Carbocation Intermediate Stability

Now we’re going to discuss the most important intermediate of all organic chemistry, the carbocation.

Carbocation Stability

Concept: Determining Carbocation Stability

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Video Transcript

Let's talk about intermediates. Remember that intermediates are high-energy compounds that can be isolated but exist at a higher energy state than the products or the reactants. That's the definition of an intermediate. It turns out that the most popular, the most widely used type of intermediate in this course is going to be the carbocation. What that means is I want to go more into depth on exactly what are the properties of carbocations because those are the most popular intermediates.
Remember that carbocations are just carbons that have three bonds instead of four, so they're always going to have a empty orbital. That empty orbital we give it a positive charge because it's missing a bond. And carbocations it turns out are going to be stabilized by something. There is something that can make that positive charge a little bit better. That's something called hyperconjugation.
What hyperconjugation says is that the electron density from neighboring sigma bonds, so adjacent sigma bonds, can share electron density with the carbocation, with that empty orbital. So if I have – the more sigma bonds I have close to it that are able to donate, the more stable my positive is going to be. Because of the fact that my positive is missing a ton of electrons, if I can share a little bit of these electrons into that positive charge, it's going to make it a lot more stable.
Basically, hyperconjugation is defined by the delocalization of charge of the interaction of an empty p orbital with an adjacent eclipsed sigma bond. It turns out that there's only one group that's really good at having eclipsed sigma bonds to p orbitals and that's R groups because R groups remember they always have hydrogens coming off of them. So those hydrogens are going to be able to eclipse with the p orbital. What that means is that the more substituted my carbocation, or the more R groups I have attached to it, the more stable.
So now I want to show you guys a chart of carbocation stability and show you guys the differences between what's a really bad carbocation and what's a good one.
A bad one would be all the way down here at primary. This would be called a primary carbocation because I only have one R group that's going to have H's that can donate electrons. Because remember an R group just means it's a carbon with some H's on it. That would be an example of a really, really bad carbocation because it only has one side donating electrons.
Now on the opposite end of the spectrum, a tertiary is awesome because a tertiary is going to have hydrogens here, here, and here. And they're all going to be able to donate electron density to that positive charge stabilizing it. So what that means is tertiary carbocations are always going to be your most stable no matter what.
Then in between, we have other ones like secondary and then these guys here. What are those? These are actually going to be ones that can resonate. This is called allylic. And this is benzylic. All these words mean is that – allylic just means next to a double bond so that's what this is. Benzylic means it's next to a benzene.
Remember when we talked about resonance structures way back that a positive charge next to a double bond could resonate and it would form another resonance structure that looked like this. That is a form of delocalization where now my positive charge isn't just stuck in one place, it can be distributed over several atoms so that's going to be more stable.
So allylic carbocations are more stable than secondary because secondary, it's good, it's better than primary because it has the ability to donate with hyperconjugation, but it doesn't have the resonance capability, so it's not as good as allylic. Same with benzylic. Benzylic is also better than secondary.
But the only thing is that tertiary is actually always going to be the best even if it can't resonate, that's fine. Because the tertiary one has so much hyperconjugation that that one's actually going to be more stable overall than even the benzylic or the allylic ones that can resonate. So you might just want to put here that these can resonate. But then this one just has – the tertiary's just the most stable overall.
Now I want to do this example. And I want you guys to think about this for a second and say which of the following alkyl halides would generate the most stable carbocation.
Now I am kind of asking you guys to think about an extra step, so I'm going to tell you what that extra step is. Just so you guys know. Maybe you've learned this already or not or maybe you remember it or you don't, but that's fine. Alkyl halide, which we're going to talk about more in a little bit, have the ability to leave. And if alkyl halides have the ability to leave the molecule completely that means that they're going to leave a carbocation behind.
What you need to do is think after this F leaves and becomes an F negative and I get a positive charge here, what kind of carbocation would that be. Do the same thing with all four of these and then figure out which of them is going to be the most stable carbocation. So go ahead and try to solve this question. Then I'll answer it. 

Carbocations are stabilized by a phenomenon called hyperconjugation.

Hyperconjugation is the delocalization of charge by the interaction of an empty p-orbital with an adjacent, eclipsed σ-bond. Basically it’s like resonance for single bonds (more complicated than that but don’t worry about it). 

Since this is only possible with -R groups, the more substituted the carbocation, the more stable

Example: Which of the following leaving groups would generate the most stable carbocation?

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Carbocation Intermediate Rearrangements

Concept: Understanding why carbocations shift.

