Show how you would synthesize each compound from starting mate...

Question

Show how you would synthesize each compound from starting materials containing no more than six carbon atoms.

Accepted Answer

Hey everyone and welcome back in today's video, we are going to show how the following compounds can be synthesized. Starting with a reactor and having not more than six carbon atoms. Now, we have three problems to solve. Let's begin with a we are forming a key tone, right? Because we are forming a key zone and we actually have the ring on that structure. It makes sense to choose our starting material, which will not require us to form that ring. And this will actually correspond to our starting region. We want to make sure that our starting region actually involves the ring. So let's throw our five member ring and this would correspond 25 carbon atoms. So we satisfy the criterion because we're forming a carbonell compound. One of the easiest ways to form a carbon compound is V a greener reaction. Even though we are forming an alcohol, we can easily oxidize that secondary alcohol and a ketone. So our goal is to actually choose an alcohol bromide and form the greener region out of it. First. If we start with our ring, we can choose a bromo substitution bonded because whenever we have grooming and whenever we treat this al kill bromide with magnesium in the presence of ether, we can easily form the green region for us. So we're going to use magnesium in the presence of ether. Now from here we're going to form the green region which essentially incorporates magnesium into our structure and we have our granular region. When we have our graveyard region, we need to choose an appropriate carbonell compound to form our alcohol. Right? So to do that we have to look at our structure because the final product will correspond to a key tone after oxidation of that secondary alcohol. We understand that we need to make sure that the alcohol that we are forming is a secondary alcohol, meaning we want to use an alga hide in our stop because whenever you treat an al to hide where the graveyard wage and you're going to form a secondary alcohol after the high troll is a stop. So if we want to get a secondary alcohol, let's choose an appropriate al to hide. And we need to think about how many carbon atoms do we want to add would rather be one too. Right. So for that purpose we are going to use ethanol. So we're going to draw our carbonell group our al the hide. Right? Right. And we essentially want to add one more carbon atom. That's our muscle group bonded to that carbon hill. So we have S. N. L. And let's recall that in the stop where essentially attacking our electrical center. Breaking the pi bon. And then this is where we get our elk oxide and iron. So let's try the result on intermediate our five member grain. Now we're here we will have two carbon atoms. This is our metal group. This is our Elcock sign. And now we need to protein ate it. So we're going to protein ate it in acidic conditions. Let's see that in the next reaction step, we're going to use H30 plus which will correspond to acidic work up. And that will allow us to produce the alcohol. Once again our five member green. Nothing changes. We are going to add two carbon atoms. And now Alcock side is protein ated into an alcohol. Now from here, when we have the protein nation step, we're essentially going to oxidize it. If we oxidize a secondary alcohol, we get a key tone in order to oxidize that we can use the majority of oxidizing agents. P CCS. Where an oxidation any strong oxidizing agents such as jones Regent. For the purposes of this problem, we can use a common region for oxidation of secondary alcohols that would correspond to sodium hypochlorite in the presence of acetic acid. That's acceptable. That's one of the oxidizing agents that you also know. So this oxidizing agent would essentially allow you to get a key tone out of a secondary alcohol. So let's throw the result on product. Okay, so we got our key zone. That's indeed our final product. So let's label a as solved right, we have our reaction sequence. Let's label this as our reaction scheme. We can also label the steps. So we can clearly understand what's happening. So this is our reaction number one. We need to add magnesium in the presence of ether reaction. Number two, addition of our carbonell compound which reacts to the green yard Regent step number three. It's really important to not make these tea together. We want to make sure that the stick work up takes place when we form our Alcock side, we get our alcohol reaction number four. The oxidation step together. Our key zone. Moving on to be, we see that we want to synthesize a key tone. We have our benzene ring. Right. And essentially because we have benzene ring in our structure, we are going to use benzene as our starting material because we definitely don't want to synthesize that. So let's see how that works. First of all, if we start with benzene, why don't we just use Friedel Crafts isolation. Right. We have our SCL group and essentially we are going to add it via a single step. So let's see how that works. Going to part B. If we start with our six member ring, dad's fencing and we want to perform the following ketone. We made something to recall how Friedel crafts isolation works. If you're starting with benzene ring, let's suppose we are drawing a general reaction scheme over here. So benzene ring is treated with your acid chloride, you want to make sure that you have an acid chloride so that we are carbon deal with chlorine. And then we have some our group were essentially forming a bond between the following two carbon atoms were taking one carbon from the ring. We are bonding it to our carbon group and chlorine is removed from the structure. Right? We're essentially creating that carbon carbon bond and we get our asset chloride. We want to make sure that we're using our louis asset aluminum chloride. That's how Friedel Crafts isolation, isolation works. And now we are getting our final product which corresponds to the following key tone that we are benzene ring. We are going to create that carbon carbon bonds. We are immediately adding our carbon deal with and our group. So now the question is, what is our region that we're going to use? Notice that we want to