Sciencemadness Discussion Board

Sequential Beckmann Rearrangement?

Leo Szilard - 27-5-2015 at 20:13

I was wondering if you can perform sequential Beckmann rearrangements. For example, with cyclopentanone as the starting material, you can get to a δ-lactam. Can this undergo another Beckman rearrangement to yield 1,3-diazepan-2-one? I can't seem to find much about this compound. Is this compound unstable? Or is there a common name?

This idea seems obvious, so I can't believe it hasn't been attempted. Maybe the acidic conditions required for the rearrangement causes the δ-lactam to open up. If this is the case, do you think the second ring expansion could be done under milder conditions (maybe with an OTos leaving group)?

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byko3y - 27-5-2015 at 21:24

Amide is not a ketone.

Leo Szilard - 28-5-2015 at 03:36

Search for "azepan-2-one". I didn't invent nomenclature.



Or "Hexahydro-2H-azepin-2-one". Or "Hexahydro-2H-azepine-2-one"

[Edited on 28-5-2015 by Leo Szilard]

CuReUS - 28-5-2015 at 03:52

I think this is the reason why what you are suggesting doesn't happen
in the beckmann-rearrangement,you initially convert a ketone into a ketoxime by reacting it with hydroxylamine.in the mechanism,the carbonyl carbon gets a positive charge.but when there is already an N atom next to it(like in an amide),the positive character will obviously be less(due to lone pair donation by N).So hydroxylamine will not react with it at all to form an oxime,and without the oxime,no beckmann :D
byko was't complaining about nomenclature.He just said it in one line ;)

[Edited on 28-5-2015 by CuReUS]

Leo Szilard - 28-5-2015 at 04:09

Thanks. Yeah I get the "amide is not ketone" comment now. Thank you! I wasn't thinking.

Nicodem - 28-5-2015 at 12:16

Amidoximes can rearrange to ureas, but the rearrangement is called the Tiemann rearrangement and is analogous to the Lossen rearrangement of hydroxamic acids rather than the Beckmann rearrangement (for this reason it requires activation, generally by O-sulfonylation, and a basic medium).

Amidoximes can be prepared from amides, but not directly with hydroxylamine as the aldoximes and ketoximes. Instead, amides are generally activated by forming the chloroimidate (e.g., with oxalyl chloride) and then acylating hydroxylamine with it.

Leo Szilard - 29-5-2015 at 16:56

Thanks! I'll check it out.

byko3y - 30-5-2015 at 01:46

As far as I can see, Schmidt, Hofmann, Beckmann, Lossen and Tiemann rearrangements all go through the same mechanism of nitrene formation https://en.wikipedia.org/wiki/Nitrene , the latter can also be made by oxidation of carbamate with persulfate, and nitrene is known to be capable of addition to carbon.
While Schmidt, Lossen and Hofmann go through isocyanate intermediate after nitrene rearrangment, Tiemann goes via cyanamide, while Beckmann goes via nitrile, thus being slightly distinctive from others because electronegative atom is absent on carbon side (but the nitrile intermediate is still positively charged).

CuReUS - 23-6-2015 at 03:37

Quote: Originally posted by Nicodem  

Amidoximes can be prepared from amides, but not directly with hydroxylamine as the aldoximes and ketoximes. Instead, amides are generally activated by forming the chloroimidate (e.g., with oxalyl chloride)

I read somewhere that NaBH4 could reduce amides to amines,in the presence of an acidic reagent.(I may be wrong,but I am guessing the "acidic reagent" is POCl3,I remember reading an article on that)So could amides be reduced by NaBH4 if the chloroimidate was formed first?