Organic 2 Test 3 Practice Q&A

[00:00:00]Welcome to uh test three practice uh for the summer of 2026. Now that we're here, um we have already essentially done the same test three practice. I liked how it worked from last semester, so I carried it over this semester, which means that if you want a more complete rundown of every topic that's covered in this practice test, I would refer you to my previous video. So, if you go to my you're on my YouTube channel, if you go to my uh profile page, you can scroll down and find the last organic two test three practice where I kind of go into this in more detail. I'm not going to go over exactly those things. Instead, I think what we're going to do here, um, I would I would direct you there if that's what you want. What we're going to do here is instead try to bridge some gaps, answer some questions, talk about this stuff in more detail where it's appropriate. that way we have uh complimentary kind of videos. So I have a few people who've hopped on uh taking some time out of their morning to ask me some questions and uh I appreciate that.

[00:01:06]So we're going to go through those questions. We're going to post this video and hopefully it will be a good uh partner to go along with the last video I posted. Um all right. So uh any questions? What are some things you guys are thinking about? Some concerns you have? Some things you want cleared up?

[00:01:25]Could you go over the percentages part?

[00:01:28]I think I got confused with that. Like I get when we use LDA, it's more of like the 100 to 0%.

[00:01:36]>> Yeah. So that's a that's a really great question. Um that's kind of one of the hearts of what we're doing here.

[00:01:43]So LDA, so if we're comparing bases, we have LDA, which looks like this. It's lithium disopylamine.

[00:01:54]So lithium has a positive charge because it's a metal and then it's an amine which is the A and di isopropyl is what the L that's what the D stands for lithium disopyl lithium diisopropyl amid and so if we look at LDA and we want to define it if we want to characterize this base which is a very common base in enolate chemistry These two big old isopropyl groups make it bulky and a negative charged nitrogen makes it very very very strong. This is an exceptionally strong base.

[00:02:40]Um so these two cate notice that these two are not uh these two things sound similar bulky and strong right we tend to think of bigger things as being stronger. They're they're not necessarily the same thing in in organic chemistry. You can be bulky and weak. You can be non-bulky and strong. Um but you can see LDA specifically has these two different characteristics. It is both bulky and it is strong. If we compare this to like an oxygen base, so we're we're talking about sodium hydroxide, sodium ethoxide, etc. We're talking about any sort of nao h n ao e t even ko k o e t heck it could be l i right like anything that where once you break out the ionic bond you can see that it's kind of this motif which is why I tell you guys anytime you see a metal strip that metal off draw the ionic because it kind of shows the heart of the reactivity of this thing.

[00:03:54]This this is generally less bulky. So sodium hydroxide is even more so.

[00:04:05]And it's it's relatively weaker. So the problem with calling something strong and weak is it's all relative, right? An O minus is more stable than an N minus.

[00:04:17]And normally sodium hydroxide is strong enough for PE for most people's purposes, right? If you're doing biochemistry or you're doing chemistry in water, which is what most chemistry is doing outside of organic chemistry, sodium hydroxide is a strong enough base to do anything you want, which is why it's defined as a strong base in general chemistry. But in organic chemistry, we can have much stronger bases because we're not reacting in water. Um, and we can talk about that more if you want, but uh, this is weak relative to LDA and it's also weak relative to our goal for this unit, which is essentially to take an alpha proton off of a carbonal. So that means the carbon one away from the carbonial because there's this resonance because you can push the electrons up into the carbonial.

[00:05:14]A base can take a proton off of a carbon even though it's really hard to take a proton off of a carbon. Carbons are not acidic.

[00:05:25]They they don't want a negative charge for the same reason a nitrogen wants a negative charge less than an oxygen.

[00:05:31]Right? Think about carbon. carbon's one further to the left compared to nitrogen, it really doesn't want to lose this proton.

[00:05:40]Um, but since there's resonance with this carbonal, we can get this to happen. And this becomes now a usable nucleophile.

[00:05:49]So I have this I have my enolate and I use a base to get there.

[00:05:59]Now these are the two most common styles of base to form enolates. um LDA since it's bulky always forms the enolate and it's strong right having this since this is such a strong base there is no problem for LDA to take this alpha proton and form the enolate it happens and when I have these two things together. We'd much rather have this negative charge on the enolate than the negative charge on the nitrogen from the LDA.

[00:06:41]This the the conversion to this is 100%.

[00:06:46]There is no starting material left over because this gets pushed all the way to this point because this negatively charged nitrogen is such a strong base that there's no question about it. If I put one equivalent of this base in here and one equivalent of my alpha of my carbonial, I will get full conversion to the enolate.

[00:07:07]Sodium hydroxide is in this category.

[00:07:11]And the thing is this isn't a complete picture, right?

[00:07:15]I'm also comparing my base to my protetonated base when I'm looking at the equivalent here when I'm looking at the equilibrium here, right? Which side is favored? Well, really depends on whether this negative charge is more stable than this negative charge. And it just so happens that this being a very unstable negative charge, we push everything this direction. This being a more stable negative charge means that there is more equilibrium here. And when you look when you look at the reality of it, when you look at this through when you do the do the chemistry roughly, you're you're not going to form a ton of this enolate in relation to the total amount of starting material you had because this O is actually slightly more stable than this enolate because you had to take a proton from a carbon even though there's resonance. So there's these conflicting factors. This is a carbon. It doesn't want to have its proton taken away from it but there is some resonance to make it somewhat acidic.

[00:08:35]Um but what we see is that there's actually a very small about roughly 1% of this enolate is formed in relation to the starting material which we still have a lot of left over.

[00:08:55]So what's happening is because the sodium hydroxide is a weaker base, it is unable to convert all of the starting material in one go to this enolate.

[00:09:07]That's compared to LDA which is much stronger and therefore is able to convert all of your carbonal into enolate.

[00:09:16]So the the difference comes with the relative strength of the bases how they and yes O you know sodium hydroxide is strong base but in relation to the acidity basicity of this enolate is what we have to compare it to.

[00:09:35]Um LDA is because there's negative charge on this nitrogen it's a much stronger base and is able to push that equilibrium to 100%. And this has ramifications for the reaction itself.

