Carbocation Intermediate Rearrangements: Study with Video Lessons, Practice Problems & Examples

Carbocations can undergo rearrangements to achieve greater stability, primarily through hydride shifts and alkyl shifts. A 1,2-hydride shift occurs when a hydrogen from an adjacent carbon moves to a carbocation, enhancing stability. Conversely, a 1,2-alkyl shift involves the movement of a small alkyl group when no hydrogens are available. Additionally, ring expansions happen when a carbocation is adjacent to small rings (3-5 members), allowing the ring to grow larger, thus stabilizing the positive charge. Understanding these mechanisms is crucial for predicting reaction outcomes in organic chemistry.

Understanding why carbocations shift.

Now I want to talk about one of the most interesting but also annoying things that carbocations do, and that's called the carbocation rearrangement. So it turns out that carbocations are going to be able to rearrange to more stable positions if they're adjacent to the carbocation and if it has more R groups than the carbocation has at the moment. That's called a shift. All right? And there are a few different ways that this can happen. So basically what your criteria are is this. Your carbocation, let's say that it's secondary, it's only going to want to move if it can move one space to the right or the left and become more stable by moving that space, meaning that it could become tertiary or that it could resonate or something like that.

Hydride Shift

All right. So let's talk about the most common type of shift first, and that's the 1,2-hydride shift.

The 1,2-hydride shift occurs when there is a hydrogen located on an adjacent more stable carbon.

Okay? So here's an example of an alkyl halide. Remember that I told you guys that alkyl halides have the ability to leave. So, my Cl could leave all on its own and make a carbocation that looks like this. Okay? Now that I have it, and by the way, the Cl would just become Cl^-. Alright? So there's my carbocation.

Now my question is, is that carbocation the most stable that it can be right now? No, it's primary. Primary isn't the best. Is there a way that if it moved one space over, could it become more stable? And the answer is yes because right now it's on a primary position. This is a tertiary position. So if we could just move one carbon over, that would make it a whole lot more stable.

Okay. Well, how do we do that? Well, are there any hydrogens attached to that more stable position? Yes, there is. There's actually a hydrogen right there. Okay? That means that I'm allowed to do a 1,2-hydride shift.

The way that we draw the arrow for this, which is what a lot of students get confused about, you have to draw the arrow from the most negative thing to the most positive thing, just like any mechanism we've ever drawn. What that means is that you never draw your arrow coming from the positive charge because the positive charge is the thing that's missing electrons. You take the electrons from the bonds of the H and you attack the carbocation with it. What that's going to do is we're just going to write here this is a 1,2-H shift. And what we're going to wind up getting is a new carbocation.

Now let's, I just want to point out some things about how this carbocation is going to move.

First of all, I'm going to circle this carbon right here. You guys see that? The green one? Sorry, the one with the carbocation before, how many hydrogens did that carbon have before the shift? It has a positive charge, so how many hydrogens did it have? Not 3. I know you're thinking 3, so you can fulfill the octet, but it's not 3 because it has the positive charge, right? So it only had 2. I'm going to draw them out. It had one here and it had one here, let's say. It had 2 hydrogens coming off and then had a positive charge, meaning that it's missing 1 hydrogen.

Now that we're moving this hydrogen over, how many hydrogens will that carbon have? Well, it's going to have the 2 original blue ones. Let me draw that in blue. Okay? But now it's also going to have this red one that I just moved over. So is that carbon going to be happy now? Yeah. It's going to have 4 bonds. It's fine.

But now I have this carbon here that used to have 4 bonds and now it only has 3 because the hydrogen moved over. That means that now the carbocation goes there. That's the way that it works. Now I've done a 1,2-shift. I have a tertiary carbocation, and that is a whole lot more stable.

Does that make sense? This is a carbocation rearrangement.

Alkyl Shift

Now I want to go on to the next type of rearrangement. It's a little bit less common, but you're still going to see it and that's the 1,2-alkyl shift. The 1,2-alkyl shift occurs when only small alkyl groups are located on adjacent stable carbons. What that means is that you only do a 1,2-alkyl shift if no hydrogens are available. Why? Because it's a lot easier to move a hydrogen over than it is to move a methyl group or an ethyl group or some alkyl group over. So what you want to do is to do the hydrogen shift first no matter what. But in the case where there are no hydrogens, then you are allowed to do an alkyl shift. Just so you guys know, the two types of shifts that are common with this are the methyl shift and sometimes you will see professors use an ethyl shift. But I've never seen anything higher than that. The reason is that the bigger these alkyl groups get, the more energy it takes to move them over. By the time you get to propyl, the activation energy to make that happen is just overwhelming. It just doesn't happen anymore.

