Oxidative cleavages can help to determine
the positions of the t...

Question

Oxidative cleavages can help to determine the positions of the triple bonds in alkynes. (a) An unknown alkyne undergoes oxidative cleavage to give adipic acid and two equivalents of acetic acid. Propose a structure for the alkyne. <

of reaction>

Accepted Answer

Hey everyone and welcome back in today's video, we are going to solve the following problem. The positions of triple bonds in AL kinds can be determined through oxidative cleavage. An unknown AL kind gives melodic asset and two equivalents of carbon dioxide as shown below, propose the structure of this AL kind. As the reaction suggests, we have an unknown AL kind which undergoes two reactions. Number one edition of ozone at a low temperature negative 78 degrees Celsius and number two edition of water. Now, the products forum here would be our melodic asset, right? We have two car book silic acid groups and two equivalents of C 02. Whenever we are considering a reaction of that kind, we have to think about the oxidative cleavage in general. You may notice that these two reactions correspond to an osce analysis reaction. Right? Let's recall that whenever we are using number one ozone at a low temperature water, of course we need to use reductive workups such as addition of the metal sulfide as well. And essentially what we're going to do, we're going to remember that also knows this of AL kinds produces car book Celik assets and sometimes C O T. So let's think why and how that happens and also how we can use that information to determine the original structure or the unknown AL kind. So let's think about two cases case. Number one suppose that we have an alkaline where we have a triple bond with two alcohol. Substations are one as well as our T, which we define as an internal AL kind because it's surrounded by two alcohol groups. Supposing that you are performing an oxidative cleavage or you're performing an osteoporosis reaction in this case, number one edition of ozone at a lower temperature. And then # two edition of that metal sulfide and water, essentially that implies that you're going to break your structure apart. Now this is not really correct to say in terms of the actual mechanism, but for practical purposes to give you a shortcut to visualize how you get your products, you're essentially going to break your structure apart along your triple bond. And the best idea is to count the number of carbon atoms that you have on each end. So first of all, notice how we have our tea bonded to this carbon. Right? So we have our tea and an additional carbon. So that carbon in green will become the part of our curb oxalic acid. And we may immediately draw it out. So one of our products will correspond to car book silic acid group C. 00. H, which comes from that carbon. And then we have our our T group bonnet said. So this is how we get our car book silic acid and then our second product will be the same. Right? We are going to look at the other carbon in blue and this is the part of carb oxalic acid. So we're going to throw our car book silic acid and then on the other end we have our one. So this is the substitution bonnet to the curb oxalic acid group. So let's throw it out in red. On the other hand, the second cases whenever you have an insert a terminal alka in my apologies. So a terminal alka and essentially implies that you're going to have to suppose we are adding one alcohol. Supposition for example. So if we are breaking this double my apologies, this triple bond instead, then we are forming something else. So on the left hand side we understand that we are forming our curb oxalic acid in the form of R one and then C. 00. H. But because this is a terminal AL kind, this carbon only has hydrogen bonnet said. So this yields carbon dioxide and you may immediately understand how we got to most of carbon dioxide. That's because we must have to terminal al kind ends to produce two modes of C. O. T. Okay, so now that we clearly see how we apply an osce analysis reaction. Once again, the second part was another example of this analysis or oxidative cleavage number one, ozone number two, dimethyl sulfide and water. So we just have to analyze our products. Whenever we see car bookseller cast group, we can go into reverse and we can understand that this carbon atom must have formed a triple bond within an alkaline. Right? Because it's part of the car book silic acid group notice how we sat in green that this carbon atom from the car bookseller cast group must have had a triple bond. So let's label our triple bonded carbon atoms. So we are having this as our triple bonded carbon atom. Then we have a chain. Right? We have siege to. So nothing happens here. But then we have another adam that belongs to carve oxalic acid. So that means these two within the structure must have been triple bonded. So we have our triple bonds here and here in between. Nothing changes. The structure is still connected via CH two. Okay, so now because two equivalents of C. 02 are produced. That means during the cleavage of these triple bonds there was no al kiel substitute print either here or here. There was actually hydrogen on each hand. Because as we previously said, if we're performing the cleavage for a terminal al kind, then C. 02 is produced for this carbon atom meaning here we are going to have two hydrogen atoms within our alkaline. Okay, so we have performed our analysis, meaning all that we have to do is just replace these H. 00. With triple bonds right? Which will correspond to that carbon. In other words, there will be triple bond on the left side of the triple bond. We will have hydrogen. We can essentially think about this hygienists and implicit Haijun. Then we will have siege two bonded to another carbon atom that will contain a triple bond and that would be pretty much it because on the other end of the triple bond we will have that carbon atom coming from C. O. T. And then there will be our hydrogen which again we can think of it as an implicit hydrogen atom. So we are ready to draw our structure. Let's see that the structure of an unknown AL kind will correspond to, as we said before, that would be a triple bond to begin with because we have our internal part of the AL kind. Right? We have hydrogen monitor that first carbon atom and this is our second carbon atom. So we can also label them. This is our first carbon atom which comes from C 02. Then we will have our second carbon atom which comes from C H. Then we will have our third carbon atom which will be seized to group our fourth one and then this will be our fifth one at this end because it comes from C O. T. Again. Right? So 12345. We have thrown carbons number one. Number two. Now we need to add carbon # three which corresponds to a siege to group so to do that we are just going to extend it and we are going to say that here we have our siege to. Now we need to draw a carbon number four which corresponds to a triple bonded carbon atoms. So we can add a triple bond and we can say that we are done because then we have our carbon number five in this position which contains hydrogen only. So if we label our carbon atoms, That would be carbon # one, which comes which actually produces C. 02. Right upon the cleavage. Carbon number two for the asset. Carbon number three for the middle carbon Over here, carbon number four, which produces this carbon within the carb oxalic acid group, and carbon number five, which produces another equivalent of C. 02. So this would be our unknown al kind. Let's redraw it more clearly so we can see how the reaction takes place and what our final products are. So our Al Qaeda would correspond to triple bond. Now we have seized two, then we have our next triple bond and we understand that this works for us, right. This structure undergoes those analysis. The two reactions are number one edition of ozone at negative 78 degrees Celsius. And then we have reductive work up with dimethyl sulfide in the presence of water. This gives us our melodic asset. So we can essentially state that we have a church O C. Right? We can draw it the way it's drawn in the problem plus two most of C. +02. So that's our overall reaction. Let's only label the final answer to this problem because we were interested in the unknown AL Kind itself here. This we got it. And we may simply conclude that the correct answer choice to this multiple choice question is a. However, if you have any questions, don't have States Leader comment down below in the comment section. Thank you for watching.

Oxidative cleavages can help to determine the positions of the triple bonds in alkynes. (a) An unknown alkyne undergoes oxidative cleavage to give adipic acid and two equivalents of acetic acid. Propose a structure for the alkyne. < of reaction>

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Oxidative cleavages can help to determine the positions of the triple bonds in alkynes. (a) An unknown alkyne undergoes oxidative cleavage to give adipic acid and two equivalents of acetic acid. Propose a structure for the alkyne. <
of reaction>