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Consider the reaction below to answer the following question(s) cooh FeClz Clz PRODUCTThe nucleophile in the reaction indicated by letter2. The Lewis acid catalyst ...

Question

Consider the reaction below to answer the following question(s) cooh FeClz Clz PRODUCTThe nucleophile in the reaction indicated by letter2. The Lewis acid catalyst in the reaction is indicated by letter Draw the structure of product _

Consider the reaction below to answer the following question(s) cooh FeClz Clz PRODUCT The nucleophile in the reaction indicated by letter 2. The Lewis acid catalyst in the reaction is indicated by letter Draw the structure of product _



Answers

Draw the structure of the ester you would obtain by acid-catalyzed reaction of the following carboxylic acid with 2-propanol:

Okay, so in this problem, we are given the reaction of CIS tube you teen plus each to. So we're gonna draw our reaction a double bond here, plus each to, and we know the double bond the H two is gonna add to both sides. So we're simply going to get this you teen product. And this beauty chain is a Cairo because on no carbon in this product, you're gonna have four different groups. So we know that butane is a Cairo. And now an ice immer of butane jarred out. We could move a medical group and just come up with this. So here is a nice summer of beauty mean, and there you have.

So what are we going to get if we react this carbon eel? They're beautiful. I mean, so of course this nice region is going to attack. We're going, Tio, lose water and you'll end up for replacing that oxygen with a nitrogen carbon double bond. And you'll end with this. I mean, he can generate an mean only when you have a primary a mean and he can lose two of these hydrogen tze with this oxygen in the case of water. But if we were to react with Dia Philomene, then of course, this is going to react, but you'll only lose hydroxide and you'll be left with this The mini um ion, Of course, it's a mini, um ion is not very stable at all, So we'll go through a tot memorization to generate this in. I mean so again. When you have secondary, it means you'll generate the Inamine we have primary. It means you'll generate the mean And if you have a tertiary, I mean, you'll have no reaction at all. Next we have this and hydride, I knew if we react this with our beetle mean, then we'll just generate our Amit. Yeah, will lose this acid and generates this. Um, I'd In this case, the same thing can occur with our secondary. I mean, because again, we're only needing to lose one of our hydrogen atoms. So will generate a tertiary Hamid and anhydride has the same reactivity as an easel Chloride. It's, of course, where their beetle mean and this pencil chloride, we'll simply generate the secondary and lied. And if we use the awful mean will generate the tertiary, Um, I'd

The first part of this problem assets to draw the mechanism of the reaction of tube you tine with one equivalent of beer to. So here's our to view time. So the first thing that will happen is let's make your to like that. So the first thing will happen is one of these pi bonds will come and attack the bro mean? And then the bro me in one of the lone pairs on the Berman will actually come and attack one of the carbons as well. And we'll kick off that second burning. So we get a transition state that looks like this. I only have two bonds there. We used one of the pie bonds to attack. The brewing is attached to both carbons, and it's positive because it has two bonds. We also have this br minus leftover. Um, now the br minus going to come in and attack one side. Um, but because this PR pyramid transition state hurt, not transitions. Intermediate is so big, it has to come in from the opposite side. It cannot attack from the same side that this bromate is on because there's no room. So this bro means on the bottom. This one has to come in from the top and attack one carbon. And then whichever one it didn't attack. Uh, we'll get the other brought me that was already there. So our products will look like this. So if we put this roaming on the bottom, this one has to be on the top now, the more proper way to draw that for an all keen would look like this. So roaming here, rooming here. And then we have our Russell groups on the second part of the question asked us to predict the configuration of the product. So basically, what they're asking is is this year see, because it's an Al Kane s o E means it stands for end gegen, which is the German word that means away from. So it's basically Trans and, uh, Z is what the German work Suzanne man, which means together is cysts. So it's easier for you to think of it like that. Um, this, in this case, the bro means are trans to each other or e. Um, so we know that because if we look at this line right here, this plane that the double bond is on, they're on opposite sides of the plane. Um, so this is and e configuration

This is the answer to chapter to problem number 65 from the Smith Organic Chemistry textbook. Uh, in this problem gives us five reactions and asks us to label the nuclear file and Elektra file and each reaction on then to draw the products. So I'm gonna go through and label nuclear files and electric files first on and then go back through and draw the products of each reaction. Um, and so I remember that Louis bases our pardon me. Louis acids are also Electra files on Louis bases when they react with something other than just a proton. Um, our nuclear files, Um and so looking at all of these, except for D, what we see is that these air of the form of a central Adam, often a metal with three halogen, is around it. I'm so, for example, aluminum. Try chloride iron, try bromide, um, and pour on, try fluoride. And so you should get used to seeing these things are common Louis acids that you'll encounter a lot throughout organic chemistry. And so definitely definition aly definition. Aly Yeah. Okay. So by definition, according to what I just said, Luis acids are going to be electrify. Lt's so we can safely label all of these very apparent Louis assets as Electra files. And this makes sense because the definition of a Louis acid is that, uh, it wants to accept a pair of electrons. And so, um, electro file means electro loving or electron loving on DSO, and they sense it's something that wants to accept a pair of electrons Would be an Electra servile so we can label of each of these as the electric file. Ah, I'll skip D um, briefly and label e. And then, since since we've labeled the Electra file obviously the the other starting material must be nuclear file. Ah, and we can see that all of these nuclear files have electrons that they are willing to donate. So whether it's bro mean or an oxygen or the sulfur, each of these have electrons that they are ready to donate to the electron except er And so, um, that can actually help us go back and decide which is which. In de eso, the water has electrons on the oxygen that it can donate. Um, and then the cycle Oh, heck, sane, um, with a positive charge. Uh, can accept electrons in that unfilled valence shell, so we can now label that as well, so electrify I'll and water will be our nuclear file. Okay, um and so next we just need to draw the products of each of these reactions. Um and so remember, um, in every case, it's going to be the electrons that can be donated, being donated to where they could be donated to, um, to form new bond. So in each case, the electrons are are going to basically do this. We call it a nuclear filic attack. Okay? And so then our products are going I to just be our two starting materials bound together at the positions at the electron movement indicated. And then, of course, we do need to keep in mind that we're going to be creating charges. So our nuclear file was gonna end with a positive charge, and our electric file will end Ah, with a negative charge seated on it. And so, at this point, it becomes pretty much just so in the blank. Almost so. A l c l three. Then we can put our loan pair and our charges. So see is gonna be exactly the same. Put in our charges. Um and so actually, for D since Indy, we started with Ah, positive charge. Um, so one thing that I had not brought up yet I'm is that the overall charge state of the starting materials and products is the same. So there's no charge in the starting materials for the 1st 3 parts of this problem. Um, and then one positive and one negative charge in the products. I mean, so the overall charge status zero in the products as well, Because the positive and negative cancel one another Indy, we start with a positive charge. And so, in order to keep the overall charge state the same, we have to finish with the positive charge. And so we don't this time have a native charge on our, uh, the what was the electric file, So I hope that makes sense. Ah, and then sew for D. You're right. Yes. So for e is going to look like this. Um, And then again, we need to account for our charges. So this, bro mean will have a positive charge, and this iron will have a negative charge. Ah, and so that's it. um, so remember, um, Louis acids are electric files. Um, also, um, something with it with an unfilled valence shell is going to be an electric file on nuclear files. Always have, Uh, maybe not. Not always. But in this problem, the nuclear files always have a pair of electrons which they're willing to donate. Um, and so that's the answer to chapter to problem number 65.


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