Orgo 2 Exam 2 Summer 26

[00:00:01]For question one, it is asking for which two products are expected to be the least abundant under kinetic and thermodynamic control.

[00:00:15]So, the way we start this is to look at the the four versions that we could get from this dying.

[00:00:30]And if we draw the mechanism for the first one, we could go and get this.

[00:00:39]We could get this version right here.

[00:00:43]Or we could get a tertiary allylic because we added the hydrogen there.

[00:00:49]Well, what if we added the hydrogen not there?

[00:00:59]And what if we added it like that?

[00:01:03]We added the hydrogen here.

[00:01:07]We could have done that, right?

[00:01:09]We generate a primary carbocation, which is very, very unstable.

[00:01:15]And then we could do the same process over again, but just looking at the other double bond.

[00:01:29]And what four versions could we get from that?

[00:01:33]We could get this one because we added the hydrogen there.

[00:01:40]Or we could have gotten this one because we added the hydrogen there.

[00:01:47]Okay.

[00:01:49]And so, when we take a look at this, which is the most stable intermediate.

[00:02:00]We would say that this one right here is the most stable.

[00:02:10]And then we could say, "Hey, this one is probably the second most stable because that the second one is a secondary allylic and the most stable one is a tertiary allylic.

[00:02:27]Now, going to these next ones right here.

[00:02:33]Looking at that one at the very top, that's a primary carbocation.

[00:02:40]Not stable at all.

[00:02:43]And a primary carbocation, not not very stable.

[00:02:46]Okay?

[00:02:50]And so if we wanted to find the least abundant, then the molecule on the bottom has a a disubstituted alkene. This one only has a monosubstituted.

[00:03:08]So, based off of that, we could assume the one at the very top would be the the least stable.

[00:03:17]Okay?

[00:03:18]And so now we have our bromine that's left over.

[00:03:24]Okay?

[00:03:26]Which is actually a bromide now.

[00:03:29]And then that could technically come in and attack there.

[00:03:32]All right?

[00:03:36]Like this.

[00:03:43]So, that's a very unlikely, if not nearly impossible, compound.

[00:03:50]And then we have this other version from this intermediate that would have given us this one.

[00:03:57]Okay?

[00:04:01]Now, the thing is when we take a look at this top one, what type of addition did we do?

[00:04:11]That is a 1 2 addition.

[00:04:14]Okay?

[00:04:19]And then we take a look at the bottom one here, what type of addition is that?

[00:04:23]That would be a 1 2 addition as well.

[00:04:28]Okay?

[00:04:29]Um So, what What do we do here?

[00:04:45]The thing is even if you added heat or you did it cold, these two right here would still be the least stable.

[00:04:59]And they're the least stable due to no resonance.

[00:05:04]And so, what we have here is that those two intermediates that now I'm just going to circle again in green, those two carbocation intermediates are going to be the ones that are going to give us the the uh least abundant product at the end of the day.

[00:05:28]And so, do we see that on the the graphic here? This one at the very bottom, do we see that as an answer choice?

[00:05:37]That one.

[00:05:39]And then do we see this top one anywhere as an answer choice?

[00:05:46]That would be that one.

[00:05:50]And so those are the least abundant products.

[00:05:54]And so the answer choice that makes sense would be answer choice D.

[00:06:06]So question two, major product of NBS. This is a reaction that we remember that we have a benzylic carbon and when you have a benzylic carbon with NBS, that's where the bromine [clears throat] is going to be added is onto the benzylic carbon. So giving us that as our major product.

[00:06:33]And the mechanism that ex- explains why that's the case because you are going to generate an intermediate that looks like this. You're going to generate a benzylic radical and that's relatively stable.

[00:06:49]And that's that lead that stability of that radical leads to that the product that you see in the answer.

[00:06:56]>> [snorts] >> Question three, which of the criteria listed is not necessary for nucleophilic aromatic substitution reactions?

[00:07:09]That would be that one. That is not a requirement.

[00:07:18]The requirement is actually ortho or para.

[00:07:29]Diels-Alder, what's the major product here?

