- 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: kinetics and substrate
Mechanism of an E1 elimination reaction. Created by Jay.
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- For the alkyl halide example (starting at8:17), why wouldn't that undergo an Sn1 reaction? It's a tertiary carbocation with a weak nucleophile, isn't it?(19 votes)
- The answer to this question lies chiefly in the Gibbs free energy equation:
∆G = ∆H - T∆S
Elimination produces more products (three) compared to substitution (two), and thus, is more entropically favored, which means that at high temperatures, ∆G is negative for elimination which in turn makes it more spontaneous compared to the substitution reaction.(4 votes)
- 10:25how come the ch3 not on the ring not concidered as beta?(12 votes)
- You are correct. There are three ß carbons: C-2 and C-6 of the ring and the C of the methyl group.
The ß hydrogen atoms on the methyl group can also be removed to give an alkene, but the product is not as stable as the one with the double bond inside the ring. That is probably why they were ignored in the video.(9 votes)
- At6:47, the protons from the acid are responsible for the -OH leaving the cyclohexanol ring. How is it an E1 reaction if the concentration of both cyclohexanol and sulfuric acid determine the rate of this step?(6 votes)
- The concentration of H2S04 is not a factor in this case because it is your solvent. You can assume that there is an endless supply available.(4 votes)
- What happens to the sulphate ion? Does it remain an ion or reform a molecule?(6 votes)
- It will stay as an ion. If there is a cation with which it can precipitate, it will do that, but this is often not the case.(4 votes)
- At4:00the guy says that the water will take the the proton from hydrogen attached to the beta carbon...
why would the water behave in such a way? we just said that oxygen is very electro negative and does not like to have a positive formal charge.
so what's the deal?(5 votes)
- Electronegativity refers to the attraction of the O atom for the bonding electrons. It says nothing about the lone pairs.
The O atom can still use a lone pair to bond to a proton. The fact that O has a formal positive charge in H₃O⁺ does not mean that it is unstable. In fact, if you combine liquid anhydrous HClO₄ and water in a 1:1 molar ratio, solid hydronium perchlorate forms.
The hydronium ion is stable or, to put it another way, the reaction
H⁺(aq) + H₂O(l) → H₃O⁺ (aq)
is thermodynamically highly favourable.
In solution, the 'aq' bit of H₃O⁺(aq) becomes important. Hydronium ions don't exist in isolation. They have a solvation shell around them that consists of other water molecules tempted in by the polarity of the hydronium ion. This lowers the overall energy of the system.
So when we have the reaction
carbocation(aq) + H₂O(l) → alkene + H₃O⁺(aq)
we are going to a lower energy situation because the alkene is more stable than the carbocation and the hydronium ion is stabilized by hydration.(1 vote)
- H2SO4 will become HSO4 minus, giving a negative charge on an oxygen. That will be stablilzed by resonance structures. Will this stable structure be able to pull of the beta-hydrogne as well or is it too stable?(3 votes)
- It is too stable. HSO4- is a stronger acid than water, so water is a stronger base than HSO4-. It will be the water that removes the beta hydrogen.(3 votes)
- what is an alpha and beta carbon?(2 votes)
- alpha carbon- carbon on which the functional group is attached, beta carbon - carbon attached to alpha carbon.(4 votes)
- At2:32,Jay says that the best place for the oxygen to get electrons would be from its bond with carbon. My question is,why carbon?Why not the 2 hydrogens it's directly linked to? That should be easier as H is way less electronegative than C,if I'm not wrong.(3 votes)
- H+ is not as good a leaving group as a carbocation. The Oxygen will grab the electrons from the carbon because the overall system will be more stable with a tertiary carbocation than with an H+. This is especially the case in an acidic solutions because there are already many protons floating around and equilibrium favors the lowering of the concentration of protons.(1 vote)
- How can we predict that a neutral molecule such as ethanol or water molecule would act as a base as shown at10:26?(2 votes)
- Any molecule that contains an atom with a lone pair of electrons, such as the O in ethanol or water, can accept a proton from an acid. For example. H2O: + H-Cl --> [H2O-H]+ + Cl-. The [H2O-H]+ is usually written as H3O+. Since the water is accepting a proton from the HCl, it is behaving as a Brønsted-Lowry base.(2 votes)
- Does the leaving group (Br) leave spontaneously like what happens in a Sn1 reactions or does it have a mediator ( CH3CH2OH)?(1 vote)
- The Bromine will leave spontaneously like in SN1.
