5
$\begingroup$

enter image description here

I have suggested a mechanism for the formation of 3-ethyl-2-methylcyclohex-2-en-1-one A

enter image description here

but the question says problems would be encountered if a similar route was used to make 2,3-dimethylcyclohex-2-en-1-one B. I think perhaps a decarboxylation step is involved? Can someone please explain how to get to B and the difficulties in the mechanism?

$\endgroup$
  • $\begingroup$ It looks like some sort of trick question. If you count the carbon atoms B has one less but it isn't clear how to remove it under the given conditions. I think the answer is that B is a typographical error. Hope this helps. $\endgroup$ – user1945827 Dec 21 '15 at 13:44
  • $\begingroup$ A similar route to B sounds to me like just removing one carbon atom from the chain to hope to arrive at B. I actually see no problem with this as long as you are in thermodynamic control … But I could be missing something obvious … $\endgroup$ – Jan Dec 21 '15 at 13:46
  • $\begingroup$ @jan how can i remove one carbon from the chain? $\endgroup$ – justbehappy Dec 21 '15 at 13:48
  • $\begingroup$ As in using a different reactant. $\endgroup$ – Jan Dec 21 '15 at 13:48
  • $\begingroup$ @Jan oh i see. As in changing the starting material completely? That was not clear from the question.. I will try it! Thanks! $\endgroup$ – justbehappy Dec 21 '15 at 13:49
4
$\begingroup$

Your starting product for A is symmetrical, but your starting product for B will be asymmetrical - one carbon less on one side. This is going to mean you may get 2 or more different products; B may not be your major product, and you'll have to separate them afterwards. I don't believe there's inherently a tricky difficultly in the mechanism for B.

$\endgroup$
  • 1
    $\begingroup$ Actually no. Under thermodynamic control (i.e. given enough time) the product B is inherently more stable than the potential side product and it will form. Reasoning: Tetrasubstituted versus trisubstituted double bond. $\endgroup$ – Jan Dec 21 '15 at 14:38
  • 2
    $\begingroup$ @Jan My two cents: you are right, but the question may not necessarily be. I think this is the answer the question wants. If OP can edit and make a note of this piece of information I'd +1. $\endgroup$ – orthocresol Dec 21 '15 at 15:41
  • $\begingroup$ Wouldn't the more favoured enolate in the starting material needed to form B (6-oxo-2-octanone, probably not the IUPAC name) actually mean B as a product is not favoured? $\endgroup$ – Beerhunter Dec 21 '15 at 21:34
  • $\begingroup$ @Beerhunter As I wrote above, under thermodynamic control B is more stable than the side product 2-ethylcyclohex-2-en-1-one and will thus be formed preferentially. $\endgroup$ – Jan Jan 15 '17 at 22:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.