- Formation of hydrates
- Formation of hemiacetals and hemiketals
- Acid and base catalyzed formation of hydrates and hemiacetals
- Formation of acetals
- Acetals as protecting groups and thioacetals
- Formation of imines and enamines
- Formation of oximes and hydrazones
- Addition of carbon nucleophiles to aldehydes and ketones
- Formation of alcohols using hydride reducing agents
- Oxidation of aldehydes using Tollens' reagent
Formation of imines and enamines
How aldehydes and ketones react with primary and secondary amines to for imines and enamines. Created by Jay.
Want to join the conversation?
- The amine is basic, won't it take a H+ from the acid and thereby become unreactive?(7 votes)
- From what I have been told, these reactions occur at a specific pH, which keeps the amine from being protonated.(12 votes)
- Does it matter if the acid catalyst was added before or after the amine attacks the carbonyl carbon? I'm thinking it won't since aldehydes and ketones are more reactive than amines and therefore even if the acid catalyst was added before the amine attacks the carbonyl group, the oxygen on aldehydes and ketones are more likely to get protonated instead of protonating the amine.(3 votes)
- That is an absolutely fabulous question! I thought about it for an hour before I figured it out!
The product is the same, but it follows a different mechanism which have written below. To understand what I have written, please draw the reaction pathway as I explain. (Sorry! I can't post an image here)
As seen in4:26the reactions will proceed in the second mechanism Jay has written to the first step, where there is a positive charge on the nitrogen and a negative charge on the oxygen.(Rate Determining Step) If no acid is present in the reaction mixture, the oxygen will grab a proton from the nitrogen, immediately, to make both of the atoms neutral.
(NYH2+ ==> NYH
Then, after the acid is added, the proton has the choice of attacking the oxygen or the nitrogen. It chooses the former because it is more electronegative. Hence we get protonated water in the compound. This water exits, as it is a good leaving group, leaving a positive charge on the carbon.
Because there is a positive charge on the carbon, elimination takes place and you have your required imine!
This is quite long and extensive, but I hope it helps!
Again, wonderful question isaacheba!(6 votes)
- H2SO4 as a strong acid will protonate the amine to CH3CH2NH3+ which is not a nucleophile anymore !! right? If yes, how the reaction will be completed??
thanks in advance .(4 votes)
- The H₂SO₄ is only a catalyst and is present in small amount.
It protonates both the carbonyl oxygen and the amine nitrogen until it is used up.
But most of the ketone and the amine are unprotonated.
The protonated ketone molecules can then react with the unprotonated amine molecules.(4 votes)
- In general, what is the difference between the "Y" group of the imine and the "R" groups?(3 votes)
- The Y means it could be R or H, whereas R means it's a Carbon group/chain(4 votes)
- Is it possible to deprotonate cyclic ring in primary amine? if so, pri-amine will also form enamine...(1 vote)
- It can happen, but only to a small extent. It doesn't matter if the proton is in a ring or not.
1° amines form the more stable imines. 2° amines form enamines.
Consider the reaction of acetaldehyde with CH₃NHR, where R can be H or CH₃.
The intermediate carbocation is CH₃-C⁺H—NRCH₃.
R= H (1° amine): The intermediate is CH₃-C⁺H—NHCH₃. It can lose an H⁺ from the N to form the imine CH₃-CH=NCH₃ or an α-H from the terminal CH₃ to form the enamine CH₂=CH–NHCH₃.
These are tautomeric forms that are in equilibrium with each other. The H on the N is much more acidic than the α-H, so the imine is the more stable form.
