Show how you would use a protecting group to convert 4-bromobu...

Question

Show how you would use a protecting group to convert 4-bromobutan-1-ol to hept-5-yn-1-ol.

Accepted Answer

Hey everyone and welcome back in today's video, we are going to learn how to solve the following problem, illustrate the use of a protecting group in the synthesis of hacks four in 10 from three bromo pro pan one. Oh when we think about this problem, we have to understand that we're given our starting material and the final product. So let's draw both of them because we want to propose the correct synthesis three bromo propane one, all is our starting material. And we understand that this is propranolol, which contains three carbon atoms with the alcohol group at position number one and the bromo substitution at position number three From this starting material, we are going to synthesize the required structure which is hex four I N one. Oh hex implies that there are six carbon atoms, there's a triple bond at position number four and the alcohol group starts at position number one. So we may simply draw our alcohol group at position number one and then we will move towards position number four. So that's carbon number 1234. And this is where the triple bond starts. This is where our fifth carbon is and we need our sixth carbon atoms. We are going to add it over here. That's our final structure that we want to synthesize. Our goal is to propose reagents for that transformation and we may notice that what's interesting is that we are still having our alcohol group in the final structure but we are losing roaming and we are making our chain longer also are producing that triple bond and one of the ways to enlarge the al kill chain as well as to add that triple bond is a lesson to reaction where we use the deep rotation stop so that the steak hygiene can be damaged, right? We can essentially lose it and we can have a lot of byproducts cycle ization of the compound and so on. So we don't want that. We want to make sure that we protect the alcohol group and let's see how that works. So if we redraw our sign material once again, we are going to immediately protect the alcohol. And one way to do it is to use try Aiso probo silo chloride, which is T. I. P. S cl followed by the addition of tri ethyl amine which acts as a base. Now, what we're going to do here, we're going to replace our acidic hydrogen with try isopropyl asylum group. And the result on structure would look like this oxygen would now be bonded to Silicon that contains three eyes. Oh, purple groups. This is our protecting group. Now we are protecting our oxygen. We essentially want to make sure that we're synthesizing our AL kind because we're trying to synthesize an AL kind. It makes sense to add three carbon atoms to our AL kinds since we are starting with three we have six total. So we may start with pro pine, which is a three member alkaline. We're going to draw that triple bond a math group. Now, what happens here is that we call that if we want to add that al kind to our chain, we want to get rid of grooming and grooming is a good leaving group. We can make use of an essence to reaction. So we are going to produce a satellite and iron out of it because there's an aesthetic hygiene here. And that's one of the reasons why we had to protect our alcohol because it's also it also has a mistake hydrogen atom. So there's an acidic hydrogen atom over here. And if we treat it with sodium a mite, right? This is a strong base. It will essentially be protein, it that acidic hydrogen to produce a satellite and the satellite will be a good nuclear pile. So when we produce a satellite, we may essentially True or the result on new profile. A satellite would then react as a nuclear file in particular. Now we are ready to react these two together and a satellite would essentially attack the electrical exposition which has a good living group followed by the loss of the living group. So this is an essence to reaction that we are performing right? And we are going to being nearly done right. If you look at the result on structure, we essentially understand that we are getting let's draw these carbons first. Right? 123. So that would be 123. Then we still have our protecting groups. So that we oh silicon three Aiso propel groups. Now instead of roaming at this position, we're actually creating a triple bond. So let's throw that triple bond. And we understand that there is a method group Bonnett said Now this is very close to our final structure. Right? If we look at the carbon atoms as five. Right? So I'm basically missing one carbon atom because we're actually attaching the triple bond, the triple bonded carbon at that position. So let's make a quick fix here. So we're going to add that one more carbon atom basically this carbon atom and then we're going to add that triple bond with a method group. So now we're good. Right that that's 123456. So we're nearly done. We're just thinking about the deep protection stuff and to deeper attack that compound. We are going to use bu four M. F. Followed by the addition of water. This is the tribunal ammonium fluoride which acts as a de protecting region. And essentially the addition of water will allow us to restore the alcohol group. And this is how we managed to get our final product going from this reaction towards our final product. So looking at this proposed synthesis just to review all of the steps that we have performed. We started with our compound. We said that okay we are going to protect that alcohol group using try isopropyl silo chloride addition of that protecting group allowed us to perform an essence a reaction of course. Since we started with propane, we had two D. Protein idealistic hygiene to make sure that we produce a satellite. A satellite is a strong nuclear file. It attacks that electrolytic position. We're creating our intermediate right. That's how we can call it because we still have that protecting group. And then the final step we needed to de protect it to form our final compound. Seeing all of these steps, we understand that the correct answer choice to this multiple choice question is C However, if you have any questions, don't hesitate, slave your comment down below in the comment section. Thank you for watching.

Show how you would use a protecting group to convert 4-bromobutan-1-ol to hept-5-yn-1-ol.

Key Concepts

Protecting Groups

Nucleophilic Substitution Reactions

Alkyne Formation

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