Intramolecular Nucleophilic Catalysis - Online Tutor, Practice Problems & Exam Prep

Intramolecular nucleophilic catalysis, a form of neighboring group participation (NGP), alters reaction rates and stereochemical outcomes. In reactions involving azide ions and cyclopentane rings, the nucleophile can perform an SN2 reaction, leading to inversion of configuration. The presence of bulky groups, like sulfur with a phenyl ring, can hinder the leaving of chlorine, affecting reaction speed. Intramolecular processes are generally faster due to reduced steric hindrance, while intermolecular reactions may face challenges from larger substituents, demonstrating the significance of steric effects and reaction mechanisms in organic synthesis.

Intramolecular Nucleophilic Catalysis Concept 1

Hey, everyone. In this video, we're going to take a look at intramolecular nucleophilic catalysis. Here, we're going to say it is a form of neighboring group participation, which we're going to abbreviate as NGP. We're going to say that NGP is just a change in the reaction rate and the stereochemical outcome of a reaction due to a neighboring group. So, if we take a look here, in the first one, we have a methyl and we have a chlorine attached to our cyclopentane ring.

Here, our nucleophile is the azide ion. We're going to say that this can undergo an SN₂ reaction, so it would come in and hit this carbon, kicking out the chlorine by SN₂. Remember, this would cause an inversion of our bond. So now, the azide ion is going to have a wedge bond. In this next one, we again have our chlorine and we're going to say the bond is dashed.

But, with this one, we have a little bit more difficulty because we're going to say that this azide ion wants to come in and kick out this chlorine molecule. And we're going to say that this chlorine molecule is going to have a little bit of difficulty leaving because we have this sulfur with this big benzene ring on the same side as it. So, there's going to be some difficulty in kicking out that chlorine group, but if we do successfully do it, we'd hit this carbon kicking out the chlorine and again our bond would switch to wedged. And in this final one, we can say that intramolecularly, that the sulfur has its own lone pairs, and it could come in, hit this carbon and kick this out. This will give us an intermediate form, where we're going to make a ring.

Now, we hit this carbon and kick out the chlorine, the bond inverts again, so this would be wedged. We'd still be connected to this benzene ring here. We'd still have this methyl group over here. The sulfur is making 3 bonds, so it'd be positively charged. Now what can happen is the azide ion comes in.

It can hit here causing this ring to open. Remember, it's a 3-membered ring, so there is some angle strain here, so we want something to come in and hit this less substituted carbon to pop it open. And remember, we're hitting the less substituted side similar to what we would see when it comes to epoxide ring openings. This is a negative nucleophile, so we'd attack the less substituted side. And here, as a result of this, the bond inverts back to the azide.

Now, we could talk about their relative rates to one another. We could say here that this one does it by intramolecular process, so it would be the fastest. And between these two here, this azide ion here, it's coming in from the opposite side to kick out the chlorine, it'd be affected by this methyl group. Yes, it would, but it would still be a moderate relative rate. This would be the slowest because we'd have to contend with this large group here.

This sulfur, this benzene ring that's in the way, would make it hard for this chlorine group to leave as easily. So we'd say that it would show a slower rate than we would typically see compared to the other two. Alright. So these are the stereochemistries that are involved. We can see that our neighboring group does have an effect on how quickly our reaction can go.

This sulfur with the phenyl group is a pretty large group, which would get in the way of things coming in leaving. Here, though, if we do it by intramolecular process, it actually can help to enhance this leaving of this chlorine group, creating an angle strain three-membered ring which can then be attacked by a nucleophile on the less substituted side.

Intramolecular Nucleophilic Catalysis Example 1

In this example question, it says, the following 2 compounds react with methanol to yield substitution products. Identify the compound that will react at a considerably faster rate than the other. Alright. So they're saying that these are going to react with our methanol. We're going to say here that methanol would come in and it's trying to attack this carbon here, kicking out the group.

So you just have to imagine how easy it is for the methanol to come in and hit that carbon. Right? So here, we could hit this group kicking it out. Come here, kicking this carbon, kicking it out. In this one here, we have this thiomethyl group that's pointing up out of the way.

Out of the way of this methanol that could potentially come in and attack this carbon kicking out the bromine. Here, the methanol would have to try to attack this carbon, but this group is on the same side as it. It's pointing down as well and that methyl group branches out its hydrogen, so they could potenti

The following compound has two stereoisomers. One isomer reacts with acetate more than a billion times faster than the other. Draw the structure of that isomer and write a plausible reaction mechanism.

Hey, everyone. In this practice question, it says, the following compound has 2 stereo isomers. One isomer reacts with acetate more than a billion times faster than the other. Draw the structure of that isomer and write a plausible reaction mechanism. Alright. So here, we have this structure and it's showing us this squiggly squiggly line for OH, meaning that it can orient itself up or down. Now, if we orient this, let's say we're going to orient it down, we're going to draw this bicyclic compound. Now, this OTs is up here, and let's say we orient this OH down. We're going to say here that acetate looks like this. It will come in, hit this carbon and kick out this tosylate group. That will give us this structure initially. Again, this group is pointing down. And to abbreviate the acetate group, I'm just going to write it as OAc. Now, we're going to say the OH is too far, it's pointing down away from this group here. So, we can't attack it. So, we're stuck with this form. This is the form that's going to be slower. The faster form is this one where I'm going to draw my bicyclic compound and I'm going to have this OH oriented up which means it's closer to this carbon here and because it's closer, it can do this. It comes in and hits this carbon, kicks this OTs out. What we get initially now is this. Okay. So the O now, here it is. It's attached right there. It still has this H connected to it. That oxygen is making 3 bonds so it's positively charged. And now, we can have our acetate come in. It's going to hit this same carbon to open up the ring and what we get here is this structure. So here goes our bicyclic compound. We'd still have our OH here oriented up. And then we'd have OAc over here. So we'd say it's this structure that's a billion times faster because the OH is up, it is in better position to attack this carbon that has the tosylate group attached to it. And then later on, to alleviate this ring strain, this angle strain that we've created, we use the acetate to attack this same carbon here releasing the hydroxyl group again and that's how we get our acetate attached to our bicyclic compound. Right? So this is what we would say in terms of the faster compound, the faster orientation, and the plausible mechanism involved with it.