[00:09:46]So I didn't ask you to do this reaction fullon. This first step of the thing is just forming the enolate. But because there's so much of this sitting around, this can still react with some of that starting material. But when I use LDA, there is none of this starting material sitting around. I it will not be able to react with another version of itself because there's just there's none of this left, right? So that's why I have you guys write these percent conversions because it helps us keep track of what's still in the reaction mixture. In this reaction, there's still a lot of this sitting around. In this reaction, there is none of this sitting around and that has to do with the relative strengths of those bases.

[00:10:32]All right, I just did a lot of talking.

[00:10:34]Did that answer your question? Do you have follow-up questions on that?

[00:10:37]>> No, that makes sense.

[00:10:39]>> Okay.

[00:10:40]>> Uh, Dr. Little, I have a question.

[00:10:42]>> Yeah.

[00:10:43]>> If we have a case where we're adding one of those like relatively weak bases to a carbonal instead of forming an enolate, does that still apply where it doesn't convert all the starting material?

[00:10:56]>> No. Um, and I mean that's a really good question. um at that point carbonial work a little bit different.

[00:11:06]So when you add into the carbonal let's do something like this.

[00:11:13]When we add in that's a that's a really good question. When we add into a carbonal we're always comparing the forward steps to the backward steps in reality. Like that's what's happening.

[00:11:36]But the two paths here that we're comparing against are this collapsing and kicking off this OM which would get us right back to the starting material or that collapsing and kicking off this OE which gets us to our transestification. product.

[00:12:09]So here this would actually be the most accurate way to draw this is this. And if we evaluate each step for its own sake, we see that what we're comparing here is a little bit different.

[00:12:39]We're comparing this product versus this product. We're comparing the stability of this esther to this esther. Realistically, the best way to look at that is we're looking at this tetrahedral intermediate. Is this leaving group significantly better than this leaving group?

[00:13:00]No, they're about the same, right? And so, actually, if you go back to test two, this question was on test two.

[00:13:11]There's there's really nothing that would say that this product is any stronger than this product or any any more stable. There's no reason to think that this is preferred over this. So if we wanted to make this, how would we do it?

[00:13:27]Well, Lhatier's principle. We'd add a lot more of this, right? So realistically, I mean, if I were to do the percentages out, which you don't have to do for this test, I would say there's, if I just did this, it would be about 50/50 starting material final product. That's where the equival uh the equilibrium would lie. So we have to do something to break that product and push this forward. And that's adding more of this.

[00:14:03]So yes, this equilibrium still exists in this, but I'm I don't I'm not going to ask you to do it and write me all the percentages because it doesn't really change kind of how the reaction works.

[00:14:15]Whereas this one, it matters because this ends up reacting with itself again in a way that um is a little bit different. But yes, there's always equilibrium in all of these reactions.

[00:14:29]I just really care that we write this one out because it helps us see why something could do a selfd doll could react with itself versus not. It's it's more instructive to the reactivity.

[00:14:41]But but yes, equilibrium exists here as well.

[00:14:48]Thank you.

[00:14:50]>> Yeah, I mean if if Yeah, that's this is a hard concept to wrap your head around.

[00:14:54]So, we can spend a little bit more time on it. If you guys have have more pointed questions, I could talk about it. Um it to be honest, it's one of those things where it's like, man, you just have to sit there thinking about equilibrium for a long time to really start to understand what's what I'm actually saying because it's It's a little tricky.

[00:15:25]>> Do you guys have any more questions?

[00:15:28]>> I have a question. Um, this will come up like as we get further down, but I guess like here obviously like I know you told us to just go for the enolate and like later we'll like go all the way through, but I know like sometimes we reduce the enolate with the um like H30 and like so there's no double bond like when when will we have to do that and like do we need to do that every time? Like if we get to the end after we've like gone through a reaction and like either done the self alol or like added to something else. If there is an enolate left, should we always reduce it?

[00:16:02]Yes. Okay. Um. Yes. So that's a great question.

[00:16:08][clears throat] When you form an enolate by nature, you see this enolate being formed. That means in the step before you broke a carbonal.

[00:16:20]Enolates equal broken carbonal.

[00:16:30]What do we know about broken carbonial?

[00:16:36]What do carbonial want to do when they're broken?

[00:16:38]>> They want to reform.

[00:16:46]broken carbonial want to reform.

[00:16:52]So the way that enolates can reform is by collapsing and attacking something.

[00:16:59]So the thing is for a lot of these like for this reaction I would have another electrophile sitting around for it to collapse and attack.

[00:17:17]That's there.

[00:17:20]But there are sometimes points in time when you get to the end of a reaction and there's no electrophile sitting around, right? Specifically, and we could just look at this.

[00:17:34]Well, that's going to be kind of an annoying one to drop. But if we get to the end of the reaction and we have an enolate, like let's say we did a 14 addition with some nucleophile.

[00:18:04]I have an enolate. I broke a carbonal.

[00:18:06]It wants to reform, but I don't have any more electrophiles sitting around. So, the only way it can reform is if I give it some acidic proton.

[00:18:29]So, yes, when you have an enolate, enolates aren't very stable. They don't want to sit around. and they don't want to remain in that state. You're you're never going to you're never going to do this.

[00:18:47]Just proteinate this alcohol, this oxygen.

[00:18:55]You're never going to form an enol because almost always this enol will will immediately turn into the ke the keto or the ald doll or the aldahhide like the keto version is significantly more stable. This is where it's going to go.

[00:19:25]So you would never form this enol because you would much rather form this keto. So what we see instead is that which gets us to the keto version.

[00:19:40]So if you find yourself with an enolate or an enol, almost certainly the motif you're doing that next step is collapsing and attacking whatever you have, whether that's an acid or an electrophile available.

[00:19:55]Does that help? As as a heristic, you see an enolate, always collapse it down to the carbonal and find something to attack.

[00:20:09]Cool.

[00:20:19]All right. Do you guys have other specific questions about this that you want to talk about? um or there things that don't don't feel that clean.

[00:20:28]>> I have a question about one of it was another enolate formation one.

