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.

(c)

Accepted Answer

Hey everyone and welcome back in today's video, we are going to propose a synthesis plan for the following compound. You can use any Elkins or psycho Elkins that contain no more than six carbon atoms as the starting material because the problem suggests us that we can use cycle Elkins and no more than six carbon atoms. We can simply start by breaking this carbon oxygen bond and we can just use a five member drain because it's a psycho Elkins, we cannot use two double bonds, we can only use one of them. So let's suppose that we are essentially starting with cyclo Pantene. Now let's draw a cycle of fencing and we have to be careful regarding where we assign our double bond within the final structure. We have two double bonds at these positions and essentially there is a way to produce one comma three. Buta dying. Right, because that's what we have. There is a way to produce these two double bonds via elimination. And you may simply be eliminating these hydrogen atoms from your structure If you have two good leaving groups or let's suppose, theologians such as bromine. If you're essentially eliminating this hydrogen using a strong base of producing a double bond fold by the loss of the leaving group and then you repeat the process and you get your one comma three beats a dying. So essentially we want to use a double bond at this position, right? Because we will then use an electro filic edition of bromine and we will be able to perform an elimination afterwards. So this was a brief discussion of our first steps to understand how we decide regarding the position of our double bond in the first structure. So we are going to draw a five member grain and as we just said, we are going to add a double one in this position. Now, because our goal is to produce two double bonds and use a double elimination. Our first step is to make sure that we add roaming in the presence of carbon tetrachloride, that's our standard solvent. And we have electrolytic edition of booming across the double bond. So in the first step we will essentially get two bromo substitue ints right bonded to these two carbon atoms, one grooming and another one adjacent said, okay, now we are going to perform that elimination stop in the elimination step. We are first of all looking at the first alpha carbon which has the leaving group, right? And the beta hydrogen over here will be eliminated. We can use a strong base such as potassium hydroxide in the presence of heat which promotes elimination. This would be our E two reaction. We are forming our double bond fel by the loss of the leaving group and we repeat the same process afterwards. So we can draw it on the same reaction scheme. You may notice how we have this beta hygiene and this position hydroxide attacks the proton. We are forming a pi bond followed by the loss of the leaving group. So this would give us two double bonds desired. And this is really good for us. We can throw our five member ring and now we have two double bonds. The region that we used was potassium hydroxide. We can state access. But that's not really important. Let's keep in mind that potassium hydroxide is in excess. We're using a solution of potassium hydroxide in the presence of heat. So we get our one comma three vita dying and we want to think about adding this whole substitution to our structure. The world. First thing that we can do is just to think about the fact that this structure is an ether right? It has oxygen which has two substitutions now because it's an ether. One of the ways to form it is to first of all dominate the a lalique position over here and then replace roaming with an alcohol group. As long as we have an alcohol, we will be able to form an ether easily. So we understand that our first step is to make sure that we add roaming to olympic position. This is the olympic position because notice how this carbon atom is adjacent to a double bonded carbon atom. Same over here. Right? So in order to ruminate the elect position, we are going to use NBS NBS is the reaction that allows us to perform nomination at the position. So if we use N. B. S, we are going to replace one of the two hydrogen atoms with bromine. So we are going to get our structure to double bonds. And then bro mean at the olympic position. Now as we previously said, we want to make sure that we're getting an alcohol. So to get an alcohol we can use a simple sn two reaction. Right? If here we use an E two reaction here we are going to use an sn two reaction. Two substitute bombing with hi Troxell group. So O. H minus would do the work. We cannot have any elimination product because if you notice these two positions, even though they contain hydrogen atoms, they're actually sp two hybridized and elimination will not take place at these positions. So we can only have a substitution product. Hydroxide attacks. This is followed by the loss of the leaving group and we are forming our alcohol. So let's draw the result on structure for this. Stop That will be our five member ring. Let's draw it a bit more clearly so that will be our five member ring. We understand that we have double bonds over here and now we are essentially getting our alcohol O. H. Instead of bromine. Okay, now if we get our alcohol, we want to form a nuclear file out of it because if we form a nuclear file out of it, perhaps we can use the volumes and ether synthesis or something similar to that. So let's see how that works to form a nuclear file out of this alcohol and in particular a strong nuclear file that can possibly participate in a nuclear Felix substitution reaction. We need to make sure that we are removed this proton. So to remove this proton, we can essentially use a strong base such as sodium hydride, let's say N A H. We call that sodium hydride is a very strong base because