Despite it being equally favorable, opening of the epoxide doe...

Question

Despite it being equally favorable, opening of the epoxide does not happen in the absence of an acid catalyst. How does acid make the reaction faster? Demonstrate this concept by directly comparing the reaction coordinate diagram for both situations A and B.

Accepted Answer

Alright. Hello everyone. So this question says two reaction mechanisms for the oxide ring opening are shown below. The reaction does not occur in the absence of an acid catalyst, explain how the acid makes the reaction faster and provide a reaction coordinate diagram to compare the two reactions. So the first mechanism shown here is the oxide ring opening without acid and the second reaction is with acid. So what's the difference? Right. Well, in the first reaction, we can see that water is directly attacking one of the carbons of the oxide ring which breaks its bond between itself and the oxide oxygen. This creates a positive charge on the oxygen that just attacked the carbon, which is then mitigated by ad pro nation step to form the final product. So now the difference between reaction one and reaction two is that in reaction to the first step is actually the probation of the oxide oxygen. So then by the time water is attacking one of the carbons of the oxide ring, that oxide oxygen already has a positive charge. So this opens the ring. And after another derotation step, we end up with a dial specifically a trans dial. So now let's go ahead and start with the reaction coordinate diagrams. And here I went ahead and just redrew the X and Y axis. The X axis is the reaction progress and the y axis is energy. So now let me go ahead and scroll down to start with reaction one that is without acid. So recall that in a reaction coordinate diagram, we're starting with the reactants on the left side and the products on the right side of the graph. So here the highest peak of the graph is determined by the rate determining step. And the rate determining step for both reactions is the step in which one carbon oxygen bond is broken and the other is formed. So in this step, the carbon atom that's being attacked has a partial positive charge. This is because it's bound to a more electron oxygen atom. So here when it comes to the step without acid, right, this first step is rate limiting. So the attack of this oxide carbon by water is rate limited. So here that particular step should have the highest peak. And so from there you go down to the first intermediate which is still higher in energy than the starting material. So then because we have another step here, we have another peak in the graph, but it's not going to be as tall as the first one because it's not the rate limiting step. So finally, we go down to the products which have a lower energy compared to the reactants. Now just to go ahead and clarify. Part of the reason why this first step is rate limiting is because the ring strain of the oxide is still present. So here on the very left side, I'm just going ahead and drawing out the oxide ring with water, right. And so my intermediate here in the middle, that is this point right here in between both peaks of the graph is going to represent our intermediates. So that's where oxygen from the water molecule that attacked carbon currently has a positive charge. And then lastly, on the very right side, we have our product. And so just to go ahead and clarify this, the reason why the final step is lowest in energy than the reactant or lower in energy than the reactant is because the ring strained from the oxide is no longer present. So it's more stable. All right. So now let's compare this with acid. Now, what's important to mention is that the protonation of the oxide oxygen in the first step of reaction too makes the subsequent step a lot easier because proton, the epoxy oxygen results in a positive charge on that oxygen. And so what that does is give a sort of incentive for the oxide ring to break because when the oxide ring breaks, you're giving electrons to that positively charged oxygen and thereby making it neutral, which is favorable, right. So once again, we have more steps here. So that's more intermediates and more peaks in the graph. This time, it's the second step, not the first that's going to have the highest energy. So let's start with our reactants right here. We have our reactants. Let me actually scroll down some more. And so our first step is going to involve the probe nation of the oxide ring. And so that transition state which represent the peaks of the diagram is going to be higher in energy than the starting material, but not as high in energy as the rate limiting step. So from the first transition state, I would go down somewhat to represent our first intermediate before going back up again to the highest point of the graph representing the rate limiting step. So the second intermediate is going to be lower in energy than the first because of the fact that ring strain is no longer present. So we get our third transition state and then finally, our products which are lowest in energy. So here let me go ahead and draw it out. So our starting material is the closed oxide ring, right? Our first intermediate is or a peroxide ring with proton oxygen. Our second intermediate is now the open oxide ring, but with the positively charged oxygen from water oops and then we have our products on the very right side. So now that we have the two reaction coordinate diagrams contrasting each other what role does the acid play in activating the reaction? Exactly. So here just to go ahead and summarize the acid, basically activates the oxide rent and it's going to activate the oxide ring towards nucleic substitution by forming a better leaving group. And when you form a better living group, you make the reaction faster. Because once again, that positive charge on the oxide oxygen that's already present, gives more of an incentive for that oxide ring to break. So that oxygen can no longer have its positive charge. So there you have it. Now weve arrived at our final answer. Here's the reaction diagram of the first mechanism without acid. And here's the diagram of the second with acid. So in this case, the acid activates the oxide ring towards nucleic substitution by forming a better leaving group, making the reaction faster. And so if you watch this video all the way through, thank you so very much for doing so. And I hope you found this helpful.

Despite it being equally favorable, opening of the epoxide does not happen in the absence of an acid catalyst. How does acid make the reaction faster? Demonstrate this concept by directly comparing the reaction coordinate diagram for both situations A and B.

Key Concepts

Epoxide Reactivity

Acid Catalysis

Reaction Coordinate Diagram

Watch next