[00:20:32]>> Um >> it was the third one. I when I originally did it, I pushed those electrons to like when I took off the H in the middle, I pushed the electrons to the right side. Um, is it would it go to the left side because like there's more like like I don't know why it would go to one side versus the other?

[00:20:55]>> That's a good question.

[00:20:58]>> I'm glad that you saw you take the middle one. It's significantly more acidic even though it's less substituted.

[00:21:06]This proton is more acidic than this one even though it's less substituted than this one. And it's because it has it can interact with both of these carbonial, >> right? Yeah. But let's let's do something.

[00:21:22]What have I drawn?

[00:21:27]>> Resonance.

[00:21:29]>> Yeah.

[00:21:32]So, which one of these resonant structures is right?

[00:21:40]>> Maybe they're both right. Is the animal a mule or is the animal a sorry is animal a donkey or is animal a horse?

[00:21:50]>> I remember your analogy but I don't remember the answer.

[00:21:55]>> Is a mermaid a man or is it a fish?

[00:21:58]>> It's both.

[00:21:59]>> Yeah. So which one is it? It's both you know. So >> um one you know this one might be slightly more stable than this one. you know it what we say is resonance contributor. So like you might have something where it's like you know 65% of the real structures explained by this one versus 35% explained by this one or something like that. But I don't really care.

[00:22:27]It's the reality is it's neither of these pictures. So drawing one versus the other is not more or less correct.

[00:22:33]If you draw the correct resonance structure of whatever thing you're drawing or draw a reasonable resonance structure of whatever you're drawing, then um that's fine.

[00:22:46]>> Okay, thank you. That makes sense.

[00:22:47]>> Yeah, that's a great question. Resonant structures are the same molecule.

[00:22:56]or question two versus 14. Why do I attack the um I don't know if you already answered this, but how come we attack the bond inside the ring instead of attacking the carbonal?

[00:23:16]>> Wait, what? For which question? This one? The one that we're just going over?

[00:23:20]>> No, no, no, no. The it's it's in the section 12 versus 14 edition.

[00:23:27]Like when I solved it, I did me and attacked the carbonal, but they the answer was wrong. And then I did ME verse and then attacked the one in the ring and then that was wrong again. Like I don't understand.

[00:23:48]>> All right. So I put these two questions next to each other for a reason.

[00:23:52]Um, it's showing something important.

[00:23:55]It's showing that when you've got a carbonal, so far this semester, I've been nice and I've only given you guys carbonial as the electrophile. So, things add into this carbonial, but because of resonance, when there's a double bond next to a carbonal, there's another electrophilic site right here.

[00:24:19]So there are some things that are going to add into the carbonal like here. There are some things that are going to add into the end of this double bond attached to a carbon. This is called a one two addition because if we number these things out, if we start at the oxygen, 1 2 3 4 1 2 3 4.

[00:24:49]A one two addition means that I'm attacking the two carbon and pushing into one. A 1 14 addition means I'm attacking the four carbon and pushing into the one.

[00:24:59]So what things do the one two addition? Well, a very small group of things will still add into this carbonal. For the most part, most things add into this this double bond attached to the carbonal. So, for this, we've got Grineyards, RMGBr, and we have lithates, which these are two harder nucleophiles. That's the kind of term that they have to describe it. These are hard nucleophiles.

[00:25:32]An Lah would be a harder nucleophile.

[00:25:35]Um, this is pretty much everything else.

[00:25:43]We'll just add into that four position.

[00:25:46]So, when I look at this, when I see a electrophile, I see a carbonal with a double bond next to it, I'm immediately thinking, I'm going to assume one for addition.

[00:25:59]I'm going to assume addition into that double bond unless there's one of these three exceptions.

[00:26:14]So, this is actually the default. I'm thinking I want to attack there. All things given equal. But if I get one of these exceptions, if I get a Grineyard, a carbon chain with a lithium on it or LH, then I'm adding into the one two position instead.

[00:26:34]>> Okay. Yeah, that makes a lot more sense, >> right? So, we have to now start looking right as this semester goes on. And one of the things that we're doing and one of the reasons why I kind of organize the class the way I do it is that we're increasingly spreading our focus further away from like when we start it's one atom attack one atom.

[00:26:59]Then it's like an epoxide where one atom attack something and now it's two atoms away. Now we're thinking about enolates where something you know like and now we're doing 14 additions where we're we're attacking something even further away. So we're broadening our scope. You see a carbonal. Now we're also looking to see if there's a double bond next to it. So it's just a a a step more thinking about what's all on the molecule that's not just the carbonal itself.

[00:27:32]So if there's no carbon or if there's no double bond next to the carbonial, do the soft nucleophiles still add into the carbonial then?

[00:27:39]>> Yeah. And that's just like everything else we've been doing like here.

[00:27:47]Like this would be a quote unquote softer nucleophile. Like if I had a like if I had that same nucleophile and my molecule looked like this, it would add here instead.

[00:28:02]But since there is no double bond here, it adds into the carbonal.

[00:28:10]So we're always looking for that 14 addition.

[00:28:19]>> Does that make sense?

[00:28:20]>> It should it should be it's now the default. Look for the 14 edition except for those few. And and the problem is that the exceptions are big reactions.

[00:28:29]Grineard, that's a huge reaction. LH is a huge reaction. So important things still add one, two, but almost anything else adds 14.

[00:28:42]Chim, what were you going to say?

[00:28:44]>> I also had another question. There's that issue I keep running into of knowing when I'm doing substitution or when I'm doing an enolate and and section addition into carbonial versus enolate. like how do I know when I'm doing one or the other?

[00:29:02]>> Yeah. So, this is this is so someone asked this other question in the chat.

[00:29:07]Um these key concepts this this stuff like I might mine like conceptual questions at the beginning off of this, but really these are things that I want you to be focusing on when you do the mechanisms and the real questions. like the test is going to look more like how you've done tests before where I'm going to just give you a bunch of reactions like this.

[00:29:32]>> Uh but what I want you to be thinking through where I want your focus to be is on these these concepts. And so I'm kind of hopefully priming you to be thinking about these things as you go into the practice.

