Imines and Enamines are the products formed by the nucleophilic addition of a primary (1˚) amine, secondary (2˚) amine, or ammonia (NH3) to carbonyls such as ketones or aldehydes.
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Imines look like carbonyls and are formed from a primary amine (or ammonia) reacting with some type of acid (H+, H3O+) with a carbonyl. In simplest terms, it looks like a carbonyl with a double bond to nitrogen instead of oxygen.
Notice that after the reaction is over, you will be left with a structure that looks like this (above in green). The 3 carbon chain coming off is simply the R group that was attached to the original primary amine before it reacted.
On the other hand, enamines form from carbonyls reacting with a secondary amine and does not have a double bond to nitrogen, but instead to carbon.
You can remember this because the nitrogen already has 2 R groups in the beginning and therefore would be positively charged if it contained a double bond. For example:
Notice that the double bond here is between 2 carbons: 1 from the carbonyl carbon that formed it, and the other is the alpha carbon.
Enamines are formed from the same mechanism that imines are formed except for the final step which is deprotonation of the iminium cation (more on this to come later).
The reagents needed to form an imine are simple and really come down to 3 things:
It is commonly seen to combine parts 2 and 3, the NH3 and the H+, so that we don’t have an extra step involved. If we did this the reaction simply looks like this:
For enamines, the amine being used will be 2˚ and therefore commonly seen is a cyclic amine such as azetidine:
Assuming protonation like we did before, we would get the reaction seen below and be ready to start our mechanism.
Now, as we go into the mechanism we will focus on forming the imine through 0˚ or 1˚ amines. However, if we use the cyclic protonated amine above, the mechanism is the same until we get to the end
Protonation is the first step where the oxygen from the carbonyl reacts with the protonated amine that we formed above. As you can see below we have already taken a proton and formed our +OH.
From here we can draw a resonance structure so that we have a positive charge not on the oxygen anymore like before, but now on the carbon of the original carbonyl.
The green resonance structure is not absolutely necessary to draw because the arrows will be exactly the same once we perform the next step, however it is a more complete picture and sometimes easier to visualize.
Next, a nucleophilic attack [1,2-addition] occurs with the lone pair from the original neutral amine to the positive charge on the carbon. From this, we get a tetrahedral intermediate with a new C-N bond and nitrogen now has a positive charge again.
Now, you may want to draw the NH3+ with one of its H’s hanging off to the side because we are trying to eliminate water and therefore do a proton exchange, which is our next step. This means the oxygen from the OH will grab one of the H’s from the H-NH2+ (this is simply NH3+ arranged differently) and form –H2O+.
Once we form water we can kick out our leaving group or eliminate water. The way we do this is to use the electrons from the nitrogen, form a double bond between carbon and nitrogen and in the process form an iminium ion. Is the product looking familiar now?
At this point we have our iminium cation which will be your checkpoint for these types of reactions. If you don’t get this structure you most likely did something wrong and should go back to your mechanism and the different steps to figure out where you want off track.
For the enamine mechanism, this will be the tetrahedral intermediate formed before we do a proton transfer and form water. From here, water will be our leaving group again that will be kicked out in our next step.
This is the general formula for the iminium cation (R2C=NR2+), which is the key intermediate for imine and enamine reactions.
The iminium cation formation will be our last intermediate before we get to our final product which can either be an imine or an enamine. From here we can use water or the conjugate base of our original NH4+ which is NH3 (ammonia), to deprotonate the final H on the nitrogen to give us our final product, an imine.
The iminium cation for enamines looks very similar. Below is an example of one when the amine used was secondary (2˚) and cyclic.
Imine general formula: R2C=NR
The product of the imine mechanism will look like a carbonyl except there is a N (nitrogen), instead of an O (oxygen) and an extra H (hydrogen) is attached. The rest will simply be our byproducts.
Also, notice that NH4+ was reproduced, but if we used water to deprotonate our iminium cation we would instead have H3O+ and NH3 as byproducts instead. Both would be correct!
For enamines our product would not have the double bond to nitrogen, but instead a double bond from the original carbonyl carbon position to one of the alpha carbons (more on this below).
All the steps above were described for when a 0˚ or 1˚ amine is used, however when a 2˚ degree amine is used we get a different last step. That is because if we look at our iminium cation (in green above) we have no hydrogen to deprotonate on nitrogen.
The question we ask ourselves then would be: Where do we have a hydrogen (H) to remove from? The answer to that is at our alpha position from our original carbonyl.
Just as before, remember we can deprotonate with the conjugate base, which in this case would be the cyclic amine (seen below), or water.
Deprotonation Step for Enamines
We mentioned the alpha position briefly when talking about enamines, but why is this important? Well, it turns out that enamines can do reactions at their alpha position just like many other compounds in Organic Chemistry, like enols & enolates.
Enamines specifically are able to use their nucleophilic alpha position to add different R groups and form alpha substituted carbonyls. This happens via an iminium salt intermediate that is able to be hydrolyzed back to a carbonyl.
Hydrolysis in this case is referring to the ability to go from our iminium salt back to our carbonyl using water and a little bit of acid. The mechanism involves 5 steps: protonation, addition of H2O, proton transfer, elimination, deprotonation. Seem familiar?
Here is a video of that mechanism being used to convert an ester back to a carboxylic acid.
The hydrolysis can be done for both imines and enamines to turn them back into carbonyls, as see below with the hydrolysis of an enamine.
This hydrolysis mechanism will be important when doing enamine alkylation and acylation problems because hydrolysis back to the carbonyl is usually the last step.
Another important topic arises when discussing enols, the alpha carbon, and ketones in particular, which is tautomers and tautomerization. The reason this is applicable is because imines and enamines can tautomerize with each other.
First, we have keto-enol tautomerization which can be acid or base catalyzed. Remember, these 2 compounds are simply constitutional isomers.
The same rule applies with imines and enamines. Notice the only difference between the 2 is the double bond and 1 H (hydrogen) switched places.
This concept of tautomerization with imines and enamines will become increasingly important when dealing with synthesis reactions where you need to add R groups to the alpha position.
4. Schiff bases & other nitrogen compounds
Now, when we say imines we should recognize that they can also be referred to as Schiff bases when the group attached to the N is not hydrogen, but an alkyl group.
Take for example the structure below, which is the general structure for an imine.
We can actually call this compound different names based on what R1, R2 and R3 is. Let’s go through some of the most common names:
Lastly, just be aware that enolates, which react very similar to enamines, can undergo a condensation reaction known as aldol condensation. It is when 1 enolate acts as a nucleophile (Nu-) and attacks an electrophile (E+) such as another carbonyl.
This forms what is known as a ß-hydroxycarbonyl which can dehydrate via 2 mechanisms to form a conjugated unsaturated carbonyl.
Here is an example: (more details on this later)
Q and A:
Why is it necessary to draw equilibrium or reversible arrows?
What do we get if we use a tertiary amine?
Can we go from an imine to an enamine?
What is the mechanism for the reverse reaction going from an imine or enamine back to the original carbonyl?
What were the 5 steps again for the mechanism?
Here is a summary of the imine mechanism. Notice the 5 steps involved: Protonation, Nucleophilic Attack, Proton Transfer, Elimination and Deprotonation.
Here is a summary of the enamine mechanism. The same 5 steps for the imine mechanism can be seen here, with the exception of the deprotonation step of the iminium cation where our conjugate base grabs a hydrogen from the alpha carbon.