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Which of the following compounds would react with Tollens' reagent, Benedict's reagent, both, or neither?...

Question

Which of the following compounds would react with Tollens' reagent, Benedict's reagent, both, or neither?

Which of the following compounds would react with Tollens' reagent, Benedict's reagent, both, or neither?



Answers

Which of the following nucleophilic substitution reactions will take place?

This is the answer into Chapter 24. Problem number nine from the Smith Organic Chemistry textbook. This problem asks us what Aldo product is formed when two molecules of beauty now react together in the presence of a base. What re agents were needed to convert this product? Each of the following compounds. Okay. S o when we, uh, react to equivalents of beauty now. Ah, in the presence of base, our product is going to be this one. Oops. Oh, no. Okay, there we go. I drew ah to too many carbons at first. Um, okay. And so we'll have an alcohol here mean move this down a little bit. So we'll have an alcohol here. Ah, and we will have on Aldo hide up here. Okay. Ah. And so now we're asked to convert this product into four other different things. Um, And so for the 1st 1 we are asked to get to here. Um, and so, looking at this, we can see that, um, what has changed is that we've reduced Ah, the ala hide to an alcohol on. And so we can do this with sodium bora hydride. S o N e B h four. Ah, and this reaction is always done in methanol. So sodium bora hydrogen methadone will affect that reduction that we need. Okay for part B were asked to get from our starting material, too. Um, material within Aldo hide on an extra method group. And so, in order to you that we're going to do that in two steps. So the first step Ah, well, use bass on this is gonna give us the elimination product. So we have this elimination product here, uh, the d hi dehydrated algal products, if you will. Um And so now we need to add that metal group and get rid of the double bond. And so to do that, we can use ah, the, uh, living cooperate. So it would be the metal form since we want to add metal group. Ah, and then has always worked that up with water, and that is going to give us our desired product here. Okay, so there we go. There is be, rather than try to cram see Indy And on that page, I'll put them on their own page eso for C. We are going to start with that intermediate that we just made and b eso We'll start from the the Aldo Condensation. Herb. Pardon me. The out all dehydration product. Um, And now if we just reduce, uh, this product that will get us to where we want to be for C eso again, will you, sodium? For hydride. Ah, and as I said before, always done in methanol. Um, and that gets us our desired product here. Okay, Um, so lastly for D Ah. Okay. Um, we are also going You start from this same intermediate from part B. So the album dehydration product. Ah. And now, um, we can get to the product for D in a single step, using a metal grin. Yeard ch three mg b r o. And, as always, will work that up with water. Okay, so there we go. There is our desired product for party. All right, on then. Here again, our a and B. And the answer to the initial question of what we get from the outdoor reaction of two equivalents Butte now. Ah, and that is the answer to Chapter 24. Problem number nine

This is the answer to Chapter 20. Problem number 47 fromthe Smith Organic chemistry Textbook. This problem asks us to draw the products of each of these reactions which involves a new organic metallic re agent. Okay, so we ate organic metallic reactions here. Okay, So So for the 1st 1 we have Grenier free agent, and we're reacting it with carbon dioxide, followed by, um, some acid for the work up. Ah, and so this is going to result in, um, the carb oxalic acid. So remember when we react Ah, grin. You agree, agent, With carbon dioxide, we end up with a carb oxalic acid instead of like, an alcohol like we typically get. Um, when we use just analogy. Hideaki Ito. So we have, um basically just replaced our MGB are with the c 00 h. So we have the car back sell gas in there. Okay, so that's a B is a more typical grin. Your reaction. So we have Ethel Griner agree, Agent um and we're reacting it with this This bring system this by cycle. Um, and there's a key tone on the right ring. And so, um, the reaction's going to occur at that key tone carbon. And so we will add our Ethel group from the Griner Free agent. Ah, and our key tone will end up as an alcohol. Okay, so there we go. That's B for see? Ah, we're reacting. Um, and ask chloride with excess green. You're free agent followed by water work up on. And so here, what we're gonna end up with, um, is a central carbon on. And so we started with one federal ring. Ah, and our green Your dory agent is ah, fennel Grenier on. And so we actually end up with three federal rings on this central carbon. Ah, and then also Anelka hole. Okay, um, for D. We're reacting in Esther with excess greener, every agent on water work up. Um, and so what we're gonna get here eyes double addition of the greenery agents. It's a metal grin yeard on. And so we end up with, um, our federal ring fromthe starting material. So there is that. Ah, and then here is our carbon. That was, um, the Esther Carbon. And so it's going to have an alcohol on it now. Ah, and we will have the two method groups that added So one method group and a second. Okay, so moving on to the next page eso for e were reacting just ah Green. You agree? Agent with formaldehyde. Ah. And so the result of that reaction is going to be, um, just the addition of one more carbon to this chain with an alcohol on it. And so we have five carbons to start. Your product will have 456 carbons on the alcohol. So one hexen all is our product. Okay, Um, so then for F, I were using ah, copper lithium re agent. So it's the metal organo. Cooperate. Um, and s o. The reaction here is going to take place at the double bond. Um, so not the key tone key tones gonna be infected. Um, and basically, we just end up adding, ah, metal unit to this change. So it's an eight carbon chain to start 2345678 nine. Now it is a nine carbon chain. The addition of that method group took place at the the AL Keen on. And so there is our key tones still intact. And so that's f for G. G. Is just another typical grin. You agree, Agent s. So it's gonna take place at the key tone in this molecule. Nothing else about the molecule, including that Al Keen eyes going to be affected by the reaction with discrim yeard. So there's the out keen still intact. As I said, um, and then here's the rest of our molecule, so our key tone has been reduced to an alcohol. Um, and we've added a method group here. Um, move that down a little bit, just so that it's easy to see. Here we go. Okay. Yeah. So we've added, Ah, I've added a meth. A group on our key tone is now an alcohol. Okay. And so that's g. And so lastly, um, we'll take a look at h S o h your organo cooper. It that we have here is gonna act is nuclear file and open this up oxide. Um, and following the rules for a pox side opening, it's going to attack at the less substituted carbon in the epoxy side. Um and so we will end up with this of hoops. So here's our C six h five unaffected by the reaction. Okay, um, and so again. The the cover lithium is gonna act as, ah, nuclear file an attack at this carbon. And so what we end up with is an alcohol here. Um, and there's our second carbon from the pox side on then. Since it was, ah, metal, copper, lithium or metal Cooperate, uh, that we were using. Um, there is our added metal group, okay? And so, um, that's the product for h. Ah. And that's the answer here. Uh, the only you know, there's no secret to these. You just need to know how these organic metallic re agents react under all of these conditions. Ah, and that's the answer to Chapter 20 problems of a 47.

