Show how you would synthesize each compound, starting with alk...

Question

Show how you would synthesize each compound, starting with alkenes or cycloalkenes that contain no more than six carbon atoms. You may use any additional reagents you need.

(b)

Accepted Answer

Everyone. And welcome back in today's video, we are going to look at the synthesis problem which reads, proposing synthesis plan for the following compound. You can use any al Kins or psycho Elkins that contain no more than six carbon atoms as a starting material. And when we look at the structure, whenever we identify our heater atom, which is oxygen, we can essentially break it apart and think about all of these Adam's substitute prints and functional groups added later. Because if we do so, we notice that we indeed can have five plus 16 carbon atoms as our starting material. So we can propose This to be our starting material because it says Elkin, right. So we are going with one method cycle band one. In this makes sense to us, we have our method group, our double bond. The problem allows us to use Elkins only. That's why we're also adding a double bond here. So this is our starting material. How can we get all of these substitutions later? Essentially, we noticed that we have oxygen bonded to it, right? Oxygen is bonded to the ring. And what is also important is that oxygen is anti to that metal group right there, pointing in opposite directions. Oxygen is pointing down metal group is pointing up anti means that we can form an alcohol using hydro operation oxidation. And this would give us anti arrangement. Let's see why. Essentially if you perform hydro operation oxidation, which requires two steps. First, adding, worrying in the presence of tetra hydro Furin as our solvent. And number two treatment of the resulting intermediate with hydrogen peroxide and hydroxide recalled that this reaction performs a syn addition of hydrogen and beach. To write, essentially, we're looking at our double bond and we want to make sure that we are forming the ante Markovic of product, meaning hydrogen is added to a more substituted position and beach to the remaining part of the molecule is added to a less substituted position. This is part one and this is a syn addition, right? And if this is in addition to the beach to group will be pointing up and hygiene will be pointing up. So to give you a better idea, to understand it properly, we can grow our intermediate and we understand that the metal group will be pointing down. This hae jin is sin To be H2 groups, right. So this is our intermediate because these are in syn addition the methyl group is pointing down and therefore now it's anti to be used to. And in the in the next step, when you're adding hydrogen peroxide, you can essentially say that nothing changes regarding configuration at this position, the bond will still be pointing up because it's not an essence to reaction, right? It's not following an essence to pattern. So you have a retention of configuration at this position and the product that you are forming would be the desired alcohol where that metal group is pointing down and the alcohol group is pointing up. Let's keep in mind that this in addition, can take place two different ways, two wedges or two dashes. So we have plus in anti armor, plus in anti armor. That's why you may think that my structure doesn't seem to be consistent with a bond arrangement. Actually, this because we're forming a pair of finance commerce. So one of them will have a wage group pointing down and the metal group pointing out just as displayed over here. So for the purposes of this problem, we are going to use in and in our next step. So keep that in mind. Okay. Okay. So we have formed our alcohol, what's next? So when we form our alcohol, we need to think about how to get that carbon chain right? We went to bond a carbon chain to oxygen, but let's not rush. And let's think about this part over here. There's a Cyclo propane ring recall that cyclo propane can be formed using the Simon's Smith region. In other words, you're just adding that cease to group across the double bond forming a ring, right? So if we just break these carbon carbon bonds, we can visualize that there must be a double bond at the beginning. So if you look at our carbon atoms that we want to re carbon atoms, we need a substrate to form. So that substrate would be 123 carbon atoms, right? And as we said, 123, and we need a good leaving group. So let's suppose, bro mean because we want to think about the substrate that can participate in a potential substitution reaction, the position of the double bond must be over here, right? Because he wants to make sure that during the attack, the double bond is further from you. Now, this is our possible substrate for a new catholic substitution reaction. And indeed, when you think about your alcohol, you can form a nuclear file out of it to do that. You want to use a strong base that can be pertinent that acidic hydrogen, one of such bases is sodium hydrate. You can use sodium A might if you wish or potassium hydroxide, but this base is even stronger, which would be protein eight hydrant and form your elk oxide. And I am. So when you form elk oxide, we can already think about our unanimous okay. So we can just draw the nan timer that would be oxygen with a negative charge that metal group, the elk oxide and on with essential attack followed by the loss of the leaving group. This is how we can add our allele chain. So let's see what we got. That would be our product destroyed five member doing method group oxygen bonded to 123 carbon atoms. So that'd be 12, three carbon atoms and we have a double bond. And our final step is to use the Simon Smith region to form that cyclo propane range. Just as we said before we call that the Simon Smith region involves methylene iodide in the presence of zinc with an impurity of copper. We can use either copper or copper, one chloride that's perfectly fine. Okay. And this is how we get the final product. So this is more like the analysis of our steps. Let's just conclude it in a concise and clear scheme. Okay. We are going to outline the correct synthesis step by step as we said before, our starting material is one method, Cyclo pant one in. And we said that in our first step, we are just going to use hydro abrasion oxidation which requires two reactions to be carried out, boring in the presence of tetra hydro Furin. And number two addition of hydrogen peroxide in the presence of hydroxide. This essentially forms us two alcohols that are in an tumors. And the one that we're going to draw will now be the one that we need to get right. We are going to draw the correct in. And for our required reaction. So that would be the alcohol group pointing down. And the metal group pointing up, we can say that we get an answer straight and the step, right? Because there are two possibilities to add, boring above or below the plane. But we are only interested in this specific in. And we said that after this stop, we are going to treat, we can essentially erased so that you don't get confused. But keep in mind that indeed there will be an an and tomorrow we'll just add it at the very end. Okay. So now we're going to use our strong base sodium hydrate. And when we format we get our elk oxide and iron, so we can just draw elk oxide. The result on intermediate is our elk oxide sold that be oxygen with a negative charge. And the metal group. Now, when we form that Alcock side, our next step is to add a little bromide because if we do so, we may notice that addition of a little bromide which essentially corresponds to the given structure. During that essential reaction, we are adding our L L substitution C oxygen and we can now draw the results on product for this reaction. There will be oxygen, bar nitty three carbon atoms. Now 123, there will be double bond and there is that metal group. So we got it. And the final step is the application of the Simon's Smith region. So we can say here we are using methylene iodide sang in the presence of copper impurity. Okay. And we said that the Simon Smith Regent allows us to form that cyclo propane ring over here. Meaning this is how we can get our final product. Let's draw our final product can essentially say that everything looks exactly the same except from the fact that now we have oxygen with 123 carbon atoms. There was that double bond which now becomes our cyclo propane ring, right? All good. And now we still have this metal group. And now the world last thing is that you have to recall, we previously said we have formed and an antimatter in the step, meaning in each step that in Antrim are basically propagated until the worry end. Okay. So that's why we said instead of constantly adding or drawing that structure, saying pleasant, anti, more pleasant and more pleasant. And we just made it more concise for us to see just the structure we are interested in. But there's also an anti performed in each step until the final step. So we said, let's just write at once at the very end. So this is our complete synthesis several steps here. And we can conclude that the correct answer to this multiple choice question is c However, if you feel that this problem was complicated for you, feel free to drop your comment down below in the comments section. Thank you for watching

Show how you would synthesize each compound, starting with alkenes or cycloalkenes that contain no more than six carbon atoms. You may use any additional reagents you need.
(b)

Key Concepts

Alkenes and Cycloalkenes

Reagents in Organic Synthesis

Functional Group Transformations

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