[00:29:44]So what are we looking for here?

[00:29:48]>> Uh when I answered that when I did the enolate.

[00:29:52]So this is the enolate. How do we know?

[00:29:55]>> I don't know.

[00:29:57]>> Okay. So the answer is when we add into a carbonal, we form a tetrahedral intermediate.

[00:30:15]>> Right? And when I look at this tetrahedral intermediate, I think if I wasn't given this context, if I just was given this tetrahedral intermediate on its own, I would think this is my kicking group. This is my only leaving group possible.

[00:30:34]Mhm.

[00:30:35]>> So, the reality is that happens and we're just going to get right back to the starting material, >> right?

[00:30:46]>> So, while this does happen, this isn't fruitful. It doesn't go anywhere. It just returns back to the starting material. So, we don't worry about drawing it.

[00:30:55]>> Okay.

[00:30:56]>> Right. So, this I don't even need to draw because it's just not going anywhere. Right.

[00:31:07]But so I have to think I still have a negative charge and I need to put it somewhere. I'm going to form the enolate in this in this situation, the more substituted one.

[00:31:24]Um, this one is similar OE.

[00:31:31]I think about adding it into the carbonial first.

[00:31:48]And I notice that I've got a kicking group.

[00:31:53]And the only things the only other thing that can be a leaving group on this tetrahedral intermediate is equivalent to what I just added. So even if I collapsed and kicked and I kicked this off, the only thing I have, what ends up happening just like before is I just reform the starting material.

[00:32:16]So, a good rule of thumb is if you start going through it and start evaluating these things, let's let's throw away the rule that the nucleophile can't be the leaving group.

[00:32:28]>> Mhm.

[00:32:30]>> Well, actually, we still need that for some things, but >> but I got the same answer you did at the end, but on the practice test, it said like the answer key said it's wrong.

[00:32:40]>> This isn't the right answer.

[00:32:41]>> Oh. Oh, okay. Why?

[00:32:45]>> Um, not sure.

[00:32:51]>> What does this look like?

[00:32:55]>> Oh, square one.

[00:32:56]>> It's the starting material. We We did the same thing where we returned right back to the starting material. That's >> Yeah, >> it doesn't help us. So, if you find yourself returning the starting material, that means, hey, this is this is wrong. This isn't right.

[00:33:10]>> Right. If you're honestly genuinely following the mechanism and taking it for what it is and you find yourself with a starting material, again, that can't be the mechanism. I'm not going to give you a reaction where you're just going in a circle.

[00:33:25]So, I think this negative charge has to has to go somewhere else. So, I form the enolate instead. Does that make sense?

[00:33:33]>> Yeah, that that makes sense.

[00:33:36]So if you find that the only leaving group when you add into the carbonal is what you added or equivalent to what you added then uh you do the enolate.

[00:33:57]So >> okay.

[00:34:03]So if what you added is or is equivalent to the only leaving group then we do the enolate.

[00:34:27]Now something really interesting happens when we look at this. Let's look at so this is OE negative let's look at the acidic version of this same nucleophile so ethanol and let's do it instead in acidic conditions so let's have some generic HA acid I proteinate my electrophile file and then my nucleifile attacks. We've done this before.

[00:35:11]There's something interesting that happens here.

[00:35:16]So, I get to that same tetrahedral intermediate.

[00:35:23]Oh, this one doesn't really make any sense. Hold on. Let's do that. Not do that here. Let's do that up here.

[00:35:34]It's a better example. All right. We're going to do what I just said. We're going to do the acidic version instead.

[00:35:45]And we're going to get to the tetrahedral intermediate here.

[00:36:10]All right.

[00:36:12]I want to compare that directly to the tetrahedral intermediate we got when we were in basic conditions.

[00:36:25]In these conditions, we said this isn't fruitful to add into the carbonal because this is the only leaving group.

[00:36:32]So, I'm going to collapse and I'm going to kick this off. But here we see something different, which is yes, this is a leaving group.

[00:36:42]But is it the only potential leaving group?

[00:36:50]>> Yes.

[00:36:51]>> No. An O can be a leaving group.

[00:36:55]>> Oh, but I thought an Oh, I see. Never mind.

[00:37:02]>> Where minus cannot be a leaving group.

[00:37:13]So this will eventually collapse and kick off this water. This cannot do that because O minus will never be a leaving group but O can.

[00:37:36]So we see that this reaction forms an enolate in acidic conditions but forms or sorry adds in the carbonal in acidic conditions but forms an enolate in basic conditions.

[00:37:50]So, but that all comes from looking at the tetrahedral intermediate and evaluating it is the only thing is the only possible leaving group what you just added or equivalent to what you just added. If that is the case, it's going to be the and so this this almost happens exclusively in basic conditions if that also helps.

[00:38:20]You're not going to form an analyte in acidic conditions.

[00:38:25]I mean, you can, but we're not going to in this class.

[00:38:29]All right.

[00:38:32]Oh, someone asked if we need to name the type of reaction on the exam. Yes.

[00:38:38]Did I forget to do that on practice test key?

[00:38:42]>> Yeah, not a key.

[00:38:44]>> Okay, I'll fix that on the key. I thought I I thought I remember doing that like right be before I hopped on and I was like, "Oh gosh, I need to fix that." Um, so I'm going to fix that. I'm going to update the the um key to have the reaction names. Um, I'm sorry I forgot to do that. That will be uh I'll update that. I'll repost that. I'm also trying to put together a uh updated so you know we did a test two retroynthesis practice. Uh I'm trying to create an updated version for test three. Hopefully that will be helpful.

[00:39:27]Um that's just it's the same worksheet.

[00:39:30]I'm just I've just added the the few things that we've added from test two to test three for synthesis and give you more retro synthesis practice with that.

[00:39:38]Um so I'm working on that and I will hopefully post that this morning and and work on a key today uh for that. So um hopefully that gives you more uh more retroynthesis practice in case that's something that's a struggle for for you guys. Um, but okay, does that make more sense? I've also, um, for those of you guys that are hopping on late, uh, I also went through these sections in more detail in the last video I did that walked through almost this exact practice test.