The number 1 or 2 tells us how many reactants are involved in rate determining step. Because the reaction is E1, a higher concentration of the alcohol will NOT speed up the reaction.(3 votes)
- [Instructor] Let's look at the mechanism for an E1 elimination reaction, and we'll start with our substrate, so on the left. Let's say we're dealing with alkyl halide. So the carbon that's bonded to our halogen would be the alpha carbon, and the carbon next to that carbon would be the beta carbon, so we need a beta hydrogen for this reaction. The first step of an E1 elimination mechanism is loss of our leaving group, so loss of leaving group, let me just write that in here really quickly, and in this case, the electrons would come off onto our leaving group in the first step of the mechanism. So we're taking a bond away from this carbon, the one that I've circled in red here, so that carbon is going from being sp3 hybridized to being sp2 hybridized. So now we have a carbocation, and we know that carbocations, sp2 hybridized carbons have planar geometry around them, so I've attempted to show the planar geometry around this carbocation. So that's the first step, loss of the leaving group to form a carbocation. In the second step, our base comes along and takes this proton, which leaves these electrons behind, and those electrons move in to form our alkene, so this is the second step of the mechanism, which is the base takes, or abstracts, a proton, so base takes a proton to form our alkene. And let me go ahead and highlight those electrons, so these electrons here in magenta moved in to form our double bond, and we form our product, we form our alkene. So the first step of the mechanism, the loss of the leaving group, this turns out to be the rate determining step, so this is the slowest step of the mechanism. So if you're writing a rate law, the rate of this reaction would be equal to the rate constant k times the concentration of your substrate, so that's what studies have shown, that these mechanisms depend on the concentration of only your substrate, this over here on the left, so it's first order with respect to the substrate. And that's because of this rate determining step. The loss of the leaving group is the rate determining step, and so the concentration of your substrate, your starting material, that's what matters. Your base can't do anything until you lose your leaving group. And so, since the base does not participate in the rate determining step, it participates in the second step, the concentration of the base has no effect on the rate of the reaction, so it's the concentration of the substrate only, and since it's only dependent on the concentration of the substrate, that's where the one comes from in E1, so I'm gonna go ahead and write this out here, so in E1 mechanism, the one comes from the fact this is a unimolecular, a unimolecular rate law here, and the E comes from the fact that this is an elimination reaction, so when you see E1, that's what you're thinking about, it's an elimination reaction, and it's unimolecular, the overall rate of the reaction only depends on the concentration of your substrate, so if you increase, let's say you have, let's say this was your substrate right here, and you increase the concentration of your substrate, let me just write this down, so if you increase the concentration of your substrate by a factor of two, you would also increase the rate of reaction by a factor of two, so it's first order with respect to the substrate, so this is some general chemistry here. If you increase the concentration of your base by a factor of two, you would have no effect on the overall rate of the reaction. So let's talk about one more point here in the mechanism, and that is the formation of this carbocation. Since we have a carbocation in this mechanism, we need to think about the possibility of rearrangements in the mechanism, and you need to think what would form, what substrate would form a stable carbocation, so something like a tertiary substrate forming a tertiary carbocation would be favorable for an E1 mechanism. Here we have a tertiary alkyl halides, and let's say this tertiary alkyl halide undergoes an E1 elimination reaction. So the carbon that's bonded to the iodine must be our alpha carbon, and then we would have three beta carbons, so that's a beta carbon, that's a beta carbon, and that's a beta carbon. So the first step in an E1 mechanism is loss of our leaving group, so if I draw the lone pairs of electrons in here on iodine, I know that these electrons in this bond would come off onto iodine to form the iodide anion, so let me draw that in here, so we would make the iodide anion, and let me highlight our electron, so the electrons in this bond come off onto the iodine to form the iodide anion. And this is an excellent leaving group. Iodide is an excellent leaving group, and you know that by looking at pKa values. The iodide anion is the conjugate base of a very strong acid, HI, with a approximate pKa value of negative 11, so HI is very good at donating a proton, which must mean that the conjugate base is very stable, so the iodide anion is an excellent leaving group. So if we lose the iodide anion, that means we're gonna have a carbocation, so we lost a bond to this carbon in red, so we're gonna form a carbocation, let me go ahead and draw that in, so this is a planar carbocation, and so the carbon, let me go ahead and highlight it here, the carbon in red has a plus one formal charge, it lost a bond. So that's the first step of an E1 elimination mechanism. The second step of an E1 elimination mechanism is the base comes along, and it takes a proton from a beta carbon, so our base in this case would be ethanol, so let me go ahead and draw in lone pairs of electrons on the oxygen, so notice we're also heating this reaction, so the ethanol is gonna function as a base, so ethanol's