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Video Transcript

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.
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.
And there's a few different ways that this can happen. So basically, what your criteria is 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 can become tertiary or that it can resonate or something like that. 

Carbocations will rearrange to an adjacent, more stable possible if possible. These have different names based on which atoms are rearranging. 

a. 1,2-Hydride Shift occurs when there is a hydrogen located on an adjacent, more stable carbon.

Concept: Carbocation Rearrangements

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Video Transcript

Alright 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, OK? 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, OK? Now that I have...And by the way the CL would just become CL negative, alright? So there's my Carbocation, now my question is that Carbocation the most stable that it can be right now? No it's primary, OK? Primary sucks, 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 it could just move one carbon over that would make it a whole lot more stable, OK? Well how do we do that? Well are there any hydrogen attached to that more stable position? Yes there is, there's actually hydrogen right there, OK? That means that I'm allowed to do a 1-2 hydride shift, the way that we draw the arrow for this this 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 so what that means is that never draw your arrow coming from the positive charge because the positive charge is the thing that's missing electrons you don't want to do that, 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 we're going to wind up getting is a new carbocation, OK? Now let's I just want to point out some things about how this Carbocation is going to move OK? First of all I'm going to circle this carbon right here, OK? You know see that? The green one? How many.... Sorry the one of the Carbocation before, how many hydrogens did that carbon have before the shift? It has a positive are so how many hydrogen did it have? Not three, I know you're thinking three so you can fulfill octet but it's not 3 because that's the positive charge, right? So it only had two, I'm going to draw them out, it had one here and it had one here let's say, OK? It had 2 hydrogens coming off and then had a positive charge meaning that it's missing one hydrogen, alright? Now that we're moving this hydrogen over, OK? How many hydrogens will that carbon have? Well it's going to have the 2 original blue ones, let me draw that in blue, OK? 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 be it's going to have 4 bonds it's fine, OK? But now I have this carbon here that used to have four bonds and now it only has three because the hydrogen moved over so that means that now the Carbocation goes there and that's the way that it works so 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? That is a Carbocation rearrangement, OK?

b. 1,2-Alkyl Shift occurs when only small alkyl groups are located on an adjacent, more stable carbon.

Concept: Carbocation Rearrangements

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Video Transcript

So now I want to go on to the next type of rearrangement, it is a little bit less common but you're still going to see it and that's the 1-2 alkyl shift, OK? The 1-2 alkyl shift occurs when only small alkyl groups, OK? Are located on adjacent stable carbons, so what that means is that.... Sorry that's the Alkyl, OK? This you only do a 1-2 alkyl shift if there's no hydrogens available, why? Because it's a lot easier to move the hydrogen over than it is to move a methyl or an ethyl group or some Alkyl group over, so what you want to do is you want to do the hydrogen first no matter what but in the case of there being no hydrogen is there then you are allowed to do an Alkyl shift, alright? Just so you guys know the two types of shift that are common with this are the methyl shift and sometimes you will see professors use ethyl shift but I've never seen anything higher than that and the reason is just because the bigger these R groups get the more energy it takes to move them over so by the time you get it to purple that's just like the activation energy to make that happen is just like overwhelming it just doesn't happen anymore, OK? So here you'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 BR negative. Alright so now I've got my Carbocation, is that able to shift to a more stable location? Well letÕs 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 about if it went to the left? Yea I mean 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 they're all the same size so 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 close to it but these are all even the same distance away as that's just the way I drew it happens to look closer, OK? So now what would the arrow look like? 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, OK? 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 fact that the red carbon here used to four bonds but now it only has three because the methyl group left, so now I just went from a tertiary I mean from a secondary carbocation to a tertiary carbocation and that's going to be a lot more stable, is that making sense? So remember hydrogen shift is the easier or hydride easy one then I'll alkyl shifts come next, methyl is before ethyl, ethyl is like your last resort, OK? We rarely see ethyl shifts but it is possible.

c. Ring Expansion occurs when a carbocation is adjacent to a 3, 4 or 5-membered ring.

