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Anea Flen rencllun Wuchuln Tolla TAFHTEHlat LAnEt Eaetel Hun Hlorkul ubunt Cururun 'EJne 01l4DIOLlFollowng compound- least-lavored praduct frum the given reacl...

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Anea Flen rencllun Wuchuln Tolla TAFHTEHlat LAnEt Eaetel Hun Hlorkul ubunt Cururun 'EJne 01l4DIOLlFollowng compound- least-lavored praduct frum the given reaclant; Which of the reiction that proceeds by an El mechanism"! assuming that unucteoC? Hint "forget abowt carbocation rearrangementsfound in tke given alkyl halide? How many unique hydrogens are{

Anea Flen rencllun Wuchuln Tolla TAFHTEHlat LAnEt Eaetel Hun Hlorkul ubunt Cururun 'EJne 01l4 DIOLl Followng compound- least-lavored praduct frum the given reaclant; Which of the reiction that proceeds by an El mechanism"! assuming that unucteoC? Hint "forget abowt carbocation rearrangements found in tke given alkyl halide? How many unique hydrogens are {



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How many alkyl chlorides are obtained from monochlorination of the following alkanes? Disregard stereoisomers.

Could This problem we're trying to figure out if are leaving group That is excellent. A compound will result in a compound that is Resident Lee. Stable or not, we're looking for are leaving the to be in one of two forms. The first is the A Lilic position. When are leaving Group is separated by it from a double bond by at least one other carbon. And so an example of this where leaking group exits is If we have this right here, this leaving group takes those electrons leaving a carbo cat eye on a positively charged carbon. On the end, this is going to be Resident Lee stable with its mirror image because the double bond I share some of those elections. So this would be Resident Lee Stable. We are all set. The other former looking for is the Benz Ilic position. Where are leaving group is at least two bonds away from a ring of carbon and for us when we're actually going to look at ah, negative examples. One example of this not, uh, of something that's not in a pencil, a position that you might think it is or a leaving group is only one bond away from neighboring of carbons. Now, if this leaving group exits the compound, we're left with a ring of carbons where one of the carbons not have a positive charge. It is a positively charged carbo cat ion, that is any double bond and this is just incredibly unstable. It doesn't work. We don't have stability. So for there to be resident stability, you need to have your leaving group B at least two bonds away from the carbon ring. So looking at some examples, our first example here of the something you don't think is a Lilic is where this chlorine right here is only one bond away from this double bond and this just won't work. If this chlorine left, you'd be left with a positively charged carbo cat eye on here. That's in a double bond. So it's not stable. It's not going to work. However, this glory is in the a Lilic position in just two bonds away from the nearest double bond. So when this leaves were left with a carbo cat ion, it isn't involved in a single bond. And this has residents Stability, just like we saw were here. So This is all set for Arbenz Ilic examples. This bro me is just too close to the ring of carbons, causing again a positively charged carbo cat ion that's in a double bond. This doesn't have residents stability. It's not stable at all. However, this bro me is to bonds away from the ring of carbons, meaning that when it leaves your left with resident.

Hello Today we're doing problem nine point seventy eight. This is another one of those challenging problems and states that rearrangements can occur during the dehydration of primate alcohol's, even though no primary carb Okada and his form that is a one to shift can occur as the carbon pro nated water bond is broken, forming a more stable secondary tertiary Kara Kara as shown in the question just them. So using this information draw the stepwise mechanism that shows how our starting material here, this beautiful group, can be dehydrated with sulfuric acid to yield our elimination products has shown here. Oh, so it says also that we will see other types of examples of this in future chapter. So this is very useful thing to learn. So just like before, I like making my strong acid h two s o for into h ages, for simplistic sakes. So a jade associations h plus and a minus and solution, and you need identify what nickel follows so we know that the alcohol is going to be our nuclear file because of the lush on rich properties of oxygen. Will pick up that proton to form pro nated water now the question stem is saying that although no primary carbo cattiness formed, we could get some sort of hydride shift occurring while water breaks. But before we do that, let's first form our first product here on our left for the continent. Base a minus. And we did identify the beta carbon beta proton so beta being read adjacent to his Arab ADA. Protons so easily weaken reformer acid catalyst by picking up proton When we make a bonding break the bond and the bond that breaks a Sigma bond making a new Al Kane making a bond there. We gotta break the bond between this water shield us our product one and water. Alternatively, what can happen is that from our pro nated water step, the questions stem is stating that we can have some sort of shift occur in which these beta protons one of these can just spontaneously shift an attack. Remember, Hydride isn't negatively charged Proton, so it's actually those two electrons in its signal bond that are doing the attacking to form. And you bond here. And when you make a buying your break upon so water gets kicked out on what that feels for you is now a stabilized second car broke down with water. So here, obviously we form the cyst. Trance, I summers. The reason for that is that we have a Sigma bond which can freely rotate. So in some time of space, this molecule exists like this. In other times, it will exist like this. So because of that, we know that once we deep probe Nate, those beta protons without conjugated base, we will form the cyst and trans ice. Immers, you make a bond, break a bond and you form your AL keen to neutralize your car broke down, which yields your assist. Trans ice summer. Or I should say transit. I summers actually.

