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14. Draw curved arrows for the following mechanistic stepMe15. Will the following cation undergo rearrangement? Why or why not?...

Question

14. Draw curved arrows for the following mechanistic stepMe15. Will the following cation undergo rearrangement? Why or why not?

14. Draw curved arrows for the following mechanistic step Me 15. Will the following cation undergo rearrangement? Why or why not?



Answers

Draw curved arrows to show the flow of electrons responsible for the conversion of the following reactants into products:

Yes. Let's draw a stepwise detailed mechanism for the following reaction. So we'll start here reactant. Yeah. And for step number one and remove a bro Ming Adam, plus charge. Then we'll have each O c h two ch three. This would come in here, and we can have two cases here. Um, over here, in case number two, we can have above the plane attack. And that would look so this here we have six black seen ring. We have oh, ch two ch three positive church each. And this would come in here and then we'd have step number three minus c each Plus, this would yield Bring oh ch two ch three. And this would be plus each PR or this could come over here. And step number two would be below the plane attack. This would yields Oh, ch two ch three plus hydrogen. This would come in here Step three minus h plus. And you would have structure here. Oh, ch two ch three also plus HBR. So here's our stepwise mechanism that shows the formation of the products and the two products that are formed. Uh,

So we're going to see through the air pushing mechanism how a meth oxide can replace Ah, chlorine and a car box that gas derivative. So first things first, um, lone pairs on the meth oxide are going to attack the central carbon to the complex acid derivative and force electrons from the doubly bonded oxygen back onto itself, forming a tetra usual intermediate. So here we have the Tetra huge RL Intermediate. We have our oxygen now with the native charge. We have our chlorine and we have our newly bonded methyl either. So when we look at the Tetra Wegelin Air Intermediate, we want to see which ah, component of it is the best leaving group. And obviously the benzene ring is not a great leaving group. The, uh, negatively charged oxygen is a bad leaving group because it would leave with a negative to charge the meth ox are the methyl ester berm. Ethyl ether is a poor leaving group because it would leave an oxygen does not hold a negative charge on its own Very well. But the chlorine is an excellent leaving group because it can hold a negative charge in solution on its own very well. So what's gonna happen is two electrons from this oxygen are going to reform the double bond and kick out the chlorine yielding your substituted. Ah, car, box of gas. A derivative. Now be. Or now with the Esther, So O c. Age three. Ah, plus C l minus. And that's all there is to it.

Bus on John Iwas us on Johnny one requires to do this. The sin we have on s the ring. But this time we have lady. Oh, I only know I dry in our core solution. I see this illusion reviews. I'll go home off course because they don't. I drive what it is called. Go home. 123 Okay, now we dough the we can use. Um, Now let high dry from the late um hydride, this girl like this one. So we have this one this small Who so I have this one and this? No, no, no. This will be more DiFelice. So they got minus here and I got hydrogen here. This farm, they don't know how dry I let me know my drive. So now this will. But this I Let's some policy gin. Yeah, that's what. And now this go. But this one and this will go do this born. And now you want this one? No. Going to You need to be whole Hey, want to? Okay, now what? This one. Unless you will have another high dry re picked the mechanism. This will go this way and this one Well that this firm. Um, so you have this gin like this one. Do 345 And this sausage, onion, hydrogen. So they will nick still to a brook alone from You're looking on this. I'm this for one on this. Most here. This will. But this you did. And now this will goto this one and this one. You got this one? I don't ride along. Pagis off. You soon understand. Along by here, we'll talk the hydrogen. Okay? And finally we got the product they require, okay? And we don't It's my you. I don't write the also Jinlong the long black hair. But you should remember that belong back here, okay?

Okay. This problem is asking us to draw the curved arrows to represent the mechanism of this direction. So let's go ahead and go to this first one. Okay? So I'm the first one. We see that we have this alcohol on the left, and we're going direct with HDL in order to produce these two compounds over here, the appropriated to alcohol as well as the chloride ion. So, in order to undergo this direction, I have to do a couple of Barrows. I have to show the movement of electrons from this auction and gonna move that there's electrons onto this hydrogen. Okay, so the reason I'm doing this is because HCL is a very strong acid versus alcohol. That is a very not a, not a particularly strong base or particularly basic. But it is basic enough to get donated by my various at a compound which is a seal so that relatively, it's actually considered to be a weak base. But that base is going to get pregnant by my HCL. It's gonna take off that proton. And after I take off that proton, I'm gonna have to move the electrons from that. H c l bon onto the chlorine. Okay, so that is going to produce the protein of alcohol as well as the chloride ion. So that that chlorine is gonna have that extra set of electrons, meaning that it has that negative charge. And then because I'm moving a making a bond to this alcohol, that means that I'm gonna add a bond. Mean that I'm taking off electrons, meaning that it's gonna have that positive charge. Okay, what about this one? Okay, so for this one, I have my triumph looming over here. I'm going to record with my car box look acid, and then eventually, I'm gonna create this compound right here, in which I have a negative charge on the oxygen and then have the positive charge on my next trip. So again, this is gonna be an asset base reaction. And so trifle mean again, It's not a particularly very basic compound, but it is basic enough to get protein aided by my carb oxalic acid. Because carb oxalic acids are relatively citic at least their aesthetic enough to get to undergo a d pronation in relation to this triumph for me. So what's gonna happen is the set of electrons on this night, June are gonna go ahead and move onto this 100 to take off that proton from my carb oxalic acid. And as I do that, I'm gonna have to move the electrons from this hydrogen oxygen bond onto that oxygen. So I'm moving electrons onto the oxygen mean that it's gonna have that negative charge get. I'm gonna have those lone pairs on the oxygen and then because I'm moving a hydrogen on to the next in that nitrogen is gaining a bond, and it's basically losing a set of electrons, so it's gonna have that posit charge. So that means I'm gonna have this protein try Fleming and my deep rooted curb oxalic acid, a k a. Car box late. I on


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