R = CH₃ (2° amine): The intermediate is CH₃-C⁺H—N(CH₃)₂. It can lose only an α-H from the terminal CH₃ to form the enamine CH₂=CH–N(CH₃)₂.(6 votes)
- At1:45can you please explain again why C (after the protonation of O) is a good electrophile? And anyway, shouldn't the ammine attack the C from both sides (from the top and from the bottom) forming a racemic mixture?(2 votes)
- First question: It is a good electrophile because when the O of the carbonyl group grasps a proton, the O will have a positive one formal charge. O is very electronegative, so it can not stand the condition of formal charge. Thus, it will take the pi-electrons from the carbonyl group, making the C positively charged (quite bully...). C is far less electro negative so C is quite okay with that. The two scenarios I mentioned are the resonance hybrid. So, C is more electrophile.
Second question: Yes, you are right. But it is not mentioned here because it does not make a lot of sense here. :)(2 votes)
- I thought there was a proton transfer from the amine to the OH to make OH2+.(2 votes)
- In an imine (Schiff base) formation the Nucleophiles are usually ammonia, amines which attack the activated or unactivated Electrophiles (Carbonyl Compd).
Can an ammonium salt e.g NH4Cl, Ammonium Succinate, Ammonium Oxalate etc act as a Nucleophile (in place of ammonia, amines) & kick-on this reaction?
I am asking because I intend to use Acetone ( a Carbonyl compd) as solvent to extract an organic ammonium salt from an inorganic salt (KCl, NaCl) mixed with it. I am wondering if Acetone will serve only its intended purpose of an extracting solvent here rather than initiating a cascade of Schiff base formation reaction.
Could Ethyl Acetate or Toluene act as a suitable alternative to Acetone in this type of extraction procedure?(1 vote)
- If I wanted to use an imine and form an aldehyde with an animo acid , I would simply follow these steps backwards?(1 vote)
- pretty much, however if you were to draw out the mechanism it may be slightly different by 1 step(1 vote)
- What happens if a 3° amine is used in this reaction?(1 vote)
- my guess is it will not react because it does not have a hydrogen to donate.(1 vote)
Voiceover: Let's see how to form imines and enamines and start with the formation of an imine we start with an aldehyde or a ketone. And you add an amine to it and you need an acid catalyst and over here on the right if your Y is equal to hydrogen or an alkyl group which is a R double prime you will form an imine. So let's go ahead an write that this would be an imine. And the Y can be other things, we'll talk about those other things in the next video. You also form water and so if you, since it reacts as [that] equilibrium if you wanted to shift equilibrium to the right, you could remove water as it's formed and equilibrium will shift this way and give you more of your imine product. So let's look at the mechanism to form imines. And I've seen two different ways to start off, the mechanism and so I'll present both ways and you can choose which one you want to use. And so one way would be to think about an acid being present right, so we'll say it's H-A plus the generic acid and you think about this as protonating your imine. So you have an imine and protons, you're protonating your imine to form this generic acid here. And then your carbonyl oxygen is going to be protonated. A lone pair of electrons on the oxygen takes this proton, leaves these electrons behind, so let's go ahead and show the results of that. So now we have our oxygen, with a plus one formal charge and we still have our prime group over here, so let's show those electrons. So this lone pair of electrons here on the oxygen picks up this proton and that forms this bond right here. We've talked about in previous videos how protonation of the carbonyl oxygen makes the carbonyl carbon more electrophilic. And so we have a good electrophile, this carbon here is a good electrophile and the imine can be a good nucleophile. So a lone pair of electrons and the nitrogen over here are going to attack our carbonyl carbon and that's going to kick these electrons in here off onto the oxygen, so let's draw the results of that. So now we're going to have our nitrogen is forming a bond with our carbon, so let's show those electrons so let's make those blue. So this lone pair of electrons of electrons on the nitrogen forms a bond between this nitrogen and this carbon here. This nitrogen is also bonded to two hydrogens. So let's draw those in. And still our Y group like that. That gives our nitrogen a plus one formal charge. And for our carbon, it's also bonded to this oxygen, this oxygen now has two lone pairs of electrons. So let's show the movement of those electrons so I'll use green here. So these electrons in green moved out onto our oxygen. The carbon is still bonded to our alkyl group so R and R prime, so I'm saying we start with a ketone here. And so that's