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I have read where it is explicitly stated that the Hofmann Rearrangement is a reaction whereby primary amides are converted into primary amines with one less carbon. I have tried to find examples where a secondary amide is used, but haven't been able to find it addressed anywhere. I have been able to find examples where a N-hydroxylamide is used, but that's still a primary amide.

What happens when a N-methylamide or an N-methylhydroxylamide is subjected to Hofmann Rearrangement conditions?

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    $\begingroup$ Simply put, secondary amides will not react. $\endgroup$ – Berry Holmes May 18 '17 at 13:54
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I think it's not possible, because if you go through the reaction mechanism, you will see that in the 1st and 3rd step two protons are needed to be abstracted by the base to proceed the reaction. So if you take a secondary amide, i.e., one of the H being substituted by any alkyl/acyl/aryl group, the base will not be able to abstract that group and hence the reaction will not progress to any fruitful product.

Pic reference: http://www.name-reaction.com/hofmann-rearrangement

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See you the isocyanate intermediate in the mechanism above? It's impossible the existence of this intermediate from a secondary amide. So, Hofmann, Curtius and Lossen rearrangements only works with PRIMARY amide.

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