Beta Hydrogen - Online Tutor, Practice Problems & Exam Prep

Elimination reactions involve the removal of beta hydrogens, transforming two sigma bonds into one pi bond. The presence of multiple beta hydrogens can lead to different products, depending on which hydrogen is removed. By identifying non-equivalent beta carbons with available hydrogens, one can predict the number of possible products, typically up to three. Understanding this concept is crucial for mastering elimination mechanisms and their outcomes in organic synthesis.

Unique β-carbons and possible products

Now I want to 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 essentially pull off beta hydrogens and then they form double bonds. If you think back to the definition of elimination, we recall that, let's say you have a single bond to a leaving group and a single bond to a hydrogen, what winds up happening is that these two sigma bonds get pulled off and turn into one pi bond. Alright? So that's the whole process of elimination. The actual definition is that two sigmas turn into one pi. Alright? So that's not bad. But the tricky part comes in with the beta hydrogens because it turns out that you will rarely have just one beta hydrogen that applies towards this rule. Often, you're going to have several beta hydrogens from which to choose. This complicates things more because if you select a different beta hydrogen to extract, that might actually yield a new product. So, this

Identify the number of unique products

All right. So this is actually a really easy question. It can 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 alpha. That means that every carbon coming off of that one is beta. Okay? Now, as you can see, I have three beta carbons. But do I have three different nonequivalent beta carbons? In order to figure that out, I have to see what they're attached to. So, I would say, okay, I have three beta carbons. Are they all attached to the same thing afterward? What you're going to notice is that all three of them are part of an ethyl group overall. 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, they're all exactly the same substituent. Now the next thing I have to do after figuring out if they're equivalent or not is see if they all have at least one hydrogen on them. And yes, actually, each of these has two. I've got two here, two here, two here. That doesn't matter. I don't care about one or two. I just care is there at least one. Yes, there is. So the answer is that I could use any of these beta carbons, but I'm only going to get one product because they're all equivalent. So the answer is I would just get one product. And I'm also going to 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 going to go up with it. And that's going to be my one product. Because remember that basically what you do is you make a double bond between the alpha and the beta. And that's going to tell you what your end product looks like. You look at your alpha, you look at your beta, you draw a double bond between them, and that's that. Now you might be wondering, if I picked a different one, I picked the one that was at the bottom. It's the same exact thing. It doesn't matter because this is symmetrical. And then maybe you're like, does it matter if I drew the top thing this way? Same exact thing. It doesn't matter. These are all equivalent, so it doesn't matter how you draw it. 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. Remember that the way you draw the product is to draw a double bond between the alpha and the beta that you pick. So go for it.

Identify the number of unique products

All right. So let's start off with the alpha carbon. This is my alpha. So where are my betas? I've actually got 3 betas. 1, 2, 3. Okay? Do those 3 betas all have at least 1 hydrogen on them? Yes, they do. Okay? There's at least 1 hydrogen coming off of each one. So now I have to figure out are they all the same or are they different. Okay? And it turns out that 2 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 it's between the two rings. Okay? So because of symmetry, even though I have 2, even though I have 3 carbons total that could participate, I only have 2 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 going to get 2 different products. So let's go ahead and see what they would look like. So I'm going to put here 2 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 they're symmetrical. So you could have also drawn it on the left-hand side. Okay? What's the other product going to look like? Well, the other product would be the same general shape, but now I'm going to have a double bond in the middle. Okay? And that's because I and 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. Alright? Now, I know you guys might be already starting to think, Johnny, which one would I get more of or would I get the same? Don't worry about that yet. We're going to have rules that decide that later, but for right now, I just want you guys to get used to just drawing these products and knowing how many different ones you could get. All right? So let's move on to the next one.