[00:40:16]Um, so if you want another explanation, more explanation, uh, you can go watch that old video as well. I'm just using this today to kind of answer pointed questions. Um, so you guys have questions on this section or any other general questions that we can talk about?

[00:40:33]>> Um, I was going to say for the section that says putting it together reaction practice, do we use the same Well, actually that wouldn't I don't know how that would apply. For the very first question, I did enolate again, but it didn't go back to the same answer.

[00:40:54]The first question, this isn't a nolate question >> because when I my default thinking should be to add in to the carbonal and when I do that I notice that I have a leaving group that's not what I just added.

[00:41:19]>> Mhm.

[00:41:19]>> Like this is a potential leaving group.

[00:41:21]This is what I added. This is not a leaving group. This not a leaving group.

[00:41:26]But that means I can collapse and kick.

[00:41:34]>> So basically my first re my [clears throat] first thought is to do a substitution and if I can't >> your first thought should be add into the carbonal.

[00:41:43]>> Okay?

[00:41:43]>> And if it reforms the starting material then start over and do the enolate.

[00:41:48]>> Okay? And the more you see that, the more you practice doing that, the more you'll quickly see before you even have to add in there, hey, this situation, I don't need to add in the carbonal.

[00:41:57]>> All right.

[00:41:58]>> Right. So, if you do it the right way, if you kind of step it out and justify it, hopefully at some point you start to make the connection of like, yeah, I know that this is going to replace this and therefore that tetrahedral immediate is going to be a little bit different.

[00:42:12]I'm going to add in the carbonal.

[00:42:15]But the default should be added to the carbonial evaluate. If you are not sure, someone said, "Oh, will we have to name the type of reaction on the exam?" Yes, I'll post those answers. Sorry. Could you go over the last two questions on the new retroynthesis question section?

[00:42:38]Um, new retroynthesis.

[00:42:43]Yeah.

[00:42:44]So two carbons away, we've traditionally always used a grineard to add into an epoxide to make this bond formation here. If I need to make that cut and it's two carbons away from whatever my head hetereroatom is, in this case a methyl gyard into this epoxide would be great.

[00:43:16]Um, but we learned recently that there's another way to make the same motif and it's using an enolate with some sort of leaving group, right? So um notice it's there's a big difference between MGBR brgme this is a nucleophile this is an electrophile. Do not get those confused because this is an electrophile and this is a nucleophile.

[00:43:50]You're you're flipping reactivity.

[00:43:53]Technically this is what's called backwards reactivity. Umpalong is the word we use. It's a German word for backwards. Um, but this the enolate stuff kind of solves some problems that the epoxide has. It allows you to do things that the epoxide can't do, such as and a Grineyard will only ever be able to add into less substituted.

[00:44:19]But in this case, we can get different reactive possibilities. Um, so there are times when they both work for this motif. Um, and there are times when just this works and there are times when just this works. Um, and it has to do with really the focus is that uh a Grineer is always going to be basic. You can't do an acidic Grineard because it's a negatively charged nucleifhile. If there's an acid in here, it will destroy the Grineard before it will ever act as a nucleifhile. You can only do basic nucleophiles. It can only add here.

[00:44:54]So, um, we can only really get this motif. But if I wanted to get something like this, if I wanted to get something like this and I do my retroynthesis and I know notice something, this Grineyard cannot add into that carbon. It can't add here. So, this can't work. So, if I start doing epoxide and I notice that my Grineyard's not working, it's adding into the wrong place. Now, I have to think the Grineyard's not going to be the way that I approach this. I have to instead envision this as a enolate with a leaving group.

[00:45:59]Right. The the alternative is this fixes a problem you could run into with epoxides.

[00:46:12]So that's that's the point. That's that's one of the useful ways of using an an enolate versus an epoxide in a synthesis.

[00:46:30]Can you go over the wedig one? The wig reaction.

[00:46:42]This one.

[00:46:44]>> Yeah, that one. [clears throat] >> All right. So, I recognize this immediately as vidig formation reagents.

[00:46:59]So if we remember this P acts like a nitrogen and it's going to do a substitution onto my leaving group. So after a couple of steps this is a base.

[00:47:10]After a couple of steps, I'm seeing that what this does is it forms a vidig on a leaving group, right? That's that's what those conditions do.

[00:47:33]But we're not done because what what do vidigs do? Well, vidigs look for carbonial to make a carbonarbon double bond. Right? So, there's a carbonal nearby. There's this vidig here. That's a nucleophilic carbon.

[00:47:50]There's an electrophile in my picture.

[00:47:52]If there's a nucleophile and electroile in my picture, they're going to interact and we're going to form a 1 2 3 4 5 six membered ring where there's a double bond between one and six.

[00:48:12]Okay, that makes sense. So, um, vidig only form from leaving groups.

[00:48:17]>> Yes.

[00:48:18]>> And they only add into ketones or aldahhides?

[00:48:22]>> Yes.

[00:48:23]>> Okay. Thanks. And can you also go over the first reaction on the last or I'm sorry, the last page of reactions right before synthesis practice. The first one on the top of that page. This Um, yes.

[00:48:40]>> All right. This is a strong bulky base.

[00:48:44]It's going to form the less substituted enolate because it's bulky.

[00:48:54]Dang.

[00:48:56]It's going to form 100% of it because LD is a very strong base. Um, so I don't have any starting material left to react with.

[00:49:06]So, so I'm looking for something for this enolate to react with.

[00:49:10]Me e that means there's a methyl uh and that's connected to a carbon with one hydrogen and one oxygen. Oxygen wants two bonds. Carbon wants four bonds. So, it makes sense that they'd have two bonds together. So this is just C.

[00:49:34]CHO is a common shorthand for just an aldahhide which makes sense when you draw it all out. A carbon connected one oxygen one hydrogen is going to be an aldahhide.

[00:49:46]Um right it's missing. If it's C H2O, well that's an alcohol.

[00:49:57]Right? Because now this carbon is connected to four things, right? But in this case, this carbon was only connected to three. Um, so enolate collapse attack the aldahhide.

