Imines and Enamines

by James

in Aldehydes, Carboxylic acids, Esters, Ketones, Organic Chemistry 2

(part 5 of  a series on the reactions of neutral nucleophiles with carbonyl compounds)

I’m in love with the organic chemistry of nitrogen – alkaloids, in particular. To my mind,  imines are just about one of the most beautiful functional groups there are.  I’d love to give you the full scoop on all the amazing things imines can do, but  today we’re going to keep it really simple: we’re going to talk about the mechanisms involved in their formation.

What are imines again? Well, they look like this:

That’s an imine on the left. To its right are its chemical cousins: oximes and hydrazones. What they all have in common is a C=N double bond, and they are all formed by the same process. Let’s talk about how that C=N is made.

The reaction starts by stirring an aldehyde or ketone in solution with a little bit of acid.[note 1] You might recall that acid makes carbonyl carbons more electrophilic – it binds to the carbonyl oxygen and weakens the C=O bond. Now comes the exciting part. Upon addition of an amine, the carbonyl carbon (an electrophile) is attacked by the amine nitrogen (a nucleophile) and there is a 1,2-addition. The next step is a proton transfer. This serves two purposes. First of all, the OH is protonated to give OH2(+), which is a tremendously better leaving group. Secondly, proton transfer frees up the lone pair on the amine nitrogen.


Finding itself free again, the amine lone pair now descends upon the neighboring carbon like a horned owl swooping down on a moonlit field mouse, expelling the OH2(+) as a neutral water molecule. The final step is a simple deprotonation of the resulting positively charged imine (iminium, we call it – most positively charged organic molecules end with -ium), resulting in our neutral product, the imine. In practice, this reaction is usually run with a dessicant like magnesium sulfate to soak up the water that is formed along the way. That’s Le Chatelier’s principle at work again: you’ll notice that the reaction is an equilibrium.  If you take your imine and then dissolve it water,  you can get back the original aldehyde/ketone plus the amine.

So breaking down the mechanistic sequence, we get: protonation/1,2-addition/proton transfer/1,2-elimination / deprotonation. If you look at it hard enough, it might seem familiar: that’s the exact 5-step sequence for the formation of an ester from a cartboxylic acid, the hydrolysis of an ester, and the hydrolysis of an amide. that we talked about last time. Behold: underneath those four seemingly different reactions you will find unity in their mechanism.

By the way, the reactions that form oximes and hydrazones proceed exactly the same way. The main difference is that  hydroxylamine and hydrazines are more nucleophilic than amines (not on the exam: the alpha effect) and the products are more stable. In your teaching labs, you may have had to characterize an aldehyde or ketone by forming a hydrazone and taking the melting point. In particular, the hydrazones formed by 2,4-dinitrophenylhydrazine (DNPH) are a brilliant funky red color and are often highly crystalline, which provides another method for analysis.

Now up to this point you might have noticed that the imine we were dealing with was a primary amine: connected to only one carbon. What happens when we are dealing with a secondary amine?  Well, they form the C=N bond too,  but there’s a key difference: you can’t deprotonate them at the end, since they have two N-C bonds (this is also called an iminium, by the way) Somehow you have to make this compound neutral. What do you do?

Here’s how you do it: deprotonate the alpha carbon. The resulting chemical species is called an enamine.  Enamines behave a lot like their chemical cousins, enols: they are nucleophilic at the alpha carbon and react with electrophiles. The key difference is that they are actually a lot more reactive. This makes then very useful chemical intermediates. Here’s an example of an enamine in action,  a process first demonstrated  in detail by Professor Gilbert Stork of Columbia back in 1963. Sometimes these are called Stork enamine reactions.

Note here that at the end of the reaction there is another deprotonation to give you back the neutral enamine. If you need to get your ketone back, they will react with aqueous acid: you go to the iminium, which then hydrolyzes through the reverse of the mechanism that I drew for imine hydrolysis.

Further reading: check out some of the amazing advances that have been discovered in the catalytic chemistry of enamines over the past 10 years or so.

[note 1] – the acid is not an absolute requirement for this reaction, particularly with imine formation from aldehydes. However, acid does accelerate the reaction, provided that the pH is not so low that the amine is protonated.

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{ 23 comments… read them below or add one }

jampaiah amanaganti August 27, 2010 at 12:47 pm

please give one simple method for the synthesis of imine and how can we remove the water formed in the reaction?


James August 30, 2010 at 2:11 pm

Depends on the type of imine, and the reactivity of the amine and aldehyde involved. For simple imines with electron rich amines and aldehydes, anhydrous magnesium sulfate [ Mg(SO4)2 ] is usually sufficient. One can also use other dessicants such as silica gel or powdered 3A or 4A molecular sieves. For stubborn imines, heating in the presence of catalytic acid with an affixed Dean Stark trap using toluene or benzene as solvent will do the trick. Hope this helps – James


jampaiah amanaganti August 27, 2010 at 12:50 pm

how to remove water in the imine formation reactions?


vishnu September 24, 2011 at 10:54 am

there are different methods which does not involve water fromation like dehydro halogentaion, azawittig reaction. it depends on the type of imine you are working on.


james September 24, 2011 at 12:45 pm

True. This post is written for people taking organic chemistry for the first time, who probably don’t come across the aza-Wittig reaction and other alternative methods of making imines.


