|Ch. 1 - A Review of General Chemistry||4hrs & 47mins||0% complete||WorksheetStart|
|Ch. 2 - Molecular Representations||1hr & 12mins||0% complete||WorksheetStart|
|Ch. 3 - Acids and Bases||2hrs & 45mins||0% complete||WorksheetStart|
|Ch. 4 - Alkanes and Cycloalkanes||4hrs & 18mins||0% complete||WorksheetStart|
|Ch. 5 - Chirality||3hrs & 33mins||0% complete||WorksheetStart|
|Ch. 6 - Thermodynamics and Kinetics||1hr & 19mins||0% complete||WorksheetStart|
|Ch. 7 - Substitution Reactions||1hr & 46mins||0% complete||WorksheetStart|
|Ch. 8 - Elimination Reactions||2hrs & 24mins||0% complete||WorksheetStart|
|Ch. 9 - Alkenes and Alkynes||2hrs & 10mins||0% complete||WorksheetStart|
|Ch. 10 - Addition Reactions||3hrs & 33mins||0% complete||WorksheetStart|
|Ch. 11 - Radical Reactions||1hr & 57mins||0% complete||WorksheetStart|
|Ch. 12 - Alcohols, Ethers, Epoxides and Thiols||2hrs & 34mins||0% complete||WorksheetStart|
|Ch. 13 - Alcohols and Carbonyl Compounds||2hrs & 14mins||0% complete||WorksheetStart|
|Ch. 14 - Synthetic Techniques||1hr & 28mins||0% complete||WorksheetStart|
|Ch. 15 - Analytical Techniques: IR, NMR, Mass Spect||7hrs & 18mins||0% complete||WorksheetStart|
|Ch. 16 - Conjugated Systems||5hrs & 49mins||0% complete||WorksheetStart|
|Ch. 17 - Aromaticity||2hrs & 24mins||0% complete||WorksheetStart|
|Ch. 18 - Reactions of Aromatics: EAS and Beyond||4hrs & 31mins||0% complete||WorksheetStart|
|Ch. 19 - Aldehydes and Ketones: Nucleophilic Addition||4hrs & 54mins||0% complete||WorksheetStart|
|Ch. 20 - Carboxylic Acid Derivatives: NAS||2hrs & 3mins||0% complete||WorksheetStart|
|Ch. 21 - Enolate Chemistry: Reactions at the Alpha-Carbon||1hr & 59mins||0% complete||WorksheetStart|
|Ch. 22 - Condensation Chemistry||2hrs & 13mins||0% complete||WorksheetStart|
|Ch. 23 - Amines||1hr & 43mins||0% complete||WorksheetStart|
|Ch. 24 - Carbohydrates||5hrs & 56mins||0% complete||WorksheetStart|
|Ch. 25 - Phenols||15mins||0% complete||WorksheetStart|
|Ch. 26 - Amino Acids, Peptides, and Proteins||2hrs & 54mins||0% complete||WorksheetStart|
|Aromaticity||8 mins||0 completed|
|Huckel's Rule||10 mins||0 completed|
|Pi Electrons||5 mins||0 completed|
|Aromatic Hydrocarbons||15 mins||0 completed|
|Annulene||17 mins||0 completed|
|Aromatic Heterocycles||20 mins||0 completed|
|Frost Circle||15 mins||0 completed|
|Naming Benzene Rings||13 mins||0 completed|
|Acidity of Aromatic Hydrocarbons||10 mins||0 completed|
|Basicity of Aromatic Heterocycles||11 mins||0 completed|
|Ionization of Aromatics||19 mins||0 completed|
|Physical Properties of Arenes|
|Resonance Model of Benzene|
|Cumulative Aromaticity Problems|
|Polycyclic Aromatic Hydrocarbon Nomenclature|
Concept #1: Aromaticity of Hydrocarbons
Hey guys! Now that we know those four tests of aromaticity and now that we're experts on counting up pi electrons, it’s time to put that information together to figure out if a molecule is aromatic or not.
