|Ch. 1 - A Review of General Chemistry||4hrs & 47mins||0% complete|
|Ch. 2 - Molecular Representations||1hr & 12mins||0% complete|
|Ch. 3 - Acids and Bases||2hrs & 45mins||0% complete|
|Ch. 4 - Alkanes and Cycloalkanes||4hrs & 18mins||0% complete|
|Ch. 5 - Chirality||3hrs & 33mins||0% complete|
|Ch. 6 - Thermodynamics and Kinetics||1hr & 19mins||0% complete|
|Ch. 7 - Substitution Reactions||1hr & 46mins||0% complete|
|Ch. 8 - Elimination Reactions||2hrs & 21mins||0% complete|
|Ch. 9 - Alkenes and Alkynes||2hrs & 10mins||0% complete|
|Ch. 10 - Addition Reactions||3hrs & 28mins||0% complete|
|Ch. 11 - Radical Reactions||1hr & 55mins||0% complete|
|Ch. 12 - Alcohols, Ethers, Epoxides and Thiols||2hrs & 42mins||0% complete|
|Ch. 13 - Alcohols and Carbonyl Compounds||2hrs & 14mins||0% complete|
|Ch. 14 - Synthetic Techniques||1hr & 28mins||0% complete|
|Ch. 15 - Analytical Techniques: IR, NMR, Mass Spect||7hrs & 20mins||0% complete|
|Ch. 16 - Conjugated Systems||5hrs & 49mins||0% complete|
|Ch. 17 - Aromaticity||2hrs & 24mins||0% complete|
|Ch. 18 - Reactions of Aromatics: EAS and Beyond||4hrs & 31mins||0% complete|
|Ch. 19 - Aldehydes and Ketones: Nucleophilic Addition||4hrs & 54mins||0% complete|
|Ch. 20 - Carboxylic Acid Derivatives: NAS||2hrs & 3mins||0% complete|
|Ch. 21 - Enolate Chemistry: Reactions at the Alpha-Carbon||1hr & 56mins||0% complete|
|Ch. 22 - Condensation Chemistry||2hrs & 13mins||0% complete|
|Ch. 23 - Amines||1hr & 43mins||0% complete|
|Ch. 24 - Carbohydrates||5hrs & 56mins||0% complete|
|Ch. 25 - Phenols||15mins||0% complete|
|Ch. 26 - Amino Acids, Peptides, and Proteins||2hrs & 54mins||0% complete|
|Ch. 26 - Transition Metals||5hrs & 33mins||0% complete|
|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|
|Aromatic Heterocycle Nomenclature|
|Cumulative Aromaticity Problems|
|Polycyclic Aromatic Hydrocarbon Nomenclature|
Concept #1: Heteroatoms
There's only one more type of molecule that we need to be able to predict aromaticity for. That’s heterocycles.
What is a heterocycle? That’s just going to be any ring that contains at least one heteroatom within it. Recall what a heteroatom is. That’s just going to be any non-carbon atom. That could be something like nitrogen or oxygen but also phosphorus or sulfur. These are all very common atoms we found within rings. When you that, it's called a heterocycle.
One of the best examples of a heterocycle that I could think of is pyridine. Pyridine was a base that we commonly used in Organic Chemistry 1. If you guys remember, it had this basic lone pair that could be used for reactions especially for acid-base type reactions. Later on in this section, we’re actually going to discuss why that lone pair is basic.
But for right now, we have to understand that heterocycles are going to present one extra complication to figuring out aromaticity which is that typically a heteroatom is going to have one or more lone pairs on it. The question is going to be “Do I count that lone pair towards the pi conjugated system?” For pyridine, would I go ahead and count this lone pair towards the total sum of electrons to determine Huckel’s Rule? Or would I ignore it?
It turns out that it's not a clean and simple rule. There’s actually a few steps that you have to go through to figure that out. It's not just that they either donate or that they don't donate. There's situations in the middle.
Let's look at what the rules are. Heteroatoms can choose to donate up to one lone pair each. That means for example, oxygen has two lone pairs but only one of them is able to be donated into the ring. Why would oxygen want to donate one of its lone pairs to the ring? Let's take a look.