[00:07:35]Wow.

[00:07:36]Let's just draw it out.

[00:07:39]I'll just go ME for the methyl.

[00:07:44]And then draw it like this.

[00:07:48]Because we want it to approach endo.

[00:07:53]And so that's going to give us a six-membered ring here.

[00:08:01]And our hydrogen is up. The O-methyl is down.

[00:08:08]And our nitryl piece right there is going to be in the ortho position pointed down.

[00:08:16]>> [cough and clears throat] >> So do we see that answer choice?

[00:08:23]F right there.

[00:08:27]Because we want the outside group and the electron-withdrawing group to be on the same side. So that's down, that is down.

[00:08:38]And then we've learned the ortho-para rule.

[00:08:41]That says we want the electron-withdrawing group to be in the ortho or para position. And for these two reagents, the only possible one would be ortho or meta.

[00:08:54]Para is off the table. That can't even form in this set of reagents.

[00:09:00]And we know meta is never a good option.

[00:09:03]So it had to be just ortho.

[00:09:08]Which of the compounds shown will be the least reactive towards a Friedel-Crafts reaction?

[00:09:17]Let's see.

[00:09:21]So if you draw those all out, you draw benzene and anisole, a phenol, and a chlorobenzene, we have seen that all those do work with Friedel-Crafts.

[00:09:32]The only one that doesn't is when you have that nitrobenzene.

[00:09:36]Nitrobenzenes are two electron withdrawing and doesn't allow the Friedel-Crafts reaction to proceed.

[00:09:45]So, that would be that one.

[00:09:49]Okay, so how many antibonding molecular orbitals does benzene have?

[00:09:54]For this one, remember that you can draw a frost diagram.

[00:09:59]Looks like this.

[00:10:01]So, we have orbitals there, there, there, and there.

[00:10:07]And then if we just go right in the middle here, right in the middle there, how many electrons does benzene have? Has six.

[00:10:15]Or how many pi electrons?

[00:10:18]There.

[00:10:20]So, these right here are our bonding orbitals, right there.

[00:10:25]And these are our antibonding orbitals.

[00:10:28]So, there's three of them, which would give us answer E.

[00:10:36]Which of the following cannot have an S-trans bond?

[00:10:41]Cannot.

[00:10:42]Okay.

[00:10:44]So, if I take a look at A and we just draw that out, we have butadiene like this. It says it's a 1,3 Then it says at carbon two we have a methyl.

[00:10:59]Like that. So, that that would be a molecule A, cyclohexane.

[00:11:12]Would be simply like that.

[00:11:16]A 1,3 hexadiene.

[00:11:21]So, hex meaning six, 2 3 4 5 6, 1,3 hexadiene.

[00:11:30]Then cyclopenta diene this and butadiene would just be this simple thing like that.

[00:11:44]>> [clears throat] >> Let's see which of the following cannot have an S trans.

[00:11:55]Trans trans trans, okay?

[00:11:58]So, the way this is drawn already, that's S trans.

[00:12:03]That's S trans. That's S trans.

[00:12:06]And this one doesn't even apply cuz there's no double bonds in it.

[00:12:12]So, these three right here can rotate around that red bond and go from S cis to S trans.

[00:12:22]Whereas this one right there, since it's in a ring, it's a locked in place. It cannot interchange between S cis or S trans.

[00:12:33]It can only be S cis because that's what it's locked in as.

[00:12:39]So, that would be the correct answer for this one.

[00:12:48]Which of the following best describes the effect of a hydroxyl group substituent on a electrophilic aromatic substitution reaction?

[00:12:58]And the correct answer for that one would be C.

[00:13:12]Question nine.

[00:13:14]Why does benzene undergo a substitution reaction with bromine whereas cyclohexene undergoes an addition reaction with bromine?

[00:13:24]So, if we just write that out so we can kind of get a visual of what it's saying is we're going to have this versus if we just have this one right here, cyclohexane, going to this product right here.

[00:13:51]All right.

[00:13:57]All right, so this one is a addition reaction because we also added a hydrogen.

[00:14:09]That's the definition of an addition.

