- 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: stereoselectivity
Stereoselectivity of E1 reactions to favor more stable Zaitsev product.
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- If the cis and trans products make up 100% of the products, what happened to the carbo cations at3:00?(2 votes)
- Jay drew those to illustrate how the resonance structures "pass" the + charge around in order to stabilize it. You have to remember that the actual molecule is a combination of all of the resonance forms, so the + charge is never truly localized on any of the carbon atoms that its drawn on when drawing out the resonance forms. The resonance form that has the + charge outside of the phenol group will have the greatest influence on the resonance hybrid, which is why the alpha carbon is considered to be the one outside of the ring.(2 votes)
- At1:40, why did Khan avoid deprotonating the beta carbon on the benzene ring. Is there a rule that involves using the beta carbon attached to the most hydrogen?(1 vote)
- So the beta carbon on the benzene ring doesn't actually have any hydrogens to be extracted by a base. Two bonds from double bond to another carbon on the ring, and two additional single bonds to other carbons means carbon has the maximum number of bonds being tetravalent with an octet. There is no more room for a hydrogen to be bonded to that beta benzene carbon. Hope this helps.(2 votes)
- How do you know when a molecule is stereoselective? Like is it safe to say that all molecules with a stereocenter have the ability to be stereoselective? Or are all molecules susceptible to being stereoselective?(1 vote)
- In order to be an E1 reaction, does heat need to be added along with the H2SO4?(1 vote)
- Can the base only attack the proton on beta carbon?
Is it the same always??(1 vote)
- [Narrator] In this video, we're going to look at a E1 reaction that is stereo selective. So first, we're gonna draw out the mechanism, figure out the products, and then we'll talk about why it's a stereo selective reaction. On the left, we have our alcohol. The carbon that's bonded to the O-H is our alpha carbon. And the carbon next to that will be our beta carbon. So this beta carbon has two beta protons. Because we know in our mechanism, in our E1 mechanism, we're gonna lose a beta proton. On the left here, this carbon's next to the alpha carbon but there are no hydrogens on that carbon. So we only have to worry about the carbon on the right. Sulfuric Acid is a strong acid. And it will protonate our alcohol. So I'll just draw on an H plus in here to save some time. So, a lone pair of electrons on the oxygen picks up a proton from Sulfuric Acid. And let's draw what we would have now. So I'll put in my Benzene ring, right in here. So let's put in those electrons. And I'll put in my carbon chain. And if we're protonating our O-H, let me draw this wedge in here. Now oxygen would have two bonds to hydrogen, one lone pair of electrons on the oxygen, and the oxygen would have a plus one, formal charge. So let me show those electrons here. So these electrons, in magenta pick up a proton from Sulfuric Acid. And we could say that it formed, that it formed this bond, right in here. And then we also know there's a hydrogen on this carbon, going away from us, in space. I'll draw in that hydrogen. And then for our beta carbon here, we know we have two beta protons. I'll put those in. One will be on a wedge. And one would be on a dash, like that. And then I'll go ahead and draw in the C-H three, right here. The next step in our E1 mechanism is loss of our leaving group. So now we have water as a leaving group. And we know water is a good leaving group. These electrons, in here can come off onto the oxygen. And we would lose water. We're taking a bond away from this carbon. So we're going to form a carbocation. So actually, let me just go down here real fast and get some more space. And let me draw in our carbocations. So we would have our Benzene ring, right here. So let me put in these electrons. And we would have, let me draw in the chain, like that. We would have a plus one formal charge, on this carbon. So after the water leaves, we got a plus one formal charge right here. We know in a E1 mechanism we need a stable carbocation. And this is a Benzylic carbocation so it's very stable. Because we can draw several resonant structures. We won't draw them all in, to save time. But just to give you an idea, I could take these electrons and move them out to here. And let me go ahead and draw that now. So if I move those electrons out I would have a plus one charge. I took a bond away from, let me use red for this, I took a bond away from this carbon. So now that's where I have a plus one charge. So plus one charge right here. And then I would have these other Pi electrons in the ring. And you could keep going and draw more resonant structures. For example, you could move these electrons into here. But I won't do that. I just wanted to show you that this is a Benzylic carbocation that is resonance stabilized. So let's go back up to here. So we can figure out our products for this reaction. And I also want you to think about the possibility of free rotation around this Sigma bond