Aromatic Hydrocarbons

Hey, guys. So now that we know those four tests of air mentis ity and now that we're experts on counting a pie electrons, it's time to put that information together to figure out if a molecule is aromatic or not. So remember what those four tests of Air Metis City were. We had the whole has to be a ring thing cyclic. We had fully conjugated we had that has to be plainer. But remember, there was that last rule that was a little confusing. The foreign plus two pi electrons Hucles rule rule. All right, so now we know how to count up I electrons, but foreign plus two is still kind of confusing. What does that mean? Well, the whole reason that we have this idea of foreign Plus two is because someone realized that foreign plus two would be an easy shorthand to memorize thes magical numbers that make a molecule extra stable and therefore aromatic. So the way it works is that end is equal toe Any integer, any whole number, right in in woman. I'm sorry, guys into jerk. And when you make end to equal and any integer, then what you do is you wind up getting these numbers that are the super stable numbers. So that would be if n equals zero, then that would be, too if n equals one and there'll be four times one plus two, which would equal six if n equals two. Then there'll be two times four plus two, which equals 10 and so on and so forth. We get these numbers that go to 6, 10 14 etcetera forever. Okay, these numbers are the Hucles room numbers. And some of my students just prefer to memorize the numbers instead of foreign plus two because they think it's easier that way. I'm gonna leave that to you if you want to just memorize +26 10 instead of foreign plus two. That's easier for you. Go for it. All I care about is that you use the right numbers on your exam. Now you might be wondering, Why are these numbers so great? Why are they so stable? They look like normal numbers. To me, that is the topic for a different video. In another video, we're going to discuss why these numbers actually contributes to stability and why they make the molecule so badass but for right now, just memorize it then. Remember that we had this other category, which was, Let's say that you meet the first three tests, but we get a four end number of pie electrons. Well, thes air, different numbers, right? Thes we're gonna be the multiples of force. These we're gonna be numbers like four, 8, 12 16. Guess what? These air magical numbers as well, But they're magical in a bad way. They suck. They make the molecules super unstable. In fact, it's really hard to even synthesize these molecules in the lab because they're so unstable, so thes molecules, they're gonna be what we call anti aromatic. Remember these air molecules that are much less stable than normal. And remember that these air called sensitive follow brass lows rule because Brussels rule said that you have the first three tests met. You're still cyclic. You're still fully conjugated. You're still plainer, but you have the wrong number of pie electrons. In fact, you have a four end number of pie electrons. So now let's talk with this third category of non aromatic recall that I stated that non aromatic molecules are simply molecules that fail one or more of the tests. So if you're not a ring, you're automatically non aromatic. Now, one thing I want to point out is that some pie electrons can actually count towards failing the rule If you have an odd number of pie electrons. So I would literally be any odd number. For example, the number e g for the number seven. Okay, the number seven doesn't fall into any of these Any of these types of electrons, right? It's not foreign. Plus two, it's not for n. It's simply left out. We never discussed it. If you have an odd number of pie electrons, you're also said toe fail Hucles rule. So that means that you would be automatically non aromatic. Okay, so just keep that in mind that you can even fail. Um, the test severe Metis ity by the number of pie electrons that you have, you might be wondering when would you get an odd number of pie electrons when you have radicals? Because radicals count on Lee one. Okay, so now I have a ton of practice for you guys. We're just gonna do one at a time. Go ahead and look at the first one. I know it's a little bit too easy, but let me know if you think that first molecule is aromatic or not. Based on the four tests of domesticity and your ability to count up for M plus two pi electrons go.

So is this molecule aromatic? Of course it is. But let's obviously go through the four tests to verify it. So, one is it a ring? Yes. To is it fully conjugated? Yep. Three. Is it cleaner? You bet four doesn't have a four end plus two number of pie electrons. It has six pie electrons. So that would be one of those magical Hucles rule numbers. This is aromatic. Awesome. So now we know how to prove something that you just vaguely knew before. Okay, so now move on to the second compound and tell me what type of Irma to see it has.

Is this compound aromatic? Actually, no, it's not. In fact, it's anti aromatic. Let's go ahead and find out. Why one? Is it cyclic? Yes. To Is it fully conjugated? Yes, it is. Positive charges can participate in resonance, so the entire every single atom can resonate. Three. Is it plainer? Yep. Four is It doesn't have foreign plus two number of pie electrons. No, it actually has to. For it only has four pi electrons, which, if you recall, is one of the really bad numbers. This is a four end number, so it's got four end number of pie electrons. So that means this is gonna be anti aromatic. Not so bad. Right? Let's move on to the next question.

What was your conclusion here? Hopefully everyone said non aromatic because we have a problem right from the get go. It is not cyclic so immediately. I don't have to keep going. I don't have to test anything else out. If it fails one or more of my tests, it's automatically non aromatic, which simply means that it's in a category that would have been counted by all the other chapters of organic chemistry. It does not belong in this chapter of organic chemistry. Okay, so we could just move on to the next compound.

So what do you think about this next one? This next one was anti aromatic. Why? Because it is cyclic. Two. It is fully conjugated three. It is plainer. Let's go ahead and draw this out really quick. Remember that it's double bond. Double bond, double bond. Negative charge. How maney pie electrons does it have? It has four end number of pie electrons because it's a multiple. Four. It actually has eight pi electrons, correct. So because it's four end, this would be another example of anti aromatic. Beautiful. So let's move on to the next question.