[00:14:12]And in the red, we did a substitution because we started with a hydrogen and we substituted that hydrogen for a bromine.

[00:14:23]So that's what's happening and taking those words and drawing it out.

[00:14:29]But the question is asking why does the benzene undergo substitution?

[00:14:34]Okay.

[00:14:36]When you read through all those answer choices, the best one that makes the most sense is B, right there.

[00:14:51]Because if we took benzene and did an addition reaction, it would look like this. And if we did an addition reaction, we would get this.

[00:15:03]Br H This is aromatic.

[00:15:08]And now we have gone to non-aromatic.

[00:15:14]And so this is very stable, very low in energy, and this would be higher in energy.

[00:15:21]So there's really no energetic incentive to like do this reaction because you would break aromaticity, which is just so, so stabilizing.

[00:15:33]So, question 10. This is basically just a definitions to see if we remember the difference between cumulated, conjugated, and isolated double bonds.

[00:15:45]Okay.

[00:15:46]And so, right here, this one would be cumulated.

[00:15:52]Those are conjugated, and those are isolated.

[00:15:57]Now, why is number three isolated?

[00:15:59]It's because we start with our double bond, but then we go one, two single, then our next.

[00:16:08]In a conjugated system, it has to be double, single, double.

[00:16:13]So, what would the correct answer be for this one?

[00:16:17]That would have to be E.

[00:16:21]Okay, question 11.

[00:16:25]Identify the structure for RZ three tert-butyl, four methyl, 1,4-hexadiene.

[00:16:36]Okay, that's a mouthful.

[00:16:40]So, let's just start with that right there. Hexadiene. So, that means six.

[00:16:45]Two, three, four, five, six.

[00:16:49]And it tells me that it's in a one four relationship. One, two, three, four.

[00:16:57]And it says we also have a methyl at carbon four. So, one, two, three, and four. So, there's our methyl.

[00:17:12]We'll add that there.

[00:17:15]And then on carbon three, it says to put a tert-butyl.

[00:17:21]So, we'll put a tert-butyl right there.

[00:17:24]And now we have R and Z that we have to figure out.

[00:17:32]So, let's do the R.

[00:17:35]And so, that means we have to find our stereo center.

[00:17:40]This would be our stereo center right there.

[00:17:43]So, we have to ask ourselves, are we going to make that a wedge or a dash?

[00:17:51]And so, let's let's break it through here.

[00:17:54]Let's just say I chose to do the a wedge. What configuration would that be?

[00:18:06]I know that we have a hydrogen there.

[00:18:09]So, now we just have to figure out our substituents here.

[00:18:14]Now, we have that group, and then we have this group, this group, and that group, which you can see are all four different groups.

[00:18:30]So, that tells us why we have a stereo center.

[00:18:33]And now, in order to prioritize things, we need to take a look at the carbons that are directly attached to that red stereo center.

[00:18:45]So, this one in yellow.

[00:18:47]In parentheses, what is directly attached to that carbon? There's going to be four carbons right there.

[00:18:58]If we take a look at this one, in parentheses, what are we going to have? We're going to have a carbon carbon carbon and a hydrogen.

[00:19:08]And then the orange one, we're going to have a carbon, carbon, carbon, carbon.

[00:19:14]Okay?

[00:19:16]So, we can say at this point that hydrogen is going to be four.

[00:19:21]And this carbon is going to have to be three.

[00:19:25]And so, now we have to break the tie.

[00:19:28]Which one's going to be one and which one's going to be two.

[00:19:32]So, we just go to the next carbons.

[00:19:36]We look at this one and this one.

[00:19:41]And in parentheses here, in red, we could see carbon, carbon, carbon.

[00:19:46]Oh, there's one hydrogen.

[00:19:51]But this one down here, we have a carbon then three hydrogens.

[00:19:58]So, that means this right here is going to take priority number one.

[00:20:04]And this would have to take priority number two.

[00:20:08]So, if we go and trace the numbers one, two, three, we are going in the clockwise direction, giving us an R configuration.