Okay. So here we've got another one. Let's go ahead and make our carbocation first. How do we make our carbocation? Kick out the alkyl halide. So what I'm going to get is a carbocation that looks like this. Are you guys cool with that? Cool. And then plus Bromine ion.Br^-

Alright, now I've got my carbocation. Is that able to shift to a more stable location? Well, let's say it went to the right. Would that make it more stable? No. It would just still be secondary. Right now it's secondary. How would it be if it went to the left? Yeah, that one on the left definitely has a lot more groups than the secondary. So now I have to ask which shift do I use? Well, do I have any hydrogens attached to that carbon? No, I don't. So that means my only choice is to do an alkyl shift. Now all three of these alkyl groups are the same size, so typically I'd want to pick the smallest one. But since they are all the same size, it doesn't matter which one I use. I'm just going to use the one closest to it, but I mean the one that I drew closest to it. But these are all even the same distance away; that's just the way I drew it; it happens to look closer. So now what would the arrow look like? The same exact thing, it would just come from the bond to the carbocation. What this means is that now I'm going to get what we would call a 1,2-Methyl shift.

And what I would wind up getting is now that I have an extra carbon coming off of that one down there and now I have a carbocation there. Why? Because the carbon here used to have four bounds, but now it only has three because the methyl group left. So now I just went from a secondary carbocation to a tertiary carbocation. And that's going to be a lot more stable. Does that make any sense? So remember, a hydrogen shift is the easier one, then alkyl shifts come next. Methyl is before ethyl. Ethyl is like your last resort. We rarely see ethyl shifts, but it is possible.

Ring Expansion

And this brings us to the last type of rearrangement which is called a ring expansion. So what is a ring expansion and why does it happen? Well, a ring expansion occurs when a carbocation is next to or adjacent to a 3-, 4-, or 5-membered ring, basically when it's next to a small ring. When you have a positive charge that's immediately next to a small ring, you can get something called a ring expansion. Now notice that the molecule that I'm using to show this to you is very similar to the molecule that's at the top of your page. At the top of your page, recall that we used a cyclohexane with a carbon and a chlorine. And we said that this molecule, once the carbocation forms, is going to do a hydride shift, a 1,2-hydride shift. But now, just by making the ring one size smaller, I'm actually going to make it do something different because notice that after this chlorine leaves, I'm going to get a carbocation that forms on this carbon. And now because I made my ring size just a little bit smaller, now instead of 6, it's 5. A 5-membered ring counts as a small ring. And remember that small rings like to do what? They like to expand. They like to do a ring expansion. So now that this positive charge is next to the 5 of a ring, I do something completely different which is that I basically grab that carbon and I pull it into the ring to make the ring bigger.

Now let me show you what the mechanism looks like for this. For the sake of showing you the mechanism, I'm going to draw 3 different carbons. I'm going to draw that this is a red carbon. I'll just make it a red dot. This is a blue carbon, and this is a green carbon. Now we know that carbons aren't circles or aren't drawn as circles. But just for the sake of drawing the mechanism, I think it's easy to do that. Now notice how many hydrogens each of these has. Red has 2. Blue also has 2. It's always going to have 2 hydrogens, but it's going to have a positive charge after the chlorine leaves. And notice that green is the odd man out, it has just 1. Well, what happens with this mechanism is that, in a ring expansion, the ring is strained. It doesn't like to have those bond angles and that torsional strain. It doesn't like to be a five-member ring. It wants to be bigger. So instead of doing a methyl shift or a hydride shift, the ring is actually going to donate its electrons to that carbon to make it bigger. Okay. So imagine that the bond between red and green gets broken and those electrons are used to pull the blue one in. Let me show you. Imagine that you took these electrons and you used them to pull blue in between both of them. So that now, instead of having red and green directly attached, now it's red, blue, green.

Well now I've got a 6-membered ring and where are these carbons? Well, let's say this is still red. Notice that red has 2 hydrogens. So is red going to have a charge now? No. Red is fine because it's got 4 bonds. Now notice that red is attached to blue. Why is it attached to blue? Well, because of this new bond that was created. How many hydrogens did blue have? Still 2. So is it going to have a charge? No, it's neutral. But what else happened? Well, the loser in this situation is green because notice that green was happy before. He had 4 bonds. But now, we just broke this bond, right? So that bond doesn't exist anymore. And green had how many hydrogens? Just one. So that means that now green with 1 hydrogen is going to have a positive charge. And that is a ring expansion.

So what happens in a ring expansion, guys, is that you take basically a smaller ring and you expand it to make a bigger ring. Do you still have a carbocation at the end? Yes, but now that carbocation is located on a bigger ring. So now, without drawing all these arrows, I just want to show you that what it's usually going to look like is this: Like a carbocation that looks like this would rearrange to form a carbocation that looks like this. Basically, the carbocation with 1 carbon and a 5-membered ring, see how I have 5+1, it's going to engulf and it's going to become 6. So 6 is just the sum of 5 eating up 1 and making it into the 6th carbon of the ring. All right. So guys, also remember this is only going to happen if you start off with a 5, 4- or 3-membered ring. Over here in the bottom example, the carbocation here would not expand because it doesn't make sense to engulf it and to make a 7-membered ring that's not more energetically stable, so that doesn't really happen a whole lot. All right. So I hope that made sense, guys. Let's move on to the next page.

