Organic 2 Final Practice | Spring 2026

Added: 2026-05-05

[00:00:00]Welcome to the final practice that we'll be doing uh for spring 2026 organic 2.

[00:00:06]Here's just a brief list of all of the topics that might be asked. These are kind of the things that I've decided are the main ideas in organic 2. These are the things that are most likely to kind of come up and and be asked in in this situation. And so we have everything from the general topics that are that are new for this unit which are aromatics. Uh what what does it mean to be aromatic? How do how do I tell if something is aromatic?

[00:00:38]How does the things attached to to rings uh to aromatic rings affect its reactivity?

[00:00:46]Um how does it affect the reactivity of deals alder reactions? So whenever we're dealing with these kind of more complicated systems, how can we see what things how things are connected to them impact their reactivity?

[00:01:00]Um on top of that, pretty much all the reactions that we have gone through or done are fair game here. Um so just a couple of things to kind of keep in mind. I think in general you guys are pretty good at this, but uh a couple of kind of trickier bits that might catch you um are understanding which reactions do one four additions and which ones do one two additions. Um being familiar with all of the different kinds of substitution aelamine enamine. Essentially all all this is is the tetrahedral intermediate stuff. So once you add into a carbonal can you follow your tetrahedral intermediates? Can you find the kicking group and the leaving group? The the students that leave the most on the table. You know, the difference between someone getting a B and an A usually is someone kind of getting lazy with the tetrahedral intermediates and not uh not being alert to when whenever you add into a carbonal, you need to be thinking, hey, is there a kicking group?

[00:01:58]Hey, is there a leaving group? I need to assign those things. If it's acidic conditions, I need to do a proton transfer. So uh really that's can be the the differentiator between a B and an A I I think is on all these mechanisms being very vigilant about when I see something do I know when I do a tetrahedral intermediate? Do I can can I always recognize that I'm doing a tetrahedral intermediate? Can I always find the leaving group, the kicking group? Do I know when I need to do a proton transfer? Uh do I know when I'm doing a one two versus a 14 addition?

[00:02:33]uh do I know when I'm doing an forming an enolate versus adding into the carbonal? Right? These are all things that were kind of benchmarks throughout the semester as things that might be uh a differentiator.

[00:02:46]Uh the last thing is another tricky point is epoxides and caroxyic acids is understanding that these these two things are a little interesting because they work differently in basic versus acidic conditions. So you can get different products depending on which kind of conditions you're in. um and to just be aware of that. Uh here's just a list of all of the synthesis, all the major bond forming reactions that we use some different ideas. Um obviously gineards are are very common, but esester formation. So doing an alcohol with some sort of asyl.

[00:03:26]So just being aware of that. Um, a big part of this new unit is electrophilic aromatic substitution. So, which group you put onto a benzene ring first dictates where the second group goes.

[00:03:39]So, the order in which those happen matter.

[00:03:43]Um, we have substitution which SN2 a leaving group and an alcohol usually um enolate and alkalhalide. That's something that was a big part of the last test. Um, so essentially doing reaction that looks like this where we're able to substitute out the leaving group. So being able to plan that, being able to understand that the cut would go here which is two carbons away from the hetereroatom. Um, so seeing those that pattern and replicating it.

[00:04:27]Um, vidig is something that you can use. Um, oftentimes there's ways around it in the kind of synthesises that synthesis that we're doing. Um, but it's still a good thing to know. Um, we didn't really talk about this one, so I wouldn't worry as much about that.

[00:04:46]Uh but we did talk about the reduction of nitro groups to am means. So that is something that is fair game that you're likely to run into. It's a very common kind of reaction where you reduce that nitro to an amine to to get an amine that you can use um for things like reductive aminations. So whenever I have an amine and I want to add a carbon group to it, um the best way to do that is to have it be a ketone or an aldahhide. Uh and then we would reduce the amine that we'd form. So this is from a couple of tests ago. This is reductive amination. It's a useful reaction to form amines because the problem is if I try to use a leaving group, this nitrogen would over add. So, one thing I thought I would try to do is uh go back to that chart that we've done every day and kind of remind you of the big picture ideas that kind of permeate through this class. Right? So, um I'll start with kind of any sort of anything attached to a metal I'm looking to be a nucleophile. So anything with a negative charge, anything attached to a metal. This is an electron donor. This is something that I'm going to assume is going to attack an electrophile of some kind, amines, alcohols.

[00:06:30]These are things that I'm looking to be sources of electrons.

[00:06:35]And then uh double bonds. But to be honest, double bonds aren't really good nucleophiles, and they're even better nucleophiles than benzene rings, which are just kind of lousy nucleophiles, but we can get them to act as nucleophiles.

[00:06:51]The thing is that these these need good electrophiles and it helps to have an electron donating group orthopera for reasons we'll talk about in a little bit. So um reasons to do with resonance and stuff like that. So having uh electron donating groups makes benzene a better nucleophile. Another nucleophile that I kind of mentioned up top that we need to be careful of or another thing that kind of fits this gap here is between nucleifile and electrophile or caroxyic acids.

[00:07:43]It's good to keep in mind that this thing in acidic conditions becomes an electrophile and in basic conditions becomes a nucleophile, right? Which is pretty Oh, did not mean to do that.

[00:08:03]Oh no. Get out of here.

[00:08:08]I don't know how to remove this thing.

[00:08:12]Well, look at me.

[00:08:14]There we go. All right.

[00:08:17]So, it's pretty, if you think about it for a second, it makes sense, right? If I add base to this, it's going to remove the proton off the caroxyic acid. Then it becomes something that's negatively charged. It becomes nucleophilic. If I add a acid to this thing, it becomes positive, which makes it more electrophilic.

[00:08:36]So um the electrophiles we need to remind ourselves that if I've got a carboation we need to look for rearrangements in a way that forms the most stable carboation. That would be ideal. So look for hydride shifts. If you ever see a carboation at any step for anything, always be looking for hydride shifts.

[00:09:08]Um, we have a big class of things with leaving groups on them. Um, so that's usually iodine, which is going to be better than bromine, which is better than chlorine, which is better than positively charged oxygen. Even florine.

[00:09:25]So florine is a good electro uh good leaving group on a benzene ring.

[00:09:33]This is really when you're in your SN1, SN2, E1, E2 world. Um so um this is leaving group chemistry and then organic 2 deals a whole lot with carbonial chemistry. And there's a couple things I want to note about carbonial chemistry. So one, there's two different places you could add into a carbonal. So this is one two and this is one 4.4 is really going to be the world of soft nucleophiles. So this is really going to be S minus caroxilates enolates any sort of negative charge that can be spread out. So either through resonance or on a big atom like S minus and then this is going to be kind of everything else. So, oh, also coupates.

[00:10:44]So, this big copper atom helps spread out that charge, makes it softer. Um, so this is going to be everything else. So, really, if it's not one of these small list of soft nucleophiles, we're just looking to do one to addition.