[00:20:19]So, I lucked out on that one, getting it right on the first try.

[00:20:24]If I would have done a dash orbital, I would have saw that we got a S and then I would have just replaced it.

[00:20:30]So, So, we got our wedge there.

[00:20:36]And then the next question that we need to address is the Z part.

[00:20:43]What does that mean?

[00:20:45]Well, let's clean this up a bit now.

[00:20:48]All right.

[00:20:53]So, now we have to take a look at the alkene.

[00:20:58]This one or this one? Which one do we need to worry about?

[00:21:02]Well, the one on the left can't be E or Z because it's only a a monosubstituted alkene. So, we just have to worry about the one on the right. We want a Z.

[00:21:13]So, that means we want the the highest priority groups Okay?

[00:21:21]To be on the Z or the same side. So, we have a hydrogen coming off here and a methyl coming off of this one.

[00:21:30]Right here.

[00:21:33]And so, if we take a look at carbons one and carbon two of this alkene.

[00:21:42]And that's that's not numbering it correctly in the IUPAC.

[00:21:53]Let's get rid of those numbers there.

[00:21:54]Okay?

[00:21:56]So, it would The numbers in the IUPAC are what? 1 2 3.

[00:22:00]So, carbons four and five.

[00:22:04]With carbon five, you have a methyl or a hydrogen. Which one takes higher priority?

[00:22:12]The methyl.

[00:22:14]And then when we go and look at the carbon four, what are options? We have a methyl versus this group right here.

[00:22:28]Then how do we take priority pick priority over this? It's not based off of the size of the group. It's about the carbon the atom directly attached to the alkene.

[00:22:39]So, if we expand that methyl out to look like this, what we're doing is we're comparing this carbon to that carbon.

[00:22:50]And which one's going to take higher priority?

[00:22:54]This one.

[00:22:57]And so, at the end of the day, looking at the orange circle and the red circle, do we notice that they are what?

[00:23:09]They are on opposite sides.

[00:23:12]So, that the way I drew that is implying that we have a E.

[00:23:19]And so, to fix that, all we would have to do is just take this methyl and replace it like that.

[00:23:29]And now, this higher priority group is going to be on the same side of that double bond.

[00:23:36]And that would give us our Z.

[00:23:40]And so, when you go to try to find the answer choice, um let's see.

[00:23:58]Just double-checking something real quick here.

[00:24:04]Hm.

[00:24:08]When you take a look at this the the zigzag backbone, do you notice how I started like this, pointed down, but all the answer choices actually start it starts like that.

[00:24:24]So, what I would have to do is I would have to take this molecule and flip it 180° on the horizontal axis. And when you do that and flip it on the horizontal axis, you're going to see that this and this are exactly the same.

[00:24:44]So, in essence, when you look at this answer choice C, which is the correct answer, you will notice that this is R and that is Z.

[00:24:58]And it matches all of this information right here.

[00:25:04]All right.

[00:25:08]All right, question 12.

[00:25:13]So, it's we're looking at this MO diagram of a molecule that we don't know the structure of, but we know that it's flat and planar.

[00:25:22]We know it's cyclic and it's fully conjugated.

[00:25:25]And how many pi electrons does it have?

[00:25:28]It has six of them.

[00:25:30]So, really all we had to do is [snorts] just do the 4n + 2 rule here.

[00:25:36]We can see that all these electrons right here are in the bonding MOs.

[00:25:44]So, that has to equal six.

[00:25:47]>> [clears throat and cough] >> And do we have a uh whole number that makes that statement true?

[00:25:56]4 * 1 + 2 = 6.

[00:26:00]So, that by definition would have to be aromatic.

[00:26:09]This problem right here, there is a uh typo that we need to address.

[00:26:17]And that is the hydrogen is not there.

[00:26:21]And you're going to have a positive charge on the sulfur.

[00:26:25]And then this carbon here on the second on the first molecule is also going to have a positive charge.

[00:26:33]So, when you look at those molecules, now we can ask which two molecules are aromatic?

[00:26:42]And we will see that >> [clears throat] >> it is those two.

