E1 Elimination Reactions. Created by Sal Khan.
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- I am having trouble understanding what is making the Bromide leave the Carbon - what is causing this to happen? and why is the Br- content to stay as an anion and not react further?(6 votes)
- A few reasons:
1. The C-Br bond is relatively weak (<300kJ/mol) compared to other C-X bonds. The C-I bond is even weaker. The kinetic energy supplied by room temperature is enough to get the Br to spontaneously dissociate.
2. Br is a large atom, with lots of protons and electrons. Leaving groups need to accept a lone pair of electrons when they leave. Br is a good leaving group because it can easily spread out this negative charge over a large area (we say it is polarizable). Due to its size, fluorine will not do this very easily at room temperature.
Another way to look at the strength of a leaving group is the basicity of it. The best leaving groups are the weakest bases. The F- is actually a fairly strong base (because HF is a weak acid), whereas Br- is pH neutral (because HBr is a strong acid)(21 votes)
- why E1 reaction is performed in the present of weak base ? reason with detail? otherwise why s1 reaction is performed in the present of weak nucleophile ?(3 votes)
- An E1 reaction requires a weak base, because a strong one would butt-in and cause an E2 reaction. In an E1 reaction, the base needs to wait around for the halide to leave of its own accord. In the video, Sal makes a point to mention that Ethanol, the weak base, just wasn't strong enough to push its way in and MAKE the bromine leave (as would happen in an E2). This is why it's called an E1 reaction- the reaction is entirely dependent on one thing to move forward- the leaving group going. A weak base just isn't strong enough to participate- if it was, it'd be a strong base, and all of the sudden the rate-determining step would depend on TWO things (the Leaving Group leaving AND the base entering), which would make it E2.
This is a lot like SN1! With SN1, again, the nucleophile just isn't strong enough to kick the leaving group out. It therefore needs to wait until the leaving group "decides" it's ready to go, and THEN the nucleophile swoops in and enjoys the positive charge left behind. A STRONG nucleophile, on the other hand, TAKES what it wants, when it wants it (so to speak) and PUSHES the leaving group out, taking its spot. This causes an SN2 reaction, because the rate depends on BOTH the leaving group, and the nucleophile.
So, to review:
- a reaction that only depends on the the leaving group leaving (and being replaced by a weak nucleophile) is SN1
- a reaction that only depends on the leaving group leaving, but NOT being replaced by the weak base, is E1.
- a reaction where the strong nucleophile edges its way in and forces out the leaving group, thereby replacing it is SN2.
- a reaction where a strong base steals a hydrogen, causing the remaining electron density to push out the leaving group is an E2.(15 votes)
- Just by seeing the rxn how can we say it is a fast or slow rxn??(2 votes)
- I believe that this comes from mostly experimental data. In E1 reaction, if you increase the concentration of the base, the rate of the reaction will not increase. The rate only depends on the concentration of the substrate. Fast and slow are relative, but the first step only involves the substrate, and is relatively slower than the rest of the reaction, which is why it is called the rate determining step.(3 votes)
- Just to clarify my understanding, the hydrogen that is leaving the carbon leaves both electrons on the carbon chain to use for double bonding, correct? The notation in the video seems to agree with this, however, when explaining the interaction between the partial negative oxygen and the leaving hydrogen, you make it appear that the oxygen only donates one electron to the hydrogen, making it seem that the hydrogen takes an electron, as it would need to do that to create a bond with oxygen. This is not the case, as the oxygen gives BOTH electrons in one of the lone pairs to form the bond with hydrogen, leaving two electrons on the carbon atoms to form a double bond.(3 votes)
- i was told in class that you could end up with HBr and Ethanol as you didn't start with any charges and since your product contains a charge wouldn't it be more reasonable to assume that the purple hydrogen would form a bond with Br and therefore remove any overall charges?(2 votes)
- How do you perform a reaction (elimination, substitution, addition, etc.) on an alkene or alkyne without a leaving group?(2 votes)
- From the point of view of the substrate, elimination involves a leaving group and an adjacent H atom. Substitution involves a leaving group and an adding group. Addition involves two adding groups with no leaving groups.(1 vote)
- Is there a thumb rule to predict if the reaction is going to be an Elimination or substitution ? Like in this case the partially negative O attacked beta H instead of carbcation (which i was guessing it would !)and resulting in elimination!(1 vote)
- You have to consider the nature of the
• leaving group
SN1/E1 reactions are favoured if you have a 3° substrate, a good leaving group, and a polar solvent.
