|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 & 8mins||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 & 26mins||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: Resonance of Fulvalenes
In this video, we’re going to discuss how aromaticity will drive some molecules to ionize on their own and form very strong dipoles for the sake of remaining stable. Let’s look into this.
Double bonds can be viewed as a loose pair of electrons that can undergo resonance movement and ionization if it hopes to create an aromatic compound. One of the most famous examples of this type of molecule is called a fulvalene. A fulvalene is a hydrocarbon that’s composed of two fully conjugated rings joined by what we call, this is a big word, an exocyclic double bond. It just means that it’s a double bond that’s not really inside of any ring. You see this a lot when you talk about carbonyls being on the outside of the ring, something like this. Exocyclic would describe that, how it’s not inside the ring. How do you count it? Do you count it as two? Do you count it as zero? That’s what this topic is about.
With these exocyclic double bonds, we find that you can use them as a loose pair of electrons. Meaning you could draw a resonance and you could split them into charges of positive and negative to help create aromaticity. Here we have two questions about this molecule that I’d like to answer. One question is which of the following atoms, notice there’s eight atoms on this molecule. Which of them would you expect to most readily react with an electrophile e positive? That’s a really huge question. You might be like “I have no idea how to answer that.” Second of all, does this molecule possess in that dipole? If so, indicate the reaction and draw it.
This is definitely a little bit beyond our level right now because we haven’t discussed it. But this turns out to be a pretty easy question. Like I said earlier, these double bonds, if they’re exocyclic, they can be seen as a loose pair of electrons. Meaning that I should ionize them in a way that’s going to make the most aromatic compounds as possible. I’m going to actually erase this example so I can get a little bit more space. I’m going to draw another version of this molecule here.
What I want to do is draw both possibilities and have you guys be the judge of which one is better. In the first possibility, I’m going to split up the double bond so that it becomes a negative charge on one side here and a positive charge on this side here. We’re going to ionize this double bond into a resonance structure of just ions.
For the blue one, I’m going to do the opposite. I’m going to put the negative chare here and the positive charge here. What we’re ending up with is a single bond. If I could erase that and the double bond, I would but a single bond with ions. Does it really matter which direction I ionize? Are both of these equally as stable? Notice that if this does happen, they’re going to have very different dipoles. The one that I drew on the left would have a dipole pointing to the right because you always go towards the negative thing. The one that I drew on the right would have a dipole going to the left because the negative is on the other side.
Which one is right? Are any of them right? Have you guys figured it out yet? The answer is definitely if this is A and this B, it’s definitely molecule B. Can you help me understand why? Can you explain it? Can you explain it to your friend? Why would molecule B be so much more stable than molecule A? Because we split the exocyclic double bond into ions to make both of the rings stable. We have ring 1 and we have ring 2. Notice that once I place the negative charge on that six-membered ring, I get six pi electrons.
Once I place a positive charge on the second ring, I get two pi electrons. Are those number good? Hell yeah, those numbers are great. Remember that those are the Huckel’s Rule numbers. They predict that you’re going to have unusual stability in those rings. Now let’s look at the other option. What if I put the plus charge on the five-membered ring? I would get four pi electrons. What about putting the negative charge on the three-membered ring? I would also get four pi electrons.
Does this look stable to you? This is like a molecule from hell. This would have two anti-aromatic rings. It would suck. Terrible stuff. You’re definitely going to go with molecule B. this is going to help us to answer our questions. Now that we have this dipole drawn, the answer is yes it does possess in it. That dipole would go to the left. In this case, we could say to the left. But you could just draw it too. Your professor would see it.
But then the first question, which of these atoms are most likely to react with an electrophile e plus? You can answer this question. Now that I drew that resonance structure, it should be pretty clear which atom likes to react with electrophiles the most. Any amount of deduction would be that if there’s an e plus around, my negative charge would be the one attacking the e. The answer is this atom right here. It would be the one that’s most likely to attack the electrophile because of the fact that it’s the most nucleophilic atom on the entire molecule because of that resonance structure. Interesting, right?
Fulvalenes are definitely a weird kind of molecule. But this entire idea of an exocyclic double bond is actually going to come up quite a bit. I want you guys to know that you could always split an exocyclic double bond into ions, into the direction that’s going to help you make your compound aromatic.
If you don’t need electrons, put the negative charge on the top. If you do need electrons, bring them down. That’s how you work with them.
Awesome guys. Now we’re going to go ahead and we’re going to move on to the next molecule.
