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3) (12 Marks) Draw the product(s) produced (only constitutional isomers, disregard stereoisomers) when the following substances are treated with NBS and heat b)...

Question

3) (12 Marks) Draw the product(s) produced (only constitutional isomers, disregard stereoisomers) when the following substances are treated with NBS and heat b)

3) (12 Marks) Draw the product(s) produced (only constitutional isomers, disregard stereoisomers) when the following substances are treated with NBS and heat b)



Answers

(a) Draw all products formed by treatment of each dibromide ($A$ and $B$) with one equivalent of NaNH$_2$. (b) Label pairs of diastereomers and constitutional isomers.

This question tells us that these two products are formed from the reaction of methylene cycle Heck sane with NBS and they're asking us to show how each of these is formed disregarding stare ice MERS. So this is our starting material methylene cycle Heche saying, um and with NBS, we know that we have a a radical will look radical that will then resonate to another position. So this is our 1st 1 that secondary a little graphical, and we will resonate this, um to get the 2nd 1 But we say right here and just add that roaming onto the radical that gives us that first product on. Then, to get the 2nd 1 we will take the radical electron and one from the double bond like this to give us a new a double blonde right there. And then the other electron from a double bond would go up on the bat metal carbon on and then are roaming will end up there so that residents form a zoo radical. Looks like this with the radical on the metal and then back of those a second product. Once the grooming add on to that uh, Mekele radical

Okay, so I want to give the Arquin that created the following products. And in the first one we have an 11 carbon chain. So on carbon two and three we have key tunes and on robert eight maybe key town as well. So who's the outer hood on carbon 11 11, 10, 98 So, um, we know because we only have one product, we started from a ring structure and broke it open. So one possibility is a 10 carbon ring. And in order for that to get a kitten on this car would have to have a metal group. So if we start with a 10 carbon ring, so you can draw cycle vaccine and then just erase the middle part. And if we put the are keen right here and the methyl group right here to speak carbon one and this will be 234 5678 9, 10, 11. Okay, so this is going to be 11 will be the AL to hide when treated with three and um dimethyl so far, so carbon three needs cartoon and um some bicarbonate to. So if you breath said we'll get the products okay for B. We have we have a nine carbon sheen with formaldehyde as a side product or a nine carbon out of 123 456789 And um we have a key tune on seven and five. So the fact that we have two products means that we must have started from one structure as opposed to a rain, like one straight chain, arcane, graphene. So we're gonna draw in total, we have um uh 10 carbons said nine of the chain and one here. So drawing that first 579 10 and we need nine carbons, so speak carbon nine. And we put the ark in right here near the carbon eight was the key to 75. Hold on. So, Okay, so carbon are benign. Is the outer hood and 123456789 10. Okay, so Carbon five would be the cartoon And same with Carbon three. Yeah, that structure.

This is the answer to Chapter 15. Problem number 30 Fromthe Smith Organic chemistry. Textbook on this problem is asking us to consider two molecules and to draw all the possible ice summers from mono coronation of each molecule on, then asking us to determine what the major product of domination would be for each molecule. And so, firstly, to draw all the coronation products of Ana chlorination product. Okay, well, so actually first. So these molecules are given in the book as ball and stick models on DH. Obviously it's it's quite hard to draw a ball and stick models. So the first thing that I did was to rewrite each molecule in just line Ah, typical line notation because it makes them a little easier to see and therefore easier to work with, in my opinion. And then you just go through and anywhere that a hydrogen khun be substituted for a chlorine. You do so so substitution on any of these three method groups would be identical products. So you only draw at once on. Then you just go to the next carbon, which is this co ordinary carbon. So it doesn't have any heart surgeons, so you can't coronate there. So then you go to the next carbon, which is this secondary. So you coordinate their proceed to the next carbon also secondary coordinate Their are the next. Carbon is tertiary, so you can add a core ng there on then. Lastly, either of these two method groups would give identical products as well. So again you just draw one on. So that is part A for molecule, eh? Part A from molecule B. You do the same thing. So I started at this metal group coronate there you can coronate at this tertiary carbon in the ring on DH, then coronate either of the secondary carbons in the ring. They're going to give again the exact same molecules. So you just want one on, then this carbon in the top right corner of the ring as I've drawn it, his coronary. And so there's no Hodgins to substitute for Corinne's there. So then again, either of these two method groups would give identical products. So you just draw one, and that's part A for molecule B on then, to complete part B, we just need to remember that a tertiary position is going to best stabilizer radical and therefore is going to most readily reacts under these conditions. Ah, and so each of these molecules has a tertiary position in them. And so for molecule, eh? It's here in the isopropyl group a tte the right end of the molecule. So that's where the bro meet will go. Ah, and then for B. It's where this Method group connects to the ring. That's a tertiary position on the ring on. And so that's where the romaine will go and be on. So that is the answer to Chapter 15 problem number.

This is the answer to Chapter 15. Problem number five from this Smith organic chemistry textbook On this problem says compounds A and B are ice. Summer's having molecular formula C five h 12 heating A with chlorine gas gives a single product of mono halogen ation, whereas heating be under the same conditions. Forms three constitutional I summers. What are the structures of A and B? Okay, um and so perhaps the best way to get started is to calculate degrees of on saturation or hydrogen deficiency index. Um, and so this is gonna be two times the number of carbons plus to minus the number of hydrogen sze on and then all of that over too. Ah, and so we get 10 plus two is 12 minus 12 0 So there are zero degrees of on saturation here. Um, so we just have al canes. They're both out canes. Ah, and and I think it's good to start with an HD I whenever you're asked to ah, come up with a structure given formula because it lets you know, um, a pretty big chunk of information about what is going to be in, uh, you're molecule. So we're told that A is gonna give us a single amano halogen nation product, whereas B is going to give us a mixture of three products. And so, for a we need to come up with a structure that is going to allow only a single product to be formed. Ah, and so the best way to do this is probably going to be like this. So if we make this our C five age 12 um uh, when we halogen ate this, uh, and and get a mano halogen nation product Ah, we are only going to get halogen ation, um, at one of these metal groups so it could be it will be at at any of the metal groups. But since they're all equivalent, we're only going to get this single product. So that's a ah, and then for be, um, we can actually just draw the un branched al cane. So there's b, um, and to check, you know, when we do a mano halogen nation on this, we will get three possible products so we can get Terminal Halogen nation, uh, or at the next carbon in, uh, or at the center carbon. Okay. And so that's that's our three products there. Um, yeah, and so that's That's the way to approach this problem. B is pretty, pretty self evident once you know that there's no unsaturated in in the molecule anywhere. Um, the UN branched five carbon Al Kane. Well, we'll give you the answer that you need. Ah, and then for a we just need to think of how we can draw this out cane in such a way that wherever the coronation occurs, the products are all going to be the same. And that's the answer to Chapter 15. Problem number five.


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