Remember what those four tests of aromaticity were. We had the whole “It has to be a ring” thing, cyclic. We had fully conjugated. We had that it has to be planar.
But remember there was that last rule that was a little confusing, the 4n+2 pi electrons, Huckel’s Rule.
Now we know how to count up pi electrons but 4n+2 is still kind of confusing. What does that mean? The whole reason that we have this idea of 4n+2 is because someone realized that 4n+2 would be an easy shorthand to memorize these magical numbers that make a molecule extra stable and therefore aromatic.
The way it works is that n is equal to any integer, any whole number. When you make n to equal an any integer, then what you do is you wind up getting these numbers that are the super stable numbers. That would be if n equals 0, then that would be 2. If n equals 1, then that would be 4 times 1 plus 2, which would equal to 6. If n equals2, then that will be 2 times 4 plus 2, which equals 10, and so on and so forth. We get these numbers that go to 2, 6, 10, 14, etc. forever. These numbers are the Huckel’s Rule numbers. Some of my students just prefer to memorize the numbers instead of 4n+2 because they think it’s easier that way. I'm going to leave that to you. If you want to just memorize 2, 6, 10, 14 instead of 4n+2, if that’s easier for you, go for it. All I care about is that you use the right numbers on your exam.
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 contribute 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 let's say that you meet the first three tests. But we get a 4n number of pi electrons. These are different numbers. These are going to be the multiples of 4. These are going to be numbers like 4, 8, 12, 16. These are magical numbers as well but they’re magical in a bad way. They suck. They make the molecule super unstable. In fact, it's really hard to even synthesize these molecules in the lab because they are so unstable. These molecules are going to be what we call anti-aromatic. Remember, these are molecules that are much less stable than normal. Remember that these are called since they follow Breslow’s Rule because Breslow’s Rule said that you have the first three tests met. You’re still cyclic. You’re still fully conjugated. You’re still planar. But you have the wrong number of pi electrons. In fact, you have a 4n number of pi electrons.
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. If you’re not a ring, you’re automatically non-aromatic. One thing I want to point out is that some pi electrons can actually count towards failing the rule. If you have an odd number of pi electrons. That would literally be any odd number.
For example, the number 7. The number 7 doesn't fall into any of these types of electrons. It’s not 4n+2. It's not 4n. It’s simply left out. We never discussed it. If you have an odd number of pi electrons, you're also said to fail Huckel’s Rule. That means that you would be automatically non-aromatic. Just keep that in mind that you can even fail the test of aromaticity by the number of pi electrons that you have.
You might be wondering, “When would you get an odd number of pi electrons?” When you have radicals because radicals count as only one. Now I have a ton of practice for you guys. We're just going to 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 aromaticity and your ability to count up 4n+2 pi electrons. Go.
Example #1: Determine the aromaticity
Is this molecule aromatic? Of course it is. But let's obviously go through the four tests to verify it.
One, is it a ring? Yes.
Two, is it fully conjugated? Yup.
Three, is it planar? You bet.
Four, does it have a 4n+2 number of pi electrons? It has six pi electrons so that would be one of those magical Huckel's Rule numbers,
This is aromatic.
Awesome. Now we know how to prove something that you just vaguely knew before. Move on to the second compound and tell me what type of aromaticity it has.
Example #2: Determine the aromaticity
Is this compound aromatic? Actually no it's not. In fact it's antiaromatic. Let's go ahead and find out why. One, is it cyclic? Yes. Two, is it fully congegated? Yes it is, positive charges can participate in resonance so the entire every single atom can resonate. Three, is it planar? Yup. Four, does it have 4 N plus 2 number of pi electrons? No, it actually has 2, 4 it only has 4 pi electrons which if you recall is one of the really bad numbers. This is a 4 N number so it's got 4 N number of pi electrons so that means this is going to be antiaromatic. Not so bad right? Let's move on to the next question.
Example #3: Determine the aromaticity
What was your conclusion here? Hopefully everyone said nonaromatic 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. So we can just move on to the next compound.