Concept: Carbocation Rearrangements

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Video Transcript

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 guys is very similar to the molecule that's at the top of your page, at the top your page recall that we used to Cyclohexane with a carbon and a chlorine and we said that this molecule once the Carbocation forms is going to make...Is going to hydride shift, a 1-2 H 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 under this carbon and now because I made my ring size just a little bit smaller, now instead of 6 it's 5, 5 counts as a small ring and remember that small ring like to do, what? They like to expand, they like to do ring expansion so now that this positive charge is next to the 5-membered 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 three 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, OK? Now we know that carbon 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 hydrogen's each of these has, red has 2, blue well it has 2, OK? 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 one, OK? Well what happens with this mechanism is that in a ring expansion, the rain is strained it doesn't like to have those bond angles and that torsional strain it doesn't like to be 5 membered ring it wants you bigger so the ring 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, Ok? 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 use them to pull blue in between both of them, OK? So that now instead of having red and green directly attached now it's red, blue, green let me we show you what this would look like well now I've got a six-membered ring and where 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 four bonds, now notice that red is attached to blue, why is it attached to blue? Well it's because of this new bond that was created, right? I just grab the blue so this is that new single bond here, how many hydrogens did blue have? Still two 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 hydrogen? Just 1 so that means that now green with 1 hydrogen is going to have what type of charge? Positive charge, OK? 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 to make a bigger ring, do you still have a Carbocation at the end? Yes but that Carbocation is located on a bigger ring so now without drawing all these arrows I just want to show or without drawing all these hydrogens I just want to show you that what it's usually going to look like is like this like a Carbocation that looks like this would rearrange to form a Carbocation that looks like this, OK? Basically the Carbocation with a one carbon and a 5-membered ring, I see I have a 5+1 it's going to engulf and it's going to come 6 so six is just the sum of 5 eating up 1 and make me it into the sixth carbon of the ring, alright? 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 into make a 7-membered ring that's not more energetically stable so that doesn't really happen a whole lot, alright? So I hope that made sense guys let's move on to the next page.

Concept: Practice Intro

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Video Transcript

Alright guys so for these rearrangement problems what I want you to be able to do is answer two 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? Alright 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 shit.

Concept: Practice 1

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Video Transcript

Alright, will 1 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, Ok? Then I'm going to draw the arrow so what I'm going to get here is a 1-2 hydride shift, OK? And that's going to give me a Carbocation that looks like this, OK? Now notice that I didn't draw the H in my final product that's because I don't need to, OK? You don't need to draw the H, Hs are implied, alright? Let's move on to the next question.

Concept: Practice 2

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Video Transcript

Alright 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 of the right or one space down it would still be secondary, OK? Now we did have the space over here that was awesome but that's too far apart, it could never be able to get over there, OK? So the answer is no this one would not shift move on to the next question.

Concept: Practice 3

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Video Transcript

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, OK? And the reason is because I don't have any hydrogens available, OK? If you used another methyl for example if you used this one that's fine they're all the same, Ok? So my final product would be well I'm going to draw here 1-2 Methyl shift and my final product would look like this with now a Carbocation down there because now the fact that now that carbon is missing that middle carbon there is missing a methyl group that is left, alright? Final question.

Concept: Practice 4

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Video Transcript

Would this shift? No it wouldn't, alright? And by the way I forgot to write right yes here, OK? But no it wouldn't shit, why? Because it's already tertiary, OK? It's already the most stable 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, alright? So remember tertiary is never going to shift, alright I hope that made sense let me know if you have any questions.

Carbocation Intermediate Stability Additional Practice Problems

Rank the substrates below in terms of relative SN1 reaction rate. (1 = fastest, 4 = slowest)

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For the following carbocations, rank them from 1-3 according to overall carbocation stability, with a 1 under the LEAST STABLE CARBOCATION and a 3 under the MOST STABLE CARBOCATION.

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When 2-methylbutene is reacted with hydrochloric acid, where does the carbocation form and why?

1. The carbocation forms at the tertiary position because of steric hindrance.

2. The carbocation forms at the secondary position because of hyperconjugation from neighboring carbons.

3. The carbocation forms at the primary position because nature says so.

4. The carbocation forms at the tertiary position because of hyperconjugation from neighboring carbons.

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Rank the following molecules in order of increasing relative rate of SN1 solvolysis with methanol and heat (slowest to fastest reacting).

A) 5 < 4 < 3 < 2 < 1

B) 1 < 2 < 5 < 4 < 3

C) 2 < 3 < 4 < 1 < 5

D) 3 < 2 < 4 < 5 < 1

E) 2 < 3 < 4 < 5 < 1

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Circle the correct answer

 

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The following statements are true. Choose from among the following three possibilities and in the space provided, write the letter of the one or more phenomena that best explain the true statement.

A. The inductive effect     B. Hyperconjugation     C. Resonance delocalization of a charged species

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The driving force for carbocation rearrangements is that the less stable 2° carbocation is converted into a more stable 3° carbocation ( True or False )

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Identify which of the following substrates will undergo an SN1 reaction more rapidly. Explain your choice

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