When compound Z is treated with ozone, followed by dimethyl su...

Question

When compound Z is treated with ozone, followed by dimethyl sulfide and washing with water, the products are formic acid, 3-oxobutanoic acid, and hexanal.

Propose a structure for compound Z. What uncertainty is there in the structure you have proposed?

Accepted Answer

Hey everyone and welcome back in today's video, we are going to learn how to solve the following problem. When an unknown compound was allowed to react with ozone followed by the metal sulfide and a small amount of water it produced acetic asset. Three metal beats and see own and three OXO pence anolik asset determined the structure of the unknown compound, in which information cannot be determined with this reaction. According to the problem, we are performing two reactions. Number one is the addition of ozone. Number two, treating the structure with that metal sulfide and a small amount of water. So if you recall in combination, these two reactions imply that we are observing and ozone knowledge is reaction. Now this is really important for us to understand because there are two cases of oz analysis. What if we start our reaction with an elk in supposing that we have hydrogen and unknown L kill group R one R two R three. To predict the products correctly. During the ozone Oz's reaction, where number one is the addition of ozone, then dimethyl sulfide in a small amount of water. We're just breaking that carbon carbon double bond and the easiest way to predict our products is this one, we can just draw the remaining fragments of that structure. So there will be now two fragments because we're breaking that bond, there'll be R two R three over here and the other fragment. If you just draw it nearby, right, you can notice that you can fuse them together into that original one double bond. So they'll be hydrogen and our one. The last thing that you want to do is just add oxygen atoms here and here and you're done now. This is the logic that we are going to utilize in this problem. It's very useful for us to think about this reaction scheme. But in addition to that, what if we produce car broke cilic assets? Because if you look at these two compounds, they produce ketones and Aldo hides, right? Al kings produce ketones and Aldo heights upon OS analysis. But if we want to get an asset, we have to keep in mind that the starting material must be an AL kind. We must have a triple bond with R one and R two groups. And if we perform analysis on that one, number 103, number two, diet muscle sulfide in water, we produce car broke Cilic assets and there will be two car broke Cilic assets, one with R one And the other one Would be the car box cilic asset with our two. And using these two models, we are just ready to analyze them. If we notice we have two car broke Cilic assets. So these two carbon atoms within the book cilic asset group, they must form an alkaline. So these two carbons in combination form a triple bond. In addition, there's that metal group, which is our one. So we can say that there must be an R1 group sticking out. This must be an internal alkaline, right? Because we don't have any hydrogen anywhere. So that would be our our one anyway are ready to go. Now, if you look at this carbon over here, it's that part of the triple bond. So you may just start drawing it step by step. So let's start by drawing a triple bond. Now we understand that there's an R1 which is a metal group sticking out. So let's add that metal group. Right? And then this is the next carbon over here within the triple bond. So if we label them appropriately, that would be carbon number one, carbon number two. So that's our carbon number one. Carbon number two. That methyl group is just our our once we can just admitted for now because we're not that much worried about the correct numbering regarding I A pack. We just want to see the relative positions of our carbons. So let's just call it carbon A. Okay, so the method group will be carbon A. Now thinking about the remaining part of the molecule, we have to be very careful. We notice that there's no Carbonell. So we can just say that the next carbon would be the siege two groups. So let's just add it. Let's just continue and go from carbon number two to carbon number three. Carbon number three has a siege two groups. So that would be our carbon number three. That's our carbon # three. And now carbon # four is actually carbonell group and that's super important to understand because this is a key tone. A carbon group with two alcohol groups meaning this must be part of an alkaline, right? It must form a double bond. So we went to find another carbonell group from the double bond. And here this is another key tone. So these two carbon atoms in particular this one and that one they form an al king. So we already have our carbon number three. That means carbon number four. If we go up, That'd be carbon # four, it must have a double bond. And this is our carbon number four. Let's label it in a different color. Carbon number four, it must be bonded to carbon number five. So this is carbon number five. Right? Because the two carbon groups form an alkaline. So you may just say that the other carbon is carbon number five. Now let's think about our substitutions. So carbon number four has an ethyl substitution and we may simply draw an ethyl substitution from carbon number four. So that's an ethyl substitution. Right? And now if you notice carbon #5 has to substitute prints and we can just add them over here. Right, So that's simple. Let's add that metal group first over here. So that be our metal group and the remaining part or the remaining substitution would be that that substitution over here, which is an isopropyl group. So if we add an isopropyl group here, we are done. We can just label our carbons for complete clarity. So we set carbon number four and five to form a double bond. So we can just say that there will be carbon number six and seven within the ethyl group. So that's our six and seven. And regarding these ones, we can just say that the metal group is at position eight. So that's our aid And this will be our nine, son and 11. So you can just say that this is nine son and 11. So you can clearly see how that works. If you simply break that double bond, you will notice that you are indeed getting the first structure when you add oxygen in the middle, then the other part is obtained on the other side. If you break that double bond, you will notice that you're getting this compound because you have to break the double bond and the triple bond add oxygen over here and make this carbon an asset. And then the remaining part just gives you that acetic asset because you only have one carbon atom, the metal group. And you're turning that carbon into the curb oxalic acid group. So this is our structure. Let's just redraw it clearly because this is our motto. So if we started step by step, we can just draw our carbons neatly double bond in this position there same metal group an ethyl group. Siege two. Then we have that triple bond ceased to triple bond and a method group bonded. So that would be our final structure. But we also have to keep in mind that there was a question whether there is any uncertainty in the structure that we have proposed and the answer is yes. There is uncertainty because there's the presence of a double bond, we cannot really say is at E or C. I. Zimmer, because if you look at the double bond, it's really important to understand that E and C configurations are relevant in general, Right? We can have two possible eyes mars either the highest priority groups are on the same side of the double bond or opposite sides. So we cannot really tell whether it's an E or C. Iseman. Right? Just as you're used to thinking about since transistors, we cannot really say whether it's an E or C. I. Is more because this method only tells us that we're breaking the double bond and the triple bond, but the original configuration cannot be determined from that. So this is really important. We cannot tell whether this L team in particular over here within this double bond has an E or C configuration and now we may conclude that the correct answer to this multiple choice question is a. However, if you have any questions don't have states, leave a comment down below in the comment section. Thank you for watching

When compound Z is treated with ozone, followed by dimethyl sulfide and washing with water, the products are formic acid, 3-oxobutanoic acid, and hexanal.

Propose a structure for compound Z. What uncertainty is there in the structure you have proposed?

Key Concepts

Ozonolysis

Functional Group Identification

Structural Ambiguity

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