Example #4: Determine the aromaticity
So what do you think about this next one? This next one was antiaromatic. Why? Because it is cyclic. Two, it is fully conjugated. Three, it is planar. Let's go ahead and draw this out really quick remember that it's double bond, double bond, double bond, negative charge. How many pi electrons does it have? It has 4 N number of pi electrons because it's a multiple of 4 it actually has 8 pi electrons, correct? So because it's 4 N, this would be another example of antiaromatic. Beautiful, so let's move on to the next question.
Example #5: Determine the aromaticity
This next compound looks a little bizarre but all the same rules apply to polycyclic molecules that apply to cyclic molecules. So it still is cyclical right? So we would go ahead and check that off. Two, it's fully conjugated there's no atom here that can't resonate. Three it's planar I haven't been given a reason to believe it's not planar and then four how many electrons does it have? You got it guys, this is again 8 pi electrons this is a bad number, we're on a bad streak here we've got a bunch of unstable compounds this is antiaromatic. Alright? Cool so let's move on to the next compound.
Example #6: Determine the aromaticity
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. Two, 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 to be able to resonate and what I see here is that this double bond can resonate, this orbital can resonate but here I have an SP 3 hybernised carbon that does not have any available orbitals 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 atoms 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. So move on to the next one.
Example #7: Determine the aromaticity
These are getting harder. So what 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 the difficulty level this is an aromatic molecule so this is actually a very famous polycyclic aromatic molecule called Azulene which we actually will discuss later on in more depth but for right now how can we prove that Azulene is an aromatic molecule? Because it's cyclic it's polycyclic for sure. Two, it's fully conjugated. Three, it's planar and then four I'm running out of space here four it has how many electrons? It has pi electrons which happens to be one of my magical Huckle's rule numbers so I'm going to say 4 N plus 2, this is aromatic. Azulene displays highly unusual stability for its level of unsaturation. So the definition of aromaticity, Azulene meets it very well. So let's move on to the next compound.
Example #8: Determine the aromaticity
What did you put for letter H? Guys, this compound is non aromatic. Why? Because it's cyclic obviously. Two, it is fully conjugated because remember that radicals can participate in resonance in fact they do all the time. Three, it's planar and then four what number of pi electrons does it have? Well it has 2 plus 1 because remember that radicals only contribute 1 electron to the pi conjugated system so that means we've got a total of 3 pi electrons. What did I tell you guys about the significance of odd numbered pi electrons? What does that mean? What test does it mean? Remember that if you have an odd number of pi electrons you actually fail the Huckle's 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. So this would be a odd number so it fails the rule so we're just going to say it's non aromatic. It does not possess any unique stability, it does not possess any unique instability, it's just a normal molecule just like the rest of the molecules in orbit. Awesome so one more question and then we move on to the next topic.
Example #9: Determine the aromaticity
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 it's that, it's really this and you're like 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 going to take myself out of the screen so 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? Well let's go through the first step. The first one is that obviously it's cyclic. Two, is it fully conjugated? Well there is an area here that is missing conjugation if you'll notice this carbon here has two H's, correct? So this carbon is not fully conjugated. That means that in order for my pi 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 pi electrons that makes sense.
Now typically for the mono segment or for the poly segment 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 can go all the way around. If I go all the way around I'm going to get stuck here so that means that this ring can't work which means that my pi conjugated system is actually limited to only 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 only around this ring and not around this ring. So it is fully conjugated but only with one of the rings meaning that three is it planar? Sure. Four, how many pi electrons are in that ring? Well since let's say this is ring A and this is ring B, right? The pi electrons in ring B don't count because they're not part of the pi conjugated system so as long as I only count the electrons that are in the ring that's actually fully conjugated which would make 2, 4, 6 electrons this has 6 pi electrons which makes it aromatic.
So actually this is similar to a benzene ring just with carbons coming out 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 are doing great with this page in fact you're probably in like snooze mode and all of a sudden you're painfully woken up and you realize that you could be thrown a question like this in your exam and get it wrong so I just want to let us know that it's not always easy you have to think you have to use what's here. Can't turn it off just yet but if you apply the four rules you should be fine. So let's move on to the next topic.
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