Let's take a look at this molecule and try to determine what Al keen we can use and what re agent we can use to synthesize this molecule. So one key way we can do this is let's look at what substitute Wint's we have on our main chain, which is our cyclo hexane. We have a chlorine and an alcohol, which leads us to believe that well, we started with a Al cane and we added chlorine and used water as a solvent to produce this molecule. Now what is this Al King? Then we can remove that plus because I already have the chlorine on the arrow. If we try to backtrack our way here, we know that the chlorine is on this carbon and the alcohol's on this carbon. Which leads us to believe that the double bond is in between those carbons. So, in fact, our re agent will be just cyclo Hegg scene and that reacted with chlorine underwater as they're solvent would give us this product. Let's take a look at this example Now we have in our keen and we add something to make this molecule. Now it is extremely difficult to well add a alcohol or a more carbons two. A molecule from an Al Keen is extremely difficult. Thio undergo this procedure. So that leads us to believe that our Al Cain had the methyl group already there and we're just adding the alcohol, which means that we started with this Al Keen Whoops didn't mean to redraw that we started with this how keen and added a free agent to produce this alcohol. But what re agent would we use? Well, there are two different methods we can go about with making this alcohol. We can use, uh, Mercure ation and Dimmer curation. Or we can use our boron re agent boron hydride re agent. Now the question is, where do we want the alcohol? Well, if we used Mercury, we would end up with an alcohol in the Markov nick aww position, which is a tertiary alcohol, because that's the most stable carbon for it. But if we used Boron, we would get the anti mark column recovery agent. And if you look at the product, it's on the anti Markov nick off re agent or the anti Marconnet called Product is what we're creating. Therefore, we need to use our boron complex whoops and that will produce our product

Kill everyone. Today we're doing Chapter fourteen Problem six. And this from asked us how many Proton and Morrissey knows what we expect from each of the following compounds. So remember you're NMR signals will depend on how many unique, chemically distinct protons exist in the molecule. So if you determine the number of protons and where where they are located in the first example, you know that terminal methyl has three protons. This center here has one full time going down on DH. This proton this Carmen here has two protons. Mr. Mill Metal has three. So we also know that, um, Proton that come off off ch tools. These two protons can be chemically distinct if the molecule contains the stairs. Jannik centre stairs Jannik center means a Carl center. Carl Center is a carbon that has for you. Next, recipients bound to it. So we know that this carbon here is a stale carbon. So therefore most like you and me do. Our analysis will determine that these two protons are actually chemically distinct and exists in separate micro environments. Bill, go on, start from one side of molecule working away two next to determine harmony complete listing protons we have. So you start from the left side. So on the left side received that we have an A Carl Carbon here found three protons and these three protons. You can say that they are free into rotates about the single bond. So we can assume that these will all be chemically identical to one another there for producing one signal in the H and M R spectrum. Next, we move onto our Kyle center and is constantly we have a chlorine and a proton. And obviously we know that this proton, being in his own unique chemical environments, will produce a separate, unique chemical shift signaling our anymore spectrum. Now let's look at the proton adjacent to it, which are the siege? Two protons. So if we see here, we see that the upper proton assist to his electoral negative chlorine atom. While the bottom proton is Trans. Does Elektra native corn? Adama, What we What we know about Electra later Adams is that they withdraw electron density it away from your hydrant nucleus. So it actually makes your haijun less shielded. And when something is less shielded, we know that it will require a lower amount of energy to cause that spin, flip and residents So it will be downfield in our NMR spectra, however, something that is transferred only further away. So it's going to be feeling less of that electron density pulling effect from this election. Negative, Adam. So they'LL be more electrons. Ah, isolated and localized around the hydrogen nucleus, therefore causing ah higher amounts of energy to cause that spin flip So is going to be more shielded to the right side of our spectrum. So we know that the top proton must be chemically nick from the bottom proton. And then once again, we know that these three protons can spin around its signal bond, generating all one signals we have one, two, three, four, five unique protons in this example a small fee. Look at the second example we see that this carbon here is a stair genic center and we have a siege to hear. There's a good chance in these two protons will unique once again. Before we do that, let's just draw out all the protons we have in this molecule. So we know that typically terminal terminal carbons with their three protons can spin around the signal bond generating some sort of blurred effect and generating just one signal for these three protons as thes will all be equivalent and well can be considered identical. So you have one terminal, one terminal carbon here and one terminal carbon here. So we will have two signals from those six full terms And these two differ from one another. But they will be identical within the same carbon we'LL see if he moved down to this carbon here we see that this proton is in its own unique little environment All myself here and once again these protons off this carbon received at this card this proton here, Sister Electra, native chlorine and also to our method group here. However, this proton here is trans tr Electra Negative, Corrine, and also trance to our method group here. So we know that these two must exist also in is unique environment. So once again we have one, two, three, four, five You need any more signals from five unique protons. So now let's investigate these ones here. So if I were to draw out all my protons identify that this carbon here is It's Theo Genic center and we have a one siege to here and one siege to here. So these four protons most likely will be chemically distinct. But let's just work through it. Confirmed that so once again. And what we know is that these terminal carbons, these method groups, at the very end they produce one signal for those three protons. So weakness right away. See if there's anyone signal this is gonna be one signal. That's because of this free rotation and kind of blurring effect you see, with those three protons. We also know that the proton instead genetic center will be unique sense on little micro environment. Right here. I will also see that we have siege to religious interests. Theragenics Center. So we know the top Proton assist to our left from negative. Adam, Hey lied And our bottom put on his trance. So these two protons once again must be unique. But now what about the second siege? To hear this speech two is now two carbons away. Ferrar bro mean while this siege to we just finished was only one carbon away from this problem. So clearly this ch to that we're investigating now is going to be in a unique environment than the siege to those two protons we just finished analyzing. So we see that the once again, these the siege to this proton on top is two carbons away from being cyst to Romi and this proton our bottoms, two carbons away from being Trans first bombing. So these tomb once again must be unique and its own unique chemical environment. So, overall, what we see is that we have one, two, three, four, five, six, seven unique and more signals from seven unique protons.


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