TOPIC 5: Olefin Catalyses
Writing about web page http://www.warwick.ac.uk/go/ch3a2
Please click on “Comments” and type a question. I will respond as soon as I can
Peter Scott
02 Jan 2012 19:19
| 
Comments (4)
| 
Report a problem
Dan Warr
Hi
On the 2012 past exam paper question 5c I’m unsure where to start, I can see there looks to be difficulty in a direct beta elimination and the incorporation of the D leads me to believe it may be a beta hydride transfer from the phosphine?
Also on the same paper, the very last question, I understand the process to arrive at each product but I don’t know why the anti-Markovnikov is dominant? I thought maybe sterics for the attack of the H2O on the alkene after the initial atack of water at the metal centre?
17 Feb 2013, 14:16
Peter Scott
Daniel
I’ve had a few question about this. These particular questions are not very useful for you. The module has been updated a liittle from last year and while I am sure we could spend time on these questions it would not really help with the exam! Basically though you are on the right track. It is C-H (or C-D) oxidative addition that is the key step. We did not look at this process as Adrian Chaplin now does it in detail next year.
Similalrly, I would not worry about the second part of your question.
You’ll notice I did not use these question in the set problems.
18 Feb 2013, 12:03
Hi Peter
Been looking at the past paper from last year and i have a couple of problems with question 1. Question 1 b, I was wondering how protic impurities actually react with the catalyst to produce an unreactive species, and what MAO actually does to “scavange” for these. And for Question 1 d, I couldnt figure out how the increasing pressure of Propene effects the reaction, is it just Le Chatelliers principles and iy shifts equilibrium to the right and produces more product?
Thank you!
25 Feb 2013, 19:23
Peter Scott
A protic impurity could be e.g. ethanol. It would react with the metal alkyl to give the alkane and a Zr-OEt group which would not be a catallyst as the Zr-O bond is too strong. MAO woudl react with the ethanol in the same way before it could get to the catalyst.
The polymer is not formed at equilibrium, but the alkene complex at the cationic Zr centre is. Hence we expect the rate of insertion to be greater at higher concentrations because more alkene complex would be formed.
25 Feb 2013, 23:55
Add a comment
You are not allowed to comment on this entry as it has restricted commenting permissions.