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What will be the product of the following reaction?CHsOHOCH;OCH;both A and Ball of the above...

Question

What will be the product of the following reaction?CHsOHOCH;OCH;both A and Ball of the above

What will be the product of the following reaction? CHsOH OCH; OCH; both A and B all of the above



Answers

What are the products of the following reactions? a. b. c.

This question asks, What is the product of the following reaction? So when we have a glucose molecule and it will react with an ethno under an acid catalyst, the heh Mia Segal will be converted to acetyl. And it is converted in such a fashion via the Replacement Oven, an America hydroxy group within Elcock See Group. And the resultant product is going to be known as a glace Gleick aside. Okay, so the black aside is going to be a type of a seagull? No. So, looking back at our answer choices, we find that, um, the only answer choice that corresponds to this Gleick side will be answer choice be So That is why choice B is the correct answer.

Suppose we want to predict the different products of different reactions of benzene is and aromatic compounds. So let's take the first reaction. What would chronic acid do to this aromatic, competent? Well, it will oxidize all of oxidizes all What oxidizes all Ben's Ilic. Hi Dragons. So do we have any Ben's Ilic hydrogen? It's and a Benzel occasion is a hydrogen that is adjacent to the benzene ring. Kind of how a Lilic was adjacent to our king. And, well, Ben, zines do have old compounds for it. It can be treated as an a little compound so we can use this hydrogen. And since there is a pencil like Hadron, this will oxidize it into a carb oxalic acid. What about this next one? Well, same thing. Do we have been so like hydrants here? Yep, we have two of them. But as long as we have one, it doesn't matter how many we have. Now we have another one there. Technically, there's three and it will do the exact same thing. Now the question is which one will become words? Let me re trial that the question is which Ben's ill Benzel a carbon will become a car back. So, Cassidy and the answer is both. As long as a substitute has it, it will be oxidized, no matter what. This is a poly substitution reaction. What about NBS peroxide heat followed by a meth oxy for I'm a fox side. And I on Well, our first reaction would make a bro ming off of this carbon because this is an Olympic reaction and bends in. Benzene can be treated as a king in a in a sense, because double bond adjacent to it bends it like and our 2nd 1 Well, we have a primary halo cane, and we have a strong base, which is also a good nuclear file. Well, this will do s and two and we end up this molecule. What about the last one? Move this down. What about the last reaction? Well, that I think Well, b R. Two with light. We know that is just radical whole odgen ation, and it will only do it to this metal group because aromatics airway to stable too. Undergo that reaction within themselves, so it needs to be on the substitution. And if we have a negatively charged nitro group, well, that's just s and two because this is a primary hail out cane that is a strong base or a basic nuclear file and would get under s A to, uh, under s and two mechanism. We get this now, what about the last reaction, or would that create? Our last reaction will take this natural group and turn it into an ami. So we keep that carbon, we keep this carbon, I'm actually going to get rid of that, uh, dash or that underlined, because it kind of looks like a negative charge, But it isn't. We'll keep that carbon, and it'll end up as a primary and me.

Okay. This problem is asking us to predict the product of this reaction. So first step, we have a key tone reacting with an alcohol. So there's a couple of different important things here. First off the key tone, and we know it's a key tone because we have a Carbondale. And on either side of the Carbonell, we have two carbons. So ketones are considered Electra Philip towards nuclear Filic edition reactions. So here's my nuclear fall. We only have one equivalent of the alcohol, right? So we're not We know we're not gonna have to have two equivalents come in. For example, we're only gonna get this one equivalent. So the electrons on the oxygen are going to come in and attack this electro Philip Carbon. Okay, we know that that carbon is considered Electra Philip because it is part of the Carbondale. We can move these electrons from this pie bond up to this oxygen due to oxygen's, um, natural. More electro negativity than carbon. Okay, so after this air pushing mechanism, I'm going to get this ch three. I'm gonna get my carbon. This carbon right here is correspondent to this formerly electric Philip Carbon and connected to that carbon. We have the connection to an oxygen with a negative charge. So this oxygen corresponds to this one? Because I simply moved those electrons onto the oxygen mean that I had to incorporate that negative charge and then over here, also connected to this carbon. I have the connection to an oxygen. So this oxygen right here corresponds to this one. That was the incoming nuclear file and connected to the auction. We have a connection to a hydrogen, which is, of course, this one And then finally, my ethel group. Okay, but because this auction has three bonds instead of just two, that means it's gonna have that positive charge. Okay. And last, but not least, attached to this carbon as well. We have the connection to my C two h five. So we have to write that right here. C two h five. Okay, So as we can see, we only added that alcohol one time, right? If we're toe, add it twice, then we would get perhaps a key tall. But instead, we're going to form this Hemi kettle product because I only have the one equivalent incorporated. Okay, But we can't just end right here. We have to get rid of this negative charge and this positive charge, if possible. And we can accomplish that by simply transferring a proton from this oxygen onto this oxygen. Okay, so I'm just gonna represent that with Proton transfer. Some teachers will teach that this occurs in trauma molecularly, but others will teach that occurs, in turn, molecularly. But nevertheless, we're simply gonna get that proton transfer. That means that the oxygen is gonna have that new proton. I mean, that it will have a negative charge any longer. And then this oxygen will have its Ethel group minus one hydrogen. So no positive charge. And then finally, my C two h five. Okay, so this will be the Hemi Keitel product, and this will be see for the answer.

