Give the structures of the products represented by letters in thi...

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Give the structures of the products represented by letters in this synthesis. Part 2:

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Hey everyone and welcome back in today's video. We are going to solve the following problem provide the missing products of each reaction. In the following synthesis, the synthesis consists of several steps. We are starting with a material that has a double bond and a triple bond in it. After performing several steps, we are getting our intermediate A and B, followed by the synthesis of the final product C. And we want to identify these three unknowns. So let's start our analysis and we may simply through our structure. In the first stop of the reaction, we are adding ozone. So that would be a reaction number one ozone at a low temperature Followed by reaction number two dimethyl sulfide in water. If you recall these two reactions combined together may be classified as an OES analysis reaction. Remember that? And arsenal's reaction implies that we are breaking carbon carbon double bonds and triple bonds regarding the al keane's. We are forming either Aldo hides or ketones and thinking about al kinds were forming car, bookseller casas. So let's think about it. We are only going to be interested in the internal structure. But of course we can think about everything as well as our side products. Right? So if we're breaking this carbon carbon double bond, we will get two double bonds. One of them will be just one carbon atom on top. It had to implicit hydrogen atoms. We may simply draw the simplicity hydrogen atoms over here. So there will be two hydrogen atoms and there's one more over here. So after we break that carbon carbon double bond, we will have carbon atom with two hydrogen atoms. And now we need to add an oxygen atom a disposition because that's the notion elicits reactions were breaking our double bond and adding oxygen to each and of the result on two double bonds. So that's our first product and that will be the byproduct because we're not really interested in it. Now thinking about the internal structure, we have a double bond. So let's draw a double bond, add an oxygen atom on top. Now we see that this carbonell group has hydrogen bonded it. So that's an AL to hide. And then let's think about how many more carbon atoms do we have bonded? So that's 123 and four. Right, so let's draw four more carbon atoms that B 123 and four. And now let's be extra careful. Carbon number four is an Al Qaeda and carbon. And when we are breaking that carbon carbon triple bond, we are going to form a curb oxalic acid because those analysis of al kinds produces car bookseller casas. So here instead of drawing a triple bonded carbon, we're simply adding a C. 00. H. Group. The curb oxalic acid group that will be our products for the internal part and the external part is a terminal al Qaeda. Right? So we are just forming C. 02 as our third product and the one and the main one that we are interested in is the internal part. That longest change. So we can say that product A. Is this structure which involves the presence of an al the hide and a cover oxalic acid. Right, so we may say that this it's our product. A. Now let's think about our next step in the next step, we're creating compound A with sodium world high trite and methanol. Now this if you'll recall is a reduction reaction. We're reducing our carbon compounds to alcohols. Now you have to keep in mind that sodium were a high tried is a week reducing agents. So it cannot really reduce any carbon compound. It may reduce al the heights and keystones, but it will not reduce assets or assets. Right? So we will leave our asset as it is. But that this alga hide can be reduced into an alcohol. So we're going to turn that carbonell group into carbon alcohol, C. O. H bond. Right? So in particular, we will get a primary alcohol. Since there's already one hydrogen. We are going to add another hygienic disposition. Break the double bond and make that oxygen atom as our alcohol group. So let's do that. Let's reduce our al to hide into a primary alcohol. Right? So if we do that, we may draw our structure B. Keep in mind that once again the car bookseller gas, it cannot be reduced because the reducing agent is too weak. Right? So if we draw our alcohol, we may start by drawing O. H bonded to that carbon atom. So now, how many carbon atoms do we have? 12345 carbon atoms. So let's draw them. That would be one, 23, four and five. Now the fifth carbon atom is our car book silic acid group. And we are done right, Because we said we cannot reduce that, but we can reduce the alga hide part into a primary alcohol. So we may say that our structure be is this one and we are ready to go. So that would be our structure B. Finally, let's think about the final reaction where we are tweeting that structure B with sulfuric acid and heat. So H two S. 04. And heat in particular when you see heat, that should generally imply elimination. And indeed because we have an alcohol group and we're treating that alcohol with superior Cassidy and heat. We may simply recall that this is an acid catalyzed dehydration of alcohols. So what happens is that your protein eating your alcohol using sulfuric acid, which promotes the loss of the leaving group. After we apply heat and we are forming a primary carbon Karen. So let's just draw that result in carbon Karen. If we remove that protein ated alcohol, right? That B. O. H. Two, we will have the loss of water and we will form a carbon Karen at that position, which is the alpha position. So once again we have five carbon atoms. Let's draw them. There will be positive charge here and now. Because this falls the elimination mechanism. The one we have to be extra careful right recall that this reaction is an E one reaction. Let's think whether or not we can form a more stable carbo Caroline and it looks like we can because there is a source of hydrogen atoms. So we will have a one comma two hydride shift. This hire a general shift over here to form a more stable secondary kabul Karen. Right? So we can say that the actual intermediate here would be a secondary carbo Karen. Let's redraw it. And now we have a positive charge here. And that makes perfect sense. Right. We can definitely do that because it stabilizes the cardboard Carolan from a primary one to a secondary one. And that's much more preferable. Now let's think about our beta hydrants because we are trying to perform an elimination and we are going to remove one of them, The one that we are going to remove is this one because the base that we're using water is not hysterically hindered, meaning we want to produce the same product, the more substituted alkaline. So if we remove this hydrogen by water, there will be an attack the transfer of the bonding electrons into a pi bond. And we are done. We will get our final product for C. Let's see what we get. We may also label this as our final step, right? There's addition of water And we have five carbon atoms. The car bookseller cast the group and this level bomb starts over here. Now, what else? We can check is that whether this structure? See we may already proposed this is our structure. See we may simply check if the molecular formula given us consistent with our structure C 512345 carbon atoms. That's good for us. Two oxygen atoms 12, that's fine. Eight hydrogen atoms three over here plus one plus one. That's five plus two. That's seven. Right? And one from here gives us eight C five H +802 is the molecular formula of this product. So that works for us. Well done. We may say that this is our product. See so we got our intermediate products A and B. As well as the final product C. We may now state that the correct answer to this multiple choice question is A. However, if you have any questions, don't have States, leave your comment down below in the comment section and thank you for watching

Give the structures of the products represented by letters in this synthesis. Part 2:

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