- E1 mechanism: kinetics and substrate
- E1 elimination: regioselectivity
- E1 mechanism: stereoselectivity
- E1 mechanism: carbocations and rearrangements
- E2 mechanism: kinetics and substrate
- E2 mechanism: regioselectivity
- E2 elimination: Stereoselectivity
- E2 elimination: Stereospecificity
- E2 elimination: Substituted cyclohexanes
- Regioselectivity, stereoselectivity, and stereospecificity
E1 mechanism: carbocations and rearrangements
Comparing the stability of different kinds of carbocations using hyperconjugation. Mechanism of carbocation rearrangements including methyl and hydride shifts. Created by Jay.
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- So... if a Hydride shift and a Methyl shift can both occur... which one is more likely to occur? The Hydride shift?(22 votes)
- I think you need to compare both shifts and see which one will produce a more stable carbocation, that is the one that will occur.(32 votes)
- Can someone explain why carbocation intermediates can only be invoked in mechanisms operating under acidic conditions? Is it because carbocations immediately react with strong bases?(8 votes)
- Around2:40he says that it is sp2 hydridized, yet he's drawn 3 H's around the C. If the C is sp2 hydridized that means it is double-bonded to another C, and should therefore only have 2 H's around it. Right?(3 votes)
- If a C is sp2 hybridized, that doesn’t mean that the C has to be double bonded to another C, sp2 hybridization requires only that there be three other atoms directly attached to the C. This is the case with the methyl cation, CH3+.(13 votes)
- Could a carbocation rearrange twice to become more stable?
For example, if I had 2,2-dimethylpentane with a carbocation on carbon 4, would it first rearrange with a hydride shift to carbon 3, and then a methyl shift to carbon 2 to create a tertiary carbocation?(3 votes)
- A carbocation can rearrange more than once, but not in this case. You are starting with a 2° carbocation. A hydride shift would simply give a different 2° carbocation. Since this is no more stable than the first, the molecule has no incentive to undergo the hydride shift, and the methyl shift will not have a chance to occur.(4 votes)
- When you discuss the hydride and methyl shifts, are methyl and hydrogen the only two compounds that can shift around in E1 elimination reactions or will there be other compounds that can also undergo the shift? Is this shift specific for only E1 reactions or can it happen in E2 or Sn1/2 reactions as well?(3 votes)
- Alkyl groups other than methyl groups can also undergo alkyl shifts, however it is more common with smaller alkyl groups. Hydride and alkyl shifts in carbocation intermediates in SN1 reactions are also common.(1 vote)
- What is the mechanism for the hydride/methyl shift? Or does it just happen?(1 vote)
- Remember that a mechanism is just a way to represent what the electrons are doing in a reaction, there are no real arrows involved. Sometimes they help and make things clearer, and other times they can just make things more confusing and so are ignored.
In this case the electrons from the C-H bond (in a 1,2-hydride shift) move to the area of positive charge represented as a curly arrow from the C-H bond to the positive charge on the adjacent carbon.
A detailed explanation, and schematic, with a transition state, can also be found here: http://www.masterorganicchemistry.com/2012/08/15/rearrangement-reactions-1-hydride-shifts/(3 votes)
- If you have a molecule undergoing, suppose, an Sn1 mechanism, and suppose there are multiple places along the molecule where carbocation rearrangement (hydride shift) can occur (each of these places forms a tertiary carbocation). Does the rearrangement just proceed all the way to the end, and then the new structure forms, or are structures from all rearrangements possible, because each is a tertiary carbocation?