One, the oxygen or whatever heteroatom already has to be sp3-hybridized to do this. That means that if it was sp2 or sp-hybridized, then that lone pair is definitely not getting donated. It’s only getting donated if the atom was sp3 to begin with.
But that's not the only thing. We have a second criteria. One, the heteroatom needs to be sp3. But two, you’re only going to donate if it helps to create aromaticity. Meaning that you're not going to donate a lone pair if it goes against Huckel’s Rule and if you wind up getting a number of pi electrons that makes it anti-aromatic or nonaromatic. You would only donate if it makes it aromatic.
We could just go back to this example of pyridine. I’ve already given you some clues. Why don’t you guys try to solve for question A, whether you think that pyridine is an aromatic compound or not? But also predict whether you think this lone pair will donate to the ring or won't donate. Basically, should this lone pair count towards Huckel’s Rule, or should you just ignore it?
Go ahead and try to use those tools and then I'll explain the logic behind it. It’s all your turn now.
Example #1: Determine heterocycle aromaticity
First of all, do you think a pyridine is aromatic? I gave that one away already so yeah, it’s aromatic. But how so? Why is it aromatic? Is that lone pair going to donate or not? Actually for two reasons it's not going to donate its lone pair. First of all, let's look at the hybridization of this nitrogen. What is that hybridization? That hybridization is sp2. I told you as explicitly that you're never going to donate a lone pair unless it's sp3. This one cannot donate so I’m not even going to consider it.
Second of all, even if it was sp3 and it donated, how many pi electrons would you then have? We already have 2, 4, 6. If I were to donate these electrons to the ring, I would get 8. For two reasons, that lone pair is just going to sit there and it's going to be highly accessible. It's not going to be involved with the ring at all. The answer was this would have 0 lone pairs donated.
Go ahead. Try to do the second problem. Try to use the same logic and predict what the aromaticity would be.
Example #2: Determine heterocycle aromaticity
Alright guys so what do we put for aromaticity? This is aromatic, good job. So this is an aromatic molecule. How so? Well notice that the nitrogen has one pair and you have to ask yourselves would that more pair donate or not? So first of all I have to look at the hybridisation? What is the hybridisation of that nitrogen? It's SP 3 so that means it's a candidate to be donated.
I'm not saying it absolutely will be but it's a candidate. Two, if I donate those electrons will it become a 4 N plus 2 number of pi electrons? I already have 2, 4 yes it will if I donate those electrons I'm going to get 6 so I would get 4 N plus 2 so it's going to be aromatic so the answer here is that aromatic and one lone pair I should keep in the same way one lone pair will donate. Does that make sense guys? We're just going straight off of the rules. Go ahead and try to apply them to the problems.
Example #3: Determine heterocycle aromaticity
And the answer for C is? Non-aromatic, very good, non-aromatic. Now why is that? Because let's just look at the heteroatoms, we've got 2 heteroatoms this time, we know that they can each donate one lone pair if the conditions are right so let's look at the hybridisations first. So let's just draw these completely, two lone pairs, two lone pairs, my question is will one from each donate? So first of all the hybridisations of both of these happen to be SP 3.
So both of them can basically qualify to donate a lone pair. Now the question is will it be beneficial for them both to donate a lone pair? And the answer is no because if you get this one donating one lone pair, this one donating one one pair, how many electrons do you get? You get eight pi electrons in total because you're going to have 2 and 2 from the double bond so that's 4 and you're going to get an extra 4 from the lone pairs. So that means that it's going to choose to not donate its electrons because that way it can stay non-aromatic instead of becoming anti-aromatic. You might be wondering, but Johnny why doesn't it just have one of the lone pairs donate and then the other one stays the way it is? Well guys, remember that if this lone pair doesn't donate, if both of them don't donate, it's not fully conjugated. So if all it did was donate the top one and kept the bottom one the way it is then this is not a fully conjugated molecule because that lone pair isn't participating in conjugation. In order to participate in conjugation the lone pair has to flip into the ring. So anyway it wouldn't be beneficial to only donate one pair since it's not going to be aromatic anyway, it's not fully conjugated. So it's kind of an all or nothing either they both donate or they both don't donate. So this would be no or zero lone pairs donate.
