Solving the following road-map problem depends on determining ...

Question

Solving the following road-map problem depends on determining the structure of A, the key intermediate. Give structures for compounds A through K.

Accepted Answer

Hey everyone and welcome back into this video. We're going to solve the following problem. The following roadmap has some missing compounds proposed the structures for the compounds A through F. Now, if you look at that given synthesis, we understand that there is some central compound A. Which we obtained from unknown compounds in each case B F E D. Right? So we will not start with A. We will actually start with B. Because we're given a known structure a set of reagents and this would allow us to get B. Then we will move on to A. Because again, we're given regions and then when we know the identity of A. We can start with pretty much any reaction F or E. And then we can get D. And finally, let's suppose we get C. So let's label our unknowns over here. We will essentially be labeling our unknowns here and below our sentences scheme. We will be thinking about our reactions in the rationale behind each reaction. So first of all, we will we are going to draw propel magnesium bromide. You may immediately recognize that this is our green yard regent. So if we think about this first reaction propio magnesium bromide reacts with ethanol. So that would be our al to hide. Right? And in the graveyard reaction, we already know that we are essentially performing that nuclear attack at carbonell, breaking the pi bon, forming an elk oxide and ion. And then the next step when Alcock said and iron is formed. We are performing the second step where we are protein. Ating the elk oxide and producing our alcohol. So let's see how that looks like this will allow us to get structure be let's start with the profile group. That would be 123 carbon atoms. And now the third carbon atom is bonded to this carbon. So we are going to draw the fourth carbon which has oxygen. Now it's single bonded because we have broken the pi bond oxygen with a formal negative charge so far right. And then the metal group. But in the second step of this reaction, when we are protein ating that Elk oxide, we're getting our alcohol. So this allows us to get structure and be wonderful. So now that we got the structure B. We understand that this structure is treated with sodium hypochlorite and A. O. C. L. In the presence of acidic asset, H O A C. Corresponds to acetic acid. And you may recall that sodium hypochlorite is an oxidizing agent. If we use an excess of sodium hypochlorite, it allows us to oxidize alcohols because this is a secondary alcohol. Since carbon oxygen over here, this carbon atom bonded to an alcohol group has to university secondary alcohol. You may recall that the oxidation of secondary alcohols produces ketones. So the structure A would be our key tone. Let's draw our parent chain and we are going to convert this into a carbon ill group. Okay, so this would be structure A. Now we got structure A and we can simply think about reaction from F. To A. And if you look at these regions were using ozone at negative degrees Celsius. In the first step. In the second step we're using the metal sulfide and water. This corresponds to an O. Cyanosis reaction and we can use an elk in to produce A. So if we draw an arrow, let's recall the general scheme of oz analysis. Whenever we are dealing with Elkins, suppose that you have an L. Key in which contains R. one r. T. For diversity, let's add high chain and then our three. Whenever you're performing of analysis, you're going to break this carbon carbon double bond and add oxygen atoms to each end of the double bond. So performing arts analysis. That means number one is addition of ozone at a lower temperature. And number two treatment of the ozone, I'd intermediate whether the metal sulfide and water, this produces two carbonell compounds. So we are just going to break that double bond and we are going to separate the chief fragments apart. That would be R one r. T. H. R. Three and let's just add T auction adam. So we have an L. To hide in a key tone. If this hydrant is another L kill group, then we get to ketones meaning to get this carbon compound, we may essentially Draw our five member chain. Let's throw it in green. That would be 12345. And we have to think in reverse we have to remove this oxygen atom. Right? So we have removed that and bonded to some other to alcohol substitute prints. Right? So in in a general case, we can just say that these two alcohol substitute prints would be our and are okay. They might be saying they might be different. It doesn't really matter. One of them might be hydrogen or both of them can be hydrogen atoms, that's perfectly fine. We will get a by product which will not be Consistent with a. So it can be anything because we're going to break this double bond and get our carbon group in an arsenal sous reaction. So let's label it at as 03 at a low temperature negative 78°C followed by the metal sulfide and water. Okay, so this works for us. We got our structure F. So this would be our structure F. Now, when we look back at our synthesis, we also want to get A from Pent one line. So we are going to draw our al Kym, as the name suggests, we have an AL kind. So that would be triple bond at position number one. And we have a total of five carbon atoms in the parent change. So that would be 12345. Notice that A has five carbon atoms in the parent and Pent 1 9 has five carbon