Intro

All right guys. So for these rearrangement problems, what I want you to be able to do is answer 2 questions. I want you to first say will this carbocation rearrange or not? Yes or no? And then secondly, if it does rearrange, what's it going to look like? All right. So basically, here are the questions. Go ahead and start on question 1 and figure out will it shift and then what it would look like if it does shift.

Which of the following carbocations would be likely to rearrange? Draw each rearranged structure below.

Molecule I

All right. Will one shift? Well, right now it's secondary. Could it become more stable by moving to another position? Yes, it could. If it moves in this direction, it could become tertiary. So what kind of shift would I get? This would be a hydride shift because I have a hydrogen in that location. So what I would do is first I'm going to say yes that it will shift. Okay. Then I'm going to draw the arrow. So what I'm going to get here is a 12 hydride shift. And that's going to give me a new carbocation that looks like this. Okay. Now notice that I didn't draw the H in my final product. That's because I don't need to. Okay. You don't need to draw the H. H's are implied. All right. Let's move on to the next question.

Which of the following carbocations would be likely to rearrange? Draw each rearranged structure below.

Molecule II

All right. So this one is pretty easy. Actually, this one will not shift. So no. The reason is because right now it's secondary. If it moves one space to the right or one space down, it would still be secondary. Okay? Now we did have this space over here that was awesome, but that's too far apart. It could never be able to get over there. Okay? So the answer is no. This one would not shift. Move on to the next question.

Which of the following carbocations would be likely to rearrange? Draw each rearranged structure below.

Molecule III

Would this one shift? Absolutely. We've got a secondary and it has this awesome site right next to it. What kind of shift would that be? This would be a methyl shift. Okay? And the reason is that I don't have any hydrogens available. Okay? If you used another methyl, for example, if you used this one, that's fine. They're all the same. So my final product would be, well, I'm going to draw here 12 Methyl shift and my final product would look like this with now a carbocation down there. Because of the fact that now that carbon is missing that middle carbon there is missing a methyl group that just left. All right. Final question.

Which of the following carbocations would be likely to rearrange? Draw each rearranged structure below.

Molecule IV

Would this shift? No, it wouldn't. Alright? And by the way, I forgot to write yes here. Okay? But no, it wouldn't shift. Why? Because it's already tertiary. Okay? It's already the most stable that it can possibly be. I know this is tricky because you were thinking, Oh, but it wants to move down. No. It's already happy. It doesn't need to do anything. All right? So remember, tertiary is never going to shift. All right? I hope that made sense. Let me know if you have any questions.

Here’s what students ask on this topic:

What is a carbocation rearrangement and why does it occur?

A carbocation rearrangement is a process where a carbocation (a positively charged carbon atom) shifts to a more stable position within a molecule. This occurs because carbocations are highly reactive and unstable, and they will rearrange to achieve greater stability. The rearrangement typically involves a 1,2-hydride shift, where a hydrogen atom moves from an adjacent carbon, or a 1,2-alkyl shift, where a small alkyl group moves. Additionally, ring expansions can occur when a carbocation is adjacent to a small ring (3-5 members), allowing the ring to grow larger and stabilize the positive charge.

What is a 1,2-hydride shift in carbocation rearrangements?

A 1,2-hydride shift in carbocation rearrangements occurs when a hydrogen atom from an adjacent carbon moves to the carbocation, enhancing its stability. This shift happens because the carbocation seeks to achieve a more stable configuration, such as moving from a primary to a secondary or tertiary position. The hydrogen atom's electrons are used to form a new bond with the carbocation, resulting in a more stable carbocation. This type of shift is common because it requires less energy compared to moving larger groups.

What is a 1,2-alkyl shift and when does it occur?

A 1,2-alkyl shift occurs when a small alkyl group (such as a methyl or ethyl group) moves from an adjacent carbon to the carbocation. This shift happens when no hydrogens are available for a 1,2-hydride shift. The alkyl group moves to stabilize the carbocation, typically resulting in a more stable secondary or tertiary carbocation. Alkyl shifts are less common than hydride shifts because they require more energy to move the larger group. However, they are still important in cases where hydride shifts are not possible.

What is a ring expansion in carbocation rearrangements?

A ring expansion in carbocation rearrangements occurs when a carbocation is adjacent to a small ring (3-5 members). The ring expands to form a larger ring, which stabilizes the positive charge. This happens because small rings have significant ring strain due to their bond angles and torsional strain. By expanding the ring, the molecule reduces this strain and achieves a more stable configuration. The electrons from the bond in the small ring are used to form a new bond, incorporating the carbocation into the larger ring.

How do you determine if a carbocation will rearrange?

To determine if a carbocation will rearrange, you need to assess its stability and the potential for achieving greater stability through rearrangement. First, identify the current stability of the carbocation (primary, secondary, or tertiary). Then, check adjacent carbons for hydrogens or small alkyl groups that can shift. If a shift can result in a more stable carbocation (e.g., moving from primary to secondary or tertiary), the rearrangement is likely to occur. Additionally, if the carbocation is adjacent to a small ring (3-5 members), a ring expansion may happen to reduce ring strain and stabilize the positive charge.