Identify the number of unique products

Alright. So what number did you guys get? It should be 2. Okay. Let me show you why. Because this is my alpha, and then let's look at my betas. I actually have 3 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 now I've got my 3 betas. Do all of them have at least 1 proton? The answer is no, okay? This one has a proton, and this one has a proton, at least 1, but this bottom one doesn't. It only has R 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 2 options. Alright? So now what I have to say is okay, are those options equivalent or are they different from each other? What do you guys 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 if this is my red and this is my blue, I'm gonna get 2 different products. So let's write that, 2 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, oops, 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 on the methyl. 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 methyl right here. It used to be on a dash and now I'm drawing it on a stick. Why? Because if you remember back to chapter 1, we talked about hybridization, this carbon right there is trigonal planar now. It has it's an sp2 hybridized, bond and or atom 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 you it looks like you see that it's a double bond but that you don't remember that it's trigonal planar. 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 trigonal planar. 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 kind of deal. Alright. And yeah, who knows if you have a free response test, your professor could actually take off a point or 2 because you forgot that it has to be trigonal planar. 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. But now I have a double bond going towards that methyl group, so I'd act

Identify the number of unique products

Last one, and I know you guys all got it right, the answer was 3 products. Okay. And why is that? Because I had 3 carbons that had at least 1 beta hydrogen and they were all different. Let me show you. So my alpha was right here in the middle. And then I had this beta. Okay? Well, are they different? Well, first of all, do they all have at least 1 H on them? Yes. Okay? Are they all different? Yes, they are because one is a methyl, one is an ethyl, and the blue one is, do you guys remember, isopropyl. So they're all completely different. That means that if I eliminate down 1 or the other, I'm going to get a different product. All right? So let's go ahead and draw out the 3 different products that I would get. Let's start off with the red one. The red one would look like this. I would get everything the same, but a double bond going up this way. Remember, between alpha and that beta carbon, the blue one would have a double bond 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 it on top. 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 alpha and between the green beta carbon. Alright? So hopefully this helps you guys, hopefully this really helps you guys understand the differences between locating different beta hydrogens. This is actually one of the things that people struggle with the most doing elimination reactions. And this is going to work for every single elimination reaction that we talk about whether it's E₁ or E₂. Okay? So let me know if you have any questions. But if not, let's move on to the next topic.

Here’s what students ask on this topic:

What is a beta hydrogen in organic chemistry?

A beta hydrogen in organic chemistry refers to a hydrogen atom that is attached to a beta carbon. The beta carbon is the carbon atom adjacent to the alpha carbon, which is directly bonded to a functional group or a leaving group. In elimination reactions, beta hydrogens are crucial because their removal leads to the formation of a double bond, transforming two sigma bonds into one pi bond. Understanding the role of beta hydrogens is essential for predicting the products of elimination reactions.

How do you identify beta hydrogens in a molecule?

To identify beta hydrogens in a molecule, first locate the alpha carbon, which is directly bonded to the functional group or leaving group. The carbons adjacent to the alpha carbon are the beta carbons. Any hydrogen atoms attached to these beta carbons are considered beta hydrogens. These beta hydrogens are the ones that can be removed during elimination reactions, leading to the formation of a double bond.

Why are beta hydrogens important in elimination reactions?

Beta hydrogens are important in elimination reactions because their removal is a key step in the formation of a double bond. During an elimination reaction, a base abstracts a beta hydrogen, and the electrons from the C-H bond form a pi bond between the alpha and beta carbons. This process transforms two sigma bonds into one pi bond, resulting in the formation of an alkene. The presence and position of beta hydrogens determine the possible products of the reaction.

How do multiple beta hydrogens affect the outcome of elimination reactions?

Multiple beta hydrogens can lead to the formation of different products in elimination reactions. Each beta hydrogen that can be removed represents a potential site for double bond formation. If a molecule has several non-equivalent beta carbons with available hydrogens, removing different beta hydrogens can produce different alkenes. This results in the possibility of multiple products, typically up to three, depending on the number of unique beta carbons with hydrogens.

What is the maximum number of products you can get from an elimination reaction involving beta hydrogens?

The maximum number of products you can get from an elimination reaction involving beta hydrogens is typically three. This is because a molecule can have up to three non-equivalent beta carbons, each with at least one hydrogen that can be removed. Each unique beta carbon with a hydrogen represents a potential site for double bond formation, leading to different possible alkenes as products.