Propose a synthesis of the carbonyl(s) using the (ii) dihydrox...

Question

Propose a synthesis of the carbonyl(s) using the (ii) dihydroxylation/periodic acid cleavage pathways.

(a) Chemical structure of a diketone with two carbonyl groups and stereochemistry indicated by wedge bonds.

Accepted Answer

All right. Hi, everyone. So for this question, let's show the synthesis of the following carbonyl compound using the di hydroxylation reaction of an alkene followed by the treatment with per ionic acid. So the product in question right is three R s3, 4 dimethyl six oxo Hain and for ease of reference, I'm going to go ahead and label all the carbons in my parent chain for future reference. So I'm going to label carbons one through seven of my parent chain with the first one, the first carbon being the carbon of the aldehyde. So now let's go ahead and recall, right, the sequence of events described in the question, the first involves the di hydroxylation reaction of an alkene. So if we take an all keen like the generic one that I'm drawing on the screen here, recall that after di hydroxylation, we end up with what's known as a 12 dial. And what I mean by that right is that by the end of the di hydroxylation reaction, the two carbons of your starting alkene are each going to have a hydroxy group attached. So we call that a 12 dial because the hydroxy groups are right next to each other, right? They're on adjacent carpets. Now, our 12 dial can react further with per ionic acid. Like the question is mentioning because recall that when you react a 12 dial with per ionic acid, which in this case is hio four. Recall that when you react a 12 dial with per ionic acid, what's happening is that the carbon carbon bond between the two carbons in the starting? All keen, right, that bond is actually cleaved. And so you're going to yield as a result, either aldehyde or ketones. So in this case, right, in my generic alkene, the bond between the two carbons that have the hydroxy groups are being cleaved to form aldehyde. So here when it comes to figuring out the All Keen that will yield the product given on the screen here, the first thing you have to do is remove the oxygens of each carbon and combine the respective carbonyl carbons through the pipe bomb. So here, right, the car boals are on positions one and six, which means that we have to draw a pi bond in between carbons, one in six to form the starting material. No, in order to visualize what this product would look like as a ring structure, I'm going to go ahead and rotate it in a moment. So here in order to visualize better the formation of the ring, I'm going to go ahead and rearrange my parent chain, so to speak, right. So here are carbons 127. And now let's go ahead and discuss what's happening with my meo groups on carbons three and four, right? Because here, right, the Meho group on carbon three is drawn as a wedge, whereas the Meho group on carbon four is drawn as a dash. However, right, it's worth mentioning that I did have to rotate my parent chain to kind of visualize how it's going to look as a ring, right? So because I had to rotate it, I had to rotate carbon four, which means that instead of drawing the Meho group on carbon four as a wedge or a dash, excuse me, I have to draw it as a wedge to take into account the fact that I had to rotate it. However, the methyl group on carbon three, I did not have to rotate. So it's going to stay at the sink, it's going to remain a wedge in this case. So now, right, I can go ahead and combine carbon's one through six via a double bond. So when I do that, I end up with a six member ring with a double bond in between carbons one and six and an additional an additional methyl group on carbon six, because I have to take into account the carbon label number seven in the parent chain of my product. Now I also have to take into consideration the two methyl groups that I have on positions three and four. And like we discussed previously, they're both going to appear as wedges. So that is what I'm going to draw right now and there you have it. So our starting material in this case is four R five s 145 tri methyl cyclo hex one E. So when we react are all keen here, the first thing we have to do is undergo the di hydroxylation reaction. Now, hydroxylation recall that it can be either sin or anti, meaning that the hydroxy groups can be either on the same side or on opposite sides. Now, a sin addition can be achieved using osmium tetroxide with hydrogen peroxide or we can also use cold dilute potassium permanganate. On the other hand, right, for anti addition, we would have to use a perox acetic acid followed by the treatment of acid in water. Now, for this question, we're going to go ahead and react with osmium tetroxide. So that's os 04 with hydrogen peroxide. Now treatment with osmium tetroxide like we said previously is a sin di hydroxylation. So that means that my hydroxy groups are going to be on the same side and that is going to result in a mixture of dyer, right? Because on one hand, on one hand, my two hydroxy groups can be depicted as wedges. And the other possibility is that both hydroxy groups can be depicted as dashes. So it's important to recall the fact that we formed two possibilities, right? Two dimes in this case. However, right, when we react either of these two ceres with per ionic acid, we end up with the same carbonyl compound pictured in the screen at the beginning of the question, right. So here is our aldehyde and here is Akito and there you have it, right? So here is your final answer to the question. And so the main takeaway from this question, right? Is that first and foremost, we can react in all keen, we can react in all keen to form a 12 dial after a di hydroxylation reaction. Now, the di hydroxylation can be either sin or anti but from there, right, your 12 dial can react further with per ionic acid to break the bond between the two carbons of your starting keen and yield either an aldehyde or a ketone on either end. Now, just for the record, right? When one of the carbons in your starting alkene is connected to two other carbons, when you disregard the bond to the second carbon in the alkene that yield a keto. And on the other hand, right, if you have only one other carbon connected, when you disregard the bond between the two carbons and your starting all keen, that gives you an all the Hyde. Now, I know that was a lot, right? But 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 (ii) dihydroxylation/periodic acid cleavage pathways.
(a)

Key Concepts

Dihydroxylation

Periodic Acid Cleavage

Carbonyl Compounds

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