[00:50:22]Then I am going to proteinate it with some kind of water something.

[00:50:37]And then I notice that heat is added to this reaction which tells me that I need to do an elimination. We always do the ald doll condensation from the hydrogen between the two oxygens because it needs that resonance with this carbonal in order to be acidic enough to take. Um, so I'll just use the water the hydroxide from the water I just took.

[00:51:03]We can do that.

[00:51:09]Oh, what am I doing? and it forms a double bond here. So, this is an ald doll condensation. So, that heat tells me we're going to do an elimination at the end.

[00:51:20]>> Okay, thanks. That's what I got. I just want to make sure that's how you did it.

[00:51:29]>> So, like the last exam, we're going to have to do a reaction like this, and then we're going to have to tell you what type of reaction it is.

[00:51:38]>> Yes. Yeah. I I forgot to put that in the um in the key, but I'm going to update that and repost it.

[00:51:46]>> All right. Thank you.

[00:51:47]>> Yeah.

[00:51:48]>> Oh, sorry. Do we have to specifically name like SN1, SN2, E1, E2?

[00:51:56]>> No, that if I ask a question about that, that would be in the conceptual part of the test and I would probably tell you what it is.

[00:52:04]>> All right. Thank you, sir.

[00:52:05]>> I'd probably say, "Hey, you're running an E2 you're going to eliminate one of these hydrogens's which one and why or something like that you know or you're running an SN2 and you get two products what are the mechanisms products like how'd you get those right >> thank you >> excuse me Dr. little.

[00:52:29]>> Mhm.

[00:52:30]>> Would you please be able to go over the first synthesis question?

[00:52:35]>> Yeah. Um, all right. So, one, two, three, four, five. I only have one benzene ring, so it's got to be a cut there. Um, and then three. Uh, I'm thinking this is 5 43.

[00:53:03]5 43. So, there's got to be a cut here.

[00:53:08]Those are the two cuts.

[00:53:10]Um, and just based off of how how the carbons layout worked at the beginning.

[00:53:22]Uh, so I the rule of thumb is usually to start with the uh carbon near or the cut nearest the hetereroatom. So that would be this cut here. So we're going to focus on that cut first. See how that works.

[00:53:41]Although you could probably get away with doing either one here.

[00:53:48]There's probably multiple reasonable answers, right? So I see that I've got a oxygen one carbon away from a double bond or one one carbon away from the cut I made.

[00:54:03]So it's got to be a Griner with the carbonal or carbonial was on carbon 5.

[00:54:08]Uh, and now I'm going to do my other cut, which looks like this.

[00:54:16]So, if I think about doing an epoxide, which is what we did would have done for test two, I see a problem. This Grineyard doesn't want to add into that carbon. It only wants to add into that carbon, and that wouldn't get us this product.

[00:54:36]So this doesn't work. We have to go we have to think of something different because our traditional epoxide which is the hetereroatom two carbons away doesn't work in this situation. So luckily we have another tool at our disposal. an enolate with a leaving group can do a substitution and that would help us out because now we can get this carbon added onto that middle carbon. We can get that carbon group added on the middle carbon instead of the end carbon. So doing the enolate in this situation solves a problem that we were having with the epoxides.

[00:55:22]Does that make sense? Yeah, it does make much more sense. So, I just wanted to verify either cut would have been fine because when I compared my answer to your answer key, you did the other cut first. So, I just want to make sure >> they both work in the situation. You could have done >> this as well.

[00:55:41]And then >> Thank you so much.

[00:55:43]>> Yeah.

[00:55:49]Wait, why would why would it be Br and not mgbr if you do it the other way?

[00:55:54]>> Well, because this is a nucleifile.

[00:55:56]There's a negative charge on it.

[00:55:58]>> No, no, no. On the one you just wrote.

[00:56:01]>> Uh, this would be, sorry, this would be an enolate.

[00:56:09]>> Oh, okay.

[00:56:10]>> Make add into that carbon.

[00:56:12]>> Okay. Right. Cuz that's where we need the carbon to be added here, not here.

[00:56:19]And I have a question about um the very last retroynthesis about how to write the cuts or what the cut would be or like yeah how how to write the cut because when it became enolate there was a cut. Well I'll let you answer.

[00:56:39]>> One two three four one one two three four. So, I got two cuts here.

[00:56:53]My first thought is to try to be closer to this header atom. So, I'm going to do that first cut first. So, since the header atom is one carbon away, I've got that and that.

[00:57:16]And then this cut gets me.

[00:57:29]This is a reasonable way to do it.

[00:57:35]>> That's not what the answer key said.

[00:57:38]>> Well, sometimes there's uh the thing with synthesis >> is that there's often multiple ways to do it.

[00:57:45]I could have also started from this other cut, the cut on the bottom, and done this.

[00:57:53]This is probably what the answer key says.

[00:58:07]>> Yeah, that's what I got at first, too.

[00:58:08]stupid. And I was like, where would do I I can't cut a double bond without getting >> Yeah, because that's not really a double bond. It's a carbonal.

[00:58:19]It's an it was an enolate, right? This is >> So this is what it looked like before you took that proton and we could definitely cut that. So if we just pretend it's that well then it becomes easy.

[00:58:34]>> Okay. So if I do cut up, you know, like I just treat it like a carbonal.

[00:58:38]>> Yeah. Treat it like the starting carbonal.

[00:58:41]>> Okay.

[00:58:42]>> Cuz that's what it was a step before.

[00:58:43]Yeah.

[00:58:48]>> Uh for the last question, can you go over the last question on the reaction practice part?

[00:58:55]Yeah.

[00:58:57]So panic What the heck is S? Me Eu. What is a SME?

[00:59:10]Well, S is sulfur.

[00:59:14]Sulfur is directly under oxygen. So, I'm going to treat this like H O M E, which is a nucleophile.

[00:59:25]KOH. This is a base.

[00:59:29]And so what we see is that just like anytime you have a nucleophile, electrophile and a base is the base is going to deproinate the nucleophile.

[00:59:45]So O minus going to take the proton from the HSM. And so now I have a negatively charged sulfur which is going to attack 14.

