Skip to main content
10 events
when toggle format what by license comment
Dec 23, 2015 at 18:25 vote accept Dimenein
Dec 23, 2015 at 0:57 comment added Jan @jerepierre I admit, I just drew a methyl group there because it’s the easiest way to introduce stereochemistry on the double bond. But yes, you are correct; that can be done better.
Dec 22, 2015 at 16:41 comment added jerepierre Although I like this analysis, the wording of 'less hindered hydrogen' makes it seem that the hydrogen from the beta-methyl would be participating, at least according to the book in question. Do you have any evidence that the more substituted alkene is formed?
Dec 22, 2015 at 14:29 comment added Dimenein By less hindered hydrogen do you mean hydrogen to be removed, leads to the more stable transition state.
Dec 22, 2015 at 14:25 vote accept Dimenein
Dec 23, 2015 at 18:19
Dec 21, 2015 at 18:41 comment added Jan @VinayakAgarwal I just realised a major blunder I had in my images which has now been fixed. You could also draw a transition state towards the terminal methyl group rather than along the chain. That would work equally. However, the double bond you create would be terminal. Therefore, under thermodynamic control (evidently, it’s a pyrolysis after all) the higher-substituted double bond should be created as shown here. I mean, I could turn that methyl into another tBu group but I’m too lazy now.
Dec 21, 2015 at 18:38 history edited Jan CC BY-SA 3.0
Fixed a major blunder in my image there …
Dec 21, 2015 at 18:25 comment added Jan Hm, we probably could remove the methyl one, too, since that is technically also beta. Let me think about this …
Dec 21, 2015 at 17:17 comment added Dimenein I didn't get it . We will remove hydrogen from methyl or from 3rd carbon?
Dec 21, 2015 at 16:53 history answered Jan CC BY-SA 3.0