[00:11:02]When we do one to addition, it's important to keep in mind that we need to get into our tetrahedral intermediate world.

[00:11:10]Uh this is we need to find a leaving group and a kicking group.

[00:11:18]We need to follow those steps. Um if it's acidic, uh we need to do a proton transfer.

[00:11:32]but only if it's acidic.

[00:11:35]Any other situation we will not have an extra proton. So we do not have to transfer it around.

[00:11:41]Um, another thing in this kind of family is epoxides. And remember that we will add into the more substituted side in acidic conditions and the less substituted and basic.

[00:12:08]All right. Then we have our non-nucleophilic base.

[00:12:21]Our non-ucleophilic bases these tend to be things these will only take hydrogens. So you have something like NAH which is used usually to deproinate O and NH to make them good nucleophiles.

[00:12:47]Um then you kind of have your bulky bases.

[00:12:52]This is like LDA which is strong.

[00:12:57]It's a strong bulky base. you have KO2BU, which isn't a very strong bulky base.

[00:13:03]It's just a bulky base. So, some things you can do with LDA you can't do with KO2BU.

[00:13:08]Um, and that's kind of where we are with non-ucleophilic bases, right? There's there's a handful of them.

[00:13:18]Some are bulky.

[00:13:21]Some are just kind of strong bases that we can use to deproenate uh hetereroatoms, but these do the less substituted substituted enolates and elimination.

[00:13:43]So these are nice to use because They will always take from the less substituted side whether it's a elimination or an enolate formation which chemically are are pretty similar situations.

[00:14:15]They're both electrophiles that get turned into nucleophiles for later use.

[00:14:22]All right. And then last category is acids. So a lot of the acids you've seen in your life look kind of similar.

[00:14:29]They've got a similar format. It usually starts with H and it lists different things like H2O4, H3, P4, phosphoric, sulfuric acid. We've got all of our H hallogen acids. These are good acids because they're conjugate bases.

[00:14:50]What they turn into, the negative charge they turn in a after you take the proton is really stable, right? So there's the old adage that acids are determined by the strength of the conjugate base. The conjugate base is simply what happens once the acid gives its proton away. So once the acid does acid things that it turns into its conjugate base. We also learned about a new um world of acids called lis acids.

[00:15:22]These usually follow the format of like a L X3 where X is a H hallogen or F X3 or sometimes it's just a metal like Fe.

[00:15:34]Uh so you'll see this what these do is they activate H hallogens.

[00:15:48]They activate hallogen let's say electrophiles.

[00:15:59]So they'll generally pry off halogen electrophiles h hallogens off electroiles to form carocations.

[00:16:12]These usually these usually can form carboations.

[00:16:19]All right. So, this is kind of the whole class summarized into a little chart. I mean, there's a lot that's kind of between the lines, right? Um, what does it mean to be a tetrahedral intermediate?

[00:16:34]You know, what's the what happens after you do 14 edition or one two addition?

[00:16:39]But these are kind of all the tricky things. This this is everything that can can get you that can um catch you off guard if if you're not paying attention.

[00:16:48]So, uh here's our big chart for the year with kind of everything involved. I've already posted an answer key, but I'll I'll screenshot this and I'll post this just in case you want this as a reference. Um hopefully it's something that could help you if you're studying for this final or if you're studying for the MCAT or or something else. Hopefully this um I would love to get feedback from you guys at some point in time if you guys found doing this chart every day or um something like this to be something that was helpful to frame and and put reactions into context. Um but yeah, it'd be good to know because I I use I spend a lot of time on it because I think it's important, but it'd be good to good to hear if um y'all's opinions validate that.

[00:17:36]All right. So, here's how we're going to do the rest of this. Um, I have a video.

[00:17:43]I don't know if too many people noticed this, but starting at this point, I have already done all of these mechanisms. This is the same final practice from this point downwards as I did last semester.

[00:18:00]So, I'm going to post that uh when I I'll post a link to that when I post this video. Um, but I'm going to just fly through some of these and tell you kind of what I see and how I think about these. And I won't necessarily do the whole reaction out. Um, but I'll I'll post a video where I show each step and and go through each step of the mechanism. Um, so we can kind of spend a little bit more time talking about how I think you should be studying and how I think you should be thinking about these reactions and like what I think about when I make them. that would hopefully help you guys see kind of my goal my goal is that you can do these reactions very intentionally um that when you get on a test you know what you're looking for you see the patterns and so in order to do that I what I want to mostly do is highlight and point you towards the the important patterns um instead of just doing a whole bunch of mechanisms. I hope by this point mechanisms are a little bit more uh you're feeling a little bit more comfortable to them at that and now the difference is we just need to be able to kind of know which mechanism we're doing. Right? So this says each reaction pair contains one reaction that proceeds significantly faster than the other. Provide the potential product of each reaction if not provided already. Then give a resonance structure that explains why one is faster than the other.

[00:19:27]All right. So, what I see here is this is an electron withdrawing group and this is an electrophile.

[00:19:41]So, it's the same thing in both both of these questions.

[00:19:45]So, what it's trying to get me to think about is that makes my benzene my nucleophile.

[00:19:57]And when benzene's my nucleophile, I have to know a couple of things about it. One, there's a whole bunch of spots I could add this electrophile.

[00:20:08]So that's why directing groups when benzene is a nucleophile are important because I have many different places that could potentially be my nucleophile. I know that this electron withdrawing group is a meta director.

[00:20:25]So I know that if if I were to add this methyl right kick off this iodine as a leaving group I would add it meta right where this is the methyl right and you can go through the mechanism um it's something that you know oh wait this is an electron donating group which is an orthopera director which means that it's going to add here and here.

[00:20:57]So I get two products on this one.

[00:21:04]>> Dr. Little, >> what's up?

[00:21:06]>> Are there specific um conditions when like an EAS reaction would not occur?

[00:21:17]>> Uh yeah. So, like this one up top would probably be hard to get happen, but I wouldn't unless I ask you which one is like, is this a viable reaction? Just kind of do the reaction if that makes sense.

[00:21:35]>> Thank you.

[00:21:35]>> But yes, the one I'm more like to ask a question like this, right? The reaction on bottom, this would be way faster.

[00:21:44]It's about relative speed more than it's about like absolute this could not happen.

[00:21:50]But the one on bottom is faster. And I hopefully it's a little intuitive why it's faster. Well, it's faster because this is an electron donating group.

[00:22:03]There's a there's these electrons right outside the ring that when we draw resonance can push into the ring.

[00:22:11]Um, so I can draw this resonance structure like this. And you can clearly see that in the first resonance structure I draw, there's extra electrons in this benzene ring. This benzene has been turned into a nucleophile because there's something that's a literal electron donor attached to it.

[00:22:34]So this is faster because electron donating groups make benzene a better nucleophile. Right? If nucleophiles are electron donors, having an electron donor attached will make me a better electron donor.