atoms. So that's perfect. We may essentially recall that this is an acid catalyzed hydration of alcohols we want to make sure that the alcohol group is added to carbon number two. Right? Even though you're still seeing a key tone, of course we will recall a keto in all sorts memorization. But because we want to add an alcohol group at position number two, we don't want to use the anti more comic of addition. Instead we want to use sulfuric acid which catalyze the reaction right? Mercury to self aid and water. This would correspond to an acid catalyzed hydration of alcohols. And let's see how that works. So essentially in the stop we're following the more cosmic of school and we're saying that hydrant is added to a less substituted carbon and the oh age group is that it's a more substituted carbon. And you may notice that if this is the case you're getting a double bond in stud and this would be an Enel. But because channels are not really stable they are in equilibrium which shifts towards a keto form. So we control our key tone which indeed corresponds to the given structure. So this would be our E. Right. This set of regions would be E. And acid catalyzed hydration of all kinds and then we can go back to our synthesis and think what else we have so far. Now we only need to determine structures the N. C. We're going to go into reverse. So if we know a that's a key tone, we want to use a structure D. And from structure D to a we have a Gilman region. So this is the Gilman Regent because we have an organic cooperate rage and right lithium with a positive charge, copper with a negative charge. And copper has two metal groups bonded to it now because they're same methyl group over here. And because we are producing a key zone in particular, we can expand this about to understand what's happening. The Gilman Reagent is Ch we two times copper lithium. Now we essentially wants to use it as our region and produce our key town because we are forming a key zone. It makes sense to use an acid chloride. So we can simply say that for this conversion, we are going to replace this metal group which comes from the Gilman reagent with a chlorine atom. Right? Because chlorine would be a good leaving group upon reacting our acid chloride with the Gilman reagent which contains that method. We already see it over here. So all that we have to do is just at chlorine here and get our asset chloride. Okay, so this will be our sl chloride for the reaction in this reaction in particular. The Gilman Reagent which is capable of transferring this l kill group to carbonell carbonell would form oxygen with a negative charge by breaking the pi bond over here and then by restoring the pi bond, we would have the loss of glory in And this is how we are adding our methyl group. Right? So this works for us. We understand that structure D. Would be an sl chloride. We can label this as D. And I was to form D. We want to propose structure C. The Regent used this Lionel chloride. So that would be S O C L. C. We call that in order to produce sl chlorides using vinyl chloride were essentially using assets. Right? So whenever you use an acid meaning, we are going to turn this chlorine into O. H. So we get our carb oxalic acid group and the remaining chain is not changed or affected in any way. So whenever you're treating and a car book silic acid with vinyl chloride, you're essentially converting it into a corresponding sl chlorides. We're converting this O H into C E L. Meaning this is our structure. See a car broke Cilic asset and we got all of our structures. We can go back and label them separately so we can clearly see what our structures for this roadmap are. Okay, let's start with A so that we go in an alphabetical order for a. We understood that we needed to use a key tone. So that is our structure A. Now for B. Structure B was a corresponding alcohol where we performed the Greenert reaction that gave us an alcohol structure C. Was a car book silic acid that we got at the very end. Right. We treated a carb oxalic acid with vinyl chloride to form a corresponding sl chloride. So C was our car book silic acid. Now for the we had our sl chloride so we can draw our carbon eagle group and cl For E. We didn't really have any regions or product. That was just a set of regions. And we said that to go from .19 to our key tone, we needed to use an acid catalyzed hydration of all kinds which corresponds to three regions sulfuric acid, mercury to self aid as well as water. And finally off, we had to devise an elk in which would undergo owes analysis to produce that tone. We said that there might be a diversity of possibilities here, as long as we have a double bond here. So that's the most important part of the structure. And then we simply said that there might be two alcohol substitute prints on each and right. So that was our proposed structure. If you wish to propose something specific, you may just add your substitue ints, those might be hydrogen atoms or L kill groups. So you may label these as our unknowns and conclude that the correct answer to the small cultures question is c. However, if you have any questions don't hesitate to leave a comment down below in the comment section and thank you for watching

Solving the following road-map problem depends on determining the structure of A, the key intermediate. Give structures for compounds A through K.

Key Concepts

Nucleophilic Addition

Grignard Reagents

Oxidation and Reduction Reactions

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