Hydrolysis of imines to give ketones (or aldehydes)
Description: Treatment of imines with water leads to their hydrolysis back to aldehydes (or ketones) and an amine.

Notes: The reaction is assisted through the use of an acid catalyst.
Examples:
Notes: Note that the third example is intramolecular.
Mechanism: Protonation of the imine nitrogen (Step 1, arrows A and B) results in the formation of the iminium ion, which undergoes 1,2-addition by water (Step 2, arrows C and D). Transfer of a proton (Step 3, arrows E and F) followed by 1,2 elimination of ammonia (Step 4, arrows G and H) lead to an oxonium ion, which is then deprotonated to give the neutral ketone.
Notes:
- Acid is helpful but not an absolute requirement for this reaction. Reasonable mechanisms can be drawn without acid.
- The “Cl” here in H3O+ Cl- is completely unimportant, just meant to show a balance of charge for H3O+. Other counter ions such as Br-, HSO3-, etc. would work just as well.
- Note that this is an equilibrium reaction and goes in this direction because of the large excess of water. It is the exact reverse of imine formation.
- There are certainly other reasonable ways to draw proton transfer (Step 3) and other species besides H2O that could conceivably act as bases in the last step.


{ 5 comments… read them below or add one }
Great outline! Excellent helping me right before my exam!
Great review, but I’m a little confused as to where the fifth carbon went on the intramolecular imine hydrolysis under examples?
It’s a typo. Fixed – thanks for spotting it!
Thanks this was really helpful!
Just one thing I’m still confused about… why can’t step 5 happen before step 4? Wouldn’t it be more favorable to deprotonate the oxygen first and then have the lone pair kick down and get rid of LG?
Sure… but what’s going to deprotonate it? : – )
This is done under slightly acidic conditions.
Compare O-H (pka about 15) with protonated carbonyl (pKa -2)