Thank you!(2 votes)
- I see - because thermodynamically, they would all be equally stable. Thanks Simon!(1 vote)
- How do you know when carbocation rearrangements will occur in terms of alcohol reactions?(1 vote)
- A rearrangement will occur if
a. The OH is converted to a good leaving group.
b. The carbocation formed by loss of the leaving group can become more stable by a hydride or alkyl shift.(3 votes)
- how we would know that a compound goes methyl shift or hydride shift(1 vote)
- If a carbocation can become more stable by a hydride or a methyl shift, it will so so.(2 votes)
- Is there such a thing as a plus 2 formal charge?(1 vote)
- Yes. The S atom in the sulfite ion, SO₃²⁻. has a formal charge of +2.(2 votes)
- [Instructor] Let's say that our goal is to show a mechanism for this E1 reaction. So we're going from the alkyl halide on the left to the alkene on the right. But the first step is loss of our leaving groups. The electrons in this bond come off onto the chlorine to form the chloride anion, and we know that the chloride anion is a good leaving group. We're talking a bond away from the carbon that I just marked in red, so that's where our carbocation will form. So let's draw out our carbon skeleton, and we'll put in these methyl groups right here. And the carbon in red is this carbon here, so that would be a secondary carbocation. So there's a plus one formal charge, and it's secondary because the carbon in red is directly bonded to two other carbons. So we have a secondary carbocation. And if you look at it, there's a possibility for a rearrangements. We could take one of the methyl groups on the carbon in magenta on the right, and we could move it over to the positively charged carbon. So we get a methyl shift here. And the reason why we would get a methyl shift is that's gonna create a more stable carbocation. So now there's only one methyl group on this carbon. There's a methyl group over here to begin with, and we just moved a methyl group to the carbon in red. So let me highlight the carbon in red here. We moved a methyl group to that carbon, so let me draw that in. So that means we took a bond away from this carbon, so that's this carbon, and so that's where our positive charge is now. So we'll plus one formal charge on the carbon that I've now labeled in light blue. And this is a tertiary carbocation. So let me highlight those carbons here. The carbon in light blue is directly bonded to three other carbons, so this carbon right here, this one, and this one. So this is a tertiary carbocation, which we know is much more stable than a secondary carbocation. Our carbocation is further stabilized by the fact that we have a polar product solvent here. So this is a very stable carbocation. At this point, you could have two possible reaction paths. You have water, which could function as a nucleophile. And if that happen, you would get a substitution product. But now, we know we need an elimination product. We have an alkene on the right here. So water is gonna function as a base, and water is going to take a proton from one of the carbons next door to our carbocation, so one of the carbons in magenta. And if you you look at our product here, the double bond formed between this carbon and this carbon, so that means we need to take a proton from the carbon in magenta on the top right. So let me highlight that carbon here. So we need to take a proton from this carbon. So let me draw in one, because we know there's one hydrogen bonded to that carbon, and water is gonna function as our weak base here. So let me draw in H2O. And water is gonna take that proton, so lone pair of electrons on the oxygen take this proton right here, and then these electrons would move in to form our double bond, and that gives us our product. So let me highlight those electrons. So the electrons are formed are double bond, like the dark blue right here. So those electrons move in here to form our alkene. And we've shown, we've shown a mechanism for this reaction. Let's look at another E1 reaction, and let's say our goal was to draw all of the products from this elimination reaction. On the left we have our alcohol, and we're reacting our alcohol with sulfuric acid, and we're heating our reaction mixture. We've already seen from earlier videos, the first step when you have an alcohol, say proton transfer, the alcohol functions as a base, and sulfuric acid donates a proton. So I'll draw an H plus here. We're gonna protonate our alcohol first. So one of the lone pairs of electrons on oxygen picks up this proton, which came from sulfuric acid. So let's draw in what we would have. So here's our carbon chain. We would have two methyl groups on that carbon. And the oxygen now has two bonds, two hydrogen. It still has one lone pair of electrons, but now the oxygen has a plus one formal charge. So our electrons in magenta pick up our proton from sulfuric acid to form this bond. Now, the reason why this step happens first when you have an alcohol is it forms a better leaving group for your E1 mechanism. Water is a much better leaving group than the hydroxide anion. So next, the electrons in this bond come off onto the oxygen, and water