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.)
d. hex-2-yne
e. hexan-2-one
f. hexanal

d. hex-2-yne

e. hexan-2-one

f. hexanal

Accepted Answer

Hey everyone and welcome back in today's video, we are going to solve the following problem using B 20 in as your starting material show how the following compounds can be synthesized. Number one B T I number to be 10 T own and number three beats. Now, starting with number one, The starting material is B- 29. So we are drawing a four member chain with a double bond at position one. That's our position one, that's our position. So there's a double bond in between. And according to the context of the problem, we are trying to synthesize beat to iron. So that's an al kind just as the name suggests. And the triple bond starts at position number two, meaning there must be an additional carbon atom over here. And because there are a total of four carbon atoms, we are going to add one more carbon atom. So that's our final material. We are ready to think about the synthesis. And regarding the synthesis, our primary goal is to use the idea of a double D hydro Hallett nation because there is no direct way to convert an Al king into an Al kind, especially looking at these two structures, the position of the double bond is different from the position of the triple bond. This makes perfect sense for us. So what we're going to do, we're going to think about how imagination first. So if we use an electrolytic edition of roaming across the double bond, first of all, recall that we are going to add two bromine atoms across the double bond by breaking that pi bond, right. So we are forming our alcohol hey light. And we can essentially draw the result on structure. One brahman is added over here. So we are going to add it. Another blooming is attached at position number one. So we are going to add the remaining blooming Adam over here. So we got our alcohol hey light and what we want to do, we want to essentially eliminate them. However, we cannot immediately form beat to iron out of it because that would require us to remove this beta hydrogen. And unfortunately, the two blooming atoms are adjacent to each other. They're not bonding in the same carbon, right? So even though we have two beta hydrants here, we can remove one of them, followed by the loss of the leaving group, we would form our double bond. But unfortunately, there, there's no grooming on the same carbon that would allow us to perform double D hydrology nation. So it wouldn't work for us, but we can achieve a structure that would work. And to do that, we are, first of all going to form a different al kind. Okay. So don't worry now what we're going to do, we're going to form an internal alkaline first. And to do that, we call that one of the regions that allow us to achieve that kind of transformation as sodium Amytal and A NH two, we are going to use an access of it followed by the addition of water. If we do so, we essentially understand that there are two beta hye Jin's of interest, one beta hygiene with respect to this bombing and another beta hydrogen with respect to this bombing, the removal of the T will allow us to form an internal al kind. But of course, first of all, we are going to form our a satellite because we're using an excess of sodium a mite, right? And that, that a satellite is then protein ated by water to form the final al kind. Okay. So we got it right. We can call this as our terminal al kind, right? And we, we are now ready to think about double D hydro elimination, which would work for us. We essentially want to add to bromine atoms at this position as we previously discussed because the removal of these two hydrogen atoms for by the loss of the living group will allow us to achieve the desired transformation and get the required alkaline. So what we're going to do here, we are going to add two equivalents of, let's say HPR two equivalents. Okay. So that B H B R two equivalents. Now, the reason why we are adding HPR hydroponic asset is because according to the Mark optic of school, you're going to add hydrogen to a less substituted position and room into a more substituted position. So if you add two equivalence, that would work, right, you would add two bromine atoms in this position. And the product that you expect here would be our four member chain with two bromine atoms at position number two. Perfect. And now just as we discussed before, because there are two different positions with beta hydrants, we are going to use this one for our double D hydro elimination because that would allow us to form a more stable elkin. Right? In the first step, the removal of this hydrogen was forming more stable elkin than this one because apparently you can notice that the formation of the double bond here would result in to al kill substitutions. While the removal of this hydrogen would only contain an elk in an elk in with one substation, right? So a more substituted Al King would be expected if we removed this hygiene. So we are going to remove both of them and to make sure that we are forming an internal al Qaeda, right? Not a terminal one, we cannot use sodium. Um it anymore, we have to use a strong base for that elimination, that will be double D hydro elimination via and E two reaction. And we need to use but assume hydroxide in alcohol in the presence of heat. And that said that's how we managed to synthesize B two. We can now add all of these steps, right. Number one, once again, addition of romaine. Number two. So D M A mine access, right. Number three, addition of water to make sure that a satellite is protein ated into that terminal al kind. Number four, for number four, we said we want to add to equivalence of hydroponic asset. And finally, number five is the treatment of the resultant L kill bromide with potassium hydroxide in alcohol in the presence of heat. Okay. Well done. So let's mark these as our final answers for the first reaction. And let's move on to the second one where we want to form B 10 C own now to form B 10 T own. Let's keep in mind that the starting material is unchanged and we can just say that the starting material is still beat one in. And here we are forming a key tone. So the product that we are forming is B 10 T own. So let's draw a four member chain that would be 1234. And the position of the carbon ill group is at carbon number two, right? So that's B 10 T own. And here we can think of what we got in the previous steps, notice that if we actually take