use an acid chloride, which will correspond to exactly same fragment. But now we're going to add chlorine to this carbon. So we need to use the following. We're going to draw that carbon steel fragment on the right, we have our metal group, right? This is our our on the left because we're creating that carbon carbon bond, we are going to add chlorine and that would be it. So, we got our region, we just want to make sure that we are adding aluminum chloride which acts as a loose asset in the mechanism of this Friedel crafts isolation. Let's label this is our final answer for this simple one step transformations. We're not even going to label any steps. We just have a single step reaction here. And finally, we want to concentrate on the world lost part. That would be C. We want we want to form a key tone. Right? We have 123456 carbon atoms. There are actually multiple ways to form ketones. Let's just think about the formation of a ketone. Whenever an alkaline undergoes acid catalyzed hydration. It's really important to make sure that it's catalyzed using mercury to sulfates. So we have more cognitive school that we obey. That's why we form ketones instead of Aldo hides rights. We're going to use an acid catalyzed hydration in the presence of mercury to sulfate. But still that we need to make sure that we have a corresponding AL kind. So the corresponding AL kind that we want to form. We essentially want to make sure that is a terminal AL kind because we want to get one product. And according to the mechanic of school, we want to have our triple bond over here so that hygiene gets added to a less substituted position at the terminal and a wage group gets added to a more substituted position. We get our total memorization. So our goal is to first will synthesize that al kind. Right? That would be our 123456 carbon atoms. So that would be hex 19. We want to form hex 19 and to do that we can choose any sort of starting material with a good leaving group and perform an essence to substitution reaction. So let's see what our possibilities are. We can use one bromo butane as our starting material. That would be a good substrate for us. We have four carbon atoms were in the limit of six. So that's fine. Let's suppose that we are starting with 1234 carbon atoms. We have Romo substitute. And the reason why we're choosing our Romo substitution is because it's a good leaving group. And what we want to do, our goal is to form an al Qaeda hex 19 to form hex one eye. And we need to add two more carbon atoms to our chain. And we want to make sure that this is a terminal alkaline. So the region that we can use can be a satellite ion right? We can perform an essence to reaction. So you can choose a satellite for this part. Now, a satellite would essentially correspond to let's let's say it's sodium a satellite, right? Because generally want to include those ions Spectators. So we have sodium with a positive charge and then a satellite would correspond to two carbon atoms triple bonded. And then we would have a negative charge on that carbon. So now this really works for us because notice how we can show that nuclear filic attack stopped, right? We have an attack followed by the loss of the leaving group. And this is how we are forming our alkaline. So let's throw it out. We will have 123456 carbon atoms. So that we 1234, Five and then six. Right? It's basically linear because we have sp hybridized carbons. So let's throw it out. That's our corresponding al kind. Okay. And we got our alkaline from here. As we said before, we can actually perform an asset catalyzed hydration. And when we do so we have to keep in mind that an acid catalyzed hydration follows the more cosmic of regional chemistry. So we're going to include that more comical ritual chemistry tells us that we're going to add ha jin to a less substituted position and the alcohol group to a more substituted position. And if we do so we get our email which is not so stable. Let's see how that looks like. And of course let's include our regions. So for an acid catalyzed hydration, we are going to use the following set of regions mercury to sulfate, Let's draw it in black so that be mercury to sulfate in the presence of sulfuric acid and water. As we said, we're going to add O. H. At position two hydrogen at position one. We're going to break that pi bonds so the other pi bond will remain intact. And we are ready to draw the result of the Enel for this one. So let's do that. Now for that email we're going to draw 123456 carbon atoms. So we have six carbon atoms. And now notice how we still have our pi bond, one of the pi bond remains okay. And now the alcohol group is added at position C. That's our final intermediate. And because it's not too stable, it easily totem arises into a more stable keto form essentially. Whenever you want to think about the result on ketone, you can think of that proton transfer because we can easily use that lone pair in the formation of a pi bond, the adjacent pi bon except the proton. And then we are transferring these electrons in the form of a lone parent oxygen. So this is how we get our equilibrium which shifts towards the formation of a more stable keystone. That's why one of the arrows is longer. And we get our final product just as follows, so that we 123456 carbon atoms. We get our carbonell group because we had a proton transfer and this is indeed the final product that we wanted to get. So for this reaction, we had two steps, we chose our alcohol. Hey light, right. And we said in the first step, we're going to treat it with a satellite ion. And then when we do so we're forming our al kind six member changes as we wanted. Performing an acid catalyzed hydration in the presence of mercury keiron and then the resultant Enel is not really stable. It totem arises into a key tone. So that will be our final sequence for this problem. We are going to label it and based on our answers, we can just conclude that the correct answer to the multiple choice question is C. Thank you for watching, and I will see you in the next video.

Show how you would synthesize each compound from starting materials containing no more than six carbon atoms.

Key Concepts

Functional Groups

Synthetic Pathways

Reagents and Reaction Conditions

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