[00:26:51]Those two molecules right there are going to be aromatic.

[00:26:55]So, that leaves us with answer choice C as the correct answer.

[00:27:12]All right.

[00:27:16]Here's our energy diagram here.

[00:27:22]And so, remember with a molecule like this, when we want to go to a frost diagram, we put the point down.

[00:27:34]And then from that, we can see, oh, we're going to have energy diagram that looks like this.

[00:27:41]So, that's what you see in the answer choices.

[00:27:45]And then we just look at this guy. How many pi electrons do we have?

[00:27:49]We have four.

[00:27:53]Do we have anything that matches that?

[00:27:57]Looks like that matches that.

[00:27:59]So, answer choice B would be the correct answer.

[00:28:07]Predict the major product for this reaction here.

[00:28:12]So, this is a electron donating group, and that directs it the nitration It's going to direct the nitration to the >> [snorts] >> ortho or para position, but we will Let's say we'll go to the um I'm going to go to the para position just to avoid the sterics at the ortho positions.

[00:28:42]And then, with that, we will do the chlorine.

[00:28:48]And so, we have an electron withdrawing group, electron donating group.

[00:28:53]Which one's going to be the most important in directing the second reaction?

[00:29:00]It would be the electron donating group.

[00:29:03]And we only have one option, the ortho position. So we would put the chlorine on the ortho position with respect to the electron donating group.

[00:29:16]And so we see that that would give us answer choice B.

[00:29:24]All right.

[00:29:27]So here we have a spider web here to fill in all the reagents.

[00:29:37]And we're going to answer this on the FRQ portion of the test, but we can still answer it here.

[00:29:47]And so let's take a look at the answers.

[00:29:56]Uh well, you know, let's just do it on the FRQ.

[00:30:07]All right.

[00:30:08]So the answer choices here are going to be G I and D A C L 1 2 3 1 2 3 one there and like that.

[00:30:44]Okay.

[00:30:46]So, those are the answer choices there.

[00:30:50]And if you guys want to go through the reasoning behind those those answer choices there, I would be more than happy to talk about that in office hours if you'd like.

[00:31:02]But you would have to refer to see the whole problem, we go back to our MCQ portion.

[00:31:11]And it's a big it's fills the full page right there.

[00:31:15]So, these are the all the answer choices that you would fill in.

[00:31:23]All right.

[00:31:25]So, let's take a look at this problem.

[00:31:29]And this problem says that we're going to have to refer back to the MCQ question number one.

[00:31:39]Because this problem is now asking a couple things.

[00:31:43]It is saying if you did all of this work in MCQ one, all of this.

[00:31:53]So, let's clean this up here.

[00:31:57]Just like that, okay?

[00:32:00]If we If we've done that work to figure out all the carbocation intermediates that are possible, and then we go back to the question.

[00:32:15]And this part right here is asking what is the intermediate that gives us that product V.

[00:32:27]And that product V as you scroll back up, is right there.

[00:32:34]What intermediate of these four, or you have to also make sure that you look at the most stable ones.

[00:32:46]Okay.

[00:32:48]Let's get rid of that for a second and look at the resonance structures.

[00:32:54]Because we could have this.

[00:33:01]That's our most stable and it has the resonance structure that looks like that.

[00:33:05]And then we could also have a resonance structure for the other one that looks like this.

[00:33:14]Let's get rid of that. We would have the double bond. Oh.

[00:33:20]It'd be right there.

[00:33:24]And then the positive charge would have to go right there.

[00:33:31]So now we have six carbocation intermediates that we need to consider.

[00:33:40]And the problem in the in the FRQ is asking which one is going to be responsible for this product right here.

[00:33:51]Okay.

[00:33:53]And the only one when you look at the six that you that the possible six that you have, uh let me see.

[00:34:09]Okay.

[00:34:11]The only way that you could get product five or V, however you want to say it, you would have to have this intermediate.

[00:34:26]That's the intermediate you would have to have in order to get product five.

[00:34:33]And then to get product two, what intermediate would we need?

[00:34:41]We would need that one.