leaves. So when we do that, we take a bond away from this carbon, the carbon in red. So that's where our carbocation forms. If I draw in my carbon skeleton here, so I have methyl groups coming off of this carbon. The carbon in red is this carbon, and that's a secondary carbocation, because the carbon in red is directly bonded to two other carbons, which I just marked there in magenta. So this is a secondary carbocation. And let me draw in the plus one formal charge on the carbon in red. Next, we think about the possibility of a rearrangement. So can we do anything to form a more stable carbocation? Well, just like the previous example, we could have a methyl shift. Let me go ahead and leave some more room in there. So we could take, let's say this methyl group, and we could move this methyl group over to the carbon in red. So let's show a methyl shift. So now there's only one methyl group on this carbon, and a methyl group moved over here to this carbon. So this is the carbon in red. So the carbon in red no longer has a plus one formal charge. The plus one formal charge goes to the carbon in magenta that just lost a bond. So now, our plus one formal charge is on this carbon. Let me highlight it. So this carbon right here in magenta. And that is a tertiary carbocation. And we know tertiary carbocations are more stable than secondary carbocations. Next, since this is an E1 mechanism, we know that a weak base comes a long and takes a proton. So in this case, we could take a proton from a few different places. We could take a proton from this carbon. So we think about our weak base coming along. Let me draw this in here, and our base is gonna take this proton, which means that these electrons are moving here to get rid of our formal charge, and to form an alkene as our product. So let's draw in the product up here, our double bond would form here. And let's draw in everything else. So that is one of the possible products. Let's highlight our electrons. I'll use light blue. So the electrons in this bond moved in to form our alkene. All right, what about if I took a proton from a different place? So let me just redraw that carbocation here, just so things don't get too busy. So I'm gonna redraw our carbocation. So the carbon in magenta is this one, so that has a plus one formal charge. Let me draw that in. And the carbon in red is this carbon. So what happens if we took a proton from the carbon in red? Let me draw in a proton here. So our weak base comes along and takes this proton, let's say. And so, these electrons would move in to here to form our double bond, to form another alkene. So let's draw that product. So we have our carbon skeleton here. And this time, the double bond forms between the carbons in magenta and red. So the double bond would form in here. So let's highlight those electrons in light blue. These electrons would move into here to form our double bonds. All right, there's actually one more product for this reaction, and let's go back to our secondary carbocation over here. So, if it doesn't rearrange, you could actually take a proton from this secondary carbocation and form our last product. Let's think about the carbon in red right here. So we're looking at the carbons next door, so the carbon in red. Well, the carbon in magenta to the left doesn't have any proton, so we can't take a proton from that one, but the carbon to the right in magenta does. So if I squeeze in a hydrogen in here, finally, our base could come along and take this proton. And if that happen, then these electrons would move into here, so that gives us our final product. Just let me draw in our skeleton here, and then our double bond will form right... Actually, let me just redraw that double bond, so thing we're getting... We just redraw the whole thing. I think I have a little bit more space that I thought I did. So let's sketch in our carbon skeleton here, our double bonded form here, and then we have our methyl groups coming off of that carbon. So the electrons in light blue moved into form our double bond. So we'd formed three products, three products from this E1 reaction. If we think about which one would be the major product, let's look at degrees of substitution. So let's go over to the alkene on the right. So when we think about the two carbons across our double bonds, there are one, two, three, four alkyl groups. So this is a tetrasubstituted alkene, so this should be the major product. This is the most stable alkene. Next, let's look at this one. So here's a carbon, and here's a carbon across our double bond. We can see this time we have only two alkyl groups. So this is a disubstituted alkene. And then finally, over here on the left, this would be a monosubstituted alkene. So here are two carbons. We have one alkyl group, so a monosubstituted alkene. And this one came from the secondary carbocation from no rearrangement, so this one is not gonna be a major product. Only a very small percentage of your products would be this monosubstituted alkene. Most of your products is gonna be your di and your tetrasubstituted alkene, So with your tetrasubstituted alkene being your major product since it is the most stable.