable.

Front 145

Explain why the electrophillic product here is cis?

Back 145

Because the H- attacks the metal first

Front 146

What is the electrophile preference for rings?

Back 146

Smallers rings are more susceptible to electrophilic attack so if there are two different rings it will attack the smaller one.