This next compound looks a little bizarre, but all the same rules apply to poly cyclic molecules that applied to mono cyclic molecules. So it still is cyclic right? So we would go ahead and check that off to its fully conjugated. There's no Adam here that can't resonate. Three. It's plainer. I haven't been given a reason to believe it's not plainer. And then for how maney electrons does it. Have you got it? Guys, this is again ate pie electrons. This is a bad number. We're on a bad streak here. We've got a bunch of unstable compounds. This is anti aromatic. All right, cool. So let's move on to the next compound.

alright guys. So I'm hoping you didn't put aromatic here because there's a big problem with this molecule. Let's go step by step. The first one is Is this cyclic? Of course it's a ring to Is it fully conjugated? No, it's not. This is an example of a molecule that is not fully conjugated. Remember, the entire perimeter of the ring needs toe have be able to resonate. And what I see here is that this toll bond can resonate. This orbital can resonate. But here I have an SP three hybridized carbon that does not have any available orbital's to resonate. In fact, to put electrons to move electrons into that carbon, I would need to somehow break a bond to hydrogen. Remember that you can't move Adams in a resonant structure. So this absolutely is not fully conjugated. So it fails the fully conjugated test. Which means I don't need to look any further. This is automatically non aromatic. Makes sense. Yeah, it does. Okay, so move onto the next one

thes. We're getting harder, huh? So what do you say about this one? So this is a very common molecule that you could see on your exam a little bit higher in the difficulty level. This is an aromatic molecule. So this is actually a very famous poly, cyclic aromatic molecule called as you lean, which we actually will discuss later on in more depth. But right now, how could we prove the as a lien is an aromatic molecule? Because it's sick, like it's poly cyclic. So for sure, to its fully conjugated three, it's plainer. And then four I'm running out of space here, for it has how many electrons has 10 pie electrons, which happens to be one of my magical Hucles rule numbers. So I'm gonna say foreign plus two, this is aromatic, as you mean displays highly unusual stability for its level of unsaturated. So the definition of your Metis ity as you lean meats it very well. Okay, so let's move on to the next compound

would you put for Letter H? Guys, this compound is non aromatic. Why? Because it's cyclic, obviously to it is fully conjugated because remember that radicals can participate in resonance. In fact, they do all the time. Three. It's plainer. And then, for what number of pie electrons does it have? Well, it has two plus one because remember that radicals only contribute one electron to the pie conjugated system. So that means we've got a total of three pie electrons. What did I tell you guys about the significance of odd number Pi electrons. What does that mean? What test does it mean? Remember that if you have an odd number of pilot Trans, you actually fail the Hucles rule test. And if you fail the test, you're just like the rest of the non aromatic compounds. You do not belong in this chapter. Okay, so this would be a odd number, so it fails the rule. So we're just going to say it's non aromatic, Okay? It does not possess any unique stability. It does not possess any unique instability. It's just a normal molecules, just like the rest of the molecules in or go. Okay, Awesome. So one more question and then move on to the next topic

so, guys, everything on this page was fairly easy. And now I just threw in a really hard question at the end. I'm just envisioning a few of you arguing with your friends on the computer saying, No, it's this No. What's that? It's really this and you're about to punch your friend, and then you're like, Let's just play the video and see what it is. Well, I'm sorry to tell you that you might both be wrong, because this is such a hard problem. I'm gonna take myself out of the screens that we can talk about it. Guys. Sadly, this molecule is actually drum roll aromatic. How in the world is this molecule aromatic? I know that was the last thing you were thinking. How could that be? Okay, well, let's go through the first step. The first one is that obviously it's sick like Okay, two. Is it fully conjugated? Well, there is an area here that is missing conjugation. If you'll notice this carbon here has to h is correct. So this carbon is not fully conjugated. That means that in order for my pie conjugation system to exist, it can't exist on this ring. because this part of the ring is not fully conjugated. Remember that I told you guys, I'll just bring myself back into the camera really quick. Remember that I told you guys that in order to be fully conjugated, you just needed to be conjugated around the perimeter of the molecule or around one perimeter. You just need to have one loop of pie electrons. That makes sense. Okay, Now, typically, for the mano site are for the poly cyclic molecules we've been looking at. We go around both rings, but notice that on this molecule we've got a problem. This carbon right here is not fully conjugated, so there's no way that I could go all the way around. If I go all the way around, I'm gonna get stuck here. So that means that this ring can't work, which means that my pie conjugated system is actually limited toe on Lee. This ring, this is the perimeter that I care about. This is the ring that is fully conjugated. So what that means is my loop or my loop of conjugation is Onley around this ring and not around this ring. So it is fully conjugated, but on Lee with one of the rings, meaning that three. Is it plainer? Sure for how maney pie electrons are in that ring. Well, since let's say this is ring A and this is ring Be right. The pie electrons in ring be don't count because they're not part of the pie conjugated system. So that means I only count the electrons that air in the ring that's actually fully conjugated, which would make 246 electrons. This has six pie electrons, which makes it aromatic. So actually, this is similar to a benzene ring just with carbons coming off of it. But it has pretty much the same stability that you would find in a benzene ring. So I'm sorry to burst your bubble. I know that you were doing great with this page fact, you're probably like in snooze mode, and all of a sudden you're painfully woken up and you realize that you could be thrown a question like this on your exam and get it wrong. So I just wanna let you guys know that it's not always easy. You have to think you have to use what's here. Okay, Can't turn it off just yet. and but if you apply the four rules, you should be fine. So let's move on to the next topic.