SN1 if nucleophile is poor base and substrate has no β-hydrogen.
E1 if nucleophile is moderate base and substrate has β-hydrogen.(3 votes)
- In6:21, he draws a double headed arrow originating from a single electron. Double headed arrows represent the movement of a pair of electron. Although he is conceptually right, I think this was a mistake. He should have drawn the arrow going form a pair of electrons.(1 vote)
- Agreed. The mechanism involves the movement of electron pairs.(3 votes)
- Is there a video on E1cb reactions?(2 votes)
- 6:20- Would EtOH take the H if the C were not a cation? In other words, is if necessary for the carbocation to exist to enable EtOH to take the H? Does the presence of the carbocation enable EtOH to take the H by reducing the strength of the H-C bond by pulling electrons towards itself? Do the electrons in the H-C bond move towards the carbocation before the H is taken, or only afterwards?(1 vote)
- No. The EtOH would not take the H if the C were not a cation. The cation must be there to make the H atom acidic enough to be removed by the relatively weak base EtOH. The cation does this by pulling the electrons in the C-H bond towards itself. The electrons in the C-H bond have already moved towards the carbon before the EtOH attacks, But they move entirely to C atom after the EtOH has attacked.(2 votes)
Let's think about what'll happen if we have this molecule. Let's name it. We have one, two, three, four, five carbons. No double bond. Five tells us pent. It's pentane, and it has two groups on the number three carbon, one, two, three. It doesn't matter which side we start counting from. We have a bromo group, and we have an ethyl group, two carbons right there. On the three carbon, we have three bromo, three ethyl pentane right here. So we have 3-bromo 3-ethyl pentane dissolved in a solvent, in this right here. It's an alcohol and it has two carbons right there. Meth eth, so it is ethanol. This right there is ethanol. Let's think about what might happen if we have 3-bromo 3-ethyl pentane dissolved in some ethanol. Now ethanol already has a hydrogen. It's not super eager to get another proton, although it does have a partial negative charge. It is polar. Oxygen is very electronegative. It has a partial negative charge, so maybe it might be willing to take on another proton, but doesn't want to do so very badly. It's actually a weak base. Ethanol right here is a weak base. It's not strong enough to just go nabbing hydrogens off of carbons, like we saw in an E2 reaction. It's just going to sit passively here and maybe wait for something to happen. What might happen? Well, we have this bromo group right here. We have this bromine and the bromide anion is actually a pretty good leaving group. It's a fairly large molecule. It's able to keep that charge because it's spread out over a large electronic cloud, and it's connected to a tertiary carbon. This carbon right here. This carbon right here is connected to one, two, three carbons. So if it were to lose its electron, that electron right there, it would be-- it might not like to do it-- but it would be reasonably stable. It's within the realm of possibilities. It could occur. Maybe in this first step since bromine is a good leaving group, and this carbon can be stable as a carbocation, and bromine is already more electronegative-- it's already hogging this electron-- maybe it takes it all together. Let me draw. Neutral bromine has one, two, three, four, five, six, seven valence electrons. Maybe it swipes this electron from the carbon, and now it'll have eight valence electrons and become bromide. What happens now? And of course, the ethanol did nothing. It's a weak base. It wasn't strong enough to react with this just yet. What is happening now? This is going to be the slow reaction. What I said was that this isn't going to happen super fast but it could happen. This is actually the rate-determining step. What happens after that? Let me just paste everything again so this is our set up to begin with. But now that this little reaction occurred, what will it look like? The bromine has left so let me clear that out. We clear out the bromine. It actually took an electron with it so it's bromide. Let me draw it like this. I'll do it in blue. This is the bromine. The bromine is right over here. It had one, two, three, four, five, six, seven valence electrons. It swiped this magenta electron from the carbon, now it has eight valence electrons. It has a negative charge. The carbon lost an electron, so it has a positive