Concept: Concept: General Reactions3m
In this next page, we’re going to talk about the nucleophilic addition of neutral amines to carbonyls and this is going to yield imines and enamines.
The difference between an imine and an enamine is simply going to be type of amine that the carbonyl is reacting with. As you can see, when you react a carbonyl with primary amine, you’re going to get an imine. An imine has a functional group of basically a nitrogen with a double bond on it and possibly an R group on the top. Think about an imine is like a carbonyl but with a nitrogen. If it would have been a carbonyl with an oxygen, that’s an imine with a nitrogen on it. Notice that when they react a secondary amine, I get a different looking compound. This amine has two R groups. Instead of getting the double bond directly on the N, I actually get it down towards one of the R groups.
We’re going to go through both of these mechanisms completely. But I just want to ask you guys, why do you think that double bond might move? Why do you think it might be towards the bottom? Because if you have two R groups on your N, that means that if I were to put the double bond on the N, I would get a formal charge. To eliminate the formal charge, I'm going to have to move the double bond down to one of the R groups. That’s the biggest difference between an imine and an enamine. In fact, they’re pretty much the same exact reaction except for the final step. What we’re going to do is we’re going to through both of these mechanisms and I do want to say one thing really quick, that both of these mechanisms are going to pass through a very important intermediate. That intermedia is called the iminium cation. The iminium cation looks like this. It's a nitrogen with a double bond and with two things on it, either Hs or Rs and a plus charge, so two of these and a plus charge. We're always going to pass through the iminium cation and then how we resolve that cation is we determine whether it's going to be an imine or an enamine. Let's go ahead and look into the mechanism of an imine.
Concept: Concept: Imine Mechanism7m
An imine is created when a carbonyl reacts with a primary amine and acid. You can also use ammonia. Ammonia would be considered as 0-degree amine. It’s the same thing. Either 0 degree or primary would both make an imine. You need an acid so I’m going to put H+ here.
For the purposes of this mechanism, I personally like it if I combine my proton with my solvent so that I don’t have multiple types of reagents I’m trying to work with. Can you guys just go ahead and assume with me that NH3 plus H+ would give me NH4+. That’s what I’m actually going to react with. For all of my reactions, for both of these mechanisms, I’m just going to use NH4+, which is fine. You can use any acid you want, whatever acid professor or your textbook gives you. I'm going to go ahead and draw it like this.
What do you guys think is the first step of this acid-catalyzed mechanism? You got it, protonation. How did you guess? Now I’m going to use reversible arrows, not forward arrows. I’m going to draw that I have a double bond H carbon. This has a positive charge. I'm going to go ahead and draw a resonance structure. I’m going to draw this structure so that I can understand what this positive charge is doing. Just so you know, disclaimer, your professor may not draw this resonance structure and that's okay because it’s still valid.
What I’m trying to say here is that there may be minor differences between my mechanism and the one that you learned in class, but again this is a valid mechanism and if you want to throw in a resonance structure, it's totally fine. Your professor isn’t going to dock you for doing this because all the arrows are going to the same place eventually. Now what this is going to do, since I have a positive charge, this is going to yield a nucleophilic attack. Let’s go ahead and take our NH¬3. It has a lone pair, so it's very attracted to that positive charge. We’re going to do a nucleophilic attack and what we're going to get is a compound that looks like this. We’re going to get OH, RR. Those are just R groups and then NH3+. Does it look familiar at all? If you guys have gone through the hemiacetal and acetal mechanism, pretty much everything has been the same up until this point.
The difference is going to happen here. We're going to consolidate a few steps. Instead of deprotonating this thing, we're just going to go ahead and start trying to get rid of that alcohol. The way we could do that is through a proton exchange. A better way to write this NH3+ would be to take the 3, change it to a 2 and then draw one H out. That’s going to allow me to do a proton exchange. This is going to yield a compound that now looks like, this OH2+, NH2, RR. Let me go ahead and backtrack and write down all the steps.
We said that this first one was protonation. The next one was nucleophilic attack. Now we just got proton transfer and now we're about to kick out the leaving group. But instead of it just leaving, we could actually use the electrons from my nitrogen to kick it out. What I could do is I could make a double bond and kick out the water. What this is going to make is a double bond N.
I just realized that my nitrogen is at the bottom. I wanted to face it up. If you don’t mind. I’m going to flip this molecule. I’m going to flip it so that now my Nh2 is at the top and it was making a double bond so I’m going to draw the double bond facing down. There's two R groups. I’m going to put the R groups facing down, positive charge. Do you recognize this intermediate? This is the iminium cation, plus we would also get water leaving. Now that is my compound, that’s my imunium cation. How do you think we can get rid of one of those Hs? We can use the conjugate base. Remember I had NH3 hanging around. I’m going to use NH3 and I’m going to deprotonate. I’m going to take away an H, give the elections to the N and what I'm going to get is my imine plus my catalytic acid and plus yeah, we did make a water. We lost a water like that. Now my product is an imine. How do I know this is an imine? Because it looks like a carbonyl except that has a nitrogen in it. Does that make sense?