Danny October 12, 2011 at 9:43 pm

I have a question. How to separate aromatic imine from the reaction mixture? Thanks!


james October 14, 2011 at 4:12 pm

Depends on your reaction mixture. Some aromatic imines are quite crystalline. The problem is that they are readily hydrolyzed. I would suggest going to chemicalforums and giving out your specific reaction conditions – there’s a lot of helpful people there (myself included) who can assist with this type of practical question.


B October 22, 2011 at 8:36 pm

Hey James, I have a question about the reaction of imines. I can’t seem to find anything online about them except for their formation!

Do you know what the outcome is when you react an imine with an acyl chloride? I can’t work out if the nitrogen attacks the C=O carbon or the oxygen attacks the C=N carbon.



james October 22, 2011 at 8:40 pm

When you treat an imine with an acyl chloride, the nitrogen attacks the carbonyl carbon. (Nitrogen is more nucleophilic than oxygen). This forms an N-acyl imine (or iminium, depending on structure) which is a great electrophile. Kinda like in this example (far near the bottom) ––Spengler_reaction


B October 23, 2011 at 11:35 am

That’s a great help. Thank you!


Ashley February 14, 2013 at 2:12 pm

Looking for the distinguishing factor of a enamine, three bonds and a lone pair? but more important: this rules. your enthusiasm rules. I love chemistry too.


james February 15, 2013 at 3:07 pm

Enamine: alkene directly attached to an amine. Part alkene, part amine, kind of like a spork is part spoon, part fork. Thank you for the last comment too


matheus October 25, 2013 at 2:02 pm

I like this website! I’m studing a sintesis of the Penicilin V and found a obstacle, but this page help me more than a book. thanks!


Vicfectious December 12, 2013 at 2:43 pm

What would be the resulting product with H2S gas reacting with imines? When tried to to bubble H2S gas on organic imines it resulted to a solid formation. What is the component of this solid? Why it does formed?


Ariya January 23, 2014 at 7:58 am


In your note you state that the acid is not an absolute requirement for the reaction.. It means that the reaction can still occur? even in standard temperature?



James January 23, 2014 at 5:10 pm

Yes, especially with aldehydes, acid is not an absolute requirement.


Jignesh September 19, 2014 at 12:56 am

Hi! I am doing reductive amination in which imine is a intermediate. I am using aliphatic aldehyde and aliphatic amine in presence of Ra-Ni. CAn you suggest me what is the suitable water scavenging agent for the reaction?


ian February 4, 2015 at 9:37 am

Dear James,
Thank you for a very useful website which I refer to regularly. Please could I ask whether the less hindered substituted alkene would be formed in the last part of your diagram for Alkylation of Enamines (Stork Enamine Reaction) because of allylic strain?
Thanks again and best regards, Ian


James February 4, 2015 at 5:31 pm

Yes – exactly! Allylic strain between the alkene substituent and the nitrogen substituent.


Emanuel May 6, 2015 at 6:52 pm

Dear James,
wouldn’t acid first protonate amino group and not ketone/aldehyde? Amino group is weak base thus it should accept proton easier than C=O? Yet in this reaction oxygen is the one accepting proton. Why is that?
Thank you


James May 7, 2015 at 12:16 pm

Yes, it protonates the amine but not irreversibly. Remember if the amine was protonated irreversibly there would be no nucleophile to attack the carbonyl carbon. That’s why pH is essential. If it’s too acidic, the amine is protonated irreversibly, if it’s not acidic enough (pH 7 or higher) then the reaction is slow. The “sweet spot” is about pH 4 or 5.


Ram July 13, 2015 at 3:28 pm

Dear James,

Thank you for your informative website! My question is with regards to imines formed in the absence of an acid catalyst. I’m assuming that the protons that eventually leave with the oxygen as water by product come from the amine group. If so, it is correct to draw the formation of the intermediate formed with an arrow showing the N-H bond disconnecting and the movement of the H to the oxygen to form OH? I’m trying to draw up the mechanism but I’m not sure if I’ve done it right. Thanks!


James July 22, 2015 at 10:38 pm

What you’re describing is a process called “proton transfer” where the protons on nitrogen ultimately are removed and protons are added to oxygen. The exact mechanism instructors use can differ considerably. In the short way, which you describe, an arrow is drawn that shows the oxygen deprotonating the nitrogen directly. A more painfully long way to draw it (but ultimately more correct) is to invoke some kind of exterior base (like water) to deprotonate the nitrogen, and then (in a second step) to act as an acid, protonating the oxygen.
It all ends up being the same thing.


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