Concept: General Mechanism11m
In this video, we're going to talk about a really important synthetic pathway called the Curtius rearrangement. This reaction does go by another name. It’s also called the reduction of acyl azide. That name really describes what's going on here. An acyl azide is a functional group that has a carbonyl in R group and then N3 on one side. That’s an azide. If you can reduce this molecule and get the carbonyl to come off, what we're going to wind up getting is we’re going to wind up getting a primary amine. This is going to require a really interesting rearrangement for that to take place. There's only two reagents in this reaction. There's heat and there's water. You might think that it's a really easy mechanism but actually it's really strange because of the fact that atoms are rearranging and we're going to have a reactive intermediate called in a nitrene that isn’t really commonly seen in Orgo. We're going to definitely make sure to handle that in the mechanism part.
What are two steps? The first step is heat. Heat is going to drive the rearrangement of this molecule here into isocyanate. Isocyanate is its own topic within Organic Chemistry. There are a lot of reactions that we can do with isocyanate. But for right now, this is just going to be the middle structure that we use the intermediate structure that we then react with again to get our primary amine. What's our seconds step? Our second step is addition of water to that isocyanate. That's going to result in a decarboxylation reaction and it’s going to liberate two gases at the end. It's going to liberate CO2 gas because decarboxylation reactions always liberate CO2 gas. We're also going to liberate N2 gas from the decomposition of the acyl azide in the first step. I have mentioned a few terms here, for example decarboxylation. That’s a mechanism that I teach in a separate video. If you want to know more about what a decarboxylation reaction is, you can always just type it into the Clutch search bar and you'll find that video. That being said, let's go ahead and just dive into this mechanism and see what it's all about.
I already showed you guys what an acyl azide looks like. When you draw it out in bond line structure, this is what you're going to get. That's the most common way that it's represented. It's N double bond N double bond N and with the formal charges accordingly. But this azide has resonance structure. That resonance structure has to be taken into account. That resonance would be that the negative charge could make a triple bond. That would already have three bonds on one side. That means I only need one more bond on the other or I’d be violating the octet. If I make this bond, I'm going to break this bond and form a lone pair on that nitrogen. That means that the resonance structure on the other side looks something like this. I’ve got a triple bond here, a negative charge here because I added an extra lone pair, and I still have a positive in the middle because it still has four bonds. Nothing changed for that middle nitrogen.
What’s interesting here is that usually these resonance structures both exist. They're both averaging into the hybrid. But in the presence of heat, one of these resonance structures is going to influence the character of the molecule more than the other. The reason is because we can get a decomposition reaction. Heat is going to make this decompose. The reason is because this resonance structure of the second one looks a whole lot like N2 gas. N2 gas is just N triple bond N. It composes 78% of the atmosphere. If 78% percent of the atmospheres is N2 gas, do you think it's stable or unstable? It’s highly stable, guys. I mean, you're breathing it in constantly. It’s not reacting with you, is it? Or not too much. This is a very stable entity. If you can become N2 gas, you're like in chemical nirvana. That's like the best thing you can do. Look at this resonance structure. It's one bond away from being an N2 gas. In heat, what you're going to get is a decomposition reaction where these electrons pick up and leave us a lone pair. What you wind up getting is now your N which has two lone pairs and nothing else, R, plus your N¬2 gas.
This molecule that I drew as a product, it's really an intermediate because it has the right number of valence electrons. Remember that nitrogen wants to have five. It has five. But this is highly reactive because it’s not filling its octet. This nitrogen only has six octet electrons. If you were to count lone pair is two, lone pair is two, bond is two, that’s 6. We need 8. This is a very reactive intermediate called nitrene. It turns out that this nitrene is going to want to rearrange. This is the Curtius rearrangement part. What you’re going to get is at the R group, the electrons from the R group actually attach to the N, a lone pair comes down and forms a double bond. This is just going to be a rearrangement. What we wind up getting is our isocyanate as a product. I’m going to have O double bond C double bond N with now an R group coming off of it. This is a molecule called isocyanate which as I alluded earlier is actually a really important molecule in Organic Chemistry. There's a lot of different addition reactions we can do to this. But for this mechanism, we're only going to do one and that's going to be the addition of water.
So you could end this here. If all you did was you added heat to an acyl azide, you’d get isocyanate and you’d be done. But we want to get a primary amine. Essentially we want to liberate this part. All we want is the NR component. We don't want the carbon. We don't want the O. How can we get rid of that? If we add water. In my second step, I add water. Can you guys predict which atom the water will be most attracted to in a nucleophilic attack? What do you guys think? You got it. We're going to attack the carbon. Carbonyl carbon, guys. We’re going to push the electrons down to the N. What this is going to give us is a molecule that now looks like this. OH, N, R, and eventually this is going to get an H because it's going to form a negative charge and it’s going to protonate. This is called carbamic acid. What’s special about carbamic acid is that it is very unstable. It's not a stable molecule, so we can spontaneously decarboxylate.
Again, for more on the mechanism on decarboxylation, search that topic individually. But we’ll just draw it as it relates to this molecule here. This is going to be our decarboxylation. Let me actually draw this H sticking out so it’s going to be easier. This nitrogen grabs the H. make a bond, break a bond. We're going to take the electrons from the single bond and donate them to bond between the O and the C. Now that that carbon has four bonds, the carbonyl carbon, we're going to use these electrons to make a lone pair on the end. It’s a weird reaction. But notice what you get at the end. What you're going to get is now a nitrogen with an R group. With how many Hs? Two. You’ve got the original H and we've got the new H that we grabbed. What else do we have? We also have a carbon with now a double bond O and a double bond O. Do you guys know what that is? The product of a decarboxylation is always CO2 gas. We just created two different gases. One of them is a greenhouse gas hurting the environment. We're keeping it in the lab though.
