Beckmann Rearrangement

[@dinofirechief40]

I need to start this chemistry series from the beginning. I got caught on the excellent math and astronomy vids and have been watching the chemistry as it comes out. I'm pretty lost, but you do a good job of making it interesting. Keep "cooking it up." Thank you.

[@benscrafield9575]

Some detail missed and some outright inaccuracy here. Firstly, protonation occurs at N first, then moves to O. Because of the big difference in pKa, this step is the RDS of the Beckmann, and this is why the reaction typically needs to be heated. Then, you suggest later that the regiochemical outcome of the Beckmann depends on the migratory aptitude of the R groups - this is kind of untrue. The Beckmann rearrangement is stereo- and therefore regiospecific in that the group that migrates has to be antiperiplanar to the OH (it overlaps with σ*). If you have a mixture of oximes, you will see a mixture of products, but each molecule reacts stereospecifically. If you took a single oxime isomer, you would see a single Beckmann product. The only time when the reaction is under thermodynamic control in the way you suggest is if the oxime isomerises faster than the rearrangement - then you’ll get the most cation-stabilising group migrating.

[@ProfessorDaveExplains]

Thanks for the insightful comments, which are correct. This is an old video, slated for replacement in the light of my new series on name reactions, which will be a step up in sophistication over these older, elementary videos. You are correct that the N is more basic than the OH group. I just simplified the mechanism by not mentioning intermediates NOT on the reaction path, but of course a complete mechanism should contain ALL species involved. On the other hand, the Beckmann usually needs heat even when the OH group is activated in other manners, like via cyanuric chloride or tosyl chloride, so it is not just the unfavorable protonation that makes the reaction slow. I oversimplified about the migratory ability as well. Indeed, in forming the oxime usually the more substituted group ends up anti to the OH group, and being anti it is also the one that migrates. However, this is not a correct description, and in my new video I am mentioning the “anti” nature of the migration. Of course, keep in mind that I am trying to summarize a huge amount of literature in 5-10 minutes to give undergraduates a basic grasp of key reactions, so I cannot do this at the level of a 100-page review.

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[@tunneltrance]

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[@PunmasterSTP]

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[@park8919]

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[@welelameka]

Protonation of carbonyl oxygen is step-1, since the reaction is acid catalyzed reaction.

[@muhammadzaimarif1969]

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[@khalilsaade2200]

Thanks professor!

[@koushikkumar8508]

Mgood lecture sir, thank you!!!

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[@Goomatora]

I really should've paid attention in high school chem

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[@nilsnickname4455]

Why does the resonance formula with the plus on the carbon be the major?

[@josephbeckmann8106]

Very interesting...

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[@rassimsimou1594]

Good

[@thechosenone5644]

Would it be accurate to say that this is like a Baeyer-Villiger oxidation but for adding nitrogen instead of oxygen?

[@mariababaeva5114]

Thank you ☀️

[@甜甜圈-h6m]

感謝~~ thank you

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