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2. Predict the product for the following transformations (show proper stereochemistry)_MePdIC, HzMeMePdIC , HzPh...

Question

2. Predict the product for the following transformations (show proper stereochemistry)_MePdIC, HzMeMePdIC , HzPh

2. Predict the product for the following transformations (show proper stereochemistry)_ Me PdIC, Hz Me Me PdIC , Hz Ph



Answers

Write the expected products of the acid hydrolysis of the following tetrapeptide:

Okay. This problem is asking us to predict the products given our rackets and re agents. So let's zoom into the 1st 1 and start. Okay, so we're given a tertiary alcohol and HBR. So how do we know that this is a tertiary alcohol? Well, if we look at our alcohol, the carbon in which it is attached to is attached to 123 other carbons. Those three carbons are representative of the tertiary nature of the alcohol. So why is this important? It is important because let's say if this alcohol left, what would happen if the alcohol left? Then we would be left with a tertiary carbo cut on tertiary carpet. Ketones are considered stable. At least they're more stable than a secondary and much, much more stable than a primary carpet cut on. So if the alcohol were to leave and we were left with a positive carbon Calyon, then we can have a nuclear file come in. And that is exactly what happens in this reaction. So first step that happens is the alcohol gets promoted by the aesthetic hydrogen on HBR that becomes a water and then that water is going to leave because water is much better leaving group than a alcohol by itself. And then we're left with a tertiary carbon cut on. Then my bromide ion, created by the deportation of HBR can come in and attack that carbon that has theater sharecropper cut on. So our product should be as such. We have our cycle Hexen with metal group and then we just replace our alcohol with a bro Me. Okay, so that should be the 1st 1 for the 2nd 1 We have a primary alcohol reacting with eso seal to and the reason we know if the primary alcohol is because the alcohol a carbon. So the alcohol the carbon in which are alcohol is attached to is only attached to one other carbon. And this is important because if we have a primary alcohol and the alcohol were to leave, then we would be left with a primary carbo cut on. We know that primary carbon Catalans are not stable. In fact, they're much less stable than a secondary cover cut on and much, much less stable than a tertiary. Carbon got on. So we cannot have that tertiary. Sorry, we cannot have that primary car because I am by itself so we would be prone to rearrangement to create either a secondary or tertiary car broke down. In this case, because we're racking with eso CEO to associate to is a mild reactant. It will not create conditions in which a primary carbon cotton will exist. So we will just be left will not create conditions in which rearrangement Wilker, in which we will get a secondary or tertiary car, broke down and said we'll just be left with a Corinne replacing the alcohol. Okay, so next up we have NBS reacting with this molecule right here. So this thing that is special that it with NBS reacting with an Al Keen is that it adds a bro Ming to the Olympic position relative to the cave by like I mean, let's say that we have our cocaine and by the way, this this cocaine is part of this cycle Hexen structure and this cycle hexane structure, it is not only associated with this one, so that could be confusing. But as far as the a little carbon we're dealing with one carbon away. So here is one end of my cocaine one carbon away will be right here. Here's the other animal Keen. The one Carbon away would be right here and same thing on this side. So one cardinal from this molecule from this autumn would be right here and one carbon away from this carbon would be right here. So these four circles are representative of the Olympic position. We have four different Olympic positions. But as far as writing out which one is the product of this reaction, we only have to represent one of them. So look to that Al King is right here. Okay? And again, I could have drawn my growing on this position, this position or this position. But I have to do or I can't do. And because we're only adding one broke me. Okay, so that is one potential product. But another potential product is the minor product. So, for example, let's say I have this one in blue. Actually, let's say I have the same molecule. Okay, so this thing that especially the NBS, is that we're prone to rearrangement of the Al Keen, for example, in the mechanism, I have a hydrogen here that hydrogen is attached to the carbon, the Olympic carbon so that hydrogen is going to get the protein in the sense leaving a radical on that carbon. So what I just drew there was a fish Coqueiro that push Fisher Caro is representative of one electron moving. So if I push one electron onto that specific carbon, then I will be left with this product with a radical there. So what's special about the NBS is that we can rearrange the molecule so that we can draw a resident structure so that we can add a roaming to a different carbon, for example. So I'll do this part in orange if I move this electron around and move it onto this single bond simultaneously moving an electron from this double bond. If I move this electron and electron right here, I'm moving to electrons cumulatively, so two electrons moving onto a single bond will create a double bond. And I can't just leave the elect the other electron from this double blind alone. So I just have to move it onto this carbon represented by a fish aguerro. So after that, I should end up with this where I moved my double bond onto this single bond and I moved my radical onto this carbon So in the mechanism by connecting my radicals together, in which I have a roaming with a radical here. If I attached my bro mean to my radical my carbon, it'll connect making the other product like this. I have my cocaine here and then attached to that carbon. I have a bro me. So as far as which one is major in which one is minor, this would be major because the radical for this one exists right here. And this radical is the one that will attack Abe roaming to create my product, which I do right here. So that radical is much more available than if it were on this one. Which is why I end up with this molecule as my major product. Okay, so next up I have a secondary alcohol directly with PBR three and again we know a secondary alcohol because the carbon in which my alcohols attached to is connected to 12 other carbons. So PBR three is a mild reactant. It will not create conditions in which a tertiary carbon combine will be the result of a rearrangement in the car piccata. So all I would have to do is replace my alcohol with a bro Mean. And this would be my product. Okay, next up, this is a grenade synthesis and then later protein ation. So here I have my al que bromide, that alcohol bromide is going to react with magnesium in an ether solvent, creating a greener. And what I'm just doing here is essentially sandwiching a magnesium ion. Sorry, Mechanized magnesium between a roaming and a carbon. Okay, that created a green erred. And now we know that Griner czar bury nuclear filic and very basic. But in this a particular problem, my water is not considered electro filic. Or at least not in the sense that I can attack it with a nuclear file. So instead, I'm going to do a acid base reaction. And just to make this a little bit more understandable, another way to represent a grin erred is like this where I have lone pairs on that carbon and then a magnesium bromide ion to the side. That is an an ionic bond. This carbon is going to attack a hydrogen on a water. Okay, so that should end up with just regular old beauty okay. And I'm going to represent the hydrogen just so we know what happened. Okay, so that is a product of that reaction. As for this one and this one, we are creating a Gilman re agent. And by doing that, I recognize that I have a roaming that Brahman is a good leaving group, and then I'm going to replace it with a lithium, so I'll end up with this. Okay, So as for the next step, I believe that this product is supposed to be in two equivalence. So if if I ended with two local events, that would mean that I would have for lithium and two of these reactions. So if I have two of these products and I'm going to erect it with this See you, I Then I will end up with a Gilman regent, which is just, Let's see, 1234 carbons. So I'm gonna write ch three ch two ch two ch. Three and again, I have two of those someone who represent that is to And then I just use copper lithium. So that should be my product for that one that generated generated a Gilman regent. OK, as for the next one. I have a Gilman regent right here, and I'm going to react it with this alcohol bromide. So what Gilman regions also do is they essentially bond carbons? Two other carbons. So I'm going to use this method group and attach it to this carbon. And in that process, I will release that, Romy. So I should end up with this, right? We just added a carbon carbon bond. Okay, so that should be the end of it.