one way to start off your mechanism and of course there is another way to do it. So let's talk about just using our nitrogen as a nucleophile straightaway so the imine is a pretty good nucleophile and so we can just show our nucleophile attacking our electrophile directly. So partial negative oxygen, partial positive carbon this is our electrophile and then a lone pair of electrons on our nitrogen right makes our imine a good nucleophile so they can just attack directly. Right, attack this carbon push these electrons off onto your oxygen. So let's go ahead and draw the result of that and you would have our carbon bonded to our nitrogen. Our nitrogen is bonded to two hydrogens and our Y group and that would make our nitrogen a plus one formal charge and then our oxygen this time would be a negative one formal charge. So negative one formal charge here. We have our alkyl groups down here, so let's once again show our electrons. So if nitrogen attacks directly, This lone pair in blue forms this bond right here. And then we can say that these pi electrons in here kick off onto our oxygen so it doesn't really matter which lone pair you make them as long as it states that one right there. And then we can show this intermediate going to the one we already talked about, if we just protonate our negative one formal charge on our oxygen here. So let's show that, so if we have our generic acid so H-A plus, we can show a lone pair picking up this proton, leaving these electrons behind. And so now that would give us this intermediate the same one that we had before. So whichever way you would like to start off your mechanism, once we've reached this intermediate we can think about the base. The imine here, a lone pair of electrons picking up this proton and leaving these electrons behind on our nitrogens. So let's show what we would make from that. Let's get a little more room down here alright. So we deprotonate and then we form this structure, a carbon bonded to a nitrogen and then we still have our Y group, we still have a hydrogen on this nitrogen. We have a lone pair of electrons on this nitrogen, so that lone pair came from right in here. So you take that proton and leave those electrons in magenta behind on your nitrogen. So we also have an O-H group bonded to our carbon here, and then we also have our alkyl group, so R and R prime. So this intermediate is called a carbinolamine. So let me go ahead and write that. So this is called a carbinolamine. And for the next step we can protonate the O-H group. So we take our generic acid once again so H-A plus and lone pair of electrons on our oxygen could pick up this proton leaving these electrons behind so let's go ahead and show that. So let's get once again some more space in here. So if we protonate our carbinolamine, we would have our carbon here, our nitrogen, lone pair of electrons on our nitrogen, our Y group, our hydrogen here, our alkyl groups. And then if we protonate the O-H, we would form water as a good leaving group. So right in here, this would be a plus one formal charge on our oxygen and let's show those electrons. So in here, lone pair of electrons on the oxygen, pick up this proton and so forming this bond right here and now you can see we have a good leaving group. So we have water as a good leaving group. So if these electrons came off onto the oxygen these electrons in the nitrogen could move in here to form a double bond between the carbon and the nitrogen. So let's go ahead and show that. So our next step here, this is where we lose water. So we're going to minus H2O and let's go ahead and show our next structure here. We have our carbon double bonded to our nitrogen this time, our nitrogen is still bonded to our Y group. And our hydrogen over here. That gives the nitrogen a plus one formal charge and the carbon is still bonded to our alkyl groups And so let's show those electrons here on our nitrogen. So I'm going to go ahead and make those magenta. So these magenta electrons move in here to form our double bond between our carbon and our nitrogen. And this is an important structure, this is called an iminium ion, so let's go ahead and draw that. So an iminium ion and then we lose water of course. So minus water at this stage. So we're almost to our final product, we would just have to deprotonate our iminium ion. And so we can do that with our imine. Could take that proton and leave these electrons behind on the nitrogen. So that's going to give us our final product where we have carbon double bonded to our nitrogen and then we have nitrogen bonded to our Y group here. A lone pair of electrons on that nitrogen and then we have our alkyl groups and so let's show those electrons here in blue moving in here, off onto our nitrogen. And then once again, if our Y is a hydrogen an alkyl group we have formed an imine. Alright, so if that's formation of an imine, let's look at an example. So here is our reaction, we're