Side note, this molecule is called one four dioxin and if you look at the wikipedia page for it, it used to say that it was an anti-aromatic molecule and I was like that's wrong so I went to some primary literature and I looked up like this scientific book that actually analyzed the bond links and through the bond links they were able to determine that this is a non-aromatic molecule so I actually edited the page and now if you go to one four dioxin it says that it's a non-aromatic molecule so I'm so nerdy that I'm actually writing wikipedia articles about molecules. Alright so just kind of a side note for you guys to know wikipedia isn't always right, it's constantly being updated thankfully you got smart people on there that are checking things but anyway so that's my good deed for the for day, I saved wikipedia from one small tiny little error. So let's go ahead and go to molecule D and see if you guys can predict the right aromaticity there.
Example #4: Determine heterocycle aromaticity
So molecule D is non-aromatic. Great job. I know a lot of you got that. Why? Because guys it is not fully conjugated. Notice that this carbon has 2 H's so not every atom can participate in resonance. If it's not fully conjugated could this ever be aromatic? No. If it's not going to be aromatic then why would these lone periods ever donate? I told you the only reason it donates is to help it to become aromatic but this was a lost cause from the beginning because not even fully conjugated. It's like me taking a straight chain and asking you if the lone pair is going to donate. It's not because it doesn't fulfill the four tests of aromaticity and this is non-aromatic and once again zero lone pairs will donate. Perfect, next question.
Example #5: Determine heterocycle aromaticity
And the answer for E was non-aromatic. So why is it non-aromatic? Maybe you're getting the hang of this by now, because guys let's analyse the heteroatom.
We said that the heteroatom will only donate if it fulfills two criteria. One, is it SP 3 hybridized? Yes it is so then you go to the second one. Two, would donating one lone help it to become aromatic? No that would make it anti-aromatic, right? That would make it have 8 pi electrons which we don't want so then the answer is going to be that these lone pairs are going to remain outside of the ring, they're going to remain horizontally positioned to the ring and they're not going to participate but here's the problem you're probably saying Johnny, why isn't this an aromatic molecule since I have 2, 4, 6 electrons and I'm not counting these but that's what I was trying to say earlier. If you don't count the electrons on the O then this is not fully conjugated it's not fully conjugated, ooh that's a G, fully conjugated.
So if it's not fully conjugated then there's no way that it could be aromatic, get it? Conjugation happens when you have a lone pair flip into the ring so then it participates in conjugation. So the answer is that once again this would be you know this would be a 4 N number if you did donate so zero lone pairs donate. Excellent, so let's move on to the next question.
Example #6: Determine heterocycle aromaticity
This was a tricky one. The answer is actually aromatic. So you might be scratching your brains how that happened. It's not that difficult. So first of all we have to figure out what does this molecule even look like? What kind of lone pairs does it have? Well nitrogen has one lone pair, does boron have a lone pair? No, boron has an empty P orbital. Boron and aluminum are special for always having that empty P orbital that's kind of like a cation, right? It doesn't have a charge but it's an empty orbital. So can an empty orbital participate in resonance? Sure, I could totally put my electrons into it and that wouldn't be a problem meaning that this molecule is fully conjugated if the nitrogen will donate its lone pair so let's see if it will. So one, we're looking at the nitrogen, what kind of hybridization does the nitrogen have? SP 3.
Two, will it help to create a 4 N plus 2 number if my lone pair donates? Well let's count it up. I've got two, I've got four, this orbital here that is just sitting here it does participate in conjugation but it doesn't add any electrons so that orbital with that orbital I still just have four from the two double bonds, now if I flip this lone pair into the ring that becomes my 6th electron, my 5th and 6th pi electrons. So the answer is that yes I would get 4 N plus 2, so one lone pair will donate.
Interesting right? So that time I tricked you a little bit with Boron because you probably weren't thinking that Boron could be part of conjugation but it has an empty orbital so that's the the same as seeing a carbon cation basically. Empty orbital that you can stuff electrons into. Alright so let's move on to problem G.