[00:22:56]Likewise, the first resonance structure I draw for this guy and you can see a problem if I wanted benzene to act as an electron donor.

[00:23:15]Now benzene has a positive charge in the ring that's going to deactivate it. It's not going to want to attack that methylide, right? If you think about the mechanism here, at some point that benzene is going to attack and it's not going to want to do that because the ring is electron deficient. There's a lack of electrons here.

[00:23:39]So, this is just trying to point to you like, can you explain to me the difference between how a benzene acts when there's an electron donating versus electron withdrawing? Not only how it directs the thing onto it, but it does this does this activate does this allow us to better do electrophilic aromatic substitutions where this is acting as a nucleophile?

[00:24:06]Now, this reaction is the opposite. I've got a nucleophile.

[00:24:12]I've got a base, which means that something on this picture is going to be my electrophile.

[00:24:18]And I know from experience that it's likely to be the leaving group on the chlorine.

[00:24:24]Now, in this situation, directing groups don't really matter, right? Because there's only one electrophilic spot on this whole benzene ring. So, I know that no matter what, this methanol is going to add into this carbon because it's my nucleophile. That's my electrophile.

[00:24:40]um and the base is going to deproinate it. So the only products I can get are substitution of this chlorine and you can see that uh having you can see that okay so this this question we have an electron withdrawing group on both which is good because benzene is my electrophile right so that's a good thing having electron withdrawing group makes my benzene more electrophilic but um the question is which position do I want this electron withdrawing group to be in right to better uh to better activate this reaction to better allow this reaction to So, I'm going to try to draw resonance structure so I can get a positive charge on my electrophilic carbon. And you'll noticed here that this resonance structure would end up with a positive charge here.

[00:25:56]But if I try to push these electrons over now, I get a positive charge here.

[00:26:03]So, there's no resonance structure I could draw that would allow me to get a positive charge where I want it to go, which is that electrophilic carbon.

[00:26:21]So, it's not like it's bad to have an electron withdrawing group in the wrong spot. It's just when this electron withdrawing group is orthopera, you'll see that I can very easily draw a resonance structure that puts a positive charge on the carbon that's my electrophilic carbon. The carbon that that oxygen of my nucleophile is going to add into after it's deeproinated.

[00:26:57]So, someone said, "On the LA review, there was a question where there was two electrophilic sites on the ring. How would you know which one to attack?" Um, that's a good question. There's there's a couple things I could go into it. I'd have to know which what the molecule is, but um, so I could imagine if I had two electrophilic spots, there's two situations.

[00:27:20]either like let's say they're both florines.

[00:27:31]I know from this that we want the electron withdrawing group to be orthopera or snar for substitution.

[00:27:44]So that means that I'd be more likely to attack the florine that is orthopera.

[00:27:58]Um it was an electron with redrawing group with a chlorine and ketone in the meta positions.

[00:28:17]Um, okay. So, it was like this.

[00:28:36]Is that right, Andrew?

[00:28:48]Yeah, the thing is, okay, so this is this kind of gets back to some other stuff that we've talked about before.

[00:28:57]Um, this is kind of a tricky question if if you're thinking about it mechanistically.

[00:29:03]So, if I just have an oxygen nucleophile, I could add here or I could add here. If I added here, it'd make it so that the only leaving group is what I just added and it could collapse and kick. Um, or I could add into this chlorine and essentially I'm going to skip a step so I don't draw it all out because I'd have too many arrows. But so this this isn't a good electrophile, but this is a reversible process. Um so and this one's not. This would push uh one direction that chlorine is not going to add back in. So here because this isn't really a reaction that could you that that produces anything just like when we did uh enolate stuff, right? If it's the only leaving group that adds in, you're going to collapse and kick it right back off. Um and this is irreversible. We're going to we're going to get primarily that addition.

[00:30:20]It wouldn't change if if the nucleophile was OE.

[00:30:24]It would change if this was OE because now you have a different nucleophile you could kick off or a different leaving group you could kick off. But if it's the only leaving group, then it's it's a reversible reaction.

[00:30:47]But yeah, changing this nucleophile doesn't matter.

[00:30:52]Um, but changing the electrophile kind of does. So that's that's a little bit tricky. I didn't spend too much time talking about that. I wouldn't worry too much about that. I'm more concerned about you recognizing how the electronic effects of having electron donating electron withdrawing group on the ring dictate the reactivity of the ring.

[00:31:14]Yeah, it's a good question though.

[00:31:18]Um, another way you could ask this if um, you know, if you had if you had a chlorine and a florine that were both right next right next to an electron with drawing group, you would end up adding into the florine because it's a better leaving group in this situation.

[00:31:46]So, SN the leaving groups are reversed from a regular substitution. Um, so with SN, that'd be another way to word it is like florine is better than florine, which is better than bromine, better than iodine, which is the opposite of like an SN2, where your best leaving groups are going to be iodine is better than bromine, better than chlorine, better than chlorine.

[00:32:22]Okay.

[00:32:23]So, do you guys have any questions about this this these kind of ideas?

[00:32:29]I kind of this is what I want to spend like I want to make sure that this is clear before we move on. Right? Electron withdrawing group makes benzene a better electrophile ideally in the orthopera position from where that leaving group is. You can only do a substitution onto a benzene ring when there's a leaving group because benzene is relatively electronrich. Otherwise, benzene acts as a nucleophile in which case it will be activated by electron donating groups on it and those have their own influences usually pushing things ortho and parah.

[00:33:03]And you can draw resonance that kind of proves those points, right? If I can draw resonance that pushes a negative or positive charge where I want it to be in order to be the reactive partner, then I can, right? This goes into all of this stuff, right? Nucleophiles, these are electron donors.

[00:33:23]So, the better I can make my electron donor or the better I can make my electron acceptor, depending on which role benzene plays, the better off I'll be.

[00:33:37]So I guess if I wanted to add this add something to this just like the caroxyic acid could go in both directions. If I have a benzene ring with a leaving group on it, it helps to have an electron withdrawing group orthopera.

[00:34:01]Right? And just to highlight the difference between benzene being an electrophile needing wanting an electron withdrawing group versus benzene being a nucleophile needing an electron donating group that needs a good electrophile. This is where we care about directing groups.

[00:34:29]All right. So the last one's the deals alder question if there's no questions here. Um and you can see that the difference is um addition where this guy's where the carbonal is up versus addition where the carbonal is oriented like this the denophile.

[00:34:57]And it's essentially just asking which one of these is better. So if I if I draw my dyene resonance, I see that I have I get a negative charge on this carbon.

[00:35:20]So, I want to line it up so I can get a negative charge on the carbon that it aligns or a positive charge on the carbon it aligns with.

[00:35:31]So, if I draw this resonance structure, I get a positive charge there, which means that this alignment would be better than this alignment.