[00:34:47]And you can see that back from the MCQ question one analysis.

[00:34:56]And then, right here, it's asking what is the major product um under thermodynamic control?

[00:35:07]So, if we come up here, look at our carbocations, what is the most stable one?

[00:35:17]We said it's this guy right here.

[00:35:20]Because it's a tertiary allylic.

[00:35:25]So, if we're under thermokinetic thermodynamic control, we're going to use this one, this resonance form. Because now the bromine or the bromide is going to attack here.

[00:35:41]And now our product is going to look like this.

[00:35:50]And that is a 1,4 addition.

[00:35:53]And now we have a more stable alkene right there.

[00:36:00]And so, that would be the major thermodynamic product.

[00:36:04]However, if your bromide if you were under kinetic control, or if it was done at a colder temperature, the bromide would have actually attacked this carbon.

[00:36:21]Giving you this as a major product.

[00:36:29]>> [clears throat] >> Like that.

[00:36:34]And that would be a 1 2 major product.

[00:36:38]1 2 major 1 4 major.

[00:36:43]So, the question asked, "What is the major thermodynamic product?" And that would be that guy right there.

[00:36:55]So, when we go to answer our question, that would be answer choice six.

[00:37:07]And you'll be able to see that when you scroll back up.

[00:37:11]And then, it asks, "Is product five, so this guy over here, is that a 1 2 or a 1 4 addition product?"

[00:37:25]And that is in fact a 1 2 addition product, so that makes that statement true.

[00:37:32]Okay.

[00:37:36]So, right here, question two, this is a Birch reduction.

[00:37:45]Okay.

[00:37:53]Whoa.

[00:37:56]A Birch reduction here, and I'm not going to go over the mechanism with for this one because I've done it in so many other videos.

[00:38:10]So, just check other videos or check in your textbook for the mechanism.

[00:38:15]But, the answer, the product is going to look like this.

[00:38:22]Remember that this is a electron donating group.

[00:38:27]And we know that electron donating groups do not reduce the carbon that's attached to them.

[00:38:34]So, the double bond has to be like that in a 1 4 relationship.

[00:38:40]And we do not reduce this carbon.

[00:38:48]Whereas if you look at the other version of the the FRQ, you can see we now have a electron donating group.

[00:39:02]So, what would the electron donating What am I talking about? The electron withdrawing group.

[00:39:12]And so, we know that that does have to be reduced.

[00:39:19]This carbon right there or that carbon has to be reduced. What do I mean by reduced? It has a hydrogen there now.

[00:39:27]And then you see the double bonds are in a 1 4 relationship.

[00:39:33]All right.

[00:39:37]Um >> [sighs] >> All right. Now, we're looking at a Diels-Alder here.

[00:39:49]All right.

[00:39:55]So, when we take a look at this guy, you could figure out if this is ortho or para or meta by just looking at it.

[00:40:08]Or if you feel like drawing the dying and the dienophile to help you see that, that's another way of doing it.

[00:40:19]But, if we just break it down and figure out um what should the starting materials look like?

[00:40:35]So, the way I like to approach that is we just go back to the most simple version and look at the mechanism.

[00:40:45]That attacks there. That goes there.

[00:40:49]That goes there.

[00:40:50]All right?

[00:40:51]And that would give us our product like that.

[00:40:58]And if you want to number everything, carbons 1 2 3 4 5 and 6.

[00:41:04]6 1 2 and 4.

[00:41:08]All right?

[00:41:10]Now, if I want to go in the reverse what would those mechanism arrows look like?

[00:41:17]So, I want to take this piece right here and go in reverse and get back to our starting materials. But, I want to show it to you mechanistically.

[00:41:28]What would I do?

[00:41:30]And so, I would have to do the exact opposite of the red arrows.

[00:41:36]All right?

[00:41:37]So, the exact opposite of the red arrows would be this has to go here.

[00:41:43]Those pi electrons would have to come there.

[00:41:47]All right?

[00:41:49]And then, these electrons would have to go there.

[00:41:54]And then, that would give us that piece plus that piece.

