Beta Hydrogen - Video Tutorials & Practice Problems
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In order to predict E2 products, we’ll have to get good at recognizing how many different and eligible β-hydrogens exist.
Recognizing Different Beta-Carbons
Elimination reactions remove β-hydrogens to create double bonds.
The number of non-equivalent β-carbons with at least one -H determines the number of possible products.
1
concept
Unique β-carbons and possible products
Video duration:
2m
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now on a dig deeper into one of the more important parts of elimination. And that's the beta hydrogen. All right, So you guys might remember that elimination reactions basically pull off beta hydrogen and then they make double bonds. Okay, if you think back to the definition of elimination, what we remember is that let's say that you have a single bond to a leaving group on a single bond to ah, hydrogen. What winds up happening is that these two Sigma bonds get pulled off and turn into one pi bon. All right, so that's the whole process of elimination. That's actually the definition. Is that to Sigma's turned into one pie? Alright, so that's not bad. But the tricky part comes in with the beta hydrogen is because it turns out that rarely will you just have one beta hydrogen that applies towards this rule. Many times, you're gonna have several beta hydrogen that you have to choose from. On top of that that complicates things more. Because if you choose a different beta hydrogen to extract that, that might that might actually make a new product. Okay, so what that means is that we're opening ourselves up for the possibility of multiple products. And in this page, what I want to do is just really practice how to determine if you're just gonna get one product or if you have a possibility of up to three products, okay. And usually that's the maximum amount of products you can get. Three, because that's the maximum amount of beta carbons you can have three. Okay, so let's go ahead and talk about how to figure that out. The way we figure that out is by counting the number of non equivalent beta carbons. Okay, remember that the beta carbon is attached to the Alfa Carbon. All right, I'm gonna go through all this again. So it's fine. And if those beta hydrogen is have at least one h, that's gonna be its own unique product. Okay, So for every beta carbon that's unique, that has its own H that could be taken off that's going to represent one possible product. And like I said, sometimes you're just gonna only have one product on Lee. One of the beta carbons will have a hydrogen on it, But other times you're gonna get up to three products, and that's what we're gonna do now. So I want you guys to look at example a and try it yourself. Try to figure out exactly how many different products you could get from a by looking at the beta carbons and see if they have hydrogen, and then I'll explain the entire question, how to do it. All right, so go ahead and get started.
For the following molecules, identify the number of unique products that could be obtained through elimination.
2
example
Identify the number of unique products
Video duration:
2m
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All right. So there's actually a really easy question. It could just be a little bit tricky the first time you're trying to figure it out. So how do we find out how many different products what I would do, first of all, is indicate that the carbon that is attached directly to my iodine or my leaving group is my Alfa. That means that every carbon coming off of that one is beta. Okay, now, a Z guys can see I have three beta carbons, but do I have three different non equivalent beta carbons? In order to figure that out, I have to see what they're attached to. So what I would say is, OK, I have three beta carbons. Okay? Are they all attached? The same thing afterwards. Okay, we're going to notice is that all three of them are part of an ethyl group overall. Okay, All three of them. This is an ethyl. This is an ethyl, and this is an ethyl. Okay, so are these equivalent or non equivalent. It turns out that they're all equivalent. They're all the same exact thing because it doesn't matter which proton I pull off. It doesn't. They're all exactly the same. Substitue int. Now, the next thing I have to do after figure out if they're equivalent or not, is C Do they all have at least one hydrogen on them? And Yeah, actually, each of these has to I've got to here to here to here. All right? That doesn't matter. I don't care about one or two. I just care. Is there a least one? Yes, there is. So the answer is that I could use any of these beta carbons, but I'm only gonna get one product because they're all equivalent. So the answer is I would just get one product, okay? And I'm also gonna go ahead and draw that. And what that one product would look like is just I could really pick any of the ones I want. I'm just gonna go up with it, okay? And that's gonna be my one product. Because remember that basically what you do is you make a double bond in between the Alfa and the beta. And that's going to tell you what your end product looks like. You look at your Alfa. You look at your beta, you draw double bond between them and that's that. Okay, Now, you guys might be wondering, Johnny, I picked a different one. I picked the one that was at the bottom. It's the same exact thing. Okay, It doesn't matter, because this is symmetrical. And then maybe you're like, Oh, Johnny, doesn't matter if I drew the top thing this way. Same exact thing. It doesn't matter. These air all equivalent. So it doesn't matter how you draw it. Okay, Cool. So with that said, maybe it's a little bit easier. Now. Go ahead and try to predict how many products and draw the products of B. Can't remember that the way you draw the product is to draw double bond between the alpha and the beta that you pick, so go for it.