[01:00:04]So the key is just recognizing is not panicking recognizing that sulfur acts like oxygen. Sulfur is going to be a nucleophile.

[01:00:12]You have a base and so all we're doing is the same steps. Step one base deproinate nucleophile. Step two nucleophile attack electrophile. In this case I know I have a one. What is up with my pen today?

[01:00:26]In this case, because I have that double bond there, my default is to think one four addition unless it's a small set of nucleophiles that do one two.

[01:00:47]Um, someone asked uh about the cuts note pretty much all of these synthesis there's multiple ways like here I'm showing there's multiple ways you can do it that would both be correct at this point what we're looking for isn't the perfect synthesis it's a reasonable argument is your synthesis reasonable is there's nothing wrong with it are you able to are you doing things and there's no protecting group inconsistencies or whatever.

[01:01:25]It's uh there's as things get more complicated, there becomes more ways to do it, right?

[01:01:36]Someone asked for the forward synthesis for this first one. It depends on which route you go.

[01:01:41]Um but the key to these forward synthesis, the two new forward synthesis things we have here are being able to add a benzilic or an alyic bromine that can be very useful. Um, and also it's taking a carbonal of some kind and adding LDA to form the enolate. Everything else remains the same.

[01:02:12]It's just that these are kind of the two new things.

[01:02:24]If I was just making this carbonal, I do not have to form an enolate. No, I just make a carbonal just like always.

[01:02:31]Nothing's different about that.

[01:02:39]And if I did it this way, I would have to form an enolate at this step. I'd form the carbonial first and then form the enolate.

[01:02:52]So it' be this Grineyard.

[01:02:55]It'd make this.

[01:03:00]Then I can PCC that.

[01:03:04]And then LDA would get me there.

[01:03:12]Uh mostly the radicals.

[01:03:15]uh someone asked about if there'll be a radical question. Mostly the radicals uh will either come in the conceptual side where I say what drives the formation to this spot and so you'd say oh it's alilic it's got resonance light forms radicals so we form the most stable radical which would have resonance or it comes here where you use this radical reaction practically in a synthesis. So, those are the two ways I could ask a radical.

[01:03:49]I'd be unlikely to just put it somewhere else, you know?

[01:03:56]Or I could always just have like a fill-in- thelank like, hey, here's the box. I don't care about the the mechanism. What's the product? So, those are kind of the three ways I could ask a radical question.

[01:04:12]All right. Um, I think I've been on for over an hour. So, um, I'll take a couple more questions. We'll wrap it up. Um, and I will see you guys tomorrow. And hopefully I can have some time to to get out that retroynthesis practice for test three.

[01:04:44]All right, give me give me your best shot. Come on. What do we got?

[01:04:50]>> Uh, I've just got a quick question. Um, it's more like a confirmation based on uh like some pattern recognition.

[01:04:57]>> When we have uh bases that take out hydrogen from a non like bulky standpoint, so anything that's not LDA or KOTBU, is there instances in which it won't like add on to itself? because I'm noticing when we have like a one and a two, that second step will add on and it won't break >> at all or whatever it's called.

[01:05:19]>> Yeah. So, yeah, you can let's say we had this reaction.

[01:05:27]So, you're saying, you know, almost always with sodium hydroxide is what we see is we form the enolate Sometimes it's I swear this thing runs out of ink and then you do a self doll, right?

[01:06:01]But here's the thing.

[01:06:05]if there was a better electrophile sitting around. So carbonial are controllable and reliable, but they're not they're not hot electrophiles. They're not like the best electrophile. They're kind of poor electroiles.

[01:06:22]They're middling electrophiles. But if I had something like this, that enolate is going to attack that leaving group before it would attack another version of itself.

[01:06:49]>> So it prefers leaving groups. Um, bottom line, if I give you a nucleophile or I give you an electrophile that you can clearly say, you can clearly like identify as an electrophile, use it.

[01:07:11]>> Okay, perfect. Thank you. I I really appreciate that. Yeah, for the first section and enolate formation question three.

[01:07:30]This does it matter if we did the elimination on the other carbonal?

[01:07:37]Is this the question you're talking about?

[01:07:40]>> No, it's on the first page.

[01:07:48]Yeah, that one.

[01:07:50]>> Well, one of these is way more um this hydrogen is way more acidic than this hydrogen.

[01:07:58]>> I was asking like if you went like the other way, like if you did like the double bond on the other side, >> that versus that.

[01:08:08]>> Yeah, I think so.

[01:08:10]>> Yeah, we already talked about this.

[01:08:11]These are resonant structures, so it doesn't matter which one you draw.

[01:08:14]>> Okay, thank you.

[01:08:15]>> Right. The the reality is it's not this or this.

[01:08:20]And if you ever get if you ever lose credit at any point in organic chemistry class because you drew a resonance structure of the correct thing, fight for it cuz you're right.

[01:08:30]>> Thank you.

[01:08:33]>> So, I have a question. When it comes to like one two versus one for addition, should we apply that outside of that section? Like for example on question the question in section um and putting it together reaction practice I got the I don't know let's count the first the second LDA or okay the third LDA question >> this one. Yes.

[01:09:09]>> Like would we we would use 14 one two.

[01:09:13]>> I'm going to answer your question with a question.

[01:09:18]What do you think I meant by putting it together?

[01:09:23]all the everything we know >> all this stuff up here >> is okay >> the framework I want you to be thinking of when you approach these questions down here >> oh okay okay >> right so yes yes when you do a reaction period no matter what I don't care what the I on the test I will not say look for a one two or a 14 addition or I will not say does this add into the carbonal I'll just add in the enolate. I will say tell me the reaction mechanism.

[01:09:58]And so the ability like I want you to prove that you know those things in a real environment without prompting on whether or not we're thinking about 1214 or enolate vicarbonel. We need to apply that to a real reaction. Those are just tools to help us understand the reaction.

[01:10:21]Cool.

[01:10:23]>> Yes. Yes, that makes sense.