[00:35:39]And that's all that's really looking at.

[00:35:41]Can you do the um can you do the resonance to show me to prove this?

[00:35:46]Someone said, can we do the polarity instead of drawing resonance structures?

[00:35:50]Uh you can if I just ask for the product, but if I ask you specifically to draw the resonance structure, go ahead and draw the resonance structure.

[00:35:57]It's a way that I can I want to see that you can do the full thing as well that you can show your work.

[00:36:08]Yeah, that's a good question. All right, before I move on, do you guys have any questions about how resonance how electron donating withdrawing groups interplays to these reactions?

[00:36:22]for the um nucleophilic like when benzene acts as an electrophile, >> it requires an electron withdrawing group to happen >> usually. But again, I'm not going to like give you a trick question where I like ask you for a mechanism and you say no reaction. Um it's going to be more like at the beginning of the test, I'd be likely to say which one of these is more likely to happen, why, you know?

[00:36:50]And so I care more that you can explain why one thing is bad than not than that you can be like, well this isn't going to happen. Well, I don't know.

[00:36:59]Um, so the thing is that okay, someone said, could you explain which two carbons are matching up on the deals alder? So let me just give a little bit more explanation to this.

[00:37:20]If I draw the resonance structure with this thing flipped, right? So I'll just flip one of the two of them. You can see that doing it so the oxygen are on the same side. This charge doesn't line up with the positive charge like this.

[00:37:45]uh but here the negative and positive charge do line up.

[00:37:52]So uh we you want to be able to prove that the resonance structures where the things line up is more advantageous or is more advantageous get this reaction to work. So that's why it adds where these oxygen where these things are are par to each other as opposed to meta to each other.

[00:38:13]Uh, someone asked, "Did we get 80% of people show up for the ACS exam?" We got close enough. Um, I'll give you guys extra credit. Uh, I've been trying to get the department to help me buy donuts. I can't I can't guarantee that I can get those. Um, I might have been going past my skis, but uh, yeah, you guys will get the extra credit. So, I'll try to put that up today so you guys can get some indication of where you're at. Um, along with the attendance. Um, so I'll do all of that.

[00:38:43]Um, someone asked again, can I explain again why the SNR can happen with electron donating group being meta? Is it just that resonance helps speed up the reaction, but isn't necessarily for the positive charge to end up on the carbon attached to the leaving group? I'm not sure I understand. It is faster.

[00:39:02]If we can get that positive charge on the right carbon, this reaction is going to be faster than this one where we can't get the positive charge on the right carbon. I mean, it's still good to have an electron withdrawing group, but it's best to have an electron withdrawing group orthopera because we can draw resonance structure that specifically makes that carbon more electrophilic.

[00:39:28]So the instead of doing an exit quiz that's that was we did the ACS. So that's kind of that's the five points.

[00:39:37]Um so we're the five points is the the exit quiz. Every semester I just kind of change I use that depending on what I want to get you guys to do. So it's my little way of of doing that. So um is the final in Fiser? The final is yeah, it's 7:30 in our normal class location.

[00:40:06]Um, aromatic, anti-aromatic, non-aromatic. Um, so there's a couple of rules. It's got to be a ring.

[00:40:16]Um, it's got to be fully conjugated, so no sp3 carbons.

[00:40:24]And then uh it has to have 4 n + 2 pi electrons. So 2 6 10 pi electrons.

[00:40:39]Here I've got an sp3 carbon here. So this is not aromatic.

[00:40:46]Here I have 24.

[00:40:52]And this since this can be involved in resonance here, this counts as a pi electron. So that would be an aromatic ring here.

[00:41:07]Those lone pairs count. This lone pair cannot be involved in resonance. There's nowhere it can go. It's an sp2 like >> nitrogen loone pair. It does not count.

[00:41:20]So this one doesn't count. This one does count. That gives me 2 4 6 pi electrons.

[00:41:27]So that makes that aromatic.

[00:41:33]Here we have a lone pair here that counts because it's part of an sp3 like nitrogen i.e. it can be pushed. We can push these electrons here. No problem.

[00:41:44]So if you're good with resonance, that's a good way to double check. Um and then we also have these electrons that can participate in resonance.

[00:41:52]So I have 2 4 6 8 pi electrons. That makes that anti-aromatic. So anti-aromatic is if you have four in pi is anti-aromatic. So if it's 48, it fits all the other rules, then it's anti-aromatic. If it doesn't fit the other rules, then it's just not aromatic.

[00:42:16]All right? And we've got this. These electrons don't count. They can't be pushed down into it. Right? This electron withdrawing group is not electron donating group. So that those electrons can't um participate. So I get 246. This is aromatic.

[00:42:34]So I've got two pairs of electrons here but only one can participate in resonance because once I do the resonance structure once I do the resonance structure I can't push another pair of electrons here so we will get um this is 246 it counts as aromatic and this one is two four and there's no electrons there. So, this is anti-aromatic.

[00:43:09]>> All right.

[00:43:09]>> What did you want us to know about aromatic and like anti-aromatic apart from just identifying whether a compound is like aromatic or not? Like are we supposed to talk about it with regards to like stability and like like are you going to like ask that? Sorry, I'm not sure if that made sense.

[00:43:26]>> You should be able to talk about it in terms of stability, right? So, not only should you be able to list something as aromatic or anti-aromatic, but I would hope that if I said, you know, which one of these is more stable, why, you'd be able to say, well, this is more stable because it's ant aromatic and this one's less stable because it's anti-aromatic, right?

[00:43:53]But I'm not going to ask you like how many kilogjles per mole stability, you know, like I'm not going to ask that if that makes sense.

[00:44:03]But yeah, you should be able to tell me that an aromatic ring is more stable than an anti-aromatic ring, for instance.

[00:44:11]>> And anti-aromatic is more unstable than non-aromatic.

[00:44:14]>> Yeah, like anti-aromatic is just not going to exist. I'm not going to and I'm not going to ask you to compare the two, but yes, anti-aromatic would be less stable than non-aromatic.

[00:44:28]>> And could you go over how we know which lone pairs count versus don't count?

[00:44:33]>> Yes. So, there's a couple of different ways to think about it. Uh, I think the best way to think about it is to think about resonance, right? This is an electron withdrawing group. So, it's not going to push its electrons into it. Right? A nitrogen part of a double bond. This electron is not part of this ring. You can't push these electrons anywhere. It's not going to go part of the ring because this is an electron withdrawing group that's only going to take in electrons.

[00:45:06]But a nitrogen that's not part of a double bond, right?

[00:45:12]it can push its electrons into neighboring atoms. So that lone pair does count, right? So when you see a hetereroatom that's not a part of a double bond, it's usually an electron donor. It can push its electrons in the ring. Those lone pairs count. But an a hetereroatom that's part of a double bond, those electrons can't be pushed into the ring. It does not those electrons do not count.

