## January 02, 2012

### TOPIC 4: Hydrogenation, hydroformylation and carbonylation

Please click on “Comments” and type a question. I will respond as soon as I can

### 6 comments by 3 or more people

1. #### George Sykes

Hi, i’ve been going through your notes, specifically the slides on Methanol Carbonylation (BP/Monsanto Process) and wondered if you could help me with a few things?

I’m finding the questions on the next slide difficult. Specifically:

Why does an anionic catalyst promote nucleophilic OA? it’s negatively charged , so surely will be less stable if attacked by a negativelty charged Cl- ligand.
What id the RDS? I’d assume from the rate equation that the RDS is the nucleophilic OA step when the catalyst reacts with the co catalyst…But i dont know why this step should be slower than the 1,1 M.I step that follows?

Also, to clarify, is the ML3X3 complex a 19e complex or have I added that up wrong? Is this why the M.I step is fast? because it reduces to a more stable 17e complex?

Thanks a lot

06 Nov 2012, 16:25

2. #### Peter Scott

In nucleophilic OA the metal is acting as the nucleophile so it being anionic would normally make it more nucleophilic. Incidentally, not sure where the Cl- features. Do you mean iodide?

You are quite right about the RDS. From the rate equation you can only say that either the OA or the subsequent 1,1-MI step are rate determining (because the latter is a rearrangement). Having said that, we know that anionic promoters such as iodide do increase the overall rate, which implies that of these steps the OA must be the RDS.

The ML3X3 complex must be 18 VE as M + L + X = 9 + 6 + 3 = 18.
Not sure how you are getting 19, but note that as drawn the complex is actually [ML2X4]- but since X- = L, that reduces to ML3X3.

The MI step takes it to [MLX4]- = ML2X3 = 16 VE

06 Nov 2012, 20:05

3. #### Rosa Sullivan

Hello Professor,

I hope you haven’t answered this somewhere, potentially you have, in which case I apologise.

With the catalyst reactions Such as the Wacker cycle, would you be expecting us to be able to recall the entire process? Or would it be more like you might give us the process but ask us to name reactions occuring, maybe fill in a missing step or something? Basically, I want to know whether I should be memorising these processes or just be familiar with them and understand the steps.

Rosa

24 Jan 2013, 13:28

4. #### Peter Scott

I will not be asking you to recall the entire cycle. In the past I’ve asked for the things you suggest such as classifying (naming) the reactions, filling in a missing structure, adding some detail etc but none of this needs you to memorize the cycle. As you say, you should be be familiar with them and understand the steps. I the end, the cycle I give you may be one that we did not study, but it will be sufficiently similar and use the same sorts of reactions.

24 Jan 2013, 14:19

5. #### Geoff Wilkinson

In question 1 part (ii) of topic 4, the answer you say as to why the wilkinsons catalyst is quicker to react with H2 than compound A is: “The presence of CO ligand stabilises the lower valence number (oxidation state) compared with PPh3 in Wilkinson’s catalyst. Explain how (see year 2 notes).”

I dont really understand the answer. Is it due to the fact that the PPh3 ligands are better sigma donors to the metal therefore the metal will have stronger backbonding to the alkene hence a lower energy barrier? What do you mean by the presence of CO ligands stabilises the lower oxidation state?

Thanks for the help!

18 Feb 2013, 11:50

6. #### Peter Scott

The presence of pi-acceptor ligands like CO lowers the energy of the metal orbitals (compare topic 1 slide 11 “classical complexes” with slide 12 “why 18e”). This means that those electrons are lower in energy, so they must have a higher ionisation energy (which corresponds to oxidising an electron). In other words it is harder to oxidise those elecrons, so the lower oxidation state of the complex is stabilised. If you were to replace those CO ligands with principally sigma ligands such as phosphines, there would be less pi backbonding overall and more sigma bonding. Both these things would put the energy of the orbitals up and so make the electrons easier to remove.

So, CO ligands make it harder to oxidise the compound, which is the same as saying that the lower oxidation state is stabilized.

18 Feb 2013, 11:55

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