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20) Draw the /.3-diaxial interactions of the given unstable compound:H;c ~C-CHa H H;c? 21 kJmolHsc H;c Hyc...

Question

20) Draw the /.3-diaxial interactions of the given unstable compound:H;c ~C-CHa H H;c? 21 kJmolHsc H;c Hyc

20) Draw the /.3-diaxial interactions of the given unstable compound: H;c ~C-CHa H H;c? 21 kJmol Hsc H;c Hyc



Answers

Draw the structure of a compound with molecular $C_{8} H_{14}$ that reacts with one equivalent of $H_{2}$ over $P d / C$ to form a meso compound.

Here I have to. Molecular molecules of CEO. So here's one carbon oxygen. Here's the 2nd 1 And both of them we see a partial positive charge on the carbon in partial negative charge on the oxygen in this interaction you see, Michael diaper bonds here between the carpet and the oxygen, we wouldn't put them carbon carbon extraction parallel because of the charges we had to oxygen down here at the bottom, they repel each other, but instead the part of positives pull themselves together.

This is the answer to Chapter 11. Problem number 21 Fromthe Smith organic chemistry. Textbook on this problem says, draw the products formed when this is satellite eye on, uh, that were given reacts with each compound and so were given six compounds. Um, OK. And so, uh, the first compound here isn't alcohol. Hey, lied. It's a primary outfield. Daylight. And so this is just going to do, um, pretty much what we would expect in this situation. Ah, and so the satellite ion will act as a nuclear file. Ah, and displaced that, bro. Mean And we will end up with these two carbon chains linked together on DSO. Our product will be this okay, And so that's a B is gonna be exactly the same. It's a different starting material, so the product won't be exactly the same. Um, but the reaction will be exactly the same. Um, the satellite ion will act as a nuclear file and displaced the chlorine. And so we will get these chains just linked together. So there's our answer for B. So then, for C sees a little bit of a trick question. I guess we've talked about this in previous problems. A tertiary alcohol, Hal, I'd which is what we haven't see. I's gonna be two hysterically hindered to undergo this nuclear Filic substitution reaction. Ah, And so instead, what we're going to get, uh, is elimination. And so, um, our satellite eye on will grab a proton so our satellite eye on will, uh, just become Well, it won't be a settling. I guess it'll be pro Pauline. Pro Pine. I see. Well, this will be the product or, ah, beauty. I'm rather So we'll get this from our satellite ion. Um, And then we will get, Ah, the, um, elimination products will get in Al Keen from our starting material. So we'll get this and we will get this ch three ch to to see double Bond ch ch three. So there we go. That's the elimination product. They're moving on to the second page. Looking at D um D is an alcohol hail ID, but it also has an alcohol in it. Ah. And so what's going to happen in D is that we will get, um, de pro nation of that alcohol. Um and then the alc oxide will complex with sodium ion so well get this, Uh, and then we'll get our starting material back, but deep throat nated and complex with the sodium ion. So, like this. Okay, s o n e and F or a pock sides. And so each of these is gonna be treated with our re agent, followed by water. Um And so, um, our ah satellite eye on, um is going to act as a nuclear file and open the epoxy side at the less substituted side. Ah, And then the water will serve to protein eight hour opened up oxide. And so our product for E is going to look like this. Okay, so that'll be our product for E on. And then our products for f will look very similar. The only difference is we have that metal group substituted onto the starting up oxide. Ah. And so our product is gonna have to reflect that. And well, s o, our product will look like this, uh, with the alcohol here. Okay. Ah, and so that's the answer to this problem. We basically just need to know, um, what kinds of reactions this re agent can do on how this region is gonna be limited for example in part. See, we had a tertiary alcohol. Hey, lied in Part D. We had an alcohol. So there's just gonna be a proton exchange on DS. Starting material will be deep throat nated and well complex with sodium ion. Ah. And then we also offer e and F need to know that this re agent can function Is nuclear filed open up oxides. And that's the answer to Chapter 11. Problem number 21.

So here we are, just looking at some chemical transformations where we have the final product as well as the re agents, and we are needing to work backwards in order to deduce the formula on the structure off are starting materials. So our formula being C five h 10 was listed up up over the top of the screen, as you can see there. And so for each chemical transformation 123 I've drawn out my starting material on this is the starting material that reacts with R B R. Two. Now the site of all reactivity is at that double bond where were simply adding a bro mean to each end of our double bond, and this trend is present in all three of our chemical transformations.

So here we are looking to draw the structure for the cyclo Al Keen with the molecular formula C six h 10 that reacts with C L two to give each of the compounds up on the right of the screen. So, as you can see in place of the two chlorine molecules, I've just drawn out a double bond in between them. This stands for all of my examples. So this is our first example where we form the compound that has gained two chlorine molecules, followed by three more AL keen cyclical molecules that go on to react with CEO to in order to generate the products.


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