Okay, This problem is asking us to show the algal products. So these are the reactions that they're giving us. So we have missed all the hydrate here. It will just start with this one. So if this Alba hide word to undergo an Al Gore reaction, my product would be a case in which I direct it with and a O. H, for example, and water. So first things first is my n a. H. My sodium hydroxide is considered to be a base to that base to go into deep protein. Ate this awful carpet, this heart alfa hydrogen Because Alfa hydrogen is considered the most acidic, um, hydrogen on this molecule. So that's gonna be probated leaving the electrons onto this Corbyn. OK, so we should get this after that immediate step in which we have Carbonell and then instead of a proton. And we have long periods there, I guess so those long periods are going to react with another molecule of this album hide. So after that reaction, we should get the movement of these electrons onto this Carbonell to form this product in which we have this all the hood which represent in red and then in the off position. So this position right here, we're going to draw its connection to this carbon, which will represent in blue. So it will be connected to this carbon, which is connected to an auction which is connected to 100 and that is connected to the rest of my chain. So okay. And then after impregnation by the water that we just created in the previous step, that we should get the pronation of the auction to get this Okay. So represented that in a more understandable form, who should get this is my final product. Okay, so this is where I'll draw in red. So this right here, this compound here is the starting material that we reacted airbase with. And then this part of the molecule is the second Carbonell, The second album. Five. Okay, moving on to Part B. We have a bending. So the spending has a key tone, and that key tone is going to undergo. My Alfa era started my little reaction. So same thing. Sodium hydroxide. That's what I'm hydroxide is going to be Protein ate the most acidic hydrogen on this molecule. And if we look to this carbon. There's no Hodgins attached to that. Because this is an SP to hybridize carbon right here way can only have the one other connection that is already connected to the key tone. So we don't have any Hodgins over here. Instead, we're gonna look to right here and on that carbon. We have three hydrogen, so we don't We only need one. I'm going to deponent that hydrogen, move the electrons up to his carbon. And then we should get the D printing version of this molecule just blown pairs. Lone pairs on this carbon. And that is the carbon in which a nuclear for look attack will occur and attack another molecule of my starting material. So these are repairs are going to attack that Carbonell moving the electrons up to here. Then we should get the fallen. And again, I'm going to represent this compound in red. And then this part part in blue. So I benzene unaffected Carbonell, and then right here is my connections. So this carbon is going to be connected to the carbon right here of the other blue molecule. So we draw that. Oh, So here's my carbon of my of this compound, and that's connected to an auction which has a negative charge now, a metal group. And then my benzene. Okay. And then after pro nation, we should just get the pro native form of this and then by pro nation, I mean with water. So we just created water by the pro nation of this sodium hydroxide. Okay, so that should be the answer felon right here. Okay, next up, we're going to do see, so see, starts with this compound. Okay, so we only have, um we have a couple options here. We can either department the Hodgins on this carbon or this one. It doesn't matter because it's a special molecule. So after continue doing my trend in blue after using sodium hydroxide, I'll Deep Throat Nate, that 100 Get the appropriate with the electrons onto this carbons. We should get this. Okay. And then that is going to attack another molecule of my starting material. So, like this, those electrons are gonna come in a technique, Carbonell moving the electrons up to the auction, forming this product so in blue will represent my start material. So my Carneal's unaffected, and I'm gonna have the electrons from this carbon be participating in the bonding. So that's gonna be connected to this structure. So this carbon corresponds to this carbon and connected to this carbon? I have my auction with a negative charge. Now I have my this one and I have this bond and then just the rest of my structure. And then after a pro nation of this auction, I should end up with the following, so that should be my product.

