Nucleophilic Catalysis - Online Tutor, Practice Problems & Exam Prep

Nucleophilic catalysis involves the displacement of a leaving group by a better nucleophile, enhancing reaction rates. In a double SN2 mechanism, a larger halogen replaces a smaller one, resulting in retention of configuration after two inversions. For example, when propyl chloride is catalyzed by iodine, the reaction rate increases significantly from 6.0 x 10^-3 to 3.7 x 10⁶ seconds^-1. This demonstrates how using a superior leaving group accelerates nucleophilic substitution reactions.

Nucleophilic Catalysis Concept 1

Hey, everyone. In this video, let's take a look at nucleophilic catalysis. Now, in Orgo 1, we learned that double SN₂ caused retention of R and S configurations, but it also serves as a nucleophilic catalyst. When I say nucleophilic catalyst, I mean it displaces a leaving group with a better nucleophile/leaving group. The general trend of the periodic table is as we head down a group, our atomic size is supposed to increase.

And when it comes to group 7A or halogens, this is the trend that we see. As we go down the group, we can see that the halogens get larger until we get down to iodine. Now, this would mean that iodine would be a better nucleophile and leaving group than bromine, chlorine, and fluorine. Bromine would be a better nucleophile and leaving group than chlorine and fluorine, etcetera. Now, if we take a look at the mechanism here, we're going to say that this blue star here means that that is a chiral center.

And we're going to start out with this alkyl halide. That halogen will be displaced by a larger halogen, which represents a better leaving group, a better nucleophile. And remember, because this is happening by an SN₂ mechanism, we have inversion of our configuration. We go from wedged to dashed. So here we have inversion going on.

But remember, we're doing double SN₂ so it's going to happen again. We're going to use now another nucleophile to come in and attack here and kick this halogen out. Now, that halogen is kicked out and leaves even better than our first halogen because it's larger, it's lower down group 7A. Now, hitting it again causes the bond to flip again. So we start out with wedged and we're ending with wedged.

So, in essence, we have retention going on. Right? So we inverted twice, they'll end up with the same type of bond we had at the end. This is the beauty of double SN₂. But now, we're also seeing that replacing the first halogen with a better halogen means that this second SN₂ would happen that much more quickly.

We've increased the rate of this exchange. And as we click on to the next series of videos, we'll see how this happens numerically. What is the numerical value? What is the increase in rate if we could somehow calculate it? Alright.

So just keep that in mind when we're talking about nucleophilic catalysis.

Nucleophilic Catalysis Concept 2

Now, remember, S&sub2; does cause retention of R/S configurations, but now we're going to say that displacing the original halogen with a better nucleophile/leaving group also causes an increase in the rate of the reaction. So here we have an uncatalyzed versus a catalyzed reaction. Here, we're dealing with Propyl Chloride, 1-chloropropane. It's a primary alkyl halide. So when I use this hydroxide ion, it'll displace it by S_N2.

So this will come in, hit here, and kick this out. This will give us 1 Propanol as our answer. Now, here this is just an estimate of what this will result in terms of rate. Here, we'd say the rate is 6.0×10-3 seconds^-1. Now, here, if we're going to catalyze this, remember what happens here is that Iodine would first come in displacing this Chlorine.

Iodine is a better leaving group than chlorine so that when the hydroxide ion comes in and displaces the iodine, it happens that much more quickly. We still get our 1 Propanol, but what's the difference in the rate? Here, the rate is 3.7×106 seconds^-1. We went from 10^-3 to 10⁶. So we increased several magnitudes.

It happens that much more quickly. So we learned in Organic Chemistry I that double S&sub2; causes retention of our configuration, but now more importantly, it's a way of catalyzing our reaction to help it happen that much more quickly. And, we can see that through these estimated rates when it's uncatalyzed versus catalyzed. We can see how much more quickly the second reaction would occur.

Predict the final product from the chemical steps provided. 

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Which of the following compounds would you expect to have the greatest relative rate of reaction?

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Using 3-methyl-1-butene as a starting material, predict the final product based on the list of reagents given?

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