Comments on: The Hofmann and Curtius Rearrangements

By: James Ashenhurst

James Ashenhurst — Fri, 18 Oct 2024 19:49:52 +0000

In reply to Aadity. We don't use NaI or NaBr for the Hofmann rearrangement because they are not electrophilic sources of halogen. We use Br2 or other sources of electrophilic Br or Cl. I am not aware of an example of using I2 or similar electrophilic I(+) reagents for doing the Hofmann but there is no reason I can see as to why it would not work, although why you'd prefer to use I versus Br I'm not sure

By: Aadity

Aadity — Thu, 17 Oct 2024 15:22:46 +0000

Is it possible to convert an amide into an amine just like Hoffmann rearrangement using NaI instead of NaBr?

By: 7 Tips to ace organic chemistry in NEET/JEE | Science tidings

7 Tips to ace organic chemistry in NEET/JEE | Science tidings — Sun, 02 May 2021 17:46:29 +0000

[…] Hofmann and curtius rearrangements […]

By: James Ashenhurst

James Ashenhurst — Mon, 09 Mar 2020 20:46:19 +0000

In reply to Prahlad. Not sure. Heating with Br2 and KOH is usually what you'd do with an amide (Hofmann rearrangement conditions), not an acyl azide. Eventually the heat should lead to a Curtius rearrangement, but I'm not sure what the effect would be. There's no reason to add Br2 and KOH to an acyl azide.

By: Prahlad

Prahlad — Mon, 09 Mar 2020 15:03:41 +0000

Sorry what happens if I heat acyl azide with aq KOH and Br2?

By: Prahlad

Prahlad — Mon, 09 Mar 2020 15:00:38 +0000

What happenes when an isocyanate is heated with aq KOH+Br2?

By: James Ashenhurst

James Ashenhurst — Thu, 12 Sep 2019 20:44:54 +0000

In reply to Peter. Shoot. You are correct. Thanks for spotting, fixed.

By: Peter

Peter — Mon, 06 Aug 2018 00:03:25 +0000

Hi, there appears to be a “typo” in the Part 1 image for the acyl azide: the “plus” charge should be on the middle nitrogen instead of the nitrogen bonded to the oxygen? Googling “acyl azide” seems to confirm what the correct charge distribution for acyl azide looks like.

By: James

James — Mon, 16 Jul 2018 19:49:46 +0000

In reply to SWAPNIL. Great question. You need a primary amide. If you try to perform it with a secondary amide you just end up replacing N-H with N-Br and it stops there, since there's no N-H for the HO- to deprotonate and no driving force for the reaction.

By: SWAPNIL

SWAPNIL — Sat, 14 Jul 2018 05:34:03 +0000

Hi James,
Thank you for sharing such a great explanation. However, I am wondering is it possible to have some change in an amide. Instead of CO-NH2 if we use CO-NHR will it result in the same rearrangement?
Thanks in advance James, you are the best!