[00:42:02]Carbons 5 6 1 2 3 and 4.

[00:42:12]So, when I take that very simplified retro Diels-Alder mechanism and apply it to something a little bit more complex, we just go back and forth with that very simplified molec- uh model right there. All right?

[00:42:32]So, what are some arrows that I've done here?

[00:42:36]This would go there.

[00:42:39]That would go here.

[00:42:42]And then that would go there.

[00:42:45]Or let's draw it a different way.

[00:42:48]Those would go there.

[00:42:51]And so when those when you look at those arrows really carefully and break the bonds that I'm showing that you need to break, you're going to get something that looks like this.

[00:43:13]So there's our dying.

[00:43:17]And what would our dienophile look like?

[00:43:20]>> [snorts] >> So if we number the carbons here, let's do that.

[00:43:26]1 2 3 4 5 and 6.

[00:43:34]1 2 and 3 and 4.

[00:43:38]And then where's the 5 and 6? It has to be the dienophile.

[00:43:46]All right, and that would be 5 and 6.

[00:43:53]What about these carbons right here?

[00:43:57]Those are right there.

[00:43:59]And I didn't number those just cuz, but I accounted for them.

[00:44:05]And so that's our diene file and our dying.

[00:44:10]And so if you do take those two compounds there and do the reaction in the forward direction, you would you would get this as our product.

[00:44:20]Now, what is the relationship between the outside group, this group, and the electron-withdrawing group?

[00:44:33]What is the relationship?

[00:44:35]Those are meta to one another.

[00:44:38]So, meta.

[00:44:41]We also see that both those, the the electra the outside group and the electron-withdrawing group, they are both pointing down.

[00:44:54]So, that makes those endo.

[00:45:01]All right?

[00:45:01]So, now I'll go a little bit faster.

[00:45:05]What would We could Let's just answer it a different way now or in a different order.

[00:45:13]We have this group and that group.

[00:45:17]That's our electron-withdrawing group.

[00:45:21]And that is on the opposite side of our outside group.

[00:45:27]So, that's going to have to make that exo.

[00:45:32]And then what's the relationship between those two circle or purple circles that I have there?

[00:45:39]That's in an ortho relationship.

[00:45:42]And so if you guys do the reactions here, the retro arrows, that you would get our two products that looks like this.

[00:46:06]>> And what would our diene a file look like?

[00:46:11]The same thing as above.

[00:46:17]Okay.

[00:46:19]So we got Okay, that covers all of that.

[00:46:26]So if we go to version G, let's just not go through the the major details here.

[00:46:37]I'll just give you the answers because it's all the same principles as before.

[00:46:46]So you would have to have this in order to get compound six.

[00:46:53]And then you would have to have that one to get compound three.

[00:47:02]What is the major kinetic product? That would have to be a one-two addition, and you will find that that's compound five.

[00:47:11]And it's asking is compound six a one-two addition?

[00:47:17]No, it's not. That's a one-four because we added our hydrogen there.

[00:47:24]That's a one-four relationship.

[00:47:27]That's false.

[00:47:31]Then I already talked about the Birch reduction.

[00:47:40]So now this was that's the last thing that we need to talk about. And this is a different looking molecule than the F version.

[00:47:51]All right.

[00:47:54]When we take a look at this one, this one is in the para relationship.

[00:48:01]And what is the We see that the electron donating group right there is a dashed meaning it's underneath the ring suggesting endo.

[00:48:20]And so our dying is going to look something like this.

[00:48:26]OME or I didn't do ME so we'll do CH3 like that.

[00:48:34]There's our dying and our dienophile has our nitro group there attached to it.

[00:48:45]We do the same thing for the next problem.

[00:48:49]We see that and that is in the ortho position.

[00:48:53]They are both pointing down making them endo.

[00:48:59]And then the dying will be a five-membered ring with the OCH3 there like so.

[00:49:11]And our dienophile is going to be the same as above.

[00:49:17]Just like that.

[00:49:24]And that should do it.

[00:49:34]Okay, as always if you guys need any help or have any concerns feel free to reach out and I'd be more than happy to help.