Current time:0:00Total duration:11:40
0 energy points
How an Sn1 reaction takes place. Created by Jay.
Video transcript
In Sn1 reaction, the S stands for substitution. The n stands for nucleophilic. So this is a substitution of a halogen by a nucleophile, and you can see that's what's happening here in the general reaction. We have a halogen, and the nucleophile is substituting for the halogen. Let's look at the mechanism for an Sn1 reaction. So in the first step of the mechanism-- I'll just start with this right here-- we get dissociation of our halogen. So these electrons in here are going to kick off onto our halogen. So you need a good leaving group. So if those electrons kick off onto the halogen, we took a bond away from our carbons. So our carbon is left like that. Taking a bond away means it is now a carbocation. It's positively charged. Our halogen gets an extra lone pair of electrons, so that would give it a negative one formal charge. And halogens are relatively stable on their own with a negative formal charge. That makes them good leaving groups, so the halogen is the leaving group in this mechanism. So in the next step your nucleophile comes along. So here's our nucleophile. And we can give it a lone pair of electrons, make it negatively charged. It doesn't have to be negatively charged. But this nucleophile is going to attack the electrophile, which is your carbocation. So nucleophile, electrophile. Opposite charges attract. This lone pair of electrons is going to form a bond with this carbon. And that will give us our product, where our nucleophile file substitutes for our halogen. So we have a carbocation in our mechanism. This is a carbocation. So rearrangements are possible, so you have to be very careful when you're thinking about Sn1 reactions. In terms of the rate of the overall reaction, this first step here, the dissociation of your halogen, is the slow step. This is very slow. And the second step, the nucleophilic attack, is very fast. So I have a slow step, and I have a fast step. And when rate studies are done on Sn1 reactions, the overall rate law is dependent only upon the concentration of our alkyl halide. So if you're running a general chemistry rate law, R, or the overall rate of the reaction, is equal to the rate constant, K, times the concentration of your alkyl halide only. So, the rate is only dependent upon the concentration of your alkyl halide. And the reasoning has to do with this mechanism here. So the first step is the rate determining step. The first step is the slow step of the mechanism. So it takes a little bit of time for that halogen to dissociate as compared to the nucleophilic attack, which is fast. The nucleophilic attack cannot happen until the halogen leaves. So when you're thinking about why the overall rate is dependent only upon the alkyl halide, maybe it'll help for you to use some numbers. If it took 10 seconds for the alkyl halide to dissociate-- obviously, that number is nowhere close to reality, but just to give you an idea-- and it took you one second for the nucleophile to attack, most of your time is spent in the slow step. And you can approximate the overall rate of reaction by the overall time it takes to do the first step or the slow step. And the second step can't happen until the first step happens. So it doesn't matter what you do in terms of increasing your concentration of your nucleophile. You can do anything you want, but that nucleophile will not be able to attack until your carbocation forms. So that's another way of looking at it. Changing the concentration of your nucleophile isn't going to affect anything if you don't have a carbocation present. All right. So that's actually why this is called Sn1. So, the 1 in Sn1 stands for unimolecular, meaning the concentration of only one of your reactant, which is of course your alkyl halide. So that's what your overall rate is dependent on, so I can write here rate determining step for the slow step of our mechanism. All right. Let's look at what types of alkyl halides undergo Sn1 reactions. So we know that a carbocation forms. So let's look at two different alkyl halides, and let's see which one will react faster via an Sn1 mechanism. So I'm going to start off with this as my alkyl halide right here, and I'm going to put a chlorine here. So, that's one of our starting alkyl halides. Let me go ahead and draw another one here. So for my second alkyl halide, I will put two hydrogens down here instead of those methyl groups above. And I'll put a chlorine right here. So if both these alkyl halides were undergoing an Sn1 mechanism, the first step would be dissociation of your halogen, so formation of your leaving group. These electrons are going to go off onto your chlorine. These electrons are going to go off on to your chlorine. So what would we get? So for the first example, I would get a carbocation that would look like this. So my carbocation would be right here. And in my second example, my carbocation would look like this. So I have my hydrogens here, and we would need to classify our carbocations. So this top carbon here, this carbon that bears the positive charge, is bonded to one, two, three other carbons. So we have a tertiary carbocation. Down here, if I do the same thing, this is the carbon with the positive charge. So it's bonded to one other carbon, so primary carbocation. So tertiary versus primary. We saw in an earlier video that tertiary