the hydrate and ion with a negative charge would essentially attack that acidic proton. So we can expand our oh age group and we can show that the negative charge of hydrate would attack forming hydrogen gas and then these electrons would be transferred onto oxygen in the form of a lone pair. And this is how we get our elk oxide and iron. So that will be our five member drink our double bonds. And now we have produced our strong nuclear file. Okay, now that we have formed our strong Nikola file, one of the ways is to see that the ether that were forming. If we go back to the original structure, the ether that we're forming also has an alcohol group. And this implies that we may use an epoxy side for our reaction. We would have an ip oxide ring opening and then we would protein eight. Our results on Alcock side to get an alcohol group due to the fact that it's not only an ether that we have, but also we have this alcohol group. We may use a corresponding ip oxide? The only question is that, what kind of an ip oxide would that be? If you count the number of carbon atoms, you notice that you have 123 carbon atoms. So let's draw an IPAQ side which has a metal group bonded to it. Okay, let's suppose that our epoxy looks like this. We know that there are two carbon atoms in a ring bonded to oxygen. And because we need that extra carbon atom, we are going to add a methyl group at any position. Now we need to make sure that we remember which position the negatively charged oxygen would attack because this region is basic. Right? There are no acidic conditions and the negative charge and oxygen essentially implies that this is a good protein except er so this is a basic structure. So in basic conditions we always attack the position of the pox side, that is less substituted, meaning this position in blue, the negative charge on the oxygen. In basic conditions would attack a less substituted carbon atom and then we would have the ring opening of that box side. And we may notice that if we draw the resulting structure for this one, we get our five member drink with these two double bonds, then we have our oxygen atom which is now bonded to this carbon atom. Right? So we can first of all slowly draw this carbon atom, there was ring opening. So at the next carbon atom we will have oxygen. Right? Because we are breaking this carbon oxygen bond and then oxygen still belongs to the adjacent carbon. So this one there will be a negative charge on the oxygen because now auction only has one bond and the same carbon atom that can that has oxygen also contains a metal group. Okay. And this really resembles our final material meaning our choice was correct. We just need to make sure that we are performing the proto nation stuff and we can use water for that. We can use hey what a molecule. So the negative charge on the oxygen would essentially attack hydrogen from the water molecule forming hydroxide. Right. So basically restoring the basic conditions and then we essentially get our final structure which indeed corresponds to our final desired product. So we get oxygen our carbon chain and now Alcock side gets protein ated. That would be our alcohol group. This is indeed our final product. Now I understand that the scheme was very Lenzi but we have finished our analysis and we are ready to conclude all of the steps for our synthesis. So let's go back to our starting material and let's draw a reaction arrow to illustrate all of the steps in one condensed form. So we can see what our answer is. Let's draw a reaction arrow actually we are forming and that's as our products. So our reaction era would go from right to left in this case because this is our product. Our starting material for this reaction was cyclone fencing. So let's go through our five member ring. Let's add a double bond at any position. Right? Because that's cyclamen scene. And let's start listing our steps. Step. Number one was the treatment of cyclops dancing with blooming in the presence of carbon tetrachloride. Here we have added two bromo substitue ints right. That was an electra. Filic edition. Number two was the treatment of the result on structure with potassium hydroxide in the presence of heat To perform that double elimination. And this gave us two double bonds in the ring. Number three. Treatment of the resulting structure with NBS to add grooming to our position. Right? Number four addition of potassium hydroxide. Once again we said hydroxide. But we need to make sure we include some iron spectator. Let's stick with potassium addition of potassium hydroxide allows us to perform an SN two reaction in this step. Okay, now, number five, the 5th reaction was addition of sodium hydride to make sure that we produce our strong nuclear file or D protein it our alcohol. So that was sodium high trite. Number six. The sixth reaction over here was the treatment of the result on elk oxide with propylene oxide. So we can say propylene oxide which corresponds to this structure. Right. That's our appropriately in oxide. And then the final step was the addition of water to make sure that we protein ate the result on the elk oxide after the ring opening. And we get our final product. So this is our complete synthesis. We can just label it that and when we label it, we have to make sure that we read it from right to left once again, right, Because this is our final product and this is our starting material. Those are the seven steps needed to perform the required transformation. Now, seeing all of these steps, we can conclude that the correct answer to this multiple choice question is D However, if you have any questions, don't hesitate to leave a 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.
(c)

Key Concepts

Alkenes and Cycloalkenes

Retrosynthetic Analysis

Functional Group Transformations

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