here. So there's free rotation around this Sigma bond. And let me go ahead and draw in one way to view our carbocation here. So I'm going to call this Benzyne ring here, a phenol group. So I'm gonna write P-H for a phenol group. And then this will be our carbocation. So we need to try to show Planer Geometry around our carbocation. And this carbon, right here, the one in magenta, is an S-P two hybridized carbon, in our carbocations. So there is an un-hybridized P orbital. We draw in that P orbital right here. And there's a plus one formal charge on this carbon. There's a plus one formal charge, that's our carbocation. And then we would have a carbon, right here. And since we know there's free rotation around the Sigma bonds over here. So this arrow that I drew. Let me just highlight it in red. I'm gonna pick a particular confirmation. I'm gonna have one of the hydrogens parallel with the P orbital. And another hydrogen over here. And then, that would meet a methyl group right here. So this is one possible confirmation. And to show you how you get this confirmation, let's go and look at a video. And in the video, I have the methyl group as being red. So it's easier to see. And the phenol group, over here. The phenol group is gonna be purple in the video. On the left, we have our carbocation. So you can see, I put in these paddles here for the P orbital. And the Geometry around this carbocation is planer. So hopefully you can see that with the bonds here. And we have our purple for our phenol group. We know that we have free rotation about this single bond. So I'm going to rotate to get to the confirmation that we saw earlier here. So this confirmation, has a carbon hydrogen bond, that's parallel with the P orbital. And can donate some electron density into that P orbital. In E1 mechanism, we take a proton. So I'm gonna take this proton here. So pretend like I'm taking it away. But now we can see the alkyne that forms. This would be the trans-alkyne, with the bulky phenol group on an opposite side of the double bond, from our methyl group. So we know that there's another possible confirmation that has another carbon hydrogen bond parallel with our P orbital. And so can donate some electron density into the P orbital there. If I took away this proton, you can see we would get the cis-alkyne for the product. So in this case, the bulky phenol group and methyl group would be on the same side of the double bond. As we saw in the video, the P orbital of the carbocation, aligns parallel with the breaking carbon hydrogen bonds. So, our P orbital would be in this direction. And our carbon hydrogen bond would be parallel with that. So the electron density from this bond can be donated into the P orbital. So we know, in a E1 mechanism, a weak base comes along, at this point, and takes that proton. So we're gonna take this proton here. And those electrons in light blue would move in to form our double bond. And this confirmation gives us the trans-product. So let me go ahead and draw in our trans-product here. So we have our phenol group. And then we have our double bonds. And then we would have our methyl group, so C-H three. So this would be the trans-product and we know that a different confirmation gives us the cis-product. So let's draw in the other confirmation. Or the confirmation that gives us the cis-product here. So here's our phenol. And then we would have our carbon, our carbocation, there's a hydrogen going away. And then, this would be, let me show the Planer Geometry around that carbocation. So we would have our P orbital in there like that. And a plus one, formal charge, on that carbon. And then we would have our other hydrogen. As now the carbon hydrogen bond is parallel with our P orbital. And so this is after we rotated it. So it's a different confirmation then the one on the left. And so this confirmation has the methyl group over here, like that. And so now, again, we have this electron density, that can be donated into this P orbital. And so our base comes along and takes this proton. And if that happens, then our electrons move into here. And we would form the cis-products. Let me draw in the cis-products. We have our phenol group. And then our double bonds. And then our methyl groups, C-H three. So this is the cis-product. So now we've figured out our products. Let's talk about why this reaction is said to be stereo selective. So we formed two stereo isomers as our product. We have a trans-isomer, which is actually 95% of our product. And a cis-isomer, which is only about 5% of our product. And we can explain that by looking at our two confirmations down here. So the confirmation on the left, has the bulky phenol and methyl group relatively far away from each other, in space. So decreased stereo hindrance, but the confirmation on the right has these two bulky groups pretty close together in space. So, that destabilizes that confirmation. And that's why we don't get as much of the cis-products. So the trans-products forms because of decreased stereo hindrance, and our trans-product is more stable. And so this reaction is said to be stereo selective. So we had a preference for the formation of the more stable isomer, which is the trans-product.