Concept #2: Resonance of Azulene
So guys azulene is a molecule that we discussed previously and it turns out that azulene also has resonance structures similar to what we just learned with fulvalene. So first of all azulene just to recap is a polisiklik aromatic molecule with a distinctive blue colour I actually went ahead and included an example of a mushroom that is actually coloured by a derivative of azulene so just goes to show how in nature azulene is actually kind of a natural dye turns things as brilliant blue and we have the same two questions here that we want to answer we want to answer which atom would most likely react to the electrophile? And does this possess a net dipole? Well this one's a little bit harder to figure out because there's no exocyclic double bond right there's no exocyclic double bond I'm just going to write that here no exocyclic double bond. So we can't just split any of these double bonds to ions that would not be right. So what is our goal? Our goal with azulene and this actually applies to many polisiklik is going to be, I'm going to put here goal to share a double bond between both rings. What that's going to allow us to do if we can share a double bond between both rings that's going to allow us to figure out an arrangement to have them both be aromatic by themselves. Awesome, so let me just once again show you that we basically have two different options of how we can do this, I'm going to have to write the smaller one at the bottom here I'm running out of room a little bit but there's really only two ways to get azulene double bond between both let me write the original structure again just give me one second you guys should all be writing this as well since I'm going to use it as an example there's two different ways we could do this. Either we could take this double bond and use that double bond to make a double bond in the middle. Now if we do that that's going to make that's going to break an octet right here so if we make that bond we have to break this bond and then we will do that and what that would do is that would give us a product that will now have a double bond in between the rings. Now the other option would be to go from the small ring to the middle so then I would go like this and I actually messed up with my double bonds ops. So this is an error guys this is the kind of stuff I'm human I actually drew an eight membered ring so let's just erase this really quick this is going to get ugly a little bit we're going to do this.
There we go, my apologies so if you want you can pause the videos so that you can catch up but go ahead and try out draw a seven membered ring not an eight. So back to the molecule at hand we would go ahead and we could go from the small ring to make a double bond but once again we have to break a bond so then we would break this one here. So lets look at what you would get in terms of charges in terms of products from both of these I'm actually going to use the space here at the bottom to draw both of the products and then we can see which one looks better. This will also give me a chance to redeem myself with that seven member ring, I thought I was doing pretty well but I guess not 1, 2, 3, 4, 5, 6, 7, no I did it again these are hard to draw man. So that's one and I'll cheat I got to take this and I got to copy it so that I don't have to mess up again. Cool, so now we're going to add N what the molecules look like after these residence structures are formed well notice that this double blond and this double bond are still the same but now I have a double bond here and here meaning that now what charge should I have where the original double bond left I should have a positive here and I should put a negative here. So that's one of the arrangements one of the possibilities, another possibility is that these double bonds are still the same, I still have the double bond in the middle but now I have the positive here because my bond left and my negative here. So we're comparing we're comparing one resonant structure versus another and you guys have to tell me which one you think is more stable do you think the red version is more stable or do you think the blue version is more stable, hint I think you should count up pi electrons and see what the aromaticity of both of these molecules would be.
So let's just say that we've got you know ring 1 and ring 2. So ring 1 has how many pi electrons? Oh sorry you couldn't see that so ring 1 has how many pi electrons? What has 2,4,6, from the double bond that is being shared 2, 4, 6, positives doesn't count as anything so this one has 6 pi electrons. So good so far and then molecule or ring B ring 2 my apologies so I have ring 1 and ring 2, ring 2 has how many pi electrons well it appears to have 2, 4, 6, this one also had 6 pi electrons. Right so that's great we really can't do better than that but lets just verify that blue is wrong so for blue how many pi electrons do we have with ring 1? Guys I have 8 pi electrons right, because I've got 3 double bonds and then I've got a negative charge at the top. Now for ring 2 I have how many pi electrons I have 2, I have 4 I have a positive charge that counts as 0 and I end up with 4 pi electrons here. This sucks, so as you can see this is a structure that is terrible this residence structure would never ever form whereas this resonance structure is actually highly favoured because of the fact that now I have 2 aromatic molecules. Now can you see where this is going now they know what my charges are can I answer the two questions? Absolutely so which atom would you expect to react with electrophile E. Alright guys so the answer is it has to be this atom, this is the only atom on the benzene ring I'm sorry not the benzene ring on the azulene that is going to get pretty much a full negative charge. Does this molecule possess a dipole If so indicate the direction? Hell yeah it does so you could just draw the dipole straight from the positive to the negative so would have some kind of slanted dipole like this and that would show that there's a dipole going towards the negative and actually just so you guys know one of the biggest reasons for azulene's vivid colour is it's strong dipole these chemical properties and physical properties are closely intertwined and actually make it the kind of amazing molecule that it is. So guys anyway so now you know that exocyclic double bond you can just split them into ions easily and you know that for polisiklik systems like this you have to make sure that when you're drawing resonance that your goal is that you want to share a double bond between both rings to allow it to make them both aromatic if you don't share a bond you're never going to make both rings aromatic at the same time. So I hope that made sense we're done with this topic let's go ahead and move on.
Practice: Which carbon in the following compound is most likely to react with an electrophile?
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