Show how the following compound can be prepared from the given...

Question

Show how the following compound can be prepared from the given starting material. Draw the structure of the compound that is formed in each step of the synthesis.

Chemical reaction showing the transformation of an alkyne to a compound with a carbonyl group, indicating synthesis steps.

Accepted Answer

Hey, everyone and welcome back to another video. Write the synthesis of the following compound from the provided starting material include the structures of the intermediate products. The starting material is prop one and we are synthesizing two R three s 23 dimethyl ox in which the two metal groups are in sin arrangement. So now we are going to think about this problem backwards. The reason why we're thinking backwards is because it will allow us to understand the strategy behind any similar problem. What if we simply think about our file product? It is any oxide because we have a three membered ring, right? We have our oxygen and our co two carbon atoms within the ring. And the two metal groups are in an arrangement because the two metal groups are in sin arrangement. We understand that this is ac oxide. One of the common ways that we know to easily form oxides is by oxidizing our alkins. So if we understand that we have four carbon atoms in our chain, we are going to draw an alkin which we can oxidize to obtain the required oxide. And let's remember that the reaction produces CS epoxy if we start with AC alk, right. So we have that retention of our spatial arrangement of the substituent. So we essentially need to use an alkene which in this case would be cys but two, right? And we can easily oxidize it into an into an epoxy using meta chloro perox benzo acid. That's the region that we are using whenever we want to produce oxides out of alkenes. And now if we start thinking about the starting material, we actually have a three member chain. But then the final reaction we used an al which is a four member chain. So we can think about the production of that alkin starting with prop one I right, or propine. So because propine is our terminal alkin, we can remember that we essentially can easily deproteinate the acidic hydrogen or remove that proton and en enlar and make our chain longer, right? So how does that work? If we start with propine, we need to use a strong base. So basically, what we're going to do here is that we're going to use sodium amide, which is our classical base whenever we want to deproteinate terminal alkins, right. So what happens is that the amide and ion sodium as anion spectator captures the protein and we are forming a satellite and ion. So if we get our satellite, which would be our intermediate product, we can also indicate sodium as an ion spectator. Afterwards, we essentially want to use an appropriate alky halite to make our chain longer. And because we want to have four carbon atoms in our chain, we are going to use methyl chloride, for example. And the reason why we're using methyl chloride is because chlorine is a good living group, but you may also use bromine or iodine, for example. So acetal will just attack the electrolytic carbon followed by the loss of the lead group. So we will have an S N two reaction in which we will produce an internal alkin. So from here, we can just bring our scheme slightly downwards and we are going to draw the product that we get, which essentially corresponds to be two I. And now we are nearly done because we, we actually want to form a corresponding out of it. And let's remember if we are forming a cyst alk, we want to use a partial reduction because we're breaking one of our pie bonds only by adding one equivalent of hydrogen. So first of all, let's add hyen to our reaction because we are performing a partial reduction. And we want to make sure that we're using Lindler catalyst. And the reason why we need to use it is because we want to make sure that the can assist if you want to form trans keys, as you know, you will add sodium and liquid ammonia. In this case, if reaction must produce the final cy oxide, we we need to have a cyst alkin and a cyst alkin will be produced by the reduction of an alk using the Lindler catalyst. Well done. So we have our reaction scheme. Let's review everything that we have. So first of all, let's label our synthesis scheme and let take a look at the sequence of our reaction. We're starting with propine. And we are using sodium amide to dep proin it and produce sodium. I set the light, the ayite anion attacks the electrolytic center of methyl chloride in an essence two manner followed by the loss of the leaving group. And we are forming B two I and internal alkin which is then reduced using hydrogen gas and blenders catalysts to produce cys alkene. And finally, since alkene is oxidized, using methylal perox Bezic acid to produce a corresponding cys ep oxide that we need it. Thank you for watching.

Show how the following compound can be prepared from the given starting material. Draw the structure of the compound that is formed in each step of the synthesis.

Key Concepts

Alkyne Reactivity

Carbonyl Group Formation

Synthesis Pathways

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