Hofmann elimination of alkylammonium salts to give alkenes

by James

Description: Tetra-alkyl ammonium salts undergo elimination to form the least substituted alkene. This is called the Hofman elimination.

Notes: For the elimination step, silver oxide (Ag2O) is a common base for this purpose, although other strong bases such as hydroxide ion (e.g. NaOH) can be used.


Notes: Note that the second example also shows the alkylation step (using MeI)

Mechanism: This is an elimination reaction. The strong base removes a proton from the carbon adjacent to the bulky leaving group on the least substituted carbon, resulting in formation of the double bond (Step 1, arrows A, B, and C)

Notes: Note how the extra steric interactions between the leaving group and the alkyl group on the more substituted carbon destabilizes the Zaitsev transition state.


{ 6 comments… read them below or add one }

Ned Silavwe March 10, 2012 at 5:26 am

I appreciate your well presented information. I just wanted to ask whether you meant, under bonds broken, C1-H and not C2-H. That is in the first (i.e, general) equation. Later on, under mechanism, the information looks correct.
Thank you.


james March 13, 2012 at 9:10 am

Thanks for the spot! Fixed.


Stanton de Riel January 16, 2013 at 11:51 am

That’s one f in Hofmann ….


james January 17, 2013 at 12:06 pm

Thank you.


Karthik May 20, 2015 at 2:08 pm

Just curious. What would happen if we tried reacting Tetramethyl Ammonium with a mild base (like AgOH). Since there is no beta Hydrogen to abstract, Hofmann elimination isn’t feasible so what would be the products?


James May 20, 2015 at 11:51 pm

Probably nothing would happen. You might get a little bit of methanol and NMe3 via substitution (SN2).


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