Catalysis Q + A

1. CatalysisQ + A Dr. Christoph SontagPhayao University

2. Coordination of transition metals Depending on the position in the periodic table, we can see a difference of the transition metals in their behavior when they form complexes: Early TM 16 el and lessare common Coordination 6and higher MiddleTM 18 el common Coordination 6 LateTM 16 el and lessare commonCoordination 4 and less

3. 4 basic steps of catalytic reactions

4. Oxidative Addition Which of the following complexes can undergo oxidative addition of Methyl Iodide ? Write the reactions down and indicate the oxidation number of the metal. Remember the oxidation addition reaction:

5. Answers (+1) (+3) Remember: Oxidative addition means(1) the oxidation number of the metal rises by +2(2) the number of valence electrons rises by 2 (CH3)I + CH3-I 16 elsquare planar 18 eloctahedral (+1) (+3) + CH3-I (CH3)I](-) 16 elsquare planar 18 eloctahedral (+4) No reaction, because Zr cannot have oxidation number +6 ! 14 eltetrahedral (-2) No reaction, because [Fe(CO)4]2- has already 18 electrons and the addition of CH3-I would give a 20 electron complex. 18 eltetrahedral

6. Oxidative Addition (2) Which of the following molecules will be MORE ACTIVE towards oxidative addition of H2 ? OR Please explain your reason. Answer In general: oxidation means electrons are given away ! For an oxidative addition, the metal has to be oxidized (from +1 to +3). This is easier if the metal center is more electron-rich ! Carbonyl (CO) is a ligand with back-bonding, that means electron density goes from the metal to the ligand (into the π* C=O bond) => in the second complex, the electron density on the metal is lower than on the first one => the first complex is easier to oxidize !

7. Oxidative Addition Which reaction do you expect between dimethyl-palladium (II) and methylchloride ? Give an explanation about the reaction. Answer Pd metal can interact with the C-Cl bond in 2 ways: A metal s- and p-orbital can overlap with the σ bond, transferring electron density from this bond into the empty p-metal orbital. A second interaction can occur between (filled) metal-d orbitals and the ANT-BOND σ* of the C-Cl bond, in this way can make the bond break.

8. Reductive Elimination The reverse reaction to oxidative addition: 2 ligands on a metal form a bond and dissociate from the complex. In this way they leave electrons from the M-L bond behind, reducing the metal by 2. Write the reaction for the elimination of hydrogen in this molecule: Indicate also the number of valence electrons and oxidation number of Ir in the start and end complex.

9. Answer This reaction is common for H-M-H to form hydrogen and for R-M-H to form a R-H molecule. It is also possible that 2 carbon ligands combine together to form a new C-C bond.

10. Oxidative Addition and Reductive Elimination Both names describe the same reaction depending on the direction, for example: Which direction is an oxidative addition and which one is reductive elimination ? Write the oxidation number of the Pt center for both molecules to confirm your idea.

11. Answer For this example, both reactions are possible and it depends on the conditions where the equilibrium will be. In this case under heating, the oxidative addition is preferred.

12. Olefin Coordination to a Metal When an olefin is coordinated to a transition metal, does the pi-bond become stronger or weaker ? Compare the C=C bond in ethylene and coordinated to Platinum: Answer A pi-bond is symmetric for rotation and therefore has the same symmetry as a metal s, pz and dz2 orbital: electron density from the pi-bond will go to the metal orbitals forming a bond. BUT: there is also an anti-bonding pi* orbital which has the same symmetry as a dxz metal orbital, that means that electron density from the metal can go into this ligand orbital and makes the pi-bond weaker. Opposite electron density from a metal d-orbital can go into the pi(*) orbital of ethylene, making the bond weaker Electron density goes from the ethylene pi-orbital into the metal p-orbital

13. Insertion / Migration This reaction type is especially important because a new bond between 2 ligands is formed.: Please write the number of valence electrons and oxidation number in this example: (“insertion” because it looks like the CO is inserted into the Mn-C bond. The “real” mechanism is a migration of the CH3 rest to the carbon of the carbonyl)

14. Answer The oxidation number of the metal does NOT change: Look carefully at the 2 molecules (which are in an equilibrium to each other !): It looks like one ligand moved away from the metal, taking its 2 bonding electrons with it.But the leaving ligand does not dissociate away, instead it makes a nucleophilic attack on the carbonyl C, forming a new C-C bond.

