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Recently, I have been reading up on the Baeyer-Villiger (BV) oxidation. The oxidation is most commonly discussed for ketones. However, the oxidation also works for aldehydes and $\alpha$-diketones (Rojas, 2015).

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The above image is Scheme 2.2 from Rojas (2015).

I have mainly two queries:

  1. Does the oxidation of aldehydes to carboxylic acids by peroxy compounds, such as peracids, proceed via a BV oxidation where the hydride is the migrating group? This would be seemingly different from the oxidation mechanism for the oxidation by $\ce {MnO4^-}$ or $\ce {CrO3}$ where the aldehyde is oxidised when it is in the gem-diol form.

  2. Why does the oxidation of $\alpha$-diketones yield the corresponding anhydride and not an $\alpha$-ketoester? This is extremely puzzling because the acyl group is terribly bad at stabilising any partial positive charges on the acyl carbon. How is it possible that such a BV oxidation can occur?

Reference

Rojas, C. M. (2015). Molecular Rearrangements in Organic Synthesis (1st ed.). Hoboken, New Jersey: John Wiley & Sons, Inc.

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    $\begingroup$ 1) Hydride migrates to leave the more stable cation. 2) Acyl migrates for the same reason. $\endgroup$
    – user55119
    Mar 11, 2019 at 2:32

1 Answer 1

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The Baeyer-Villiger oxidation is conducted under acidic conditions such as an alkyl peroxide in the presence of a mineral acid (e.g.; H2SO4) or a peracid (e.g.; peracetic acid or m-chloroperbenzoic acid). In the reactions described in the diagram, the reactions of aldehyde 1 or α-diketone 7 are initiated by proton of the carbonyl oxygen followed by addition of peroxide (an alkylperoxide is illustrated). Ultimately, the oxygen of the RO-moiety is protonated and is the effective leaving group (ROH) in the rearrangement. In both sequences, the red route is favored; the blue route disfavored. Cation 3 is a more stable species than cation 5 because of hyperconjugation with the alkyl group or resonance stabilization if R'=aryl. Cation 3 affords a carboxylic acid while cation 4 would lead to formate ester 6.

In the case of peroxyhemiketal 8, acyl migration leads to cation 9 which is more stable than species 11, which has a positive charge adjacent to a carbonyl group, which is destabilizing. Thus, anhydride 10 is formed in preference to α-ketoester 12.

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  • $\begingroup$ Thank you for your answer. I suppose migration of the acyl cation is favourable due to stabilising interactions from the lone pair on the carbonyl oxygen? $\endgroup$
    – Tan Yong Boon
    Mar 12, 2019 at 4:15
  • $\begingroup$ Because you haven't quite answered my query with regards to the favourability of acyl migration $\endgroup$
    – Tan Yong Boon
    Mar 12, 2019 at 4:16
  • $\begingroup$ I appreciate your hesitancy about accepting the existence of an acyl anion. I agree. But in the rearrangement there is neither an R' nor acyl anion per se. As either of these electron pairs become partial bonded to oxygen, a positive charge begins to develop in the transition state leading to 9 and 11. The positive charge developing in 11 is not good. Clearly, the activation energy to 9 is lower than to 11. It is a competitive, irreversible process. PS: An acyl cation is not involved! $\endgroup$
    – user55119
    Mar 12, 2019 at 13:30

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