Suggest a synthesis for the following molecules beginning with...

Question

Suggest a synthesis for the following molecules beginning with organic molecules containing three or fewer carbons. [You will need to use a protecting group in these syntheses.]

(a) Chemical structure of a linear carbon chain with two hydroxyl groups and one double bond.

Accepted Answer

All right. Hi, everyone. So this question is asking how would you synthesize the following compound using organic molecules with no more than three carbons. Hint a protecting group will be required for the synthesis. Now, the end product that we have on the screen here is pep three I 17 dial. So for ease of reference, I'm going to go ahead and label the carbons of the parent chain. So I'm going to label carbons 127. And in addition to that, right, what I want to point out is the fact that we have to build a seven carbon molecule from smaller fragments of no more than three carbons. So what we're doing here and this is a key word, right? Or key phrase, maybe, but here we're extending the carbon chain, right? So what we have to do is we have to add new carbon carbon bonds as we go along. Now, with respect to that, I want to point out the fact that we have an all kind within the carbon chain itself right here. We have an internal all kind. Now, the reason why that matters is because recall, we can go ahead and undergo s in two reactions with acetal light or acetal light ions. Because recall that if we take a terminal all kind like acetylene here recalled terminal, all kinds are actually relatively acidic or rather the hydrogen of the terminal, all kind is relatively acidic. So if we were to subject this terminal all kind to a base, right? Or we add a base to it, then we get the acetal light ion. Now, the acetal light ion recall is an excellent nuclear file. So here we can use the aceta light ion as a nu file and an sn two reaction to extend the carbon chain and Oula our all kind here. So if we were to go ahead and examine the Ocu groups on either end of the triple bond, then we can figure out what we have to add to our acetylene here in order to alate it on either side. So here, right, carbons five through seven of the parrot chain constitute one potential fragment because carbon is five through seven have three carbons exactly which is our maximum here. So in order to add carbons 5 to 7 of the parent chain, we can react our aceta light ion with for example, the following. So if we take for example, three promo three bromo sorry propan 10, then our aceta light ion is going to attack carbon three of the parent chain and that's going to result in a new bond between our sp hybridized carbon and carbon three of our three bromo propan 10. However, this is not exactly the case because yes, this is a reaction we want to occur. But we have to take into consideration the fact that we have a hydroxy group as well on the same molecule. Now recall that acid based reactions occur much faster than substitution reactions. So therefore, we have to keep in mind that even though aceta light is a nu file, it is also a base. So in this instance, right, acetal light would behave as a base and dep proton, our hydroxy proton. So this is where our hint comes in handy, right? Our first step is going to involve a protecting group because we have to make sure that acetal light cannot react with our hydroxy proton. In other words, we have to protect our hydroxy group from an acid based reaction. So here, right, we have to recall the role of silo ethers when it comes to the step of the process, right, recall that silo ethers are excellent in protecting groups for alcohols because we can install them easily and we can remove them easily when we're done. Right. So the first step we have to do is protecting our hydroxy group to make sure that we're not getting any unwanted reactions. Right. So here I'm redrawing R three bromo propan 10. And the first thing I have to do is create a silo ether to make sure that I'm getting no unwanted side products, right? No unwanted acid based reaction. So our first step here is the addition of try isopropyl silo chloride or T IP SCL. However, right, recall that in addition to adding T IP SCL, we also have to include a base to facilitate the removal of the hydroxy proton. So we can substitute it with a bond between oxygen and silica. So the basin question is going to be try a for a me. So here, like we mentioned previously, our bromine is going to remain untouched. But instead of having a bond between oxygen and hydrogen, we're going to have a new bond between oxygen and silica in our silo ether. And the reagent in question is tri isopropyl silo chloride. So here the silicone atom is substituted with three isopropyl groups. And now, right, my alcohol has been protected because now when I introduce acetal light, my aceto light ion has no choice but to behave as a nuclei because in this case, there is no hydroxy proton to remove. So now I can go ahead and proceed with the next step, which is the addition of a sea delight. So at this point, right, we've extended the carbon chain from three carbons now to five carbons incorporating the two carbons of my all kind here. So here is my triple bond and here is my silo ether group still. So at this point, we can move on to the next step just after I move my intermediate. So you can see it better so far, so good, right? But at this point, we have five carbons of the seven in total that we need for our final product. So at this point, we have to recognize that are other terminal hydrogen of our all kind here must be al coated as well because we have another r group on the other end of the all kind itself or rather on the left side. So here, right, notice how we have only two carbons with one oxygen at the end, right on carbon one. So in this case, recall that we can use an oxide opening to get an additional alcohol because here, right after we deproteinate the last terminal all kind hydrogen that we have here, we can go ahead and react it with a two carbon dioxide oxy, followed by protonation to give the alcohol group shown, right. So first we can go ahead and react with the base like sodium amide to get another aceto light ion and our protecting group is still intact. So here I'm going to make more room on the side here for my isopropyl groups before we proceed. But now we have an additional nuclear file that can participate in another substitution reaction. So here recall that the oxide ring opening is twofold, right. The first step involves the addition of oxy followed by protonation with HDO plus because what happens in this case, right is that my aceta light ion is going to attack either of the two carbons of the epoxy which causes the ring to open and that causes the formation of an oxide ion. Now that oxide ion is derotated or excuse me pronated in the subsequent step to ensure the formation of an alcohol by the end. So here, right, we're increasing the carbon chain by two this time given the structure of oxy rate. So after that's done, here are the two carbons from oxy and the alcohol from oxy after potent nation. And on the right side of the alk kind is my silo ether still. So at this point, right, at this point, I have the carbon chain that I need, right? Because here I have seven carbons in total. So at this point, our final step is going to be detection because now my silo ether has run its course, right. I've completed my synthesis and now I don't need my protecting group there any longer. So my last step is going to be the removal of my silo ether and recall that I can do that using a fluoride salt such as te tribute ammonium fluoride because the fluoride and ion is what I need to remove our silo ether group and bring back my hydroxy proton. So it's Tetro Butto ammonium fluoride in the presence of water. And here I've run out of space. So I'm going to go ahead and actually draw this pointing downwards so that I have space to show you the full product. So again, I'm adding my fluoride salt with water and I get my completed product by the end and there you have it. So at this point, we finally made it to the end. So let's go ahead and recap all of our steps, right? Our first step was the protection of the hydroxy group of three BROMO propan 10 to ensure that the aceta light ion doesn't participate in an acid based reaction from there. After protection, we added aceta light to increase the carbon chain by two. And then we went ahead and detonated the other hydrogen of our terminal alkin, they yield another aceto light ion which we then subjected to oxide ring opening, which generated an additional two carbons on the parent chain and an extra hydroxy group after pot. And from there, right, our final step was detection using a fluoride salt in water to yield our final dial at the end. So we made it to the end, right? And if you stuck around, I thank you so much for watching and I hope you found this helpful.

Suggest a synthesis for the following molecules beginning with organic molecules containing three or fewer carbons. [You will need to use a protecting group in these syntheses.]
(a)

Key Concepts

Synthesis Strategy

Protecting Groups

Functional Group Interconversion

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