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2} Draw all hkely aubetttutlon and elmlnatlon products In the followlng reactlon Label the produdts & SNI, SNZ,EL, or E2 Label the major productNeCHacdolleCkcle...

Question

2} Draw all hkely aubetttutlon and elmlnatlon products In the followlng reactlon Label the produdts & SNI, SNZ,EL, or E2 Label the major productNeCHacdolleCkcle the moct ctable alkene below

2} Draw all hkely aubetttutlon and elmlnatlon products In the followlng reactlon Label the produdts & SNI, SNZ,EL, or E2 Label the major product NeCH acdolle Ckcle the moct ctable alkene below



Answers

Draw structural formulas for the alkenes formed by the acid-catalyzed dehydration of each alcohol. For each part, predict which alkene will be the major product. (a) $2-$ Methyl-2-butanol (b) $1-$ Methylcyclopentanol

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.

So here we have just drawn out the structures off our Al Keane's, which is our starting material. Eyes ableto undergo are acid catalyzed hydration, which gives each of the following alcohols as the major product. So estates it in the paragraph of the top. Well, taking it out Kane and then we are most likely using water as well as an asset. So h two s 04 to generate a single bond that then has an alcohol group and then a proton attached where the proton will come from. Our asset on the alcohol group will come from our water so below I've drawn out a Siris of starting materials that would allow us to access the products that have been listed. So for some of the examples, you've got more than one starting materials. However, it would generate the exact same product containing alcohol

This is the answer to Chapter 10. Problem number 29 fromthe smith Organic chemistry. Textbook on this problem asks us to draw the products formed when each Al Kane is treated with B H three, followed by, ah, hydrogen peroxide, um, and base. Ah, and were told to include the stereo chemistry at all stereo genic centers on. And so I realized I actually forgot to include the base in each of these, so Oh, h minus which minus and oh, each minus. Okay, so there we go. All right. And so the thing to remember here is that this is effectively the anti Markov Markov addition of an alcohol and a hydrogen across the double bond on and it's it's anti Markov. Markov s o. The hydrogen is going to go where there are fewer hydrogen sze in the alcohol will end up our there or more hydrogen sze in this starting material. Um and so this this problem doesn't ask us anything about mechanism, but you should bear in mind that the boron is gonna gonna add first, um, and then the hydrogen peroxide and bass are going to replace the boron with the alcohol, but we don't need to draw any of those intermediates s o for a. Our answer is going to look like this ch three ch two, ch two because the alcohol went to where there were fewer alcohols. Um, and then C h 20 h. So we're making the primary alcohol there. Okay, um, and there's no ah, no stereo chemistry to account for here. Um, since there's no stereo centers, Okay, so then looking at B same same idea. The hydrogen will go where there are fewer hydrogen sze, um, and the alcohol will end up on the carbon that had more hydrogen is to start with. So, um, in this case, what that means is that the alcohol was going to add to the bottom carbon here, Um, and so the alcohol can come from one side to give us this. Ah, And, um, this is a sin addition. Um, so we then need to put our hydrogen on the wedges. Well, and that's gonna put this ever group on the dash here. Okay, um and so we need to draw the other possible stereo I summer that could be formed here as well. Um, and that is going to be, um, same basic structure, but we need to put the alcohol in the hydrogen on dashes and the method or Sorry, the Ethel Group will therefore be on wedge ch two ch three age. Ohh. Okay. And then lastly for part C of this problem, Um, same kind of idea, as in B. So the orientation of this method group is not gonna change. Of course, it's not involved in this reaction. Um, and then we can have, um our oxygen, which will add to this carbon with the most hydrogen, is to begin with so we can have our oxygen and our hydrogen on wedges, which will put our method group on a dash there. Or we can have the opposite configuration. So again, this metal group is unaffected. It's not participating in this reaction. S o. The obviously configuration would be. This metal group is now wedged Hydrogen on the dash and the O. H on the dash. Okay. Ah, and so that's the answer here. We basically just need to remember that Ah, hydra operation oxidation is a two step process. It adds an alcohol in a hydrogen across double bond. On that, it's an anti Markov Markov process. Yeah, and that's the answer to Chapter 10. Problem

Here we're just looking to draw the structure of an Al Keen that undergoes are acid catalyzed hydration to give our indicated alcohol products so that Al Keane's are listed over on the left on our alcohol products are listed on the right. So, as you can see in each example, we have re activity at the site off the double bond. And that generates our alcohol product, where we have the addition of a proton to one end off the double bond on the addition off a hydroxy group to the other end of the double bond, which gives us our alcohol functional group in the final product. So as you can see that in some cases we have multiple starting materials. However, most of the time, due to Markov niqabs rule, we do too only generate the one major product


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