All right. So this is a pretty easy example where I'm already starting off with a triple bond, so half the work has already been done for me. All I really need to do is deprotonate it because notice I have a terminal alkyne already. So I don't have to worry about making a triple bond, which is a more advanced part of synthesis. I already have the triple bond. All I need to do is react it with a strong base to get that H off of it, and that's what I'm going to do in my first step. So my first step is NH2− I'm going to grab that H and I'm going to wind up getting an alkynide. Okay? So the alkynide synthesis is the first part. I'm going to go ahead and draw that alkynide down here. And what it's going to look like is just the same thing with a negative. Now I react that alkynide with a leaving group. And in this case, it's a primary leaving group. What do I get when I get a reaction between a strong nucleophile and a primary leaving group? Well, this is the kind of question that we would use my flowchart for because any time you have a leaving group and a strong nucleophile, that's a perfect flowchart question. So you guys might already know the mechanism, but I'm just going to go through this again in case maybe you forgot. By the way, this flowchart that I keep talking about is something that I introduced in the substitution and elimination chapters. Okay? So if you're wondering about this flowchart, just go ahead and look at those videos because we're using that flowchart for a lot of different things. Okay? So, basically and I by the way, I will include and I by the way, I will include a copy of the flowchart in this lesson so that in case you didn't see those, you can still get the copy of the flowchart in this lesson. Okay? But instructions on how to use the flowchart are going to be in those other lessons. Okay? So basically, what we've got is a negatively charged nucleophile, so that would be on the left side of the flowchart. Then 2, what we have is, is it a bulky base? No. So then I would keep going to 3. What kind of leaving group do I have? Primary, so that equals sn2. So this is going to be an sn2 mechanism. Okay? So I would go ahead and draw my alkyl halide, and I would do a backside attack. And what that's going to give me at the end of the second step is now a triple bond with an ethyl group on it. As you can see, I just made my molecule bigger. Now this last step is a reagent that you are supposed to be able to recognize, and this comes from the hydrogenation section. This would be Lindlar's Catalyst. And do you guys remember what Lindlar's catalyst did to triple bonds? It turns them into cis double bonds. Okay? So I don't need to know the mechanism for this. All I need to know is that this is going to become this. Double bond and then ethyl. So notice that now both of my large groups are on the same side of that double bond, making it cis. So congrats guys. You just did your first multistep synthesis. There's going to be a lot more to come in terms of you have to do your own practice, but I'm just trying to show you a really common synthetic pathway. Now, this next one is also very common. Go ahead and try your hardest, and I'll go ahead and give you the answer. Okay? Go for it.
Alkynide Synthesis - Online Tutor, Practice Problems & Exam Prep
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Predict the major product.
Video transcript
Predict the major product.
Video transcript
So this question starts actually one step further back in the synthetic process where in the previous example, you had a triple bond. But in this example, you aren't even starting with a triple bond, so we're going to have to make the triple bond first and then react it with everything else. So let's go ahead and look at this. I have a double bond with I2. Do you guys know what happens when I react to any type of diatomic halogen with a double bond? It's a reaction that we learned from the addition section of organic chemistry. This would actually be a reaction called halogenation. Now if you don't remember what that is, that's okay, but you do need to know it. What it is, is basically that you would get 2 vicinal halogens together. So, if you guys remember, the mechanism was just kind of like this. I'd have a double bond, I and I, and I would do these three arrows where one arrow goes to the eye, one arrow kicks out the other eye, and one goes back. So what I would wind up getting is a bridge like that. And then the other eye would come in, I negative, would come in and attack. Okay? So at the end of this whole process, what I wind up getting is iodine and iodine. All right? So that's the end of the first step. This is the halogenation step. Okay? So now I have a vicinal dihalide. That's what this is. And do you guys know any way to go from vicinal dihalide to a triple bond? Because that's what I'm trying to do. I'm trying to make triple bonds. And yes, remember that we can take vicinal dihalides or geminal dihalides, and you can react them with excess base to give us triple bonds. Not only that, we can react them with 3 equivalents of base to give us alkynides. So how does that work? Well, I would take this and I would react it with 2 equivalents of base, which in this case is going to be H-. And what that's going to do is it's going to do a double elimination, so I'm going to wind up getti
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Video transcript
I hope you guys gave this a fair shot because this is also a very common synthetic pathway. You should know by the end of this lesson how to turn trans into cis and cis into trans. That's a huge part of this chapter. So let's go ahead and get started. Since there's no way to directly turn this into one or the other, I'm going to need to go to the triple bond first. I'm going to need to take it up to a triple bond and then I can hydrogenate it to whatever type of double bond I want. So if I'm making this a triple bond, then I'm going to start off with a diatomic halogen again. Let's try Cl2 this time. What that's going to do is it's going to give me vicinal dihalides. Okay? Then my second step is going to be to make this into a triple bond, and I can do that using one of my strong small bases in excess. Now notice that this is not a terminal triple bond. What this is going to make is actually an internal triple bond where the triple bond is inside of 2 different carbon groups. So could I ever get an alkynide out of this? Remember, an alkynide has that negative charge. No. Because this doesn't even have a hydrogen to remove. There's no hydrogen here. So when I take my base in excess, I'm just going to end at the triple bond. I'm not going to go all the way to the alkynide. Okay? So now I have that triple bond. What could I do to make that into a cis double bond? It's not that hard. All I have to do is use Lindlar's catalyst.
Okay and there were a whole lot of other ways to write it. I'm going to let you guys look that up depending on what your professor wants. But I'm just going to write 'Lind Lars' because that's another acceptable way to write it.
Okay, and it would be 'Lind Lars' and H2. H2 and Lind Lars. And what that's going to give you is that's going to turn that into a cis double bond, and now you've just made the conversion from trans to cis.
And now, you guys should for practice, really before you get to your exam, should know how to go the other way as well, to go from cis to trans. It's not that hard. Okay? So I hope that made sense to you guys. This is just a small taste. At the end of the day, you're going to have to practice, you're going to have to do homework, but I'm just trying to show you, give you hints of stuff that I've seen come up over and over again, and stuff you absolutely can't get to your test without knowing how to do. All right? So let me know if that helped, and let me know if you have questions. Let's move on.
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