Where did the border off the falling 55 reaction until if the process is superficial or enter official. So this signal bond breaks and they're five happens on the right in five petals on the left, which is right. It's 55 and they form it, you borrow. And that's a newborn, which was for Let's go in the Nichols, True for sex it then then the Littles. That's, Ah, dramatic transition state. So the reaction will not be introducing any additional Lords toe homer, so it's, ah, superficial.

Okay, um, in this problem, we are talking about the mechanisms and predicting the products of alkaline reactions. Specifically, in this case, we are going to be reacting al canes with hell agents, and that's we're going to be going over today, and we're also going to be using stereo chemistry. So for part A, we have this sigh click structure that has an elk in, and we are reacting it with diatonic probing. Remember that our mechanism arrows always go from electron rich to electron poor. This is the idea of a polar reaction using an electoral file and a nuclear file. So we're going to have this double bond. That's electron rich going to our bro Ming and picking up one of the bro means. And then are we? Pardon me, we will return our two electrons of this diatonic roaming bond onto one of our bro means So now we are going to form this ring that has a essentially a a pox side with this bro. Mean Adam And then we have a leftover negatively charged roaming an ion. So these electrons are going to attack are a pox. I'd ring and open it up and this would be our product. We have this ring here. We have a trans relationship between our roaming atoms where 11 would be appointing above the plain. One would be below, and that would be our product for our first reaction. Now, for our second reaction, we have something very similar. But now are all keen is not internal. It is a terminal, Elkin. So we're going to have a terminal, a pox side. So again, we have this benzene ring attached to this Al Kane, and we are reacting it with di atomic chlorine. So again, we're good for electron, rich to electron, poor are double bond is going to pick up one of these chlorine atoms and return the two electrons onto our chlorine. That's how we're getting an anti on. So we're going to have this benzene ring attached to an a pox side. And instead of an oxygen in this case, we're going to have a positively charged chlorine atom and then our leftover and ions we're going to have these electrons attack are a pox. I'd ring, causing it to open up. And this would be our product. Here we have this ring with this Al Caine structure with two chlorine atoms. And now, finally, we have a very similar, um, reaction as in part B, We start with this Al Keen in this case, we have it to be internal. Remember, we go from electron bridge to electron poor Are double bonds going to pick up one of these electrons or partly one of these chlorine chlorine? Return the two electrons onto our chlorine atom. We're going to form an a pox side with our positive charge. Chlorine atom, our chlorine. An ion is going to attack one of the sides of Europe. Oxide. Returning these electrons onto are positively charged chlorine. And then we're going to produce this product. And as we saw in part A, we have a trans relationship between our two chlorine atoms, and that is the end of this problem. I will make sure that all of our reactions are in one screen for you. I hope this made sense. If you have any questions, please comment or feel free to post a question I'm happy to answer. Um, I hope that this helps you in your current studies in organic chemistry


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