[00:45:38]One more thing to note um which is what I drew here is this oxygen is part of only single bonds right there's no double bonds attached to this oxygen so its lone pairs count but once one of the lone pairs gets involved with the ring the other lone pair can't right now this oxygen has become one that's connected to a double bond so that that lone pair cannot be involved in resonance and therefore it is not part of that the pi system is what you call it does not count. So you can only count one set of electrons.

[00:46:17]So long long story short if a hetereroatom is part is not part of double bonds its lone pair counts but only one lone pair counts. If a hetereroatom is part of a double bond like this one its lone pair does not count.

[00:46:34]Does that help?

[00:46:38]Yes. Thank you so much.

[00:47:01]How's that?

[00:47:04]>> Perfect.

[00:47:09]All right. So, instead of going through all of the mechanisms here at length, I'm going to briefly talk about what I'm asking here and what other ways I could ask this question.

[00:47:23]Um, what other things I could do to make this question more interesting? What am I trying to probe at in this question?

[00:47:32]So here I want you to see that this is going to form a grineard. So I want you to be able to draw the grineard recognizing that that magnesium is going to insert here and form a grineard.

[00:47:45]Right?

[00:47:52]And notice something. I've got a what's called an alpha beta unsaturated ketone.

[00:47:56]Right? I've got this double bond right next to a carbonal.

[00:48:01]And so what this question is really trying to probe at is do you know whether or not a grineyard is going to add into this position or this position?

[00:48:10]And so that goes back to the chart we did at the at the front at the top, right? One for additions. Well, Grineard is not one of the one of the things that adds into the the end into a one four editionish spot. So, it's going to be a one two addition.

[00:48:30]So, you can very very easily imagine how I might change this on a test. You know, maybe maybe I give a caroxilate instead or I give something that fits a soft nucleophile and add it here or maybe I give a different hard nucleophile.

[00:48:44]Right? So this is really probing at seeing if you understand the difference between a a one two addition and a 1 14 addition. Right? In this case, a grineer does a one two addition. But the point of this question was to see if you you recognize that it's not going to do the 14 addition.

[00:49:00]Right? So challenge yourself. Try to think of situations where you could draw a similar reaction that does a 14 addition instead. Right? So um that's the case right?

[00:49:16]All right. So this question pH of four to five. So water is usually pH of 7. So pH of four to five. This is acidic.

[00:49:27]So you can just use an acid here.

[00:49:30]Doesn't matter what acid. And all this is trying to get you to do is to do the the trick past this is that ultimately this nucleophile is going to form a cycle. So, do you understand how to do a cycle? Do you understand what happens when something that's part of the same chain reacts with itself?

[00:49:49]Try different reactions in a cycle.

[00:49:52]Maybe see if you can make this reaction cyclic. If you're struggling with a reaction, I would challenge you to try to draw starting materials and try to draw reactions similar to this. How this reaction is going to form an amine. But how this so it's going to form a cyclic amine.

[00:50:13]But how could I make it form an So could you change the starting material to form an enamine?

[00:50:30]That might be a different way to think about this question, right? So, uh, what other things can you see? What am I trying to get you to do? Why did I ask this question? Well, I asked this question because students struggle with making cycles. Students struggle, especially making cycles and then still following tetrahedral intermediates, right? You're likely to to make the cycle and say, "I'm done because I made the cycle." No, you're not done. You just add into a carbonal.

[00:50:58]Follow tetrahedral intermediates. when do I form an amine versus an enamine?

[00:51:02]These are good things to think about.

[00:51:04]Um, how could I change a starting material to to get those two different products? That's this is the kind of thinking that could get you that I think will will improve a letter grade. If after or as you're doing these problems, you're not just doing them. You're thinking about why am I asking this question? What is happening?

[00:51:26]All right. Per paratouine sulfuric acid.

[00:51:29]Someone asked about this question.

[00:51:32]I got up to where we kicked off the water group and we're left with a carbonal. Why wouldn't you take a tertiary hydrogen and make a double bond to even out the charge?

[00:51:43]>> Sorry, some people ask the dumbest questions. I'm confused.

[00:51:47]>> Sorry. What's going on?

[00:52:02]for the question on top of page four with par. I got up to where we kicked off the water group and we're left with a carbonal.

[00:52:23]Yeah. So you got here, right? Where this methanol you you proteinated because paratium sulfuric acid is just an acid. So you treat it like an acid. Then you proteinate it. Then methanol adds in and it collapses kicks and you get this product. Right? And you're asking why don't I do what a enemy does and just deproinate this to form.

[00:53:13]Something like that.

[00:53:16]uh because that forms the an enol and an enol is less stable than the keto form. Um so it's not going to do this in this case.

[00:53:33]So it's just that's just one of the things you have to know is that when alcohols add into ketones and aldahhides you form what's called an acetal where it adds both.

[00:53:54]There's reasons for it. It has to do with the stability of the relative products and actually the best reasoning is has to do with the fact that nitrogen is a much better nucleophile.

[00:54:07]So you don't need to add multiple equivalents to get addition.

[00:54:13]But yeah, nitrogen's going to do something differently when it ends up with that positive charge. It's going to form the enemine.

[00:54:23]So that's just one of those things to know from test two.

[00:54:30]Um so yeah, this doesn't happen. What it does is it forms it adds again.

[00:54:41]And yes, it does repeat. You get to the acetal, right? Eventually, you get to the acetal and then you're going to collapse and kick and get back here, right? But go to the end point. Go as far as you can go before you start repeating steps.

[00:55:01]Because what happens is this is the furthest you can go.

[00:55:05]At this point, you have to start going backwards.

[00:55:08]you form the acetal. That's the the conclusion is it adds twice.

[00:55:15]So you're right that it it yes this is a oh I don't know why I made this a benzene rule.

[00:55:21]You're right that this is a tetrahedral intermediate and that technically this could happen.

[00:55:27]Uh but we don't need to show it because we've already shown that right that just that goes backwards for the first time.

[00:55:42]Uh, someone asked about deals alder requiring like heat. Uh, no. It's just very common to heat up deals alder. So, it doesn't really change anything. It's just how you get deals alder to work.

[00:55:57]So, yeah, that that's nothing special.

[00:56:00]Um, all right. So, isopropanol. This requires you to know some very basic naming. So I'm not likely to ask you a naming question on this test, but knowing kind of what an isopropyl group is, what a methyl group, what an ethl, what a pro what a propel, what aut basic things are are good to know. We know that sodium hydide is a base. Um, and we we have one electrophile in this picture. And so being able to do this reaction where the base deproinates the nucleophile first and then the nucleophile attacks and then you do tetrahedral intermediate. That's all that's looking for. And one of the things that's tricky about this is that it it instead of forming a ring, this is ring opening.