This is really interesting. We’ve just made CO2 gas. We’ve made N2 gas here and here. But most importantly and what your professors will be most interested in is that we made a primary amine. That primary amine also lost a carbon. This is going to be an interesting synthetic reaction that we use when we’re trying to get an amine and we’re trying to lose a carbon. I’ll show you guys why you wind up that in a second. This is a really great reaction to use. It looks complicated but it’s actually used more often than you’d think in Organic Chemistry. That being said, that's the whole mechanism. I hope that made sense. Let's go ahead and look down with this question. This question does draw from our aromaticity chapter, electrophilic aromatic substitution. But that being said, if you have covered those chapters already, go ahead and try to remember, brush up on those reactions. See how you could perform this synthesis of making aniline. Basically, what I’m trying to do here is I’m trying to make aniline. We've never learned an EAS technique to make aniline before. I want you to think about how you could use the Curtius rearrangement to make aniline. Then I'll show you the answer, so go for it.
Example: Propose a Synthesis5m
Alright guys. So, maybe you didn't know it and that's okay But, what I do want to emphasize here is that this pathway that I'm showing you is one of the more important pathways for aromatic synthesis and it's one of the fastest ways that you can make aniline because if you recall I've never taught you an EAS mechanism for aniline. So, we have to get creative, when we want to make it so the first thing you do is you'd add an R group, so that was already done for you, we already have an R group there. So, if you wanted to add an R group, hopefully you could remember some reactions that you could use to add R groups, for example, friedel-crafts acylation, okay? But now, if the R group is there, what can you do? Well, we want to turn this into an acyl acid to rearrange, so the first step to doing that would be a carboxylic acid, let's go ahead and do a sidechain oxidation with KMnO4. Now, I'm already noticing, I'm going to run out of room on this box because KMnO4 you could write it but some professors want to see every single reagent, so let's write that out, it's going to be in base with Heat slash acid, okay? So, KMnO4 is the short way to put it but now I just included all the reagents. So, what that's going to do guys is that's going to add, I'm not going to draw every single structure I'll just draw it one at a time, that's going to add a carboxylic acid here, remember that you always pretty much can oxidize any sidechain to a carboxylic acid as long as it has at least one hydrogen, which it did, perfect, so the second step would be, let's figure out a way to make this in a acyl acid, we have to put an N3, so the next step would be to use SOCl2, okay? SOCl2 is a very common reagent to turn carboxylic acid into acid chlorides, okay? So, now I've just kind of heightened the reactivity of my carboxylic acid derivative, this reaction, the next two reactions that you're going to find come mostly from your carboxylic acid derivatives section of organic chemistry and if you want to brush up on any of these reactions, feel free to go look through there, if anything even if you haven't learn these reactions yet just know this pathway because this is the web you usually going to use a courteous.
So, now I've got the SOCl2, that's a great leaving that chlorine is a great liter group. So, now I can react it with NaN3, okay? What, that's going to do is you've got N3 negative, right? It's just going to do a nucleophilic acyl substitution, now that's not the correct mechanism, there should be a tetrahedral intermediate, I know, but just letting you know let's be a bridge mechanism, eventually the chlorine gets kicked out and I get something that now looks like this N3. So, guys now that you have N3, what could we use as our last reagent to form aniline? guys, we can use heat and water. So, I'm going to put heat and water how so, how does that make aniline? Well, guys remember, what's going to happen, what's going to happen is that two of these nitrogen's are going to leave. So, you're just going to instead of being N3 it's going to rearrange this 2n, right? Then your R group is going to pick up and attach to the N. Remember, you're going to get a double bond all that stuff, eventually you're going to decarboxylate taking the carbon off. So, all you're going to have left is the nitrogen attached to the benzene ring, which is aniline, okay? So, this is a very common way to make aniline in organic chemistry 2, maybe not so much in the lab but synthetically you'll see this pathway come up sometimes, just to help you guys out in terms of a shortcut. I'm not sure if you guys have noticed yet, there is one, which is that if you ever want to predict what the amine is going to look like at the end all you have to do is just look at your acyl acid, take your R group, take your R group and add the NH2 to it, that's always to do. So, in this case notice that my R group was benzene ring. So, in this case all I would do is I would take my benzene ring and I would add NH2 to it and I would say that's going to be my product, okay? Now, obviously that mechanistically, that's terrible, but as a short but if you're, you know, all your exams are time. So, you have to be shrewd about how you use your time, this is a great shortcut that you can use, you can just say, the curtius rearrangement just adds an NH2 to whatever R group I already had, okay? Awesome guys. So, I hope that made sense, I hope you found this video helpful, let's go ahead and move on to the next video.
Give the appropriate reagent to achieve the indicated multi-step retrosynthesis.
What is the product of the following reaction?
Propose a synthetic pathway from the indicated starting material to the designated product
What is the structure of the final product?
Provide the major product for the following transformation.
Provide reagent to complete the following chemical transformation.
Provide the full mechanism for the following reaction
Which of the following is not an intermediate of this reaction?
Predict the product of the following reaction.