carbocation are much more stable because of hyperconjugation. So they're more stable. They're more reactive. They're more reactive, because this is the one that's more likely to form. It's more stabilized. So I think experimentally it turns out that this example of tertiary versus primary is something like 50 million to one, just to give you an idea of the difference in terms of the stability of it. So since tertiary carbocations are more stable than the others, they're most likely to react via an Sn1 mechanism. So tertiary alkyl halides by extension-- so this over here with the chlorine, this is a tertiary alkyl halide-- these are the most reactive via an Sn1 mechanism, because the tertiary carbocation that results in the mechanism is the most stable. So if we're going to rank the structure of alkyl halides, a tertiary alkyl halide is the one that's most likely to react via an Sn1 mechanism, followed by secondary, and then followed by primary. And then you could put a methyl one in there if you wanted to. So tertiary fastest via an Sn1 mechanism. Let's look at a few problems, and let's see if we can do some mechanisms here for some simple Sn1 reactions. So if I have this as my reactant, tert-Butyl chloride, I'm going to react that with sodium acetates, which is a source of acetate anions like that. You would say to yourself, OK, where's my leaving group? Chlorine is my leaving group. And what is my nucleophile? Well, the negative charge for my lone pairs of electrons here are my oxygen. The negative charge on my oxygen is going to function as my nucleophile. So in the first step, dissociation, leaving me with a tertiary carbocation. This is where my nucleophile is going to attack my positively charged carbon like that, so I'm just going to do this here. And the end result is we get a substitution of the nucleophile. So nucleophilic substitution, like that. So that would be our product. All right. Let's do another one. So you start with tert-Butanol here. And I'll go ahead and put in our lone pairs of electrons reacting tert-Butanol with hydrochloric acid, so we have HCL right here. So in this reaction, OH is not the best leaving group. So if those electrons kicked off onto the oxygen to give us OH minus, it's not quite as stable as a halogen, as a halide anion. So if you have some protons present, that's going to form a better leaving group. If we first start with an acid-base reaction, if this lone pair of electrons grabs this proton, kicking these electrons off onto the chlorine, let's see what that would give us here. We would form water as a leaving group. So we've seen this before. So this lone pair of electrons did nothing. This gives us a positive one charge. Now if the electrons kick off on the oxygen, we have water as our leaving group, which is more stable. So this is what's going to happen. So now these electrons are going to kick off onto the oxygen, something we've seen in many mechanisms. And now we form our carbocation. So now we get a positive charge right here. And remember, we had our chloride anion over here, which is negatively charged. That's going to function as our nucleophile. The lone pair of electrons attacks our carbocation. And we end up with our product, which is tert-Butyl chloride. So you have to think about acid-base chemistry in these mechanisms as well. Let's do one more example. Let's look at tert-Butyl chloride as our reactant again. So, here we go with tert-Butyl chloride reacting with methanol. So this is going to react with methanol. So I'm going to go ahead and draw methanol in here. Now, I haven't really talked about solvents in this video. Actually, we're going to get into solvents in much more detail in a few videos. But this is actually going to be our only solvent for this. So methanol is our only solvent. It's also going to be our nucleophile. This is called a solvolysis reaction, so a special type of substitution reaction called solvolysis where your nucleophile is also your solvent. So let's see what happens. We know that we're going to get these lone pair of electrons to kick off onto our chlorine as the first step to form our tertiary carbocation. So our solvent is our nucleophile. So it's not negatively charged like it is in most reactions, but this lone pair of electrons on the oxygen is going to function as our nucleophile. So that can still be a nucleophile, which will attack our electrophile right here. So let's draw the result of that nucleophilic attack. So a lone pair of electrons and oxygen bonds with that carbon there. And let's see, there was one hydrogen bonded to that oxygen, and then there was a methyl group on this side. And then there was a lone pair of electrons that did not participate in that bonding, so put that lone pair right there. That gives this oxygen here a positive one formal charge. So in the next step of our mechanism, we have an acid-base reaction that happens at the end. So another molecule of methanol comes along. So like this, another molecule of methanol comes along. This time, instead of acting as a nucleophile, it's going to act as a base. So this lone pair is going to take that proton, kick these electrons in here off onto your oxygen. And we end up with our product. So let's see. We would draw our oxygen like that, and then we would have lone pairs of electrons on our oxygen. So this is an example of solvolysis. In the next video we'll worry about the stereochemistry of an Sn1 reaction.