not a strong base, but it can take a proton, so let me go ahead and draw in a proton right here, and a lone pair of electrons on the oxygen is going to take this proton, and the electrons would move into here to form our alkene, so let me go ahead and draw our product, let me put that in here, and let me highlight some electrons, so the electrons in blue moved in here to form our double bond. So a couple of points about this reaction, one point is, when you're looking at SN1 mechanisms, the first step is loss of a leaving group to form your carbocation, so when you get to this carbocation, you might think, well, why is ethanol acting as a base here? Why couldn't it act as a nucleophile? And the answer is, the ethanol certainly can act as a nucleophile, and it would attack the positively-charged carbon, and you would definitely get a substitution product for this reaction as well, so if ethanol acts as a nucleophile, you're gonna get a substitution reaction, an SN1 mechanism. If the ethanol acts as a base, you're gonna get an E1 elimination mechanism, so here, we're just gonna focus on the elimination product, and we won't worry about the substitution product, but we will talk about this stuff in a later video, 'cause that would definitely happen. Alright, something else I wanna talk about is we had three beta carbons over here, and if I look at these three beta carbons, and I just picked one of them, I just said that this carbon right here, let me highlight it, I just took a proton from this carbon, but it doesn't matter which of those carbons that we take a proton off because of symmetry, let me go ahead and draw this in over here, so this is my carbocation, let's say, and let's say we took a proton from this carbon, so our weak base comes along, and takes a proton from here, and these electrons have moved into here, that would give us the same product, right? So this would be, let me go and highlight those electrons, so these electrons in dark blue would move in to form our double bond, but this is the same as that product. Alcohols can also react via an E1 mechanism. The carbon that's bonded to the OH would be the alpha carbon, and the carbon next to that would be the beta carbon, so reacting an alcohol with sulfuric acid and heating up your reaction mixture will give you an alkene, and sometimes, phosphoric acid is used instead of sulfuric acid. So we saw the first step of an E1 mechanism was loss of a leaving group, but if that happens here, if these electrons come off onto the oxygen, that would form hydroxide as your leaving group, and the hydroxide anion is a poor leaving group, and we know that by looking at pKa values. Down here is the hydroxide anion, it is the conjugate base to water, but water is not a great asset, and we know that from the pKa value here, so water is not great at donating a proton, which means that the hydroxide anion is not that stable, and since the hydroxide anion is not that stable, it's not a great leaving group. So let's go ahead and take off this arrow here, because the first step is not loss of a leaving group, the first step is a proton transfer. We have a strong acid here, sulfuric acid, and the alcohol will act as a base and take a proton from sulfuric acid. And that would form water as your leaving group, and water is a much better leaving group than the hydroxide anion, and again, we know that by pKa values. Water is the conjugate base to the hydronium ion, H3O+, which is much better at donating a proton, the pKa value is much, much lower. And that means that water is stable, so the first step, the first step when you are doing an E1 mechanism with an alcohol is to protonate the OH group. So here's our alcohol, and the carbon bonded to the OH is our alpha carbon, and then these carbons next to the alpha carbon would all be beta carbons. We just saw the first step is a proton transfer, a lone pair of electrons on the oxygen take a proton from sulfuric acid, so we transfer a proton, and let's go ahead and draw in what we would have now, so there'd be a plus-one formal charge on the oxygen, so let's highlight our electrons in magenta, these electrons took this proton to form this bond, and now we have water as a leaving group, let me just fix this hydrogen here really fast, and these electrons can come off onto our oxygen, so that gives us water as our leaving group, and let me go ahead and draw in the water molecule here, and let me highlight electrons, the electrons in light blue, in this bond, came off onto the oxygen, which forms water, and we know water is a good leaving group. We took a bond away from this carbon in red, so that carbon would now be a carbocation, so let me draw in our carbocation here, so the carbon in red is now positively charged, so let me draw in a plus-one formal charge on that carbon, the next step of our mechanism, we know a weak base comes along and takes a proton, one of the protons on one of the beta carbons over here, so let's just say it's this one, and I'm just gonna draw in a generic base, so a generic base right here, which is gonna take this proton, and then these electrons are gonna move in to form our product, so let me draw our product in here, and let me highlight those electrons. So the electrons in, let's use green this time, the electrons in green moved in here to form our double bond, to form our alkene. So I just put a generic base, let me go ahead and talk about the base for a second here. I just put in a generic base, sometimes you might see water acting as a base, sometimes you might see HSO4-, right, the conjugate base to sulfuric acid acting as the base, different textbooks give you different things, I don't think it really matters, but one of those acting as a weak base, it's probably water, takes this proton to form your alkene. Sometimes this reaction is called a dehydration reaction since we lost water in the process.