|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 & 56mins||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|
|Aromatic Heterocycle Nomenclature|
|Cumulative Aromaticity Problems|
|Polycyclic Aromatic Hydrocarbon Nomenclature|
Concept #1: Heterocycles - Which lone pairs react?
So far, we haven't learned any reactions that aromatic compounds undergo because they're so crazy stable. They don't react with anything really. However, you might be asked to react an aromatic heterocycle with a strong acid.
In that case, this is going to be an acid-base reaction and nothing more. Nothing actually happens to the aromatic ring. We’re just doing an acid-base reaction with lone pairs on the outside of the ring. Let’s see how this works.
Heterocycles, as you have seen previously, often have multiple lone pairs that are available to react with acids. The question that we have to ask ourselves and the question you're going to be asking yourself in the exam is which lone pair do I react with? Because it can be very confusing. These lone pairs don't just say “Me, me, me!” You have to think about it. You have to conclude which lone pair is going to want to react with the acid.
Let's take this example. Here I have a molecule called imidazole. It has two nitrogens with lone pairs. It's got this nitrogen with the red lone pair and this nitrogen with the blue lone pair. I'm reacting this with a strong halohydric acid, HX. This could be HCl. It could be whatever. We know that this lone pair is going to be attracted to which atom? That one of the lone pairs would be attracted to which atom on the HX? You've got this incredibly strong dipole. You’ve got a partial negative and a partial positive.
These lone pairs our basic. They're going to be tempted to attack the H and do basically a proton grab. They’re going to be attracted to the proton. The question is will the H attach to the red N or will it attach to the blue N or will it attach to both?
How do we solve this question?
Concept #2: The Sp2 Rule
So the rule that we are going to use for this is that acids can only react with the lone pairs that are not necessary for aromaticity like we just said if a lone pair is making a molecule stable, why would it make sense for that lone pair to react with an H and then make the molecule non aromatic that molecule wants to stay aromatic so that lone pair that is being donated to the ring is not available at all to be reacted with H X. So that leaves what's left over right what can we actually react with, well S P 2 hybridize long pairs are basic because if you remember the S P three long pairs are the ones that are able to donate S P 2's were never allowed to donate. So that means if you have an S P 2 hybridize long pair that is a basic long pair because that one is not required to maintain aromaticity on the ring meaning that if we were to drop the final product here would you draw the H attaching to the red nitrogen or to the blue nitrogen and the answer is that to maintain aromaticity the final answer to this question would simply be this with a hydrogen here and a plus charge and probably an X negative hanging by pretty close.
So this looks like it's a reaction it looks like maybe you're supposed to add an X somewhere guys it's really a lot less complicated than that it's literally just an acid base reaction but you have to pick the right heteroatom that's the only challenge if you pick the blue lone pair you would have been wrong because that would have made a non aromatic compound no notice, let's just double check is this product still aromatic? What do you think is this molecule still aromatic, well what we have is a ring? Yes Is it plainer? Yes Is it fully conjugated? Well it's only fully conjugated if this long pair donates. If that long pair donates does it have a huckel's rule number of electrons? It has 2, it has 4 that positive charge doesn't count towards anything it doesn't add electrons and now it's got 6 so this molecule is still aromatic that means I drew it right that means I drew this reaction correctly. So now I'm going to give you guys a practice problem it's a little bit it's like the next step of this question where there's a little bit more to think about but I believe in you, think about everything you know about acids and basis to try to predict what the exact product of this reaction is, hint there's only one correct answers. So go ahead and do that and then I'll answer the question.
Example #1: Draw the acid/base reaction product
So let's just start out of by drawing all the lone pairs and what we notice is that oxygen since it only has room for lone pair here that means it must have a positive charge right because oxygen is only neutral with two lone pairs so if that threw you off I'm sorry but you just have to write that in yourself on this one. So then what else this nitrogen has a blue lone pair and this nitrogen has a green lone pair. So basically it's red versus blue versus green only one of them is going to be the winner who is it going to be. Well I mean you could use the rules that I gave you above which is going to at least narrow it down to two, so let's just talk about that for a second we said that you're only going to use electrons that are not contributing to the ring. Well right now is this ring or is this molecule this structure is it aromatic? Well is it siklik? Yes is it fully conjugated? If this nitrogen donates it is, is it plainer? Yes if this nitrogen donates the green lone pair would it be a huckel's rule number. So we would have 2,4, and then 6 would be addition of the green lone pair, what about the red lone pair would that one count? No because it says P 2, how about the blue lone pair would that one count? again as P 2 we never count those S P 2 lone pairs because they can't contribute no matter what this is aromatic right now.1:48
So you want to make sure that the product of this thing is also aromatic, so is there a long pair that I can definitely cross out that we're not using. I heard over half of you guys say green. I'm just going delusional at this point but I'm pretty sure I heard that so green is not available I'm going to cross that one out that means we have a choice it's either red or it's blue that's the one that we that's what we have to decide between are we going to use the oxygen to take an H or we going to use the nitrogen to get an H, and the answer to this one is actually pretty simple. It just has to do with stuff we learned in the acid and base chapter a very long time ago but It could simply be answered as which of the atoms is more basic is it nirtogen long pair more basic or is it oxygen long pair more basic what do you think? You got it you always hear nitrogen is more basic than oxygen so obviously we have a choice between two S P 2 long pairs you're going to pick the more basic one the one that's more likely to react with the hydrogen. So that being said that means that my final molecule would look like this and B R hanging around and this still has a positive charge and it wouldn't look like the H being on the oxygen so of course the mechanism would have been this.
Giving the H to the nitrogen meaning that that nitrogen was more basic than the oxygen so its obviously going to be the one that attacks the hydrogen. On top of that the big mistake you really can't make is that you can't say that the top nitrogen would have taken it because that means that you don't have any understanding of aromaticity and you're just lost but hopefully not at this point. So let's just go to the next topic.
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