15. Insertion / Migration A carbonyl ligand often can react with another alkyl ligand on the same metal to form an acyl rest.Which reaction would you expect in this Pd-complex: Answer We can expect a migration of the CH3 group to the carbon of the CO ligand, forming anacylgroup: Remember that this reaction is reversible ! An acylgroup can dissociate to form a CO and an alkyl ligand – but this direction is less likely than the above reaction.

16. Insertion / Migration Describe the migration of a hydride ligand to a carbon of a neighboring olefin ligand: Draw the path of the hydride ligand into the picture. Check the number of valence electrons and the oxidation number of Rh in both molecules. Does the oxidation number change ? Why ?

17. Answer We can think of this reaction as a nucleophilic attack of a hydride ligand on a carbon of the olefin: Hydride (H-) ligand uses its 2 electrons to form a H-C bond.At the same time 2 electrons of the olefin pi-bond form a C-Rh bond. This reaction is a re-arrangement of bonding electrons and the oxidation number of the metal does not change.

18. Catalytic Cycle Write the number of valence electrons of the Rh-complexes.Also indicate the oxidation number of the metal.

19. Answers The cycle starts with the 16-electron complex RhCl L3. (Rh(+) has 8 valence electrons + 4x 2 el from ligands) 16 elRh + (1) (2) (7) 18 elRh 3+ 18 elRh 3+ (3) (6) 16 elRh 3+ 16 elRh 3+ (4) (5) 18 elRh 3+

20. Explanations • RhClL3 is a square planar 16-el complex. For this geometry this is a stable configuration • Oxidative addition of a H2 molecule: the H-H bond breaks and 2 new Rh-H bonds are formed. The hydrogen ligand counts as hydride (H-), so that the oxidation number of the metal rises from +1 to +3 and the whole complex gets 2 more valence electrons (from the H-H bond). • From the octahedral complex, one ligand L can dissociate, taking 2 electrons with it. The oxidation number of the metal does not change but the total number of valence electrons is reduced. • The free coordination place can be filled by an ethylene molecule. Important is that the coordination happens by the 2 pi-electrons of the double bond. • One of the most important steps is the hydride migration to a carbon of the ethylene, forming a new C-H bond and change the C=C bond to C-C • The new free coordination place can be filled by the Ligand L which dissociated in step 3. • In the last step, a hydride ligand forms a new C-H bond (nucleophilic attack on carbon) and the reduced alkene dissociates away from the metal. This step is the opposite of the oxidative addition (step 2), the metal is reduced because 2 electrons (one from the Rh-H, another from the Rh-C bond) remain on the metal.

21. Hydroformylation An important catalytic cycle is a reaction where carbon monoxide can be added to an olefin, forming an aldehyde. H-H Write the number of valence electrons and the oxidation number of Cobalt for each step.

22. Answer The cycle starts with a tetrahedral Co(CO)3H complex, with Co(+) and 16 valence electrons. 16 elCo(+) 18 elCo(+) H-H (3) 18 elCo(+) 18 elCo(+) (1) (2) 18 elCo(+) Different from the previous cycle, we have here altogether 3 migration reactions: The important steps are the migration of a hydride to the olefin (new C-H bond) (1) and afterwards the migration of the alkyl ligand to a carbonyl, forming a new C-CO bond (2). The final product can be released (by reductive elimination) if a hydrogen molecule makes a oxidative addition (3)

23. Answer cont. Consider the final step (3) in detail: A hydrogen molecule is coordinated to the metal, breaking the H-H bond and oxidizing the Co(+) to Co(3+): H Co(3+) + H-H H +

24. Olefin Hydrogenation Well-known as “hardening” of natural fatty acids with many double bonds to make the product more stable towards oxidation and influence the viscosity (for butter, margarine): Write the names of the basic steps in this cycle. For which transition metals is this process more likely, the early, middle or late metals ?

25. Answers • The first step is the oxidative addition of a hydrogen molecule, forming 2 M-H bonds. • Then a hydride ligand can migrate to a carbon of the olefin, changing the M-|| bond into a M-C bond. • The second hydride ligand can make a nucleophilic attack on the other carbon of the former C=C bond, releasing the alkane molecule (reductive elimination) • Finally a new olefin can coordinate to the metal center. (4) (1) oxidative addition coordination(addition) reductiveelimination (3) migration (2)