8
$\begingroup$

So from the general reaction we can see that the carbonyl is reduced while the Grignard is oxidized to an alkene.

So for this hypothetical reaction (does this really occur?) we are asked to provide a plausible mechanism. This is what I got, but the prof. wants something better - why transfer negative charge from oxygen to an $\text{sp}^3$ carbon?

enter image description here

Click on the image to enlarge, if you wish.

Here are his hints:

Remember, prior to Grignard reaction the $\ce{Mg}$ center swaps out an ether solvent molecule for a reactant oxygen. This interaction will lead to the reaction but in this case, due to steric interference a hydride is delivered instead as the Grignard is oxidized to an alkene. In this way the Grignard reagent is almost reacting in the way $\ce{NaBH4}$ would in the presence of this ketone. Try to get all of this action working together in a single, concerted step.

So ... I get the first part just fine. The $\ce{Mg}$ center has an octet; it's solvated by two ether molecules in solution usually. Okay and in this case the carbonyl oxygen helps solvate the $\ce{Mg}$ center. And then hydride transfer occurs ... but how exactly do we get the hydride to leave without creating a carboanion? And is it really that important that everything be concerted?

This is my second attempt ... if this is all concerted in real life then I must have hit the orbital alignment lottery! All the stars must have been aligned ... !

The only thing I see about my second attempt is that in the end the $\ce{Mg}$ center doesn't have a complete octet - perhaps I should have shown another ether molecule popping in to help $\ce{Mg}$ out with its octet fantasies.

enter image description here

Improve this question
$\endgroup$
5
  • $\begingroup$ are there two methyl groups missing in the third step and the final product? $\endgroup$
    – bon
    Feb 24, 2015 at 18:28
  • $\begingroup$ @bon yes, sorry. Just imagine they are there; I don't think that makes a difference to my question. $\endgroup$
    – Dissenter
    Feb 24, 2015 at 18:29
  • $\begingroup$ ...Gets me dubious. Do those images need to be that large? $\endgroup$
    – M.A.R.
    Feb 24, 2015 at 18:51
  • $\begingroup$ @MARamezani I've fixed all the images with respect to chemistry. As far as the size of the images goes ... I don't know how to adjust them. $\endgroup$
    – Dissenter
    Feb 24, 2015 at 18:52
  • 1
    $\begingroup$ You can optimize what I did, and it wouldn't hurt to look at this: meta.stackoverflow.com/questions/253403/… :) $\endgroup$
    – M.A.R.
    Feb 24, 2015 at 19:09

1 Answer 1

Reset to default
7
$\begingroup$

This reaction can be viewed as an example of a pericyclic (concerted) Ene reaction. The ene reaction is actually fairly common in organic chemistry. The following diagram details the mechanism as a concerted reaction with a cyclic 6 membered transition state involving 6 electrons (an "aromatic" transition state).

enter image description here

A note on the Ene reaction...

The Ene reaction is really just a variation on the Diels-Alder theme. The Diels-Alder reaction is a concerted [4 pi + 2 pi] reaction, whereas the Ene reaction is a [(2 pi + 2 sigma) + 2 pi] concerted reaction.

enter image description here

Improve this answer
$\endgroup$
7
  • 1
    $\begingroup$ you know everything! Not even the text covers ene reactions. $\endgroup$
    – Dissenter
    Feb 24, 2015 at 18:58
  • 1
    $\begingroup$ You can do this reaction with alkenes too; check out [books.google.co.uk/… Modern Methods of Organic Synthesis] $\endgroup$
    – J. LS
    Feb 24, 2015 at 19:14
  • 2
    $\begingroup$ The simple answer is yes, but to be clear I need to make it a bit more complicated and maybe use some new words. Transition states with 4n+2 electrons are aromatic when they proceed with no orbital symmetry sign inversion (suprafacially in a 1,5-hydrogen shift). 4n electron transition states are aromatic when they involve an orbital symmetry inversion (e.g. the antarafacial process in the 1,7-hydrogen shift). The 4n+2 process becomes antiaromatic when there is an orbital symmetry inversion and the 4n process becomes antiaromatic when there is no orbital symmetry inversion. $\endgroup$
    – ron
    Feb 24, 2015 at 19:21
  • 1
    $\begingroup$ See here for example. $\endgroup$
    – ron
    Feb 24, 2015 at 19:22
  • 1
    $\begingroup$ My comment above "The simple answer is" is inadequate. It takes a chapter in a book to work through the major topics. Perhaps it would be better if I said that reactions involving 4n or 4n+2 electrons can both proceed through an aromatic transition state, but that the trajectories followed will be different. Positive overlap in the HOMO describing the bonds in the starting material that will break must be maintained in the bonds that will form. To maintain this positive overlap can require very different motions (conrotatory-disrotatory, suprafacial-antarafacial, etc.). See the above link. $\endgroup$
    – ron
    Feb 25, 2015 at 1:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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