3
example
Identify the number of unique products
Video duration:
2m
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Alright, So let's start off with the Alfa Carbon. This is my Alfa. So where are my Betas? I've actually got three betas, one to three. Okay, Do those three Betas all have at least one hydrogen on them? Yes, they dio Okay, There's a least one hydrogen coming off of each one. So I have to figure out. Are they all the same or are they different? Okay. And it turns out that two of these are exactly the same. These on the corners because they're both secondary and they're both coming off the tips of those squares. Okay, but one of them is different than the others. And that's this one because it's tertiary and its in between the two rings. Okay, So because of symmetry, even though I have to even though I have three carbons total that could be that could participate. I only have two different types of carbons. I have the blue ones and I have the green one in the middle. Does that make sense? So that means that even though I could use any of these beta protons that I want altogether, I'm only gonna get to different products. So let's go ahead and see what they would look like. So I'm gonna put here to products, okay? And what I would get is let's draw the blue one first. The blue one would just look like this. Oops. The blue one would look like this. I have my squares. And now I went ahead and I draw a double bond between one of the alphas and one of the Betas. Okay, It doesn't matter which one you pick, because there's symmetrical, so you could have also drawn on the left hand side. Okay. What's the other product gonna look like? Well, the other product would be the same general shape, but now I'm gonna have a double bond in the middle, Okay? And that's because I e I wanted to draw that in green. Sorry. Not sure if it makes a big deal to you guys, but I just like showing that the green one would be the one that would create the green product. And the blue one would create the blue product. All right, now I know you guys might be already starting to think, Johnny, Which one would I get more of or what? I get the same don't worry about that yet. We're gonna have rules that decide that later. But right now, I just want you guys to get usedto just drawing these products and knowing how many different ones you could get. All right, so let's move on to the next one.
4
example
Identify the number of unique products
Video duration:
3m
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Alright, so what number did you guys get? It should be two. Okay, let me show you why, because this is my alpha and then let's look at my betas actually have three betas, I have a beta here, A beta here and a beta here. Okay, now I know that it might have been tricky to recognize that this top one was a beta, but it is, it's a methyl group coming directly off of the alpha. So that by definition is a beta. Alright, so I've got my three betas, do all of them have at least one proton? The answer is no, okay, this one has a proton And this one has a proton at least one. But this bottom one doesn't, it only has our groups coming off of it, it only has carbons. So that means that this one cannot participate at all. That means now I'm stuck with just two options. Alright, so now I have to say is okay, are those options equivalent or are they different from each other? What do you think they're super different? One of them is on the ring, one of them is on a branch completely different. So what that means is let's say this is my red and this is my blue, I'm gonna get two different products, let's write that two products and let's draw these products, the Red one would look like this, remember it's between alpha and beta, so it would be a double bond between groups between alpha and beta and then I would still get everything else the same here and here, but now I would actually draw the beta, that's that's on the metal, I would draw it on a stick. Why is that? Because it turns out that now, I'm not talking about the chlorine by the way, the chlorine left. Okay. But I'm talking about this method right here. It used to be on the dash and now enjoy it on a stick. Why? Because if you remember back to chapter one, we talked about hybridization, this carbon right there, is triggering a planer now. Okay. It has it's an sp two hybridized um bond and or adam and that means that everything has to be on the same plane. So it's actually wrong to draw that methyl group on a dash because it looks like you see that it's a double bond, but that you don't remember that it's triggered a planer. Okay. And no one wants to look stupid. Right? So that's why you're going to go ahead and apply the rule of it being trivial planet. It's kind of like if you were drawing a triple bond and you draw it with a zigzag pattern, everyone knows duh, you have to draw a triple bond straight because it's linear, same, same kind of deal. Alright. And yeah, who knows if you have, if you have a free response, test, your professor could actually take off a point or two because you forgot that has to be triggering planner. Let's see the other product, the other product would be that I have The Blue one, which would have been that these groups are the same. Okay, But now I have a double bond going towards that mental group. So I actually draw it like this. Okay. And once again you would draw it out Like this because this is supposed to have a 120° bond angle so it wouldn't make sense to draw it straight up. Like I had it before you have to draw it out. Okay. And these are two different products for elimination. Crazy, right? But now hopefully is getting better and it's it's coming a little easier to you. Okay, let's do one more practice problem and then we'll be done with this topic.
5
example
Identify the number of unique products
Video duration:
2m
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last one. And I know you guys all got it, right. The answer was three products, Okay? And why is that? Because I had three carbons that had been at least one beta hydrogen, and they were all different. Let me show you. So my Alfa was right here in the middle. And then I had this beta. I had this beta and I had this beta. Okay, so those three different betas well, are they different? Well, first of all, do they all have it? Least one h on them? Yes. Okay. Are they all different? Yes, they are. Because one is, um Ethel. One is an ethyl, and the blue one is. You remember Isopropyl. So they're all completely different. That means that if I eliminate down one or the other, I'm going to get a different product. All right, So let's go ahead and drop the three different products that I would get. Let's start off with the red one. The red would look like this. I would get everything the same but a double bond going up this way. Remember, between Alfa and that beta carbon, the blue one would have a dull bon This way and then, lastly, the green one, which I'm gonna take myself out of the picture so you guys can see it. Or maybe I won't. There we go. Oh, gosh. What just happened? Okay, so we had a little technical difficulty, so I'm just gonna go ahead and draw out on top. Um, what I would have is the same compound, but now with a double bond facing this way. Okay, So notice that I just drew it down because I was drawing between the Alfa and between the green beta carbon. Alright, so hopefully this helps you guys, Um, hopefully this really helps you guys understand the differences between locating different beta hydrogen. This is actually one of the things that people struggle with the most doing elimination reactions. And this is gonna work for every single elimination reaction that we talked about, whether it's e one or two. Okay, so let me know if you have any questions, but if not, let's move onto the next topic.