this terminal al kind, we can perform an acid catalyzed hydration and form this ketone very easily. So let's just rewrite all of the steps that we had previously and understand how we achieved the, how we produce that terminal Al Qaeda. Okay. So we needed several steps. Number one, what is the addition of grooming when we added booming across the double bond? We said we eliminate that the T bromine atoms from the structure using sodium a mine protein ated a satellite to restore that terminal Al Qaeda. Okay. And this is where we can start thinking regarding what's different now and why are these steps exactly the same. It's basically because if you think about it, this terminal al kind can now undergo an asset capitalized hydration, right? So that would be a judge to recall that whenever you're dealing with al kinds, you want to use three regions, water still fairy cast said and mercury to sell faith, keep in mind that for all kinds, it's very similar to what we expect in Calkins. We are going to add hydrogen right to a less substituted position and the alcohol group to a more substituted position. So let's see what we get. We get our four member chain and our position number one has two hydrogen atoms, there's also a double bond, but now position number two has an alcohol group. So once again, position one gets hygiene position to gets an alcohol group and this is an Enel which is unstable. So, an Enel, if you recall undergoes total memorization into a more stable keto form. So this protein is easily transferred between the double bond and oxygen. So essentially, if we show one of the possible ways for this proton transfer, it's just that the lone pair forms a carbon steel who are basically oxygen, carbon double bond, right, we're going to form a pi bond out of these electrons. And then this double bond will essentially attack hydrogen. So this is how it gets attached to this carbon. And then these electrons are transferred in the form of a lone parent oxygen. And that's how we get our key tone and that's it. Now, this is why it's extremely useful for us to use this al kind because we can simply form our product immediately using the three steps we used in problem number one. Meaning the first step in this problem is the addition of water silvery cassette and mercury to sell fate. So this would be our re transform reaction. Number two, this is how we achieve our sentences. So just to review additional blooming followed by sodium amid water, which gives us our terminal al kind. And when we got our terminal al kind, we had to perform an asset catalyzed hydration to form email that totem arises into a more stable key tone, right? Well done. So if we move on to the third one, we can once again think about the same terminal al kind that we managed to form before. And let's see why this is once again, very useful. Think about the starting material beauty one in according to the problem we are trying to form buttes now, so that the four member chain 1234, right? And position number one has. And Aldo Haydn group notice that we can first will form our terminal al kind because if that terminal al kind undergoes the formation of an alcohol write or email, it will try to summarize into an allied as long as we are forming an anti more cosmic of addition. So let's see how that works. Okay. Now we will use the same three steps as before. Number one addition of romaine number two sodium a mine access number three, water to reform a satellite. My apologies to reform the terminal al kind. And as you already know, we are forming our Terminal al Qaeda. Okay. So this is our structure 1234. Now this is why this intermediate is so good for us because we're thinking I had previously was that if we use that intermediate and we perform an acid catalyzed hydration, we essential form an email which then forms our key zone. But what if we add an alcohol group at a different position? Yes. Previously, we edited over here, but now we can add it over here which will allow us to get the required al to hide. So let's see how that's done. We call that one of the ways is to use hydro operation oxidation mechanism. And with all kinds, we are using two reactions. So we can say reaction number one would be addition of decimals boring. So that B S I two B H, right? And the second reaction would involve the presence of hydrogen peroxide and hydroxide. As you already know, this reaction allows you to add the alcohol group to a less substituted position and hydrants, the more substituted position because it follows the anti mark optical regional chemistry. Okay. So what we are forming is our intermediate which looks like this, we are going to add an alcohol group at carbon number one. So let's add it or for simplicity to be consistent, let's suppose that this is our carbon number one. Okay. So we are going to add an alcohol group. And what we understand is now there will be double bond between carbons number one and two. Well done. So now we have our email and once again, let's keep in mind that channels are not very stable. So there's an equilibrium in which it totem arises in Savannah Aldo hide because once again, we have that proton transfer. So the lone pair forms a pi bond, the double bond at tax, there's the capture of this proton and then there's a transfer of these electrons in the form of a lone pair. And that's how we get our al to hide. 1234, right, 1234 position. Number four has our carbon group and there is this implicit high general, That's how we get our found product. And we can just say that step number four is nice. I M L boring number five, addition of hydrogen peroxide and hydroxide. And that's it. This will be our answers to this problem. I know that this problem was a bit Lenzi. So thank you for sticking with me throughout the video and the correct answer choice. This multiple choice question is b however, if you have any questions, don't have states leave your comment down below in the comments 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.)
d. hex-2-yne
e. hexan-2-one
f. hexanal

Key Concepts

Alkene Reactions

Synthesis of Alkynes

Carbonyl Chemistry

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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.)d. hex-2-ynee. hexan-2-onef. hexanal

Key Concepts

Alkene Reactions

Synthesis of Alkynes

Carbonyl Chemistry

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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.)
d. hex-2-yne
e. hexan-2-one
f. hexanal