[00:56:44]Are you comfortable opening a ring? Are you comfortable doing those things? um are you comfortable that this is going to add into here and the best leaving group is going to be this and we're ultimately going to kick that off. So again, likewise, if this is something that's troubling you, if you're if you got to the end and you didn't feel like it felt weird to to open a ring up, things it didn't work for you, um then maybe try to come up with other questions that have ring opening in them. um try to create a question that has what what makes it so I can open this ring up. Well, it's that ultimately I have something that's a part of the ring that turns into a leaving group, right?

[00:57:28]This this oxygen here is the is a leaving group connected to this esther. Ultimately, it's going to get kicked off. Could I do another leaving group? Could I do a different situation to to open up a ring? Maybe I could use a different nucleophile. Maybe I could use LH or a Grineyard and see if that works too. So, um it's helpful to to really note everything about these reactions as you go through them and to note what specifically gave you an issue. What did you have problems with?

[00:58:00]Was it that you didn't feel like you could do a ring opening? Was it that you that you attacked a carbonal and didn't follow your tetrahedral intermediate rules afterwards? Was it that you didn't you formed an enol versus an enolate versus adding a second time? Right?

[00:58:16]These are good things to know and it's good things to then note, right? Like we just did. Well, okay, I wanted my instinct was to take this hydrogen to satisfy that positive charge. But that's not the case here. When is it the case and when is it not the case? Maybe that's something I write down and at the beginning of the test, it's the first thing I write on the sheet of paper so I don't forget that, right? Because this was something that was a red flag for me, right? Understanding what to do after I've reformed a carbonal if I still have a positive charge, right? So noting specifically, I would rather you do just these problems very intentionally and write notes and think about what's being asked and what's tricky about it and why it's being asked and see where you missed is way more important than just grinding through a bunch of problems, right? Because this I write these specifically to kind of key you towards what could be on the test.

[00:59:14]>> Could I ask one question about the last one you just did with the isopropanol?

[00:59:18]>> Yes. Um, for when you open up the ring, why is the O minus protetonated? And like would you read add into the other carbonal? Sorry, I'm not sure cuz you didn't draw it out. So, I don't know how to explain it, but on the answer key like after you open up the ring, you have the O minus at the top.

[00:59:40]Um, and then like at the bottom you have the ketone connected to the isopropanol group or Yeah. Like would you read into that ketone?

[00:59:54]>> Oh, so that's an esther.

[00:59:57]>> Oh, oops. Yeah.

[00:59:59]>> So, you're saying would I do this?

[01:00:02]No, like with the excess isopropanol >> add into the esther again.

[01:00:10]>> Yeah.

[01:00:11]>> Okay. So, let's do that. That's a good question, right? That's a question worth investigating.

[01:00:30]I'll just put our group here. I just broke a carbonal. What does it want to do?

[01:00:41]>> The tetrahedral intermediate.

[01:00:44]>> Okay. Well, do I have a good kicking group?

[01:00:46]>> Yeah.

[01:00:47]>> Do I have a good leaving group?

[01:00:48]>> Mhm.

[01:00:50]>> What's the other isopropanol? Right. And so what do I get?

[01:00:53]>> The same thing. Okay.

[01:00:55]>> Right. So if you start doing it I like I think it's great that you asked about that right if you start doing something just like the one up here if you if you this is tetrahedral intermediate you are not wrong to do this but you just get back to old material that we've already formed. Yeah I mean we could add that isopropanol in indefinitely and we would just get the same thing over and over and over again.

[01:01:19]>> Understood. Thank you.

[01:01:21]>> Yeah. Cool.

[01:01:23]All right. What is this question probing at? Well, I've got an ethl that's a nucleophile, but is on on this guy. And we know this is something that I've highlighted as a soft nucleophile, right? This something that is more likely to add 14.

[01:01:43]Well, I've got a carbonal and I do have a double bond right next to it. So that's telling me that I'm going to add I'm going to do a 14 addition, right?

[01:01:56]What would you how would this reaction change if it was a Grineyard? How would this reaction change if it was a different kind of nucleophile? Right?

[01:02:03]That's the whole point of this question is to see if you can recognize when we're going to do a 14 addition versus a one two addition. Right? So taking that note, understanding that trying to challenge yourself to think about different nucleophiles.

[01:02:17]All right. Well, what's this question looking at? Well, this is clearly some sort of nucleophile. I've got an acid and I've got an epoxide, right?

[01:02:30]So, what's important here? What am I thinking about? Well, again, this goes back to one of those things at the very beginning. Um, in acidic conditions, we're going to add we're going to add into the more substituted side. So, you know, you did this problem, you checked it against the um against the answer key, you either got it right or wrong. How could I change it right? Well, maybe I have a base instead of an acid. You know, there's all sorts of things, right?

[01:03:13]Well, not really. There's two things, right? I could either ask this in acidic conditions or basic conditions if we're doing an epoxide. Be able to do both.

[01:03:20]Know know the difference between both.

[01:03:22]Know when to do one versus when to do the other. Be able to recognize what an acid is versus a base of the starting material.

[01:03:28]>> I have a question about the stereochemistry.

[01:03:31]>> So when you're attacking the acidic side, do you push what was then back now that's now forward?

[01:03:41]>> Yes. So anytime you add into This is kind of that's a great question.

[01:03:47]I'm glad you brought that up. So, let's say this is a leaving group.

[01:03:52]Anytime a nucleophile adds into a leaving group that already has stereochemistry, it flips it does a backside attack and inverts the stereochemistry, right? The old orgo one idea, right? So, anytime something already has stereochemistry, the only place this nucleifile can approach is from the backside. And so you get inverted stereochemistry here. Since the backside of this, right, what you'd be doing is kicking this off.

[01:04:22]The backside is a wedge. It does push It does push the So when it opens up, it's going to push that ethyl group forward.

[01:04:38]The nucleophile is going to be back now, right? It it it backside attacks everything. That's a really good question.

[01:04:47]The important thing is that it's going to be opposite of what it just kicked off.

[01:04:56]But yeah, good question.

[01:05:02]Um here's here's another enolate question or here's an enolate with an epoxide. Right? So, LDA I'm going to form this enolate and this is it basic or is it acidic?

[01:05:30]Well, it's tempting to say, well, there's an acidic wash here, but that's a whole different step, right? that acidic wash has doesn't exist yet. LDA is a base.

[01:05:41]We're we're in basic conditions. We have a negatively charged nucleophile. So, this is going to add into the less substituted side.

[01:05:52]Right? So, here's just another question probing at can you tell me the difference between basic conditions and acidic conditions? Can you tell me the difference? LDA is bulky.

[01:06:03]What if I use sodium hydroxide? How would that have changed this question?

[01:06:08]Um, you know, what what if I use an acid instead of a base? You know, all of these kind of questions like how how can I change these things to what am I probing at? Right? I'm telling you right now, I'm probing at you need to know that this is bulky. So, it's going to take a hydrogen from this side.