This has a lot of steps but if you've watched the hemiacetal or acetal mechanism, this is extremely similar. If you haven’t watched those mechanisms yet, then when you get to them it will seem very similar to this. I'm just trying to show you guys how the mechanisms are very, very related. Also, I think I messed up on an arrow up here. This is supposed to be reversible. Sorry about that. I get too excited with this mechanism. But you have to make sure to go back if you make mistakes, always change it before you submit your exam. We're done with this one. Let’s move on to enamines.
Concept: Concept: Enamine Mechanism7m
Alright guys, so enamines are created when a carbonyl, we're just going to take our simple ketone again this is acetone, reacts with a secondary amine in the presence of acid, so I'm going to put acid and now it turns out guys that a lot of the secondary amines that are used for the enamine process are actually cyclic because if you put an N in a ring, it's a secondary. So let's use a cyclic secondary amine. These are very very common in organic chemistry. So a cyclic secondary amine it could have not been but I'm just using it just because. Also if you guys don't mind let's go ahead and assume that my cyclic amine plus my proton is going to make the conjugate acid which looks like this. Oh wait, which looks like it's two H's. Cool, so in my first step I'm going to take my protonated version or my acidic version of my amine and I'm going to protonate. What this is going to give me is a positively charged oxygen which after you draw the resonant structure will have a highly electrophilic carbon. What happens now? You got it, nucleophilic attack. So I got my nitrogen, my square, I'm going to attack. Now if you guys don't mind, so that we can avoid what happened last time, I'm going to draw the nitrogen facing up, so I'm going to draw that I have my nitrogen with my square and that's attached to 2 carbons and at the bottom I have my O H. On top of that this nitrogen still has an H so there's a positive charge. Now this is the part that we did our proton transfer so now, see everything is the same guys. We do a proton transfer, my O is going to grab the H, give its electrons to the N, I'm going to get a molecule that looks like this with my O H 2 leaving group and now guys I'm in the perfect position to make my iminium cation because I can take the lone pair from the N, make a double bond and kick out my water that's actually an elimination reaction.
So now that I do that what I'm going to get is my iminium cation. My iminium cation looks like this. So notice that now by doing that my iminium cation has a positive charge. So how do you think we can get rid of that positive charge, this is going to go right here iminium cation. So how can we get rid of that positive charge? Well for an emine I could just deprotonate it. Can I deprotonate this nitrogen? Guys, this nitrogen has no hydrogen's on it so how do I get rid of that positive charge? Well actually the only way to get rid of the positive charge is to get rid of the double bond. So this is where your elimination or your deprotonation reaction actually turns into an elimination reaction. What we're going to do is instead of deprotonating the N, we're going to deprotonate one of the H's next to the carbonyl and that's the old carbonyl. This is also called an alpha position because the position next to the carbonyl, see this would have been the alpha position of a carbonyl. So we're going to basically take away an alpha hydrogen that's going to be very important when we talk about reactions at the alpha hydrogen for carbonyls, that's a huge, alpha carbon reactions are a huge part of organic chemistry. So now I've got my iminium cation, what can I deprotonate it with? With my neutral amine so I'm going to take my neutral amine, I'm going to take with it H, I'm going to make a double bond and kick these electrons back up to the N making my enamine. And my final enamine looks like this, nitrogen with a double bond here plus I get my catalytic acid and I got a water again, that left. So let's go ahead and write down all the steps because I know that I didn't do that in this one. So first we had protonation, then we had nucleophilic attack, then we had proton exchange or proton transfer, then this is what we call elimination this is elimination. That's an elimination reaction and then this was deprotonation. So now we have our enamine. Now I do want to point something out guys which is that notice that this double bond here I drew to the right but I could have also done it to the left right, like why didn't I draw it to the left? Just because I decided not to, they're both the same but if you were to have an asymmetrical ketone originally then that means you have two possible enamines that could occur. You could have an enamine with a double bond to the right and an enamine with the double bond to the left and if you have an asymmetrical ketone that you're starting with you should be, you will be responsible to draw both of those enamines on your exam because they could both happen. Awesome. Okay guys, so that's it for this mechanism. It looks terrible but it wasn't so bad. Let's move on to the next topic.
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Write a detailed mechanism that explains the following reaction of 6 to furnish piperidine 7 and cyclohexanone 8. Show all intermediates! Transfer protons intramolecularly!
Complete the following reaction by drawing the structure of the principal major product. Indicate relative stereochemistry where necessary. If there is no reaction, write NR.
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Reaction is incorrect. The correct product should be a .
Reaction is correct because we formed a .
Predict the product when cyclopentanecarbaldehyde reacts with phenylhydrazine (PhNHNH2) in the presence of an acid catalyst.
What product will result from the reaction shown?
d. amino acid
Predict the product for the following reaction sequence.
An iminium can be turned into an enamine in one step. Draw the mechanism of this below. Draw all the arrows to indicate movement of the all electrons, write all lone pairs, all formal charges, and all products.
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Indicate the order in which the intermediates would appear during the conversion of 1 into 2.
1) 1 → I → IV → II → 2
2) 1 → III → II → 2
3) 1 → V → III → I → IV → 2
4) 1 → V → III → II → 2
5) 1 → II → I → III → IV → 2
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What is the product of the following reaction?
Pick the corn chips scent.
Which carbonyl compounds cannot form an enamine with amine 1?
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What combination of reactants can be used to form the product shown?
What combination of reactants can be used to form the product shown?