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In electrophile, $\mathrm{E}^{\oplus}$ attacks the benzene ring to generate the intermediate $\sigma$-complex. Of the following, which $\sigma$-complex is of lowest energy? $\quad[\mathbf{2 0 0 8}]$ (a) (b) <smiles>FC(F)(F)c1ccccc1</smiles> (c) <smiles>O=[N+]([O-])c1ccccc1F</smiles> (d) <smiles>O=[N+]([O-])c1cccc(F)c1</smiles>

So for this question, here we have another mono coronation here. So here we have our two metal butane and we have our chlorine gas and some heat. So that will, through a monarch coronation, which is adding one Korean to any of the carpets we have here. Your products would include be it have one like this. Listen, what would be the one that we'd see the most? Because it's on a tertiary carbon right here. Um, but we would also have things like this where we could have one on the secondary here. We would also have them on our primary, so we would have one here, one here and one here. So this one would be these two that I'm drawing now on our primaries would be we didn't see these, most likely because they're closer to this tertiary carbon. Where is our primary? That's all. The way over here by its loans himself is because we wouldn't see this one as often because further away from deter sherry carbon attached to a secondary carbon that's been attached to a tertiary carbon. So it like an extra step away so we wouldn't see this product as much as we would see these other two primary product. So here are all of the, um, products that you could expect to see from my model coordination as we are adding one corinne to any of the carpet. Um, so now are any of these Cairo so pretty much and all of the primaries are not the reason. Why is because we have one to hydrogen. So we have I didn't hear coming on. You have a hydrogen here coming off. So have 1234 Same thing here. So 12 12 and then 34 something here. And the definitions in Cairo is that there's four different things attached. And because we have to same Adam to hydrogen, these are not Cairo. Just go ahead. Not for this one. This one would be Cairo. The reason why this one would be Cairo is because this one has a metal. It has a isopropyl. We have three carbons over here. You have hydrogen hiding, and then we have a chlorine. So this one would be Cairo. So go ahead and mark that with red is a Cairo center here. Um, this one would not be karma sector And that's because we have one Michael here and one Michael here. So we have two of the same thing again. We have an ethyl over here, Rev. According. If one of these was like a nice appropriate or Romeo or iodine or even a hydrogen, then it would be. But because it's attacked, it has two of the same structure, so a metal, it's not a couple. So now we have figured out which ones that Cairo we have to figure out which of them are optically active. So we have this one that we have to look. So is this optically active? Well, we can't really tell. Technically, yes, it would be in one way or another, but it really depends on if it's wedged, or that we could have it wedged in front like this where we could even have it on a dash going away. So but because we don't have anything to show us, which is which, the wedge or dash? We would say that it's optically inactive because we can't tell if it's gonna rotate. You know, this way or this way. That's what the whole legend dash will tell. So any time that you see a wedge or dash, that's a good option to look out for or an s to see it optically active. But we can still have Chirau Ali, even though there's no rotation show.

This question asked for the product of each of these reactions disregarding stereo ice MERS. So starting with part A, we have a terminal al keen with NBS. So we're gonna have a couple of different residents contributed that radical. So we'll start off by putting it on that a Lilic carbon right there And that will be a ch now and then we will rev innate it to put it on the end in status is what are residents will look like. We'll take the radical and one electron from a double bond to form a new double bond and leave the remaining electron behind on that, uh, end carbon to give us a another inter meter. Another radical intermediate that looks like this. And then the bro mean Congar Oh, on either one of those radicals from either of those rope in its forms. So we'll get to products. The 1st 1 will look like this from that first residence form that I drew. And the 2nd 1 will look like this from that second residence form. So there are two products for part A or part B. Again, we have NBS, um, and because Berman is so selective. We're only going to pick one side, whichever one will give us a more stable, more civil residents contributors. So I'm gonna put the radical on this end carbon over here and then when I resonated, I'll get a turkey or radical instead of a secondary radical where it's just what I would get if I put it on one of those metal groups and said so our residents form will look like this but her she really radical and the one at the end. And then again, the Bruning Congar Oh, on either one of the carbons that had radicals and our intermediates. So we'll get these two products right here. So there's the two for Part B. And then for part C, we have in our cane with br two and light. We know the bro Ming, a super duper selective. So we only have to put We're only gonna draw one product here. We're gonna put the burning on that tertiary carbon in the middle there so that product will look like this. That's the only product for C, the only major bottom for C and then for part D. Now we have a ring with steel to and light. Um, and even though seal is on as selective in this case, we only have one type of carbon. So, um, we only have one product here that looks like that for part E. Because we have sealed to without the UV light within all cane, there's something conduce You have to have, um, something to initiate the radical reaction. And we don't in this case of this one is going to be no reaction for Part E and then for part F, because now we do have the UV light. So we have this metal cycle plantain and seal to, um so here we're going. Teoh, look at each type of hydrogen we have. We know that chlorine does prefer, um, trashier over secondary or primary, but because it's not nearly selective, Hasbro mean we're going to go ahead and draw all three possible force are impossible products here because we will get all of them to some degree. So there are four possible products for a part

Let's complete the following reactions by drawing the products and indicating all stereo ice MERS that are formed for a we have cyclo hex to and own. This is gonna react with C h 32 c u l i and h 30 plus to produce If you lie summers here we would have yes, three metal cyclo ex unknown and we would have over here are three metal more space. Here are three metal. It's like low hex unknown. Those would be our two products could be again starting from cyclo pecs too and own react This with ch three m g p r and H 30 plus will produce to a spurs Here we would have our one methanol se clo hex to and all and we would have s one method cyclo, hex two and all. Most br two products for be Do it for C starting from probe io fan on and says Parola Dean reacting with he trace amounts of acid we will produce two products will produce first of all product here which is zed one dash one fennel probe one and one ill parole the dean and will also produce e 1-1 Fennel probe. Yeah, one and one ill Parola Dean, Those would be your two products for C and for D. We're starting from three. Helped unknown react This with N A, B, H four and H 30 plus produced two products. First one is s kept and three, all with second product is are helped and three Oh.

In this problem I can diabetes vaccinated. I need to see you three Plus S. or two in pageants of ash tool will give to any age S. Or three plus C. 02 This is A. And any edge S. A. Tree plus. I need to see you three will react to form To any to S. A. three plus As 2-plus co two and I need to S. 0. 3 Plus S. Will react to form Any to Ash 203. So this is being and this is C. And and to any too As two or 3 plus. I do will react to form any to ash for 46 plush to any I. So according to the option option. Beach, correct here. Oh