[01:06:27]You need to know that that makes this basic conditions and that means it's going to add into less substitute side of the epoxide which then gets proteinated.

[01:06:37]Cool.

[01:06:40]Um there's a lot of things here. I'm going to kind of hit the highlight points. I'm not going to spend all day on this because we've already been over an hour.

[01:06:53]Um, and I already have a video talking about every every one of these reactions. So, the important part here is understanding directing groups, right? These two things are meta to each other. So, how do I get things meta to each other?

[01:07:07]Well, I need to put on I know that I need to have the metadirectionirector on first, right? Since this is a metadirectionirector and this is an orthoperadirector, it would be helpful to have the metadirectionirector on first. So when I add the bromine, it adds that in meta.

[01:07:25]This is going back to test three stuff that you guys pretty did pretty good on this. Uh right when this adds in, it is the only leaving group around. So therefore, we don't do that. We instead form the more substituted enolate because this is not a bulky base, right? And then this does the selfd doll, right? This is test three stuff. And since it only forms about 1% of this, it can still add into another version of itself, etc. Right?

[01:08:00]This is the same thing, right? This thing matches this thing. So, we do the clays and condensation again. We're going to form the enolate or the enolate here.

[01:08:11]So recognizing that in these situations from the enolate versus adding in um this the only nucleophile I have is here. The only electrophile I have is here.

[01:08:23]So nucleophile we deproenate nucleophile and attacks here. Right? I'm skipping a step. And what's important is that we can push the electrons into that nitrogen. Right? Which helps stabilize it. Right? We need whenever we're doing an SN like this, we want to push it into into an electron withdrawing group and then we push it right back out. We get the substitution.

[01:08:59]So, this is the one place that it probably would change from the video I post. And one thing that I want you guys to note and and do differently if I ask you this mechanism, I want you to show this intermediate. I want you to show this intermediate. I want you to prove that we've pushed electrons into an electron withdrawing group before we kick it back out, which is something that we've showed in class and I've done several times. I don't want you guys to go robot mode off of the video I posted because that is the one update is that I I want you guys to be able to show this complex here in the middle.

[01:09:45]Um, this is a caroxyic acid question, right? So, I've got a caroxilic acid.

[01:09:50]This, even if you don't recognize what it is, there's a big tell. What is K?

[01:09:59]What is K?

[01:10:04]Any takers? Have I lost you?

[01:10:06]>> Metal.

[01:10:07]>> Potassium.

[01:10:08]>> It's a metal. It's potassium. Whenever there's a metal connected to something, I'm thinking this electron donor, this is probably a base.

[01:10:19]Um, which tells me that caroxyic acid with base is going to be a nucleophile. So that's good to know. We deproenate the caroxilic acid. It acts as a nucleophile. Oh, someone asked about the previous question.

[01:10:49]For the previous question, don't you need a base to form the enolate before addition?

[01:10:56]I don't know what you're talking about.

[01:11:08]I also have a question about the one. Um, it's like the page above it and it has NAOT and ethanol.

[01:11:20]>> Yes.

[01:11:21]for this one. I know that you form the self ald doll, but does ethanol like play a part in any of this? Like what is that there for?

[01:11:34]>> It's a solvent.

[01:11:36]>> So you don't have to like add ethanol into the selfd doll?

[01:11:40]>> No, just like water didn't do anything here or uh water doesn't do anything here.

[01:11:51]Uh, methanol doesn't do any. Actually, methanol does do stuff. It's just a solvent. THF, right? THF doesn't do anything. It's just a solvent.

[01:12:04]>> Okay. That just like confused me for cuz I thought that you would have to like add it in since that's also like a compound that could be added in if that makes sense.

[01:12:14]>> Okay. So, let's let's play that game, right?

[01:12:21]Ethanol looks like this.

[01:12:27]And if we add that in is well, okay.

[01:12:31]>> No, I mean it's like the second step like after you did the selfd doll adding it into that product that the selfd doll makes.

[01:12:41]>> All right, let's do that.

[01:12:53]So this is the product you make.

[01:12:58]>> Yeah.

[01:12:59]>> Like after you do the teddy or media, all that stuff. You're saying why does this not add in here?

[01:13:07]>> Yeah.

[01:13:08]>> Or ethanol or whatever. Because I it's the same reason we don't do that more often is that it's It's going to be the only leaving group here and we'll just kick it right back off.

[01:13:25]>> Okay. Sorry, I keep asking the same question.

[01:13:28]>> No, it's it's a good question. It's hard to wrap your head around and if you do it like any form of tetrahedral immediate only leaving group, then it's not really worth doing.

[01:13:37]It doesn't like in the equilibrium it it much prefers this. So, we just stop here.

[01:13:43]>> Okay. Thanks. Yeah, the only time that's different is with the acetal formation.

[01:13:50]>> Got it.

[01:13:52]>> That's a good question. All right. This one's a fairly straightforward Fredal crafts al isolation which we've shown in class before. Um, this one is a twist on something you've seen before. So um we've done addition reactions where this adds in and you form a positively charged thing that can do a hydride shift, right? It forms a carboaton. So this is a very similar reaction to if I'd given you like right it's just a different way to get to a carboation or it's a different way to form a carboation and so um we get the same thing but these these are essentially synonymous questions. So are there other places you've seen carboations? How where else can we form a carboation?

[01:14:53]A benzene will add into a carboation as a as a poor elect a poor nucleophile.

[01:14:57]How do I like how can I think of different versions of doing this? Um, that's kind of how I want you guys to approach this. I want you to really try to get to why I'm asking this question, right? Well, I'm asking this question to see if you remember this basic orgo one question of what happens with a double bond in an acid? Well, I form a carboation and then can you do the hydride shift? Right? That's how you'd get perfect scores by not missing that stuff.

[01:15:26]Now, we're kind of reaching time. Um there's some synthesis stuff to talk about, but really I think we we've spent some time on it. Um all we're adding really is two things.

[01:15:41]um it's directing group of benzene.

[01:15:47]So adding a directing group first will direct where the next thing goes.

[01:15:57]Um and then being able to substitute um are being able to reduce a nitro group to an amine which we can then use to do reductive aminations which I think everybody forgets about.

[01:16:30]So there's a lot of practice doing that.

[01:16:32]Everything else is like gineards and enolates and all of that stuff thrown into synthesis. So um it's all here. Um there's all practice here. I would suggest going through it.

[01:16:45]Um but it's the same stuff we've been building on this whole time. Um but now we're just you know doing a freed crafts to put on a carbonal before we do a grineard right um yeah if you're protecting you can just do PG unless you're protecting a ketone or an aldahhide um but yes so protecting groups can be helpful these things um it it's a little bit more complicated but um if you if you follow your fundamentals all we've really added is adding something to a benzene before we start doing reactions to it instead of me just giving you the benzene. So hopefully it's not that big of a step, but I'd be happy to talk about any questions you have about synthesis for the next few minutes before we hop off.

