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

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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?

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  • $\begingroup$ 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. $\endgroup$ Commented Dec 21, 2015 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
    Commented Dec 21, 2015 at 13:46
  • $\begingroup$ @jan how can i remove one carbon from the chain? $\endgroup$ Commented Dec 21, 2015 at 13:48
  • $\begingroup$ As in using a different reactant. $\endgroup$
    – Jan
    Commented Dec 21, 2015 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! $\endgroup$ Commented Dec 21, 2015 at 13:49

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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.

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    $\begingroup$ 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
    Commented Dec 21, 2015 at 14:38
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    $\begingroup$ @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. $\endgroup$ Commented Dec 21, 2015 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
    Commented Dec 21, 2015 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
    Commented Jan 15, 2017 at 22:25

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