Using hex-1-ene as your starting material, show how you would ...

Question

Using hex-1-ene as your starting material, show how you would synthesize the following compounds. (Once you have shown how to synthesize a compound, you may use it as the starting material in any later parts of this problem.)
g. pentanoic acid
h. pentanal
i. undec-6-yn-5-ol

g. pentanoic acid

h. pentanal

i. undec-6-yn-5-ol

Accepted Answer

Everyone. And welcome back in today's video, we are going to learn how to solve the following problem using beat one in as your starting material show how the following compounds can be synthesized. Number one, pro pandemic asset number two, propane, I'll number we have for I N three. Oh Now starting with reaction number one, the starting material is beat one in. So we're going to throw a four member chain with the double bond at carbon. Number one. And our goal is to form propane awake asset. So that would be three carbon atoms in the parent chain, the car book silic acid group 123, right? So that's pretty weak acid. The first thing to understand is that the parent is shorter now from four carbon atoms, we end up with three only and this implies a bond cleavage were essential ones to break some structure apart. We call that one way to do that is to use an analysis reaction or but as some permanganate in the presence of basic conditions, we want to use the source of hydroxide. What happens in such a reaction? Essentially, we want to think about our starting material of course. So we need a triple bond and actually we are going to choose a terminal Alcott and we will soon understand why. How many carbon atoms do you be one season that al kind? Because the starting material has four, the terminal Al Qaeda must also have four carbon atoms, 1234, we are going to choose be it 19. And whether you're using all this analysis, I will remind you that you can also use number one Ozone at negative 70°C, followed by the treatment with dimethyl sulfide or potassium permanganate and sodium hydroxide. What happens in this reaction is that you're performing that carbon carbon triple bond cleavage and on the terminal and you're forming carbon dioxide, which is your byproduct. And on the other hand, you're forming the assets. So you're adding the code book silic acid group at this carbon which gives you three carbon atoms as desired. But this analysis allowed us to get a four member chain. And the last thing is to think how you can get an alkaline out of an alkaline. And the standard weight status is using double D hydro elimination. You can just add bro mean across the triple bond and you will end up with in this case, actually, let's use hyperbolic acid, right? This would be appropriate for D hydrology nation. My apologies. So if we use to equal Evelyn's off hydrodynamic asset, we are going to get our alcohol hail I Bob. So that would be 1234 carbon atoms, we control 1234. Now saying that we're adding the equivalence of hydrochloric acid is not really correct. According to the more cognitive school, we want to make sure that the two bromine atoms are at positions one right carbon. Number one and two. This is really important because we want to perform an elimination reaction. Whenever we perform an elimination reaction, we essentially can get our al kind. We are removing one of the beta high transforming a double bond and then removing the other one forming the triple bond Using an E two mechanism. So basically, if you think about it, there's one beta hydrogen that is anti co planer to this, bro mean and there is another over here which is anti co planner to theater, bro means so the strong base will d protein. It this hydrant forming a double bond for by the loss of this, bro mean, now we will have a double bond and then we will perform the same step for this hydrogen D protein ating it forming a triple bond for by the loss of the leading group, this will form our al kind. But it's also important to protein ate the possible a satellite. And I and so we are going to say that the region's needed our number one sodium A my, that's our strong base that can be pertinent as hygiene. Then step # two. So we can actually say access here. That's very important. Step. Number two is the protein nation stuff. We can use water for that. And the reason why we need that step is because an excess of base would also d protein it, that acidic hydrogen over here. So we want to restore it to actually get our al kind instead of a satellite. Okay. So now we understood that. And now the question is how do we get that one comma two dye bromo substituted alcohol. Hey, light out of an L king. And the simple answer is using rumination. If you add blooming across the double bond, in particular roaming in the presence of carbon tetrachloride, you will get two bromine atoms attached across the double bond one over here, another one over here and you will get it. So we got it. We were going into reverse and we can just conclude that the reaction steps are number one edition of romaine and the presence of carbon tetrachloride. Number two treatment of the result on alcohol. Hey, light with sodium Amytal Access # three edition of water for the protein ation stop off a satellite. Number four, oxidative cleavage of that triple bond that be potassium permanganate in the presence of sodium hydroxide. We are done. So these will be our four steps for the required transformation. And let's move on to the synthesis of pra penile. Number two, the starting material is still the same. And here we are going to form a penile that would be an al to hide. So let's draw an allied group three carbon atoms broken. All right. So that'd be 123. The reaction here is very simple. If you think about oxidative cleavage, we already said that if you perform most analysis, the general reaction scheme looks like this, you can have He double bond with different substations, I suppose HR one R two R 3. And you're essentially going to break the molecule apart. You're going to break that carbon carbon double bond. The reaction conditions are number one ozone at low temperature, negative 70 degrees Celsius is what you expect to see in textbooks. And number two is the treatment of ozone ID with the metal sulfide, you're going to break it apart and you will add two oxygen atoms at each end. So one product will be R one. Hi Jen double bond and now you're going to add auction over here. So that's how you get your al to hide. Plus your next part of the double bond has to al kill substations R three and R two. And then you're going to add auction as well. So using this model, we can notice that also knows this is a perfect choice for us. We can just say if we break there's double bond Using condition No one ozone at negative 70 degree Celsius. Number two, dimethyl sulfide, one of the products form will indeed be Propane. All right, because if we break it, we have 123 carbon atoms, 123, there is this implicit Haijun, which is indeed over here. And the byproduct also would be warmed that be formaldehyde, right? Because we have two hydrogen atoms, don't worry about the byproduct. The most important part is to get pro canal itself. So it's sufficient to propose and arsenal's reaction. Now, for number three, we