Master Organic Chemistry Reaction Guide

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.


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.


  • 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.


Comment section

26 thoughts on “Hydrolysis of imines to give ketones (or aldehydes)

  1. Great review, but I’m a little confused as to where the fifth carbon went on the intramolecular imine hydrolysis under examples?

  2. 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?

    1. 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)

  3. I’m not sure if this question belongs in this topic. But I’m just wondering is an iminium ion a stronger electrophile than an oxonium ion? It is in the Strecker synthesis, but I don’t exactly understand why.

  4. I don’t quite get it. Isn’t H2O a better leaving group then NH3 or in other words, isn’t the latter a better nucleophile? Why does then reaction proceed towards the ketone? Just because water is in excess?
    Thank you for the clarification in advance…

  5. Hello,

    I want to transfer carvone semicarbazone (an imine of carvone) back to carvone using this reaction. Which concentrations are suitable to do this? Is it reasonable to use 0,1 M HCl for example?

    In other words: at wich pH has the reaction maxiumum efficiency?

    Kind regards
    Jeroen Vandervelde

  6. Hi
    Thanks for the nice illustration.
    My query is after removal of water is the resultant imine air stable?

  7. Is the imine or ketone more thermodynamically stable? It would be nice to know when I am pushing more or less uphill against the thermodynamically favored molecule.

    In my lecture, I was given a reaction of 1,3-diketones with hydrazine to make pyrazoles (no acidic catalyst). Does this mean the excess of hydrazine is implied, or is it the aromaticity of the pyrazole driving the product in the absence of a catalyst?

    1. To clarify on the follow up for anyone interested:

      Imine vs Ketone thermodynamic stability uncertain but not particularly relevant, as reactions are controlled by excess anyways. In the formation of the imine, a drying agent or molecular sieve will remove water in the equilibrium as it builds up as side product.

      The reaction I supplied is called the Knorr pyrazole synthesis, and actually my lecturer just forgot to add an acidic catalyst (more likely assumed I should know it would be there!) Imines formed from hydrazine substitution are hydrazones.

    1. The nitrogen has to accept a proton at some point in order to become a good leaving group. It’s possible to use NaOH/H2O (i.e. mildly basic conditions) but still that relies on the solvent acting as a proton donor.

  8. Hi, I am confused because what if the imine had an R group and not an H. Would it just be protonated twice? thank you

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