Hofmann Rearrangement of Amides to Amines

by James

Description: When amides are treated with chlorine and base, a rearrangement occurs, resulting in the loss of carbon dioxide and the formation of an amine.

Notes: Other halogens can be used for this reaction, such as Br2; so can the sources of reactive hydrogen NBS and NCS.


Notes: Note how Cl2 or Br2 can both be used for this reaction. In the third example, the first step makes the amide from the acid chloride. Then, the Hoffmann occurs in the second step.

Mechanism: The first step of this reaction is attack of the lone pair of nitrogen on Br2 (Step 1, arrows A and B) followed by deprotonation (Step 2, arrows C and D). Then, rearrangement (Step 3, arrows E and F) gives a carbocation, whose resonance form (arrow G) is an isocyanate.
Under these reaction conditions the isocyanate is not usually isolated. Instead, hydroxide ion attacks the isocyanate (Step 4, arrows H and I) to give the conjugate base of the amide. Then, protonation (Step 5, arrows J and K) followed by decarboxylation (Step 6, arrows L, M, and N) give the free amine.


Notes: It’s perfectly reasonable to show the deprotonation of the amide occurring before attack on Br2. There are also other reasonable ways of showing the decarboxylation/ protonation in steps 5 and 6; this is just one suggestion.


{ 4 comments… read them below or add one }

Lauren Benning

How do the last two reactions in your examples work if there is no amide to start with? Where does the nitrogen come from?


Andrew Ross

Would you be willing to post an example where Pyruvic acid (2-oxopropanoic acid) and N-hydroxymethanamine react under acidic conditions (H3O+) to form Peptin (N-methylacetamide)? The electron flow of this reaction is quite confusing.



Ah the Beckmann rearrangement. Very closely related! I should put it up here, but in the meantime… there’s always Wikipedia



Why do you use bromine to get amines?


Leave a Comment