Show how a protecting group might be used to make these reacti...

Question

Show how a protecting group might be used to make these reactions successful.

(b)

Accepted Answer

All right. Hi, everyone. So this question is asking us to provide a reaction scheme that includes protecting groups for the following reaction to be successful. The reaction scheme itself is going to take four steps. So first notice how our starting material contains both an oxide and an alcohol in our product. The alcohol is still intact, but the oxide ring has since opened to produce a dial, not only has the oxide ring opened, the carbon chain has also increased. So with that information, we get a sense on what reactions are needed in the sequence because number one, the oxide ring is going to open and the carbon chain gets extended. Recall that both of these steps can be accomplished at once because the outermost carbon of the oxide is relatively unhindered. We can use an sn two type mechanism to not only open the oxide ring but also extend the carbon chain using a carbon nucleo file recall also that a very common example of a nucleic carbon atom is in the acetyl ion. So the acetyl ion must match the R group that's being added to our carbon chain. So that's one step. But because the acetyl ion contains a triple bond. We're also going to have to reduce that triple one considering there are only single bonds in our product here. So as of right now, we have two steps, one is going to extend the carbon chain and open the oxide using an acetyl nucleo file. And the other is going to reduce that triple bond on the acetyl ion. But what about the first or what about the remaining two steps? Well, they have to do with the fact that the hydroxy group on the left side of the structure is preserved from start to finish, meaning it stays the same. So because of that, we have to make sure that this alcohol group is protected from any unwanted side reactions. With that, we have our remaining two steps because the first step in our synthesis must be the protection of the alcohol group. And our last step is de protection to remove that protecting group. So let's go ahead and put this into practice, starting off by redrawing are starting material. But here is the oxide and here is the hydroxy group. So as mentioned before, if we want to prevent any unwanted side reactions involving this alcohol group, our first step is to protect it. Now, one reaction that comes in handy is the formation of a silo ether to protect that hydroxy group because by converting our alcohol into a silo ether that silo ether is going to resist various different reagents and it can be removed relatively easily once it's no longer needed. So first lets take this alcohol and reacted with chloral tri isopropyl silent. So that's silicone with three isopropyl groups attached to it and one cory, this is also done in the presence of tri amine, which is necessary to neutralize any HCL formed during the reaction itself, just open up some space here on the side. So for our first intermediate, our carbon chain is going to stay the same and our apoc side is going to stay the same as well. But now the proton of the hydroxy group get replaced with the silo group. So oxygen is now connected to silicone and silicone still has three isopropyl groups. So at this point, we can introduce our acetyl ion without interference from the alcohol because now that's protected. So as mentioned before, right, the R group of the acetyl ion must match the R group that's being added on to our final product. So in this case, the acetyl ion in question, should contain a four carbon chain with the triple bond in between carbon one and two here as well as a methyl group on the third carbon, the terminal carbon should have a negative charge. And we can use sodium as a counter ion though I think it would be better to draw this arrow pointing downwards just for the sake of space. So let me go ahead and do that. Right. So scrolling down here, I am going to go ahead and redraw that same acetyl with sodium as a counter ion. So thats a four carbon chain with a metal group on the third position, the acetyl nucleo is going to attack the less hindered carbon of the oxide in an SN two type mechanism. And because this is an SN two type mechanism, a celly is going to attack the apo side from the back considering the oxide ring is facing the front. So after the resulting oxide ion is pro needed, we get an additional hydroxy group. But because there's no triple bond present in our final product from here, we can carry out catalytic hydrogenation to reduce this all can sorry all kind down to an all can. So the next step is going to involve two equivalents of hydrogen gas to reduce both pi bonds of the all kind in the presence of a palladium catalyst. So now we have the carbon chain that we see in our final product though, before we draw the product of this step, I recognize that now I actually did not draw an additional bond between the oxide carbon and the acetyl carbon. There we go. So now we have the six carbons from the starting oxide as well as five more carbons from the R group of ace ali or the acetyl ion. All right. So now that I have the carbon chain with 11 carbons in total let me go ahead and add our silo ether and our hydroxy group. So, because our final product is a dial, our final step is going to be the removal of our protecting group or the protection. So the nice thing about protecting groups is that they're actually very easy to remove once they're no longer needed Syo ethers have a very high affinity for fluoride. So our final reagent is going to be tetra buttal ammonium fluoride in the presence of water, that's going to remove the silo group and replace it with our proton for the alcohol. So now we have our dial like so, so now the product that we made matches with the one required for our transformation. So now I'm going to recap all of our steps here in list format to make the final answer a bit easier to see. So our first step involved the protection of the alcohol group using a silo either. So our re agent for the first step was Chloro tri isopropyl cy with tri ethel amine. After that, our second step involves the nucleic attack using an acetyl ion. So here sodium was our counter ion and our ace aide ion contained a four carbon parent chain followed by an additional methyl group. On the third position. After that, we used two equivalents of hydrogen gas in the presence of palladium to reduce the all kind down to an all can. And then our fourth and final step was de protection with Tetra Buttal ammonium fluoride in the presence of h2o this removed the silo ether and produced a dial and there you have it. So with that being said, if you watch this video all the way to the end, thank you so very much for doing so. And I hope you found this helpful.

Show how a protecting group might be used to make these reactions successful.
(b)

Key Concepts

Protecting Groups

Reactivity of Functional Groups

Deprotection

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