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Leah here from leah4sci.com and in this video we're going to look at Huckel's Rule in detail followed by a shortcut to save you time when working out aromaticity problems. In the last video, we looked at conditions for Aromaticity where a molecule has to be cyclic, planar, conjugated, and follow huckel's rule to be qualified as aromatic. The first three can be identified just by looking at the molecule but huckel's rule requires solving for the equation 4n + 2 is equal to pi electrons. When you're under pressure of an exam, you don't have that kind of time you ideally want something fast that you understand so you're confident with your answer and able to move on to the next question. If we analyze these 2 molecules using the conditions for aromaticity, we'll notice that they're both rings, the first and last carbon are attached to each other. They're both planar since every carbon has a pi bond, every carbon is sp2 hybridized making the entire system planar, you can resonate around both of these rings make both of these systems fully conjugated. But if you analyze the energy of these molecules, you'll notice that benzene is a very stable, not very reactive compound and cyclobutadiene is very unstable to the point where it actually react with itself because it's so unhappy as it is. How do you figure this out out of a laboratory setting? This is where we have to solve for 4n + 2 equals pi electrons. When we say 4n + 2 equal pi electrons, this is a slight simplification. Most of the aromatic compounds you see will have the pi electrons but you'll also see molecules that have lone electron pairs that participate in resonance and are therefore part of the conjugated system. So instead we should change this 4n + 2 is equal to electrons where we're specifically referring to all resonating electrons. We'll start with benzene since we're most familiar with it and set up the equation 4n + 2 is equal let's see, one, two, three, four, five, six. We have six pi electrons all capable of resonating. Let's solve for n. To solve for n, we subtract 2 from both sides giving me 4n + 0 or 4n is equals to 6 minus 2, 4 then we divide both sides by 4 to isolate n, divide both sides by 4 and this gives me n is equal to 4 over 4, a number divided by itself is equal to 1, n equals to 1. What's the significance of 1? nothing, except for the fact that it's a whole number. Now let's look at our other example of cyclobutadiene which we already said is so unstable it will react with itself. Let's see how huckel's rule justifies that by setting 4n + 2 is equal to 1,2,3,4. 4 pi electrons. Subtract 2, gives me 4n + 0 or 4n is equal to 4 minus 2 or 2. Divide both sides by 4 which gives me n is equal to 2 over 4, 2 over 4 can be reduced to 1 over 2, it doesn't matter which one we use because fact of the matter is 1 over 2 is a half, half is a fraction and that means this disobeys huckel's rule. This helps us understand why this molecule is so unstable and why it is not considered aromatic. Let's take a look at a few more. I want you to pay attention to this pattern. If you look at this molecule, we have a total of 8 carbons that appear to be an aromatic ring. The molecule is cyclic, carbon 1 and 8 are attached to each other. The molecule is all sp2 hybridized atoms with pi bonds all around the system, the way it's drawn right now it appears to be planar. The pi bonds can resonate over the entire ring making it conjugated but doesn't obey huckel's rule. 4n+2 is equal to 8 resonating electrons. Solve for n, subtract 2 on both sides giving me 4n is equal to 6 divide both sides by 4 which gives me n is equal to 6 over 4, you can simplify this to 3 over 2 or 1 and a half doesn't matter, it's not a whole number, it's a fraction and that means this molecule disobeys huckel's rule, is unstable, and not considered aromatic. Take a look at this interesting system. We have not one but two rings recognizing that resonance can go outside of the first ring making the entire system conjugated, that means it's potentially aromatic. Let's see, is it cyclic? yes! it's a bicyclic compound but that's still qualifies as cyclic. All the carbons are sp2, we have pi bonds everywhere, that makes it planar. We shown that it can resonate over the entire system making it conjugated and when we apply huckel's rule, we apply it to the entire resonating system which means we set 4n + 2 equals 10 electrons. If we solve for n, we subtract 2 on both sides which gives us 4n is equal to 8. Divide both sides by 4, cancels out the 4 and gives us the n is equals to 8 over 4 which can be simplified to 2 over 1, no this isn't a fraction because any number over 1 is that number, is a whole number in this case 2. That means a 10 electron system does obey huckel's rule and is aromatic. Why did we go through all these problems? Let's go back to all of this again. When we had 6 electrons, we get a whole number, it was aromatic. 4 electrons, not a whole number, anti aromatic. 8 electrons, also fraction, anti aromatic. 10 electrons, whole number, aromatic. What if we write this out as a pattern. We know that 6 is aromatic, we know that 10 is aromatic. We know that both 4 and 8 were anti-aromatic but what if we continue? If we do the math, you'll know it is that 2 gives you a whole number, then on the right, 12 is a fraction, 14 gives a whole number, 16 is a fraction, 18 is a whole number and the pattern continues. Don't take my word for it, prove it to yourself by solving 4n + 2 equals all of these numbers. This specifically refers to the number of electrons. In Organic Chemistry the faster you get through a problem the quicker you move on to the next one, the more likely you succeed on your exam so let's make this even simpler. Unless we're dealing with radicals electrons are always in a pair. We have electrons pair in a pi bond, and we have lone electron pairs. So let's ask ourselves, if we have 2 electrons, how many electron pairs do we have? 1. If we have 4 electrons, 2. 6 electrons, 3 and so on. This is how you can identify huckel's rule without working through the math. First determine if your molecule is cyclic, planar, and conjugated. If yes, count the resonating electron pairs and ask yourself, do I have an odd number of electron pairs? If yes, this molecule obeys huckel's rule. Do I have an even number of electron pairs? If yes, this molecule does not obey huckel's rule. For example, let's say I showed you this molecule and ask you to determine if it was aromatic or not. Looks a little scary, but let's break it down. This is why it helps to have a system to follow so that no matter how terrifying it is, you know exactly what to do. Is this molecule cyclic? Can I trace the path around the entire molecule so that the first carbon attaches to the last carbon, absolutely yes! Is this molecule planar without building it on the molecule. Recognize that every single molecule atom has one pi bond, no charges, that means they're sp2 hybridized, and if they're all sp2 hybridized, the molecule is planar. Next we look for conjugation. With every single carbon having a pi bond, we can assume that they can all resonate between each other because there's not a single sp3 carbon to block that resonance to break up the conjugation, which leaves the final question, does it follow huckel's rule or not? Let's see. Instead of doing the math or counting the electrons let's keep it simple by counting the number of pi bonds. 1,2,3,4,5,6,7,8,9,10,11. We have 11 pi bonds or you can think of it as 11 electron pairs, an odd number of electron pairs does follow huckel's rule so we want to say, yes, this is aromatic. Let's prove it. But, I want you to trust that this work so that you don't have to do it on your exam. We set it up as 4n + 2 is equal to 22 electrons. 11 electron pairs, 11 pi bonds means 2 per bond for 22 electrons. Minus 2 on both sides gives me 4n is equal to 20, divide both sides by 4, divided by 4, n is equal is equal to 20 over 4 which you should recognize can be reduced to 5 over 1 which is 5. We don't care that it's equal to 5, we care that 5 is equal to a whole number, making this molecule aromatic. What if I give you a molecule that does not meet all 4 criteria? You've heard me use the term Anti-aromatic or non-aromatic which is it, and how do we tell the difference? That's exactly what we'll be discussing in the next video. First, give this video a thumbs up, make sure you subscribe to my channel so you don't miss any new videos, and then visit my website for the common aromatic compound cheat sheet, aromatic practice quiz, and this entire series. leah4sci.com/aromaticity.