-1
$\begingroup$

Is an allylic rearrangement possible in formation of Grignard reagent? Going by the following single electron transfer mechanism, I think it's possible. But, I haven't found any references.

enter image description here

$\endgroup$
2
  • $\begingroup$ The discussion in the comments here might help:chemistry.stackexchange.com/questions/128087/… $\endgroup$ Commented Feb 20, 2020 at 16:53
  • $\begingroup$ Going through the comments, I feel the above SET mechanism might be incomplete. But, I would like to know if any product spread has been observed. References would be helpful. $\endgroup$
    – Shashank
    Commented Feb 20, 2020 at 18:13

1 Answer 1

2
$\begingroup$

I do not have references to the rearrangement of Grignard reagent going through single electron transfer mechanism during its formation. However, it is understand that Grignard reaction of an allylic magnesium halide with an electrophile can give a new bond at either carbon of the allylic carbanion, so two isomeric products are possible (Ref.1 & 2). Once formed, allylic magnesium halide is going to be in fast equilibrium with its other isomer.

For example, when generated from either (E)- or (Z)-bromide, (E/Z)-2-buten-1-ylmagnesium bromide (1; $\mathrm{R'} = \ce{CH3}$) would be in fast equilibrium with its other isomer, (2R/2S)-3-buten-2-ylmagnesium bromide (2; $\mathrm{R'} = \ce{CH3}$), favoring 1 in this case (Refs. 1-3). The authors of Ref.3 claims that regardless of starting material (whether it is crotyl bromide or $\alpha$-methylallyl bromide), the resulting major Grignard reagent is 1. When this 1/2 mixture is reacted with a ketone, two isomeric alcohols, 3 and 4, would formed via intermediates I and II:

Allyl Grignard

Increasing the steric bulk of $\mathrm{R'}$- and $\mathrm{R}$- destabilizes I compared to II, favoring 4 over 3 (Ref.1 & 2). For example, when $\mathrm{R'} = \ce{-C(CH3)3}$ (tert-Bu-), the product ratio of 4/3 is $8$ if ketone used is $\ce{Et2C=O}$ (diethylketone) while it is $\gt 100$ if ketone used is $\ce{((CH3)2CH)2C=O}$ (diiso-propylketone; Ref.1). This second instance can be compared to the ratio obtained for the reaction when $\mathrm{R'} = \ce{CH3}$, which is equal to $0.5$.

The allylic Rearrangements have also discussed in earlier paper by Roberts and Young (Ref.4). The suggested mechanism for this rearrangement is discussed in details in Ref.5.

References:

  1. Marc Chérest, Hugh Felkin, Claude Frajerman, “The reaction between ketones and t-butylallyl magnesium bromide. A sensitive measure of steric hindrance in the neighbourhood of the carbonyl group,” Tetrahedron Letters 1971, 12(5), 379-382 (https://doi.org/10.1016/S0040-4039(01)96446-9).
  2. Michael B. Smith, In Organic Synthesis; 3rd Edition; Academic Press: New York, NY, 2011 (Hardcover ISBN: 9781890661403).
  3. J. Eric Nordlander, William G. Young, John D. Roberts, “Nuclear Magnetic Resonance Spectroscopy. The Structure of Butenylmagnesium Bromide,” J. Am. Chem. Soc. 1961, 83(2), 494-495 (https://doi.org/10.1021/ja01463a062).
  4. John D. Roberts, William G. Young, “Allylic Rearrangements. XVI. The Addition of the Butenyl Grignard Reagent to Some Simple Carbonyl Compounds,” J. Am. Chem. Soc. 1945, 67(1), 148-150 (https://doi.org/10.1021/ja01217a051).
  5. Hugh Felkin, Claude Frajerman, “Competitive rate studies for some reactions involving the allyl, butenyl, $\alpha,\gamma$-dimethylallyl, and propyl Grignard reagents. the mechanism of the reactions of allylic Grignard reagents with electrophilic substrates,” Tetrahedron Letters 1970, 11(13), 1045-1048 (https://doi.org/10.1016/S0040-4039(01)97903-1).
$\endgroup$

Your Answer

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

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