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2. Consider the reaction below and complete parts (a)-(c): (Completein the next pages)HBr100PCWhen the molecule below trealed with HBr at high temperatures_ the #2-...

Question

2. Consider the reaction below and complete parts (a)-(c): (Completein the next pages)HBr100PCWhen the molecule below trealed with HBr at high temperatures_ the #2-addition product predominates, rather than the 4-adduct: Draw this product and explain this result: Use appropriate resonance structures t0 illustrate

2. Consider the reaction below and complete parts (a)-(c): (Complete in the next pages) HBr 100PC When the molecule below trealed with HBr at high temperatures_ the #2-addition product predominates, rather than the 4-adduct: Draw this product and explain this result: Use appropriate resonance structures t0 illustrate



Answers

Addition of HBr to a double bond with an ether (-OR) substituent occurs regiospecifically to give a product in which the -Br and -OR are bonded to the same carbon. Draw the two possible carbocation intermediates in this electrophilic addition reaction, and explain using resonance why the observed product is formed. (FIGURE CANNOT COPY)

Drew an energy diagram for the addition off HB are toe 1 19 That's the energy versus reaction. Coordinate. Starting materials spirals through this point on that girl. Let one curve on your diagram. Show the information off one Brumer contain the curve on the left would respond to the formacion off one bomber contain, which is at this point, and the curve on the right will respond to the formacion off to rumor labour the positions off all reactions, intermediates and products. The positions for the reactors and products have been ordered labeled. This position is the intermediate reading through to vomit contain Um, this position is the intermediate leading toe. One Brumer contain the higher image of tropical time. Intermediate belongs to this curve. This image is hide in this one, and that's why the same curve has high in energy. Transition state Rich is this one. The curve explains why my karmic works and that to Mama Pimpin is that major product

Find the second up in pain RDX with HBR and produces it. Prearranged bromide full of the girth airs and proposed mechanism for the whole election for convenience number carbons in the starting line Encyclopedia pain with the full of the error this burg breaks. So now there is no bond between carbons one in three and the bump between carbon once and two is firm. So is it you born the charge forms? I'm comin for you because it lost turtles. Electrons are pushed from this double burn, so it becomes a single barn and the bun between h nbr breaks informs B R minus which accept electricals and this hydrogen EDS. So this problem, that's one of those hard pigeons or vinifera. And on the second step, bromide, the maids air prayer off electoral for the grab A cattle and forms it you born

Okay. This problem is asking us to propose resident structures for this compound and this compound in order. Explain why my one Naftali is the major product. Okay, so let's go ahead and do this. So first up, I'm gonna start with this one on top. So the drawings and structures is basically moving. Carbo Kat ends around. So what I'm gonna do is move this carbon cut onto right here. And I do that by moving electrons around. Okay, so what I'm gonna do is draw a double headed arrow, which is representative of a rest in the structure. Okay, so I have my two rings. I have my alcohol unaffected. I moved this O'Kane to this position, so I moved on to that single bond, which would make a double bond. And when I do that, I'll have to make a positive charge right there. So I moved my secondary carbon headline over, okay. And then, of course, I have my unaffected, aromatic city in that Benzedrine. Okay, Now, I'm gonna go ahead and draw the extraditions form, which is breaking the aromatic city of this benzene ring in order to, in order to relieve this carbon of its posit charge, and that will make a positive charge on that carbon. Okay, so the product of that is this have my two rains alcohol unaffected my cocaine and the previous one. And then I have my new all keen right there, my other okay. And, uh, previously formed the benzene. And then, of course, I have my positive charge right there. So as we can see so far, all we're doing is moving that posit charge around by moving electrons around. Make a those L Keaton's. Okay. Now I'm gonna move the electrons from the sucking onto this single bond making a double bond. Okay, so double headed arrow indicating a new resident structure. Okay, so I have my two rings, My alcohol right there. Okay. Looking. And then new Arquin posit charge and, um, in effect, Ducky. Okay, so now I have to relieve that carbon of its posit charge. I'm going. I'm going to do that by moving the electrons from this arcane onto that carpet. Okay, so when I do that, it will make a positive charge right there, and it will. It will relieve this carpet of It's part of a charge. So let's go ahead and do that. Kids are having two rings, and then I have my alcohol, my Hokkien ok'ing. And then I have my key in there, there. And now I have a posit charge right there. Can we move this around anymore? Can we move that posit charge throughout the, um, the two rains? Any more than that, the answer is no. Because, as we can see, if our to move this one over would be back to where we started in the previous one, and then I can't move this one over because this posit charges just too far away. This movement of the selkin onto this carbon would un effect that, but it would not be effective in relieving that carbon of its bother charge. So those are all my resting structures for this 1st 1 So before I get into the details of why I wouldn't have those, they made your product. Let's go ahead and start off doing the resting structures of this compound. So, in order to relieve this carbon of his posit charge, I can't move this, um, walking around because if I moved on to this single bond to make a dope on here. That would be negligible in terms of relieving that carbon of this part of charge so as to go to the other side, which is moving the out king from this benzene over. Okay, so I'm moving the electrons from that bending over, which effectively breaks the aromatic city of that benzene. So I have my two grains. I have my alcohol over here. I have my Newell King, which I just had moved my old all keen on affect jockeying, unaffected docking. And now I have my new positive charge right there. See? So I just moved to be positive. Charge over. Okay, next up, I'm gonna move that posit charge over again by moving this cocaine. That's double bond. The elections in the double bond over to right here that will relieve this carbon of its positive charge and move the positive charge to right there. Okay, so that will look like this. I'll have my two rings, my alcohol, my unaffected looking unaffected Arquin on affected Arquin and new. Okay. And then we're gonna have to move the positive charge over here. Okay, Next up. I'm going to move that posit charge over. So when I move the elections from that single sorry double bond over to there, relieving that carbon of his posit charge and creating a new posit charge. So, like this, I have alcohol unaffected. Opinion infected, Arquin unaffected are keen and new working and then deposit charge right there. Okay. And finally, after he relieved this carbon of its positive charge. And I do that by moving the electrons from this Al cane over there. Okay, I'll do that. This will be the product. Okay, so I have my two Grinch. My alcohol, my unaffected. Okay, Unaffected Arquin unaffected docking and my new Elke. And now I have a positive charge right there. Okay. So as we can see, we have the same amount of resident structures. We have one. I'll start with this. Top one will do it. In green, we have 12345 And then for this one, we have 123 for five. So the same amount. But the reason that this one is going to be my most stable one in order to eventually form I won after all is because in this one, the arresting structures of this one, we only have this one and this one in which we have our aromatic city. Right, because in my benzene, aromatic city is we'll cover that in later chapters. But basically, aromatic city is a super stabilizing component of different organic compounds. So, for example, benzene. It is a very stable organic compound because it is the presence of these double buns help to stabilize it. Okay, it's in the conjugated pipe system. It's called aromatic City. So we have that aromatic compound and this one and we have it in this one. But if we look at this one, we only have it in this one, right? We only have this benzene. So benzene is basically a cycle Heche saying with three double bugs in it. And if we look at all these, we only have Benji and occurring in one of them. So one of them compared to 2 11 in this one. So that is why the formation for the the presence of aromatic city is why, uh, this compound are These resting structures are so favorable. So that is why it is contributing to the formation of one Ethel