charge and it's somewhat stable because it's a tertiary carbocation. Now let's think about what's happening. And I want to point out one thing. In this first step of a reaction, only one of the reactants was involved. This rate-determining, the slow step of reaction, if this doesn't occur nothing else will. But now that this does occur everything else will happen quickly. In our rate-determining step, we only had one of the reactants involved. It's analogous to the SN1 reaction but what we're going to see here is that we're actually eliminating. We're going to call this an E1 reaction. We're going to see that in a second. Actually, elimination is already occurred. The bromide has already left so hopefully you see why this is called an E1 reaction. It's elimination. E for elimination and the rate-determining step only involves one of the reactants right here. It didn't involve in this case the weak base. Now that the bromide has left, let's think about whether this weak base, this ethanol, can actually do anything. It does have a partial negative charge over here. It does have a partial negative charge and on these ends it has partial positive charges, so it is somewhat attracted to hydrogen, or to protons I should say, to positive charges. But not so much that it can swipe it off of things that aren't reasonably acidic. Now that this guy's a carbocation, this entire molecule actually now becomes pretty acidic, which means it wants to give away protons. Another way you could view it is it wants to take electrons, depending on whether you want to use the Bronsted-Lowry definition of acid, or the Lewis definition. Either way, it wants to give away a proton. It could be this one. It could be that one. It has excess positive charge. It wants to get rid of its excess positive charge. So it's reasonably acidic, enough so that it can react with this weak base. What you have now is the situation, where on this partial negative charge of this oxygen-- let me pick a nice color here-- let's say this purple electron right here, it can be donated, or it will swipe the hydrogen proton. Then hydrogen's electron will be taken by the larger molecule. In fact, it'll be attracted to the carbocation. So it will go to the carbocation just like that. Now in that situation, what occurs? What's our final product? Let me draw it here. This part of the reaction is going to happen fast. The rate-determining step happened slow. The leaving group had to leave. The carbocation had to form. That's not going to happen super fast but once that forms, it's not that stable and then this thing will happen. This is fast. Let me paste everything again. So now we already had the bromide. It had left. Now the hydrogen is gone. The hydrogen from that carbon right there is gone. This electron is still on this carbon but the electron that was with this hydrogen is now on what was the carbocation. That electron right here is now over here, and now this bond right over here, is this bond. We formed an alkene and now, what was an ethanol took a hydrogen proton and now becomes a positive cation. Let me draw that. So this electron ends up being given. It's no longer with the ethanol. It gets given to this hydrogen right here. That hydrogen right there. And now they have formed a new bond and since this oxygen gave away an electron, it now has a positive charge. And all along, the bromide anion had left in the previous step. The bromide anion is floating around with its eight valence electrons, one, two, three, four, five, six, seven, and then it has this one right over here. That makes it negative. Then our reaction is done. We had a weak base and a good leaving group, a tertiary carbon, and the leaving group left. We only had one of the reactants involved. It was eliminated. And then once it was eliminated, then the weak base was then able to take a hydrogen off of this molecule, and that allowed this molecule to become an alkene, formed a double bond. This is called, and I already told you, an E1 reaction. E for elimination, in this case of the halide. One, because the rate-determining step only involved one of the molecules. It did not involve the weak base. We'll talk more about this, and especially different circumstances where you might have the different types of E1 reactions you could see, which hydrogen is going to be picked off, and all the things like that. And we're going to see with E1, E2, SN1, and SN2, what kind of environments or reactants need to be there for each one of those to occur in different circumstances.