5
$\begingroup$

One of the ways of preparing Grignard reagent is: $\ce{R-X ->[Mg/ether] R-MgX}$

However let's assume that a carbonyl group is present along with a halide in the substrate. Will $\ce{Mg/ether}$ react with the carbonyl group in a way similar to that by $\ce{Zn}$ in Clemmensen reduction? Or would $\ce{Mg/ether}$ react with the carbonyl group in any way at all?

$\endgroup$
2
  • $\begingroup$ How do you mean by similar way to ZN in clemmensen? I don't quite understand, could you explain a bit more? $\endgroup$
    – Buraian
    Aug 21 at 15:58
  • $\begingroup$ @Buraian metals like Na,Mg,Zn can release electrons which ends up reducing or reacting with carbonyl compounds in many cases. However the reaction and end product varies $\endgroup$
    – Ashish
    Aug 21 at 16:12
12
$\begingroup$

It is known that $\ce{Mg}$ will reduce ketones. The articles (Ref.1 and 2) in the References section confirm this fact.

If you want to form the Grignard reagent of a molecule that also happens to contain a carbonyl group, then the usual practice is to protect it first. A cyclic ketal/acetal, which are stable under basic conditions, is often the first choice for this.


References:

  1. Ji Young Kim, Hak Do Kim, Min Jung Seo, Hyoung Rae Kim, Zaesung No, Deok-Chan Ha, and Ge Hyeong Lee, "Reduction of ketones to corresponding alcohols with magnesium metal in absolute alcohols," Tetrahedron Lett. 2006, 47(1), 9–12 (DOI: https://doi.org/10.1016/j.tetlet.2005.10.140).
  2. Magnesium - Organic Chemistry Portal
$\endgroup$
1
  • $\begingroup$ I have added few words to make some sense. If you don't like what I did, please feel free to roll back. Good finding though. $\endgroup$ Aug 20 at 16:50
-1
$\begingroup$

I think that the formation of the Grignard will be inhibited. The formation of a Grignard is an autocatayltic reaction. It is well known that one of the best ways to get a Grignard synthesis going is to add some of an existing solution of a Grignard reagent.

I reason that as soon as any of the Grignard forms then it will be consumed by the reaction with the ketone. As a result the reaction of the alkyl (or aryl) halide with the magnesium will be so slow that for all intents and purposes we can say "nothing happens"

It will be like trying to make a Grignard in ether that has a trace of water, I once tried to make methyl magnesium iodide and failed. I then discovered that the ether was not as anhydrous as I thought it was. I am sure that if something is present that destroys the Grignard as soon as it forms then the rate of the reaction of the magnesium with the alkyl or aryl halide is so slow that as a synthetic reaction we can regard the reaction as being "useless".

When I tried to make the Grignard in the slightly damp ether, I tried adding iodine, I tried ultrasound and even these things were not able to get the synthesis of the Grignard going. I have always found that if I exclude water strictly from the reaction (bake all the glass equipment overnight before use and sodium dried ether) that the synthesis of a Grignard never fails. What I do is to stirr the magnesium under nitrogen for a long time. I then add a little ether and some of the alkyl halide and I leave it with no stirring for a while. This normally gets it started.

$\endgroup$
3
  • 2
    $\begingroup$ I don't think "nothing happens". I think starting materials will be consumed but nothing useful comes out of the reaction mixture. $\endgroup$
    – Waylander
    Aug 20 at 20:52
  • $\begingroup$ Mg in ether , benzene and possibly other solvents couples ketones to give pinacols [1,2 diols]. I would imagine were a halide present in the molecule the results would be very exciting putting two exothermic reactions together. A similar reaction is the acyloin condensation that couples esters using Na metal. I suspect the acyloin reaction can happen readily electrochemically. The Reformatsky reaction of an alpha bromoester with zinc and a ketone reacts as an insitu Grignard attacking the ketone but not the ester. The idea of insitu Grignard reactions is intriguing. $\endgroup$
    – jimchmst
    Aug 22 at 2:33
  • $\begingroup$ Having run the pinacol reaction of acetone as an undergraduate, I know that the reaction of magnesium with acetone in benzene requires the use of a mercury salt to help things along. The mercury(II) chloride we used would have reacted with the magnesium to form magnesium amalgum which will be more reactive to the acetone than the magensium turnings would have been. I know of a greener way of doing a similar reaction which uses aquerous ammonium chloride to activate the magnesium. I think without the mercury or another activating agent that the reaction will not go. $\endgroup$ Aug 22 at 4:35

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.