are going to synthesize. Actually let's once again read through the same starting material. And according to the problem, we went to synthesize helped for I am 30. In other words, what she wants to hear, you want to understand that the triple bond starts at carbon number four, right? Meaning there must be three carbon atoms to the left. 12, see how many we have 123, right? Because it says have four iron, we need one more. So let's add that 1234. Okay. So the triple bond starts at carbon number four. And because it's helped, we need to have a total of seven carbon atoms. So that's 123456, seven. Okay. So in addition to that, we have alcohol group at position number three, meaning we are going to add an alcohol group at carbon. Number three, that's carbon number 23 has alcohol for contains the triple bond. Okay. So this is what we want is in the size. And what is extremely useful for us is basically the first reaction where we formed be it 19. Let's understand why. Keep in mind that based on reaction number one, we formed Beat 1 9. Using the following steps. We said We are going to add # one roaming and the presence of carbon electrical ride. Number two use sodium Amytal access. Number three protein ate a satellite to form Be it 1 9. Now, the question is why is this really important? It's because this is part of our final product. Okay. So let's see why. 1234 carbon atoms, right? 1234. So basically this is part of our final material that we want to synthesize because we're making our chain longer were essentially adding 123 carbon atoms. We can form a terminal, we can form a satellite out of it using an a deep rotation step. So to remove that acidic hydrogen, we are going to use sodium a mine. We already know that this is what we use whenever we want to form a satellite. And now when we form a satellite, let's call this reaction number four. And then when we form a satellite, we want to perform an essence a reaction where a satellite would attack a carbon chain that has three carbon atoms. So if we draw three carbon atoms, 123, we also need to add a good leaving group to position number right. And now the question is, does this really work for us if we perform that reaction, a satellite would attack this position? And that said we would have no way to add an alcohol group, meaning this substrate doesn't really work for us. However, because we're producing an alcohol, we can think of a nuclear attack at a Carbonell group, right, which well done which could then produce an alcohol upon acidic workups. So let's change our option for number five. And instead, let's think about the nuclear attack on propane island stud if we use propane al that be Three carbon atoms says Nita. So that'd be 123. Let's see what's happening here. So in the first step, we're getting a satellite and then it basically attacks the Carbonell group forming Elcock side. And that's why this carbon compound is much more useful than the previously suggested because the previously suggested one doesn't really have any Heathrow atom such as oxygen. That's what we need, right? To form that alcohol, we're only adding that part of the chain and there will be no way to add an alcohol group. So instead we're going with a carbon compound and this gives us our Alcock side, right? We are basically getting our Elcock said, let's draw it. That would be our triple bond With two carbon atoms to this position. And now here we will have 2 3. And we understand that the world first carbon will now have oxygen with a negative charge. We just want to protein it that and the way to protein ate it is to use a civic work up. What are in the presence of an asset, right? And then we will get our alcohol. So let's not worry about showing the final product because it's already thrown over here. Let's just enumerate our steps. Recall that the first stop was the nomination stop. The Electrolytic edition of Grooming In the presence of carbon tetrachloride. # two was the treatment of the al kill bromide form with the excess of sodium a mine. So that would be N A NH two access pro tin ating a satellite. That would be HT oh This is where we form beat 19 and then step number four, we once again worm a satellite, Meaning we can say we're using NANH two. So technically, you may also think about it and this protein nation step is not so important. If you get your satellite, you're ready to go, you may essentially even make it shorter. So it's perfectly fine if you include water here and show that. Yes, you want to isolate that product first and then form a satellite. When you form a satellite, you're going to use reaction number five where you add broken al and then after you add propane al, you want to have reaction number six where you're adding water in the presence of an asset. Now, to be honest, when you think about this reaction, you can just make it shorter because if there's no necessity to isolate that you can just exclude step number to my policy, step number three and four, right? So these are not so important because after we perform step number one and two, you immediately move on to our satellite that participates in the further reactions. So you can include them. If you want to isolate the product first, that's perfectly fine. But if you don't include them, you just do this problem more wisely, right? You just see shortcut, there's no necessity to isolate that product. So you can just keep going, whichever route you choose is perfectly fine. It's up to you okay, depending on what you want to obtain. If you want to make sure that you get that product first and if there's a requirement to get that product, yes, you will definitely include steps number three and four. But if you're just interested in getting your file product more efficiently, you can just exclude steps three and four. So that will be your final answer for reaction number three. Now, I know that this problem was a bit Lenzi, but let's conclude our findings. The correct answer choice to this multiple choice question is c however, if you have any questions don't have states, leave your comment down below in the comment section and thank you for watching

Using hex-1-ene as your starting material, show how you would synthesize the following compounds. (Once you have shown how to synthesize a compound, you may use it as the starting material in any later parts of this problem.)
g. pentanoic acid
h. pentanal
i. undec-6-yn-5-ol

Key Concepts

Alkenes and Electrophilic Addition

Oxidation and Reduction Reactions

Functional Group Interconversion

Watch next

Using hex-1-ene as your starting material, show how you would synthesize the following compounds. (Once you have shown how to synthesize a compound, you may use it as the starting material in any later parts of this problem.) g. pentanoic acid h. pentanal i. undec-6-yn-5-ol

Key Concepts

Alkenes and Electrophilic Addition

Oxidation and Reduction Reactions

Functional Group Interconversion

Watch next

Using hex-1-ene as your starting material, show how you would synthesize the following compounds. (Once you have shown how to synthesize a compound, you may use it as the starting material in any later parts of this problem.)
g. pentanoic acid
h. pentanal
i. undec-6-yn-5-ol