Okay, This problem is asking us to predict the product of this reaction. So first up, we have a key tone reacting with this, I mean, and that is going to be in either solving. But the either solvent isn't necessarily important for the thinking about this problem, because it is simply a solvent. It doesn't participate in the reaction. So I'm not gonna think about either solvent for now. But I am going to think about Mikey Tone and Miami in. So let's go ahead and do that. So ketones, as we know, are considered electro Philip because we have the natural differences in electro negativity between this auction and this carbon. So what I'm gonna do is draw a resident structure of this Keto. I'm going to simply move the electrons from this pie bond up to this oxygen. So the results of that will be this residents form in which I have my metal group, my carbon, which is, uh, this carbon and also attached the carbon. We have the connection to my oxygen so that oxygen we moved electrons onto it. That means that it has a surplus of electrons. That means it has a negative charge and we moved electrons away from this carbon. So that means it has a lack of electrons. That means it has a positive charge. And we also have the connection to that metal group. Okay, so this is my residence structure. As we can see, we have a positive charge on that carbon, right? Any time in organic chemistry. Generally, if I have a positive charge that is indicative of a lack of electrons. That means that it once electrons in order to relieve itself of that positive charge. Right? Because if we have a positive charge, that is, at least in this case, that means it is Electra Filic, right file. Or Philip means love electron means electrons. That means it wants to have electrons. Right? And I can accomplish that by simply moving the electrons from this nitrogen onto this carbon, right? That will proceed in a nuclear Filic edition reaction in which I'm moving the electrons onto this carbon to relieve itself of that positive charge. Okay, so as I do that I'm going to get this product in which I have my method group attached to this carbon which was formerly the electric full of carbon. I have the connection to my auction with the naked church. Because this one, at least this bond didn't change it all. We just had to form that residence structure. Okay, so that is single bonded. I also have a connection to my nitrogen. That nitrogen has the connection to two hydrogen Acela's my ethnic group. So because this nitrogen has four bonds, that means it has a positive charge. And then last but not least, I also have this method group. Okay, so all we did was simply making residents for him. Or we could have done this at the same time. We could have moved these electrons onto this carbon as we do as we moved the electrons up to this oxygen. So we formed this Tetra Hydro Intermediate. Okay, but as we can see, we have a negative charge and we have a positive charge, right? So if we can get rid of both both of those charges, I'm gonna go ahead and do that, and I can accomplish that by simply using a proton transfer. Okay, so in the in the textbook, at least for this, um, the set of problems, it shows that we're gonna go ahead and use H plus. So h plus a Zoe do that didn't get simultaneous steps in which we both transfer a proton using this h plus. And at the same time, we're gonna go ahead and transfer this proton onto this oxygen. Okay, so that might be a little bit confusing, but what simply happening is we're gonna have to proton transfers in which we overall want to make a good living group in this oxygen, because in, um, this, I mean, our I'm information. It's a reversible reaction. So as you can see, my reactions are a key tone and enemy. The product is going to be simply an Emmy in which I'm basically replacing this oxygen with this nitrogen. I'm making a double bonded, um, nitrogen to a carbon. Okay, so with that sense, I know that I have to get rid of this oxygen, okay? And I can accomplish that by making it into a good leaving group, which is water as opposed to alcohol. So I need to have two protons on the oxygen. Sorry, that's a little bit confusing, but simply we're transferring this proton and the aesthetic environment and also this extra proton attached to that nitrogen. So first up, using this acid, usually we're in a ph of approximately 4.5 to 5.5. The acidic environment is gonna be acidic enough to protein. Protein ate the oxygen. So the very first step of this overall to step pronation step is the pro nation of the oxygen. So we're gonna get the O. H. We're also gonna get the nitrogen, which is positively charged, because we still have that four bonds to it. And then we all said that method group, so this is positively charged. Okay, But again, we want to eventually get rid of that oxygen because this is a reversible reaction. I eventually want to create my I mean, or I mean, okay. And I can accomplish that by moving or transferring this extra proton on this nitrogen onto this oxygen because I want to form that I mean, so using a proton transfer room, I'm simply gonna accomplish that. I'm gonna move this proton onto this oxygen, So let's go ahead and do that. I'm gonna get my o h two because it has the extra proton. That means that it has three bonds that means that it has a positive charge on that. I'm gonna have my nitrogen, which is three has three bonds because I have the hydrogen. I have my ethel group, and I have the connection to this the rest of the structure. And then last but not least, I have my method group. So as you can see, we went from an overall neutral charge. We added my acid because we're in a civic environment, and then we got that positive charge, and we got this positive charge onto another Adam. Okay, so now we are at the state that we wanna be, right, Because we have this oxygen with the positive charge this is going to want to relieve itself of that positive charge. So we're gonna accomplish that by moving the electrons away from this carbon and onto this oxygen. And then we're going to do this, in which we move the electrons from this nitrogen onto this carbon to relieve that carbon of its hypothetical positive charge. So, uh, two steps happening, but this is going to be the product. I have my two method groups attached to this carbon. I have my double bonded nitrogen because again, this was formerly single bonded. But we had to add the extra bond in order to relieve that carbon of its hypothetical positive charge. And then I have my nitrogen and I have my ethnic group and my hydrogen. Okay, but because the hydrogen is a next ra bond, it has four bonds. Now, this is a positive, Possibly charged. I need to get rid of that positive charge by simply using water two d protein ate that hydrogen or d pregnant that nitrogen. Right? So I'm gonna get this as my eventual product in which I have my nitrogen double bonded to my carbon. And then I have my ethnic group as well. So this is my final product. It is in. I mean, it is simply a carbon double bonded to a nitrogen. But again, this whole entire process is reversible. So even though this might seem relatively unfavorable, we can, of course, get back to my reactant with not much problem at all. Okay, So any information


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