All right. So which of these Al keens do you think is gonna be faster to react with via Electra Filic Substitution with HBR. So you have meth Oxy on one That's para Chu and Al Keen and you have a nice, roomy other. That's power to Malki. So you might be thinking. I don't understand why this is a problem. If I have an electron donating group that's attached to a pie system, the pie system will have more electron density. It will make the AL keen more electron rich, So it'll reactive protons faster. Yeah, that's very valid, right? That's super valid. So by extension, if you have an electron withdrawing group on the on the benzene ring, it's gonna be sucking electron density out of the pie system. Um, so therefore, the AL canopy ends will be a weaker um, we'll have less electron density, so it won't be as basic for that proton. That's that's a super good conclusion. But how do you prove it with residents? Right. So, um, let's just say that I don't like how I'm gonna write this up with the H is on the left of the sea because we show accurate connectivity here. What if I tell you, Let's physically donate those electrons into the ring and let's push the electrons around. I'm gonna push that pi bonds to the other side to that carbon carbon bonds. I'm gonna push this onto the Al Keane. I'm gonna push this double bonds onto that terminal carbon. So by doing of that, I will arrive at a structure that is electron stabilized by this positive charge here and now, you'll notice that I've pushed the electrons onto this carbon that now gives me this carbon ion structure. So you'll notice that by resonance, this carbon is electron rich, right? Which means it'll B'more brought instead basic. Um, it'll be more basic in general for the proton in HBR, which means you will arrive at a faster benzel of carbon carry on in the intermediate. Um, also having that that group para means that even before I get to this resident structure, where, um, I have the, um, the oxygen on the meth oxy already in a double bonds. Let's just draw, um, mechanism where I've already taken the proton and you'll be like, Matt, what are you going to show us right So let's say I decided to do this. Um, that will give me a ring that has the benzo carbo carry on. Here's the H. We added, right. And then I could totally I'm not gonna draw all of them, because that's a little over two right now. But I could totally push my electrons, like, miss all forced troop I bonds at once. That will get me to a structure where I can. So I moved this pie bonds here. That's totally fine. And then I moved the other double bond to here. So that gives us a structure where that carbon caddy on is directly on the same carbon as an electron donating groups that I could make that most stable resident structure by pushing those down, which gives me an analogous carbo caddy onto something. Um, that I drew already up here. Right. So, um, that would give me the positive charge on the oxygen. Um, so actually, just to save me some time here, maybe I'll just draw it right here. Um so that gives me this, which is the most stable residents form for that. For the Ortho start for the Para meth Oxy styrene um, these Sylvie's showed that that, um that furthest carbon in the AL Keen is more basic for that proton by residents stabilization with that meth Oxy group. So now you might be able to say, OK, I think I understand what Matt's gonna draw now, right? So by the same extension, um, let's say that I decided to do all that same chemistry with that nitro group there instead. So I take the proton. Um, Now I'm gonna have the nature group para to where that benzo carbo Kalyan is gonna form. Right? So here is the H. I added, Here's the carbon caddy on. So now I'm gonna Porsche electrons just like I did before. And now you're going to see that I'm going to get to this incredibly high energy structure where I have the car will carry on on the same carbon as the nitro group. Um and this is super high energy. Um, and that will destabilize, um, the benzo carbo carry on in general. So this is the sorry. In the high stabilisation, it's gonna be high energy, less stabilization. Um, so this is, um, via residents argument. Why the meth oxy meth, oxy structure is mawr will react faster with HBR for Elektra Filic substitution than the nitro group will. So meth oxy more electron density. You could stabilize the carbon Canion better than the parent Nitro Intermediate, where you have the Carver County on on the same group as a missing Adam. As a Nitra group destabilizes cover cannons by withdrawing electron density, high energy structure, um, destabilizes the cover countdown so therefore less favorable.


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