Formation of enamines from ketones/aldehydes and secondary amines

by James

Description: When ketones are treated with secondary amines and heated with an acid catalyst, a condensation reaction occurs, resulting in loss of H2O and formation of an enamine.

Notes: Note how the product has both an alkene and an amine, hence the name “enamine”. It is crucial that secondary amines are used. Ketones are most common, but aldehydes can also be used.


Notes: Note that in example 3 the enamine forms preferentially at the less substituted carbon. This is for steric reasons (for more information, do a Google search for “allylic strain”).

Mechanism: Protonation of the oxygen of the ketone (Step 1, arrows A and B) makes the carbonyl carbon a better electrophile. Next, the secondary amine attacks the carbonyl carbon in a 1,2-addition (Step 2, arrows C and D) and then a proton is transferred to the oxygen (Step 3, arrows E and F). The nitrogen lone pair then forms a π bond with the carbon, expelling water in a 1,2-elimination (Step 4, arrows G and H) leaving a positively charged nitrogen. Since the nitrogen contains no protons to remove, base removes a proton from the adjoining carbon (“α carbon”) which then forms a new C–C π bond, resulting in the enamine (Step 5, arrows I, J, and K).

Notes: Acid assists this reaction in two ways, making the carbonyl more reactive in step 2, and also allowing for the formation of OH2, which is a great leaving group.

There are other ways to reasonably show proton transfer. Here is an alternate (probably more accurate) mechanism.


Note that H2SO4 was chosen here just to avoid use of generic “H+” but many acids can reasonably be used here (e.g. TsOH, H+, etc.). It’s probably more reasonable to show H2O as a base here, but note here how the catalytic acid is regenerated.


{ 2 comments… read them below or add one }

Ari Stoner

What is the mechanism for the reverse–hydrolysis of enamines?


Andres Vanegas

Is there an H missing in HOSO3 at the end of the mechanism?


Leave a Comment