Concept: Concept: Protection of Aniline Derivatives6m
In this video, we're going to talk about the challenges of doing EAS reactions aniline because it’s the most activated benzene. Strongly activated ring like aniline, remember I told you guys that NH2 is the most activating group.
Unfortunately, even though it’s activating, so it wants to react, it's so active that it opens up the ring to unwanted reactions. Let’s say that you have your aniline and you’re trying to do a nitration on it. You've got your nitric acid. You’ve got your sulfuric acid. Everything's going great. You're excited for this reaction. You’re thinking you're going to get a mix of para and ortho products because remember that aniline is an o,p-director so we wouldn't expect too much meta director. You’re thinking this is going to be great.
But we've got a problem. This aniline is so active that we're going to get some polysubstitution going on, meaning multiple reactions taking place. We may not get the desired point that we want. Actually, this is not going to work. Unfortunately, even though according to everything we learned so far this should be a perfect reaction. But it turns like I said, aniline is just too reactive. We're going to need to do something called protection. We're going to need to protect the aniline before we can keep reacting with it. How do we avoid this?
What we do is we do something called acetylation. Acetylation is a reversible reaction that makes the aniline less reactive, still activating but less reactive so that we can get a single product or the one that we’re desiring the most.
This is the way that acetylation works. Acetylation takes an acid chloride with some kind of base, any kind of base because it’s a base-catalyzed mechanism. In this case, I’m using pyridine. Pyridine is a very popular base in this chapter. It’s aromatic itself, but you could use any other base. It doesn't have to be pyridine.
What that's going to do is it’s going to react with my aniline to add the acetyl group. An acetyl group is a two-carbon acyl group. You're going to add the acetyl group to my nitrogen. This is a mechanism that at this point, you don't need to fully understand and I'm not going to fully teach. But you can imagine that it goes something like this. This is like the abbreviated mechanism. That our nitrogen would attack the carbon and kick out the chlorine. That's not right. Don't show that to your professor. But then again, your professor is not going to ask for this mechanism at this stage of the course.
Now we've got our acetylated or protected amine. What did we learn about this group? This nitrogen is still donating. It's still activating because it has a lone pair. But it now has that carbonyl that’s next to it that makes it slightly less activating. Instead of being strongly activating, now it’s just moderately activated. How's does that help? Because now that it's moderately activated, I can run my reaction on it. I could then at this point, now I could do my nitric acid over my sulfuric acid. I could get my reaction.
Now at this point, I could get it. Let's say that it adds mostly para which it would at this point because of sterics. It’s going to add mostly para. What you would wind up getting is now a nitro group in that position. The cool thing about this acetylation is that we don't want to keep the acetyl group. We’re just doing that so that we can get a moderately activated ring. Now that we’ve acetylated, we can easily remove it with base. Base is going to hydrolyze this thing off. I'm going to actually write that down. It’s hydrolysis. It’s going to hydrolyze that acetyl group off and we’re going to wind up getting the aniline again but now with my substituent that I wanted. It’s a little bit annoying, guys.
But now for the rest of this course, anytime you're working with aniline, you're always going to have to think protection. You have to protect the aniline before you can react with it because what's going to happen is you’re going to get unwanted reactions. Your professor isn’t going to be happy. We're always going to do this acetylation process whenever we’re working with aniline.
That said, we have a problem down here where I want you guys to synthesize this compound. Go ahead and synthesize the target compound using the four reagents. As you can see, we have four. Try to think about everything we’ve learned so far. This is accumulative question based on several things that we've already learned. Go ahead and try to figure out what those four reagents are. Then I'll come back and we’ll answer it for you guys.
Concept: Example: Synthesize the target molecule4m
Alright guys, so this is a very typical multi-step synthesis problem that you would face in this chapter and there's a lot going on. The first thing that we notice is that first of all none of substituents on the first molecule are in the second so we either have to figure out how to remove them or how to transform them. In the case of the nitro, do we know any ways to transform a nitro group into an amino group? Hell yeah we do. So we'll come back to that in a second. Now we also are adding a chlorine. Do we now have the tools to add chlorine to a benzene ring? Yup that's EAS. So you're like why does this take four steps? Well if you're getting annulene, if you're working annulene you're going to have to protect it.
So let's start off at the beginning. What do you think the first reagent should be? Like for example can my first reagent be an EAS chlorination? Should I do that first? No guys, you can't because if you chlorinate right now you have a metadirector you are you going to wind up getting a chlorine here. Is that what we want? No we need to transform it to an O, P director first meaning that I have to reduce it first. Now you can use any of the reducing agents that I've mentioned in the past so all of them be correct, I'm going to use the stannous chloride, I told you guys that's my personal favourite so I've got my tin 2 chloride, I'm sorry it's S N C L 2 and water. What that's going to give me is my annulene so I'm just going to draw it on the side. So now I've got my annulene in place, can I just do a chlorination? No I can't. Before this topic you would have and you would have thought you're doing great but now because of this topic we know that you need to acetylate or protect the amino group before we can keep reacting with it. So what we're going to do is we're going to add C, I'm sorry, yes that's true, C H 3 C O C L over pyridine and what that's going to give us is now a molecule that looks like this where my nitrogen has an acetyl group on it.
So now that we've got that, now I can do my reaction now I can actually do my EAS reaction and what's that going to be? My actual EAS reaction is C L 2 over F E C L 3 so let's go ahead and add that here, C L 2 F E C L 3, that's going to go ahead and give me a I'm just going to draw it over here, oops that's right where my head is, I'm going to move that. That's going to give me a benzene ring that now looks like this I've still got my acetylated or my protected nitrogen but now I've got a chlorine in the para position. Why did it add para? Guys because that acetyl group is sterically bulky, it's definitely going to favour the para mostly and now when I want to finally get to my end product all I have to do is use N A O H or any type of base to do my hydrolysis and regenerate the original annulene. Alright guys so this is a very typical type of multi-step synthesis that you could be responsible for on your exam or on your online homework or even just in your textbook homework and now that we understand this idea of protecting annulenes, it's something that you can't forget. Awesome guys, let's move on to the next topic.
Show how you would use the same sulfonyl chloride as used in the sulfanilamide synthesis to make sulfathiazole and sulfapyridine.
What would happen in the synthesis of sulfanilamide if the amino group were not protected as an amide in the chlorosulfonation step?
Which of the following molecules would not add an ethyl group with Friedel-Crafts alkylation?
Predict the proper order of the reagents that would yield the product in the given reaction below.