[01:10:24]>> So that I I kind of prefaced that in class the other day. Hey, the first part of this test, these this key concept section, >> this is not going to look like a test question necessarily, but what it is is it's trying to get you to think about the ways the common ways people might mess up or might miss or like the key points that are important for this chemistry before you do the practice.

[01:10:48]That looks like the test practice. That looks like the test. Does that make sense?

[01:10:53]>> Yes.

[01:10:56]>> Um, sorry, I was a little late to the party. So, I understand that um if you get LDA, you will get 100% of the enola and 0% of your original uh uh molecule.

[01:11:09]And then if you have um a less bulky base, you'll get 99%.

[01:11:14]You'll get 99% >> nothing to do with bulk bulkiness, but yes. or strength, sorry. Yeah. Um, you'll get 99 and then 1% enolate. But that other one down where it's a complete equilibrium, it's 50/50. Could you explain to me why it makes a complete equilibrium reaction?

[01:11:31]>> This one?

[01:11:33]>> Yes, sir.

[01:11:34]>> Because I'm comparing these two products. I mean, this is roughly the same as that, right?

[01:11:44]In terms of energy, OM negative, OE negative.

[01:11:50]>> Oh, is it because you don't form an enolate? You just go through the the substitution and then that's you basically >> Yeah, but you have to evaluate these things on you can do a substitution where it's not 50/50.

[01:12:07]>> So, >> for enolates, you have two choices. It's 100% or 0%. It's one to 99%. For these, I'm not going to ask you the equilibrium, but you should understand them.

[01:12:18]>> Like I'm not asking you to write 50% 50% on this. I was just doing that as a as a example.

[01:12:24]>> Oh, okay. Okay. Thank you.

[01:12:26]>> But like if I ask you for the equilibrium here, you should be able to tell me these things are roughly equal in in energy.

[01:12:34]>> Okay. Got it.

[01:12:36]>> Uh Okay. So, I'm starting to answer questions that I've already gone over.

[01:12:39]Um, here's a new question. Can you go over reductive amination? All right, that is something that is fair game for a synthesis. When you see, and I'll put this on the uh retroynthesis or test three practice. Um when you see that a nitrogen has made a carboncarbon bond the this doesn't work very well we would rather for various reasons and I talked about it you can go back to those lecture notes if you want to here this is a better option but the problem with this is that you can do the mechanism it doesn't form the amine directly, it forms the amine, right? But this is safer. This is safer. This can over add. It has all sorts of other problems. So, this is a safer reaction. But the problem is we have to do one extra step, which is to reduce this amine down to the amine.

[01:13:45]And you can use any reducing agent, but sorium sodium borosyanide is the safest because it's a really weak reducing agent. You don't need a strong one.

[01:13:54]Um, yeah. So, this is this is what's called a reductive amination and it helps us make carbon nitrogen bonds.

[01:14:08]>> You would add H30O plus with that as well, right? It would be like like >> sure like a like a LH reduction any reduction.

[01:14:16]>> Yeah.

[01:14:17]>> Yeah. But in synthesis I, you know, the like quenching step, the step two where you add a proton is implied. You're not going to lose points for that.

[01:14:30]>> Okay, thank you.

[01:14:32]>> Uh, someone asked if there'd be protecting groups. Um, yes, protecting groups won't go anywhere. They're they're they're here.

[01:14:41]Um, like to be honest, they're just a tool, right? If you start if you run into issues, use a protecting group. Um, like if it if you're if you look at it and you're like, well, I got two electroiles. I should protect one of them, you know. Um, it's it's not like a hurdle. Just use a generic PG or acetile.

[01:15:11]uh when there's heat, how do you know where it goes?

[01:15:15]Uh in this unit, if there's heat, you're going to have something that looks like this. And I don't care how substituted this right side is.

[01:15:24]The only proton you will ever take is the one between the two oxygens because there is resonance here. It is an alpha proton. You will only ever take this proton because O is not a good enough leaving group to leave on its own. You need the resonance with this carbonal in order to get that O to leave.

[01:15:49]You can use LH instead of sodium borocyide, but you have to be careful.

[01:15:53]If you have a carbonal sitting around, it will reduce that carbonal too.

[01:16:09]and you go through how you draw the connection of the molecules by your goalpost method when there's a reaction using sodium hydroxides so the double bond is near the more substituted carbon those are kind of different ideas um the goalpost is really just anytime oh yeah here's here's Jake here's your question on sodium hydroxide reacting with a different electrophile right it doesn't always do the selfd doll um but yeah I mean this Let's do this one.

[01:16:40]This a pretty simple one. I know it's not sodium hydroxide, but it's the same idea. So, all the goalpost method is is when I have an ald doll, when I have an enolate adding into another carbonal, this can get kind of tricky to draw.

[01:17:00]There's a lot of carbons here. There's a lot of carbon counting you need to do.

[01:17:04]So when I know I've got an enolate adding into a carbonal, I just go ahead and draw that. That's what I call my the goalpost, right? This kind of shape like that.

[01:17:18]And it's not a true goalpost, but there's an extra carbon here, right? It goes down. That just gives me a framework that allows me to not miss carbons. cuz really what what I see most often is this carbon between the enolate and the carbonial added into gets dropped a lot. So kind of drawing this motif to start with helps make sure that I don't lose that carbon.

[01:17:43]to all the goalpost method is is allowing me to have a stable structure, a reliable way to draw ald dolls and clins and any sort of enolate adding into a carbonial reaction because those are commonly ones where people will miss carbons. So the the goalpost method isn't ne isn't some magic bullet of understanding reactivity. It's just a way to draw this intermediate that allows me to kind of keep track of the carbons.

[01:18:22]Um, all right. I think I'm going to wrap it up now so I can get working on that retroynthesis worksheet. It's been an hour and a half. Um, I appreciate you guys for coming on. Um, good luck tomorrow. I think there's a LA help session sometime today. Um, I have office hours tomorrow morning, so if you have more questions, you can come see me in person. I hope this was helpful. Um, if you want a more thorough walkthrough of all the concepts in here, you can watch my old video. Like I've said multiple times, um, that goes through practically the same thing. I'll update the names on the practice sheet and I will get that back to you guys. All right, I'll see you tomorrow.