going to start with a ketone here, so cyclohexanone and react it with a primary amine. So the nitrogen is bonded to one carbon. So this is a reaction with a primary amine and then we're going to use sulphuric acid as our catalyst here. So to figure out the product of this reaction since it's kind of a long mechanism so it doesn't really makes sense to run through the entire mechanism but we could think about part of the mechanism here. We know that the nucleophile is going to be our amine and it's going to attack our carbonyl carbon here. So we know that in our mechanism we lose one of these protons on our amines so let's say it's that proton right there. And we know we're going to lose water so minus H2O and this going to think about taking us to our iminium ion steps, so we're going to have our ring. And so we've already lost water so now we're going to have a double bond to our nitrogen. And our nitrogen is going to have an ethyl group on it here and then we still have one proton on our nitrogen. So we still have the hydrogen attached here at this point. Which gives us a plus one formal charge on our nitrogen. So this is our iminium ion right here. And then in our last step we know that our amine can come along and act as a base. So a lone pair of electrons on the nitrogen could take this proton right here and leave these electrons behind on our nitrogen to give us our imine products. So we go over here and we draw our product, double bond to our nitrogen and then we have our ethyl group and lone pair of electrons on our nitrogen. So if I follow those electrons, so these electrons in here in blue ended up on our nitrogen to form our imine. So a few things that I want to point out here for the iminium ion, we still have a proton attached to it. So deprotonation for our last step will form an imine. And this is once again done with a primary amine right here. And so that's a little bit different from what we're going to talk about next, we're going to talk about formation of an enamine, and formation of an enamine is, it starts off the same way in terms of the mechanism but it's this iminium ion step that changes a little bit. So let's look at reacting a ketone with a secondary amine this time. And so let's look at this one right here, so here we have a secondary amine and our nitrogen bonded to two other carbons so a secondary amine this time. And once again, we can kind of run through the mechanism. just thinking a little bit about what happens without going through each individual step, We know the nucleophiles going to attack here and we know we're going to lose a proton. So this is the only proton we have left to lose and so we lose that one, we're going to lose water in the mechanism to form our iminium ion, so let's go ahead and draw that out We're going to have our ring, we're going to have a double bond to this nitrogen, we've already lost the proton in red so now we have two alkyl groups attached to our nitrogen and a plus one formal charge on our nitrogen here. And so in the previous reaction, let me go back up to here to the previous reaction again. Once again we still had a proton on our iminium ion at this step and so we're able to deprotonate here. But that's not the case with what we have for this ion. We don't have a proton on our nitrogen here. And so we can't deprotonate in the same place, we have to pick a carbon adjacent over here. So here's a proton over here on this carbon and our base could come along. So let's go ahead and draw out our amine base with a lone pair of electrons right here. And a lone pair could take this proton that would push these electrons in here and then push these electrons off on to the nitrogen to give us our products, so let me go ahead and draw this out here. So now we're going to have a double bond here and then we have our nitrogen with these ethyl groups and a lone pair of electrons on our nitrogen. So let's follow some electrons. So these electrons in here once you deprotonate, move in here to form our double bond and then these electrons in here move out onto our nitrogen like that to form a different product. So this called an enamine, so let me go ahead and write this out here, an "enamine." And the name come from "-en" come from the double bond. So we form double bonds, that's the "-en" portion and then the "amine" portion comes from this part up here. We have an amine so this is an enamine. And so formation of an enamine happens because once again we don't have the same iminium ion that we had before because of the fact that we started with a secondary amine. So that's why it's important to recognize the kind of amine that you were reacting with your aldehyde or a ketone. And if it's secondary, you're going to form an enamine here. And enamines are useful synthetic intermediates, So we'll talk much more about them in later videos. But once again the mechanism is the same until you get to this last step here.