Master Organic Chemistry Reaction Guide

Beckmann Rearrangement

Description: The Beckmann rearrangement is a reaction of oximes that can result in either amides or nitriles, depending on the starting material. (Oximes derived from ketones give amides; oximes derived from aldehydes provide nitriles)

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Real-Life Example:

Org. Synth. 1939, 19, 20

DOI Link: 10.15227/orgsyn.019.0020

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Comment section

43 thoughts on “Beckmann Rearrangement

  1. Regarding the nitrile formation, what if the oxime is treated with acetic anhydride and sodium acetate (as in the Wohl degradation reaction of carbohydrates)? How would the nitrile be formed? Would a Beckmann rearrangement occur? Would acetic anhydride be used to make OH a good leaving group?

    Wohl degradation reaction:

    1. That’s essentially a Beckmann rearrangement as well. What happens is you form OAc on the oxime, and then there’s a 1,2 shift of H- from the carbon to the nitrogen and it will break the N-O bond. After that , deprotonation of N-H gives you the triple bond.

  2. In one of the Beckmann rearrangement questions the reagent given was NH2OH.HCl , PCl5 and heat. What does PCl5 do?

  3. In a version of this reaction in which ring expansion occurs, does it do so with the same regioselectivity as say the B.V. oxidation? By that I mean I’m under the impression that the oxygen added in expanding a ring will generally add to the more substituted side (though that may be incorrect too). Will the NH group do the same, assuming the molecule allows it to?

    My specific example is a benzyllic ketone, located on a 7-membered ring. But even just a clarification of the generalities would be amazing.

    Thank you! This site is the only thing keeping me afloat in my class, I recommend it to everybody

  4. By this mechanism, ketone always give amide while Aldehyde give nitirile ?
    and in this reaction when R1 is ethyl and R2 is methyl or something else (When R1 and R2 is different) which R group is shifted and why?

  5. Since the group migration step is similar to that in Curtius, Schmidt, etc : Is it necessary that an aldehyde will always give a nitrile?

    In case that the Hydrogen to be migrated is not at Trans-position with respect to OH- , wouldn’t that give formic acid derivative of amine?

    1. Oximes can interconvert their stereochemistry (cis and trans) so it would still give an aldehyde unless there was some special constraint which prevented the adoption of that stereochemistry.

    1. What makes a good leaving group? Good leaving groups are weak bases. HO- is not a good leaving group, but once you get into the territory of AcO(-), H2O, etc. then the reaction will proceed nicely.

  6. I think you missed a very imp. part(Stereochemistry of the reaction)–that the alkyl substituent “trans” to nitrogen migrates.

    1. None, but with the HCl salt, you’ll need an equivalent of weak base to neutralize the HCl salt so that the lone pair on the nitrogen can be freed up.

  7. Let’s say we’ve got methyl group on carbon 2. Which carbon would do the rearrangement? C2 (more substituted) or C6? Thank you!

    1. Migratory aptitudes generally increase with carbocation stability. Carbocation stability decreases as electron withdrawing groups are added. One can think of the transition state in the Beckmann as a 3-center, 2-electron bond, which is electron deficient.

  8. It’s very bad practice to depict intramolecular proton transfers if they go through 3 or 4-membered ring transition states, as you did in step 2 of the oxime formation mechanism (4-membered ring transition state). You should instead depict two proton transfer steps involving hydroxylamine as a proton shuttle.

  9. Hi,
    I have a derivative of benzaldehyde that is converted into benzonitrile by forming an oxyme with hydroxylamine in formic acid and sodium formate. I think the formic acid is the solvent, but what would it be the role of the sodium formate?

    1. Not sure. Might be to buffer the acidity of the solution because when the pH is too low, hydroxylamine will be completely protonated and won’t be nucleophilic enough to perform the addition to the aldehyde. Are you starting with hydroxylamine hydrochloride?

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