Propose a synthesis of the carbonyl(s) using the (i) ozonolysi...

Question

Propose a synthesis of the carbonyl(s) using the (i) ozonolysis pathways.

(a) Chemical structure showing two carbonyl groups with wedge bonds indicating stereochemistry.

Accepted Answer

All right. Hi, everyone. So for this question that suggests the synthesis of the following carbonyl compound using the Ozols reaction of an alt key. Now, the product in question is three R 4 s3 dimethyl six oxo ethanol. So for ease of reference, the first thing I'm going to do is label all the carbons of my pair chain. So here I'm going to label carbons with carbon one being the carbon of my aldehyde. But with that being said, let's go ahead and talk about the OS analysis reaction right now, recall that after OS analysis, what happens essentially is that the bond between the two carbons in your starting all keen is broken. So if I take an here recall that ozone allys takes place in two steps, right? The first is the addition of ozone followed by treatment with a mild reducing agent like dimethyl sulfide and like I said previously, right, when you go ahead and you react first ozone at a very low temperature to prevent over oxidation, right? Followed by treatment with dimethyl sulfide. What happens is that the two carbons in the pipe bond essentially break, right? The pi bond is broken and both carbons of the starting alkene are oxidized to yield either aldehyde or ketones. So here in this case, we're going to get two ketones and just for the record, right, there's kind of a cheat code to figure out whether you get an aldehyde or a keto. And in order to determine that you have to figure out how many other carbons, one specific carbon is bound to when you disregard the alkene itself. So here, if I look at the carbon that's connecting to R one and R two, the first thing I have to do is disregard the bond between the second sp two carbon of my all key. Because when I do that, what I'll notice is that this carbon, which I'm going going to label a here, carbon A is connected to two others. When we disregard the alkene, it's connected to R one and R two. Therefore, when it's oxidized, it's going to yield a keto. Now, if carbon A after disregarding the alkene, right, if it was connected to only one other carbon, then that would yield an aldehyde. But with that being said, let's go ahead and talk about the given product because here, right, our carbonyl compound has two carbons in the same parent chain. And recall that when that happens, that must mean that our starting material was a ring because when you cleave an all keen inside of a ring structure, what happens is that you expand or rather you open up the ring and yield two car bottles in the same pair of jean. So to figure out what the all key must have looked like, right? What you have to do is remove the oxygens of the carbons and instead unite the two carbonyl carbons with a pi bond. So in this case, what we're doing here is we are combining carbons one and six via a pie. So to better visualize this, I'm going to go ahead and redraw our product. And here I'm going to relabel my parent jane. So here, right, what I did was that I rotated the parent chain so we can visualize how the six member ring is going to form. Now because I had to rotate the parent chain, I have to consider what's going on with the stereo chemistry of the methyl groups on positions three and four. Now, in this case, I did not rotate carbon three, which means that the stereo chemistry there is going to stay the same and the Meho group there is going to remain as a wedge. Now, the same cannot be said for carbon four, right? Because in order to rotate my parent chain into a ring structure, I did have to rotate carbon four. Therefore, right, the met group on carbon four is going to be a wedge instead of a dash. So here right now that I've drawn what the ring structure is going to look like I'm going to go ahead and finish off the ring structure. So first, I'm going to go ahead and draw my six member ring that results after the carbonyl carbons at positions one and six combine, right? And I'm going to draw a pi bond between them. Now, my methyl groups are going to stay the same because in this case, they don't react. But I can't forget the carbon labeled number seven in my final product, which in this case is going to be another methyl group coming off one of the sp two carbons of my starting. And that results in my final product being four R five s 145 tri methyl cyclo hex one E. So here, right, when I react my all keen with ozone at a temperature of negative 78 degrees Celsius in the first step, followed by treatment with dimethyl sulfide. In my next step, I end up with the carbon product shown at the beginning of the question and there you have it. So here is the answer to the question right here is the final synthesis. And the bottom line here is that oz analysis is also referred to as the oxidative cleavage of an alkene, right, essentially the pi bond between the two carbons of the alkene is cut, right? Those two carbons are separated from each other and each respective carbon is then oxidized to yield either an aldehyde or a keto. And additionally, right, when the double bond is inside of a ring, you end up with two car bottles in the same parent chain as your final product. So if you stuck around to the end, thank you so much for watching and I hope you found this helpful.

Propose a synthesis of the carbonyl(s) using the (i) ozonolysis pathways.
(a)

Key Concepts

Ozonolysis

Carbonyl Compounds

Synthesis Pathways

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