If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content

Video transcript

- [Instructor] Let's look at the mechanism for an SN2 reaction. On the left we have an alkyl halide and we know that this bromine is a little bit more electronegative than this carbon so the bromine withdraws some electron density away from that carbon which makes this carbon a little bit positive, so we say partially positive. That's the electrophilic center so this on the left is our electrophile. On the right, we know that this hydroxide ion which we could get from something like sodium hydroxide, is a negative one formal charge on the oxygen which makes it a good nucleophile. Let me write down here. This is our nucleophile on the right and on the left is our electrophile which I'm also gonna refer to as a substrate in this video. This alkyl halide is our substrate. We know from an earlier video that the nucleophile will attack the electrophile because opposite charges attract. This negative charge is attracted to this partially positive charge. Lone pair of electrons on the oxygen will attack this partially positive carbon. At the same time, the two electrons in this bond come off onto the bromine. Let me draw the bromine over here. The bromine had three lone pairs of electrons on it and it's gonna pick up another lone pair of electrons. Let me show those electrons in magenta. This bond breaks and these two electrons come off onto the bromine which gives the bromine a negative one formal charge. This is the bromide anion. And we're also forming a bond between the oxygen and this carbon and this bond comes from this lone pair of electrons which I've just marked in blue here. Those two electrons in blue form this bond and we get our product which is an alcohol. The SN2 mechanism is a concerted mechanism because the nucleophile attacks the electrophile, at the same time we get loss of a leaving groups. There's only one step in this mechanism. Let's say we did a series of experiments to determine the rate law for this reaction. Remember from general chemistry, rate laws are determined experimentally. Capital R is the rate of the reaction and that's equal to the rate constant k times the concentration of our alkyl halide. And it's determined experimentally this is to the first power times the concentration of the hydroxide ion also to the first power. So what does this mean? This means if we increased the concentration of our alkyl halide. If we increase the concentration of our alkyl halide by a factor of two, what happens to the rates of the reaction? Well, the rate of the reaction is proportional to the concentration of the alkyl halide to the first power. Two to the first is equal to two which means the overall rate of the reaction would increase by a factor of two. Doubling the concentration of your alkyl halide while keeping this concentration, the hydroxide ion concentration the same should double the rate of the reaction. And also, if we kept the concentration of alkyl halide the same and we double the concentration of hydroxide, that would also increase the rate by a factor of two. And this experimentally determined rate law makes sense with our mechanism. If we increase the concentration of the nucleophile or we increase the concentration of the electrophile, we increase the frequency of collisions between the two which increases the overall rate of the reaction. The fact that our rate law is proportional to the concentration of both the substrate and the nucleophile fits with our idea of a one step mechanism. Finally, let's take a look at where this SN2 comes from. We keep on saying an SN2 mechanism, an SN2 reaction. The S stands for substitution. Let me write in here substitution because our nucleophile is substituting for our leaving group. We can see in our final products here, the nucleophile has substituted for the leaving group. The N stands for nucleophilic because of course it is our nucleophile that is doing the substituting. And finally the two here refers to the fact that this is bimolecular which means that the rate depends on the concentration of two things. The substrate and the nucleophile. That's different from an SN1 mechanism where the rate is dependent only on the concentration of one thing. The rate of the reaction also depends on the structure of the alkyl halide, on the structure of the substrate. On the left we have a methyl halide followed by a primary alkyl halide. The carbon bonded to our bromine is directly attached to one alkyl group followed by a secondary alkyl halide, the carbon bonded to the bromine is bonded to two alkyl groups, followed by a tertiary alkyl halide. This carbon is bonded to three alkyl groups. Turns out that the methyl halides and the primary alkyl halide react the fastest in an SN2 mechanism. Secondary alkyl halides react very slowly and tertiary alkyl halides react so, so slowly that we say they are unreactive toward an SN2 mechanism. And this makes sense when we think about the mechanism because remember, the nucleophile has to attack the electrophile. The nucleophile needs to get close enough to the electrophilic carbon to actually form a bond and steric hindrance would prevent that from happening. Something like a tertiary alkyl halide has this big bulky methyl groups which prevent the nucleophile for attacking. Let's look at a video so we can see this a little bit more clearly. Here's our methyl halide with our carbon directly bonded to a halogen which I'm seeing as yellow. And here's our nucleophile which could be the hydroxide ion. The nucleophile approaches the electrophile for the side opposite of the leaving group and you can see with the methyl halide there's no steric hindrance. When we move to a primary alkyl halide, the carbon bonded to the halogen has only one alkyl group bonded to it, it's still easy for the nucleophile to approach. When we move to a secondary alkyl halide, so for a secondary you can see that the carbon bonded to the halogen has two methyl groups attached to it now. It gets a little harder for the nucleophile to approach in the proper orientation. These bulky methyl groups make it more difficult for the nucleophile to get close enough to that electrophilic carbon. When we go to a tertiary alkyl halide, so three alkyl groups. There's one, there's two and there's three. There's a lot more steric hindrance and it's even more difficult for our nucleophile to approach. As we saw on the video, for an SN2 reaction we need decreased steric hindrance. So, if we look at this alkyl halide, the carbon that is directly bonded to our halogen is attached to only one alkyl group. This is a primary alkyl halide and that makes this a good SN2 reaction. The decreased steric hindrance allows the nucleophile to attack the electrophile.