6 comments by 3 or more people

1. Matthew Taylor

   In the frontier molecular orbital picture for 1,2-MI towards the end of topic 3, im having difficulty understanding the diagrams you have drawn. i see how you have both the HOMO alkene/LUMO M-L and the LUMO alkene/HOMO M-L interactions drawn but are we saying that these are both happening at the same time? In both are the two new bonds forming at once or is the last one being formed afterwards?
   A quick explanation of these diagrams would be very helpful
   Thanks
   Matt.

   13 Dec 2012, 16:40

2. Peter Scott

   Hi Matt,
   Sorry for late reply. Been away.
   Yes, these are both happening at the same time, although the extent to which that is true depends on particular circumstances e.g. the availability of metal d electrons.

   24 Dec 2012, 07:09

3. Pratik Gurnani

   What decides which way round an alkene will bind to a metal centre to then go and undergo a 1,2 Migratory insertion. I’ve seen in a few examples eg butene the branched and linear products are made. Is this just sterics or will both products always be made?

   Thanks
   Pratik

   03 Feb 2013, 13:23

4. Peter Scott

   In the case of simple alkenes it is mostly sterics although secondary alkyls are destabilised by inductive effects as the C atom is delta negative. Also for example benzylic alkyls are stabilised so styrene would tend to insert (at a less bulky metal, or an electropostive metal) so as to give secondary insertion i.e. M-CH(CH3)-Ph

   05 Feb 2013, 10:01

5. Robin

   I’m confused about selectivity of the 1,2 MI step in the MMA catalytic cycle in the topic 4 questions. I’m a bit confused as to why it’s produces the more substituted markovnikov product as illustrated in the cycle.

   28 Feb 2013, 09:07

6. Peter Scott

   The electronic default is usually the Markovnikov product but steric pressure can change this. Here there is little steric pressure as CO is such a small ligand. Also, don’t forget that the TS will not have sp hybrid C atoms but be more bent than propyne.

   In the observed product the polarity of the corrdinated alkyne matches up nicely with the polarity of the Ni-H bond i.e. Ni(delta +) and hydride (delta -) so assuming that the MI is synchronous (which we know it to be) then this leads to the preferred transition state. The fact is that selectivity is >95.5% in the real catalytic reaction.

   28 Feb 2013, 09:09

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