[01:17:43]Okay. Well, I do talk about the retroynthesis and stuff in the other video that I'm going to post. Um, so if you want to hear me talk about that more, um, I can there, but let me just real quickly, someone said, can you explain what we could what we should break first? Right. So, um, I see I see a very clear Grineard here, right? Like immediately I see a cut one carbon away from an alcohol. I always want to kind of listen to the cuts that are closest to the hetereroatoms, right? That's that's usually a good hint.

[01:18:33]So that that kind of simplifies my stuff. Um, and Then I see that this is orthopera.

[01:18:47]Um, and this is a meta director. So it' probably be best to have that meta director on first. There's actually probably a million different ways you could do this, but I can imagine a situation where um I have this on before I do the nitro uh group. But what's important is that they're meta to each other and that I would have a meta director on before I added the amine.

[01:19:18]So, uh, yeah, it's really it's this is the same that we've been doing all semester. Um, if you're having trouble with this part, then you need to go back to do some of those old recitations where you're just grinding through some um, retroynthesis, right? Uh but then this new stuff you just have to think about well these two things are meta to each other so I need to have a metadirectionirector.

[01:19:41]Um yeah so here this is the electroile so benzene would add into this electrophile.

[01:19:49]Someone asks about that.

[01:19:53]Um, so um, a couple of questions. If we focus on the content in the final practice, will we be good for the test or should we review other content as well that's not mentioned in the list? Um, I mean, if you can do the whole final practice, it's it's a pretty good, you know, I don't know what else Besides this, well, even this, I mean this, we've covered almost all of this in the final in that final practice below, right? We had a caroxyic acid question, a 14 addition, one two addition, um an EAS, an SN, uh epoxides, uh using bulky bases, using lis acids, using regular acids.

[01:20:50]I mean, this is organic, too, you know, and I don't Um, I don't I don't know if there's a lot that I could surprise you with that wouldn't be covered in in a lot of this list if you're just if we boil it down to the fundamentals. Yeah. Recognize the different ways I could ask these things, right? Okay. Well, I didn't have an acidic caroxyic acid question in the practice final, but it's fair game, you know, right? Because I gave you a basic caroxyic acid final or question. Do you know the difference between the two? Can you tell me that when one acts one way or one acts the other way? So, I'll say the the key, and I've said this over and over now, the key to success isn't necessarily being able to do the practice final. It's being able to understand what the practice questions are asking and understand if I can flip that same concept on its head. Can I do the other version? Do I understand what's happening enough that I could write a question of my own or explain why this question works to somebody else? explain the core concepts that make this question this question. Um, that's the important bit. Not just doing the practice final, but understanding why I'm asking the question.

[01:22:03]Um, someone said, "Is the length of the test similar to the length of the practice test for time management planning?"

[01:22:10]Uh, the length of the of the test is, to be honest, is not that much longer than a regular test. It's it's a regular test with an NMR question or two at the end and maybe one or two more like um maybe one or two more questions to do with oh there's one thing I have not mentioned explicitly um this is actually a fairly common kind of question um in in regular in regular or in organic classes. I just haven't given it to you yet because I want you guys to be really good at mechanisms before I challenge you with this. But there will be a question at the end of the exam that um that is a mechanism that you haven't necessarily seen before but I give you the final product.

[01:23:08]So, I give you starting material, final product, and I say, "How do I get there using a mechanism?"

[01:23:19]Um, and so that's just a challenge for you guys.

[01:23:23]We're pretty generous with the grading on that. Uh, but it's it's can you do a mechanism you haven't seen before? If I give you the product, can you give me reasonable arrows that could get me from point A to point B?

[01:23:38]Um and then outside of that all like almost all the conceptual stuff is about electron donating, electron withdrawing groups, how they interplay with electrophilic aromatic substitution versus SN um aromaticity, you know, those are the kinds of things that will be conceptual.

[01:23:59]All this kind of stuff is where I drew the conceptual stuff from directly.

[01:24:11]Um, someone asked, could I make a list of the leaving groups?

[01:24:15]I did. This is the leaving groups. SN is leaving groups on the aromatic ring.

[01:24:20]Benzene and SN2 is the leaving groups for SN2.

[01:24:24]Um, yeah. So, the test isn't isn't that much longer than regular test.

[01:24:38]Um, it's about as long as finals I've given before, and to be honest, most people are done within an hour and then sit there and stare at it for another 30 minutes to 45 minutes.

[01:24:49]>> Um, Dr. Little, if you have time, can I ask one question about the last H&MR that you put on practice >> down here?

[01:25:01]>> Yeah, the very last one.

[01:25:04]Yeah.

[01:25:05]>> Um, the benzene ring on that is like built kind of weird. How would you know where like the car ketone carbonal group goes?

[01:25:17]>> Um, yeah, there's a couple of different correct answers here based off just this NMR alone, right?

[01:25:25]So, we know we have a tri substituted benzene.

[01:25:29]We know that we have two doulets and one singlet. So the only way we could have a tri substituted benzene with two doulets and a singlet would to look like that.

[01:25:41]Um there's no other way I could I mean this is similar like you got to have one here that's all by itself and two that are next to each other. So anyway you orient that. Um I have one oxygen.

[01:26:03]I haven't done the degrees of unsaturation. Um I haven't looked at this problem in a year. Uh I believe we have a carbonal, right?

[01:26:16]>> Yeah. in the um practice or sorry in the like answer key there's it's the degree of unsaturation is five and you have a carbonal >> probably there >> um I think you put it on like the lone one for the carbonal >> either one of those >> doesn't matter >> yeah the the only thing I'll say is the only thing that I would really take points off for here is recognizing that this is the most down field.

[01:26:48]Uh, and so you want it next to the singlet. You want the carbonial next to the singlet anyway you could get it. But which one of these it would be hard to tell. I'll give credit for both.

[01:27:01]>> Okay.

[01:27:02]>> Yeah.

[01:27:02]>> Thank you.

[01:27:03]>> That's a good question.

[01:27:07]All right. Do you guys have any other questions before I um uh before I head off? I I'll I'll I'll I'll link both videos on my canvas, but I also have a YouTube channel that has all these publicly available. I think a lot of people have been probably watching them this weekend and stuff.

[01:27:25]Um, which is why I wanted to do something slightly different. I didn't want to remake the same video. So, I want to kind of give a different focus uh to this. So, uh I hope this has been helpful. Uh if you guys have more questions, well, I I probably have to head off here soon. So, uh, thank you guys for coming. Uh, let me know if this this was good for you guys, if this was helpful. I'll post this in case you guys want to see this chart in its entirety. If if let feel free to email me, let me know if you find this helpful or not. Um, if it's not helpful, I' I'd rather know so you guys I don't waste time doing it for future classes. Um but yeah, I will catch you guys uh tomorrow and good luck with everything.

[01:28:15]Thank you.