Example #7: Determine heterocycle aromaticity
So guys you might have not realized how kind of tricky this problem is but remember that whenever you're dealing with eight membered rings, what do you have to think about? Or just actually any large annulene, you have to start thinking about planarity, right? And we said that an eight membered annulene, eight annulene, a cyclooctatetraene likes to fold like a taco so if this folds like a taco then that nitrogen donating electrons wouldn't seem to help much because it's not going to be aromatic anyway but then remember we also talked about another rule that said but if the taco can get enough electrons to be aromatic it will flatten out again kind of like a tortilla, alright. So let's see what goes on here, we've got this molecule that's got 2, 4, 6, 8 pi electrons.
Yeah that triangle could count, they're all fully conjugated but then we've still got this lone pair in the nitrogen and I'm wondering what's going to happen with that lone pair, is it gonna donate? Is it not going to donate? Well if it donates first of all let's just look at the rules. Is it SP 3 hybridised? Yes. Sorry, so is it SP 3 hybridised? Yes. Two, if it donated into the ring would it give you the right number of electrons, the Huckle's rule number? Yes, it would it would give us 10 pi electrons which would be a 4 N plus 2 number so what did I say happens to an eight member ring when it has the right number of electrons? It flattens out like a tortilla so it turns out that this actually would be aromatic because of the fact that those electrons can cause it to take the right confirmation and now all of those orbitals will be able to communicate with each other and conjugate with each other. So these last ones are getting tricky guys so just kind of do your best and then I'll explain them along the way. Here's another tricky one try to do your best with each and then I'll explain it.
Example #8: Determine heterocycle aromaticity
So what was the answer here guys? This nitrogen was definitely a little weird compared to the other hetroatom we looked at now if you just looked at the positive charge and didn't think about long pairs you could probably confuse yourself on this question but the fact that this nitrogen with 2 hydrogens on it has no lone pairs should be a giveaway that this thing can not participate in residence why? Because I said that a hetroatom can only participate in residence if it donates one one pair this one has no lone pairs to donate, so the answer is that this is going to be nonaromatic.
So let me just walk you through this, first of all we always go through this thing of saying what hybridisation with a lone pair donate, but I can't do any of that because I don't even have a lone pair. So now let's look at the rest of the molecule well the rest of the molecule you're right it has 6 pi electrons so you're thinking maybe aromatic but it's not fully conjugated once again not fully conjugated because this nitrogen doesn't have any empty orbitals it has literally 4 sigma bonds and in order to participate in resonance we have to break a bond to a carbon or to a hydrogen so that doesn't make sense so this is nonaromatic because it cannot participate in resonance with every atom the nitrogen will be excluded, super tricky. So let's move on to the last question and then we'll move on to another section.
Example #9: Determine heterocycle aromaticity
Alright guys in the final question I'm going to go ahead and take myself out of the frame so that we can have plenty of space to draw but as you see there's actually three heteroatoms to consider here and we're going to have to do you know all of the different criteria with you know all three of these so first of all they each have lone pair so it struggles in I'm just going to raise that, each of them have a lone pair. We have to take all their hybridisations and all that stuff, so the hybridization of the first one that I drew is S P 2 then I have S P 3 and then I have S P 2. So right off the bat how many of these lone pairs are available to donate? The answer is only one only one of these long piers is available, the others are not available because they're S P 2 hybridize so that means I don't have to think about those so that means the first question was you know I got it 1 S P 3 I'm going to put here 1 S P 3.
So now my second question is about pi electrons, about would it make it aromatic if I added those pi electrons in. So let's start counting up our pi electrons so I know that for sure I've got 2 pi electrons with that double bond 4 with that double bond now the pi electrons on the other nitrogen's do they count should I go ahead and say 6 and 8? Absolutely not guys remember that we said these nitrogens cannot contribute so I should not count those lone pairs so so far all I have is 4 so what if I add these lone pairs that's exactly right they will help to contribute to huckel's rule number of pi electrons meaning that they will donate and this will be aromatic. So we will put here 2 I know there's not a lot of space I'll just add it in over here, one lone pair donates. Crazy right so hopefully this has taught you guys how to navigate molecules with multiple heteroatoms it's not that hard if you use my system barely any thinking involved you just have to be consistent about how you apply the rules. So that's pretty much I threw the hardest problems I could at you these should be probably harder than anything you'll experience, let's go ahead and move on to the next topic.
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