Okay. This problem is asking us whether we can make this compound using a mixed Aldo conversation. So first up, let's draw out the first story material. So we have a benzene connected to a carbon connected to a double bond connected to an eventual Carbondale. Okay, so the trick is in order to determine what the reagents were needed to make. The album conversation product is to identify your Okay, so here's my okay. And the side of the carbon that is closest to the side of the carbon eel. So he's milking. Here's the carbon that is closest to the side of a criminal. I'm going to split the molecule down that side. Okay, So down that carbon, that is why I'm putting it. And then the results of that on the side of the car meal. That is what I was originally started with. So I started out with this. So that is one of the components of my eventual conversation reaction and then the other side. So my left side has a benzene, has a carbon. And then where the end of my cocaine was this carbon. This government here that was originally an auction, the auction. Okay, so to draw that out, it would be in Alba hide. So the question is, can we make this product using, um, these regions? And the answer is yes, because we can get this in pretty good, you know, because we only have one opportunity for D pronation. And this in these molecules, for example, on a ski tow. This one here, If I were direct, both of these, actually with any which started hydroxide, I would only get deep Throat Nation on this key tone. And that's because this key tone has available hydrogen. It actually hasn't on both sides. Whereas this been Zelda head does not have any available Hodgins ready for deportation. Right? Because if I don't even have any Hodgins on this Alfa Carbon because it's connected to SP to hybridize carbon. So I don't have an opportunity for deportation of this carbon. But I do have an opportunity for deportation of this one. Which is why this would undergo an album conversation. So for part A, that one will, and I'll get have used these two products these two regions get for party. Let's see. Okay, so I start off with this. Well, I end up with this. So benzene connected to and a king connected to a Carbonell. Okay, so can we make this compound using an elbow conversation and mixed out our conversation? Let's see. So identify on the side marking the carbon that is closest to my Carbonell. So this one Okay, someone has put that down the middle of the carbon, and on this side is what I have the carbon. Carbonell said That's what I started out with. And then on this side, I just have to same thing, draw the line. And then whatever that carbon was this governor here that was originally an oxygen. So I have my benzene. I have my methyl group attached to my joking. And then this used to be a carbon which was eventually connected to the carbon. But now, since we're breaking it down, it's an oxygen. Okay, so we're starting with these two products. Start these two reagents to see if we can make this region right here in all races to make it clear. Okay, So can we make this? The answer is no. Because let's see if I were to react, these both and if both of these were present. My solution indirect on both with sodium hydroxide. So a hydroxide ion. I have two opportunities for deportation because right here on my key tone, this key tone right here I have, um, available hide regions on this bet on this key tone of my benzene. I have also Hodgins coming off that are also available to D probation. So I have to opportunities for a deep rogue nation. And in reality, I kind of have three. Because there there's also hydrogen is on that side. Okay, so this would not be a very good mixed elbow conversation because I have the opportunity for the pronation on both molecules. So part B would be No. It would not be a very good product of an elbow concision because the two products needed the two regions needed would be it would be to multiple different products. Okay, next up, I have this product. What? You have a key tone on a cycle of vaccine and then have this. Okay, So same thing is, before I identify the side of my cocaine that's closest to my carbonell. And then I draw a squiggly line down that bond, So whatever is on the side of my Carbonell. That is what I originally started out with. So I have my cycle vaccine. And then, like he told on top. Okay, so that's one of my reagents needed and then the other side. So this said we're here. If I use all this and then what is what was originally a carbon? So this carbon right here, it used to It is now on oxygen. Okay, so, like this, So draw that all out. What was originally a carbon is now an oxygen. Okay, so if that is there, then this used to be annulled ahead. So just straighten this out. I'm going to rearrange like that. Okay, So can we make the mixed Alva conversation product using these two products using these two regions? The answer is no. Because Santana's before, If I use sodium hydroxide and in a region in a flask containing both of these, my sodium hydroxide would d protein ate both of these because I have available Hodgins here. I have available Hodgins here. One of these is going to behave as a nuclear file. The other is going to give us a Dr file and Vice versa would get multiple mixed products, so B and C would not be very good. Next Alba conversation reactions, whereas a would be


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