Acid-Base Catalysis - Online Tutor, Practice Problems & Exam Prep

Acid and base catalysis are crucial in organic reactions. An acid catalyst donates a proton (H⁺), enhancing reaction rates by forming a carbocation intermediate, which is then attacked by a nucleophile, such as water, leading to the formation of a carboxylic acid from an ester. Conversely, a base catalyst removes a proton from an alpha carbon, generating a negatively charged enolate ion that can react with an alkyl halide in an alkylation reaction. Understanding specific versus general catalysis is essential, as it influences the mechanism and efficiency of these reactions.

Acid-Base Catalysis Concept 1

In this video, we're going to talk about acid and base catalysis by first discussing the acid catalyst. Now here we're going to say the acid catalyst increases the rate of reaction by donating a proton to a reactant. So remember our proton here is just an H+. So here we're going to start out with our ester. And what we're going to have here is the breaking of a pi bond towards oxygen, and this will cause one of these lone pairs to be used to grab an H from this acid, and then a keeps the electrons.

So now we're going to have this connected to the oxygen. This carbon gave up its electron in the Pi bond, so now it's positive. And because it's positive, it becomes a target for nucleophilic attack by this water molecule. This water molecule comes in here and hits here. Now we're going to have this water molecule attached to that formally carbonyl carbon.

This whole process is the slow step in which we have the formation of our carbocation, and then the nucleophilic attack. So here, oxygen's making 3 bonds, so it's positively charged. Oxygen will prefer 2 bonds, so this a that left earlier comes back and deprotonates the water molecule. So now we're going to have an OH group here. Alright.

What happens next is we can protonate this Alcoxyl group, this OR group. So this oxygen will grab an H+ from this acid, keeps the electrons. Again, this is a slow step in terms of this process. So now this has been protonated, now it's a good leaving group. Because it's a good leaving group, this oxygen decides to make a double bond kicking it out.

That oxygen now is making 3 bonds so it's positively charged. And because it's positively charged, it'll attract this A minus which comes and deprotonates the oxygen to make it a neutral carbonyl oxygen. And here we've created our carboxylic acid at the end and we have our floating around. So this is a way of us going from our Ester which is a carboxylic acid derivative to the whole process to get back to our parent compound which is our carboxylic acid. Right?

So, this is just one example in which we can utilize an acid catalyst.

Acid-Base Catalysis Concept 2

Now, when we talk about a base catalyst, we say a base catalyst increases the rate of reaction by removing a proton from a reactant. So here, we have an alpha carbon because it's the carbon next to a carbonyl carbon. It is weakly acidic. So if we utilize a base, it can come in and remove this hydrogen. Carbon holds on to the electrons.

So carbon here has a lone pair so it's negatively charged. And what we can do here is we have this RX on the bottom, this alkyl halide, and now we've brought it over here because now we can finally react with it. So, this is going to use its lone pair hitting this R group, kicking out the halogen. So, what we've done here is we've added a carbon to the alpha carbon. This is an alkylation reaction.

This is the slow step in this process in which we're going to do this attack on this R group to kick out the halogen. So here, we've seen how a base catalyst can help to basically increase the rate of a reaction by first removing a proton from the beginning reactant.

Acid-Base Catalysis Concept 3

In this video, we're going to say that there are 2 types of acid-base catalysis and it depends on the strength of the catalyst. Now, we have specific catalysis versus general catalysis. Under specific catalysis, the proton is transferred before the slow step and it's catalyzed by a strong acid or base. Here, we will say that this happens in more than one step, so it's a non-concerted step. So, it doesn't all happen at one time.

If we take a look here, we have our alpha carbon here and we have LDA which is a strong bulky base. It will deprotonate this alpha carbon so it removes this H. The carbon holds on to the electron so now we've formed an enolate intermediate. That enolate intermediate can then attack this alkyl halide hitting this R group and kicking out the halogen. Here, this is our slow step.

And what we've done is we've alkylated our alpha carbon. Now, when we're talking about general catalysis, we're going to say this is transferred during the slow step. So the proton is transferred during the slow step, it's catalyzed by a weak acid or base, and it's a concerted step meaning it happens all at once. Here, we're going to use a hydroxyl group within an aqueous environment. So, we're going to say here that this is still my alpha carbon.

What happens here is this hydroxyl group will deprotonate the alpha carbon. At the same time, we can break the bond and that bond can be used to attack this R group here and kick out this halogen. This is all happening at one time. So we'd say here that we have the CH2 group still here. The bond between this carbon and the alpha hydrogen is breaking at the same time that the bond between the H and the OH is forming.

And then we'd say here, we are forming this bond here to the R group as it's breaking its bond to this halogen, this leaving group. This represents our transition state. And then, what happens at the end is we're basically again just alkylating our alpha carbon. So, the process the end result is the same, it's just that we have specific catalysis versus general catalysis. And it really has to do with our proton being transferred before or during the slow step, whether we're using a strong or weak acid or base.

Alright. So keep this in mind, the 2 types of acid-base catalysis that are possible.

Acid-Base Catalysis Example 1

Everyone, in this example question, it says, draw the mechanism for the acid catalyzed hydration under specific and general acid catalysis. Alright, let's work on specific catalysis first. Remember, this happens in a non-concerted step, meaning it happens in more than one step.

So let's start with our specific catalysis. We have our structure here, and here's our OH group. We're going to react it such that this π bond will break and grab an H, then the A flies away for now. We're going to do this by Markovnikov's rule, which means that the H will go to the less substituted alkene carbon, the less substituted alkene carbon. So it will go to this one here that's a CH₂, making it a CH₃, which would mean this carbon here becomes positive. Because that carbon is positive, that is the carbon we are going to attack with this hydroxyl group. We'll use one of its lone pairs and attach right there.

So let's draw this cyclic portion first. The oxygen is connecting to this carbon here. So, let’s count: 1, 2, 3, 4, 5. We're going to make a 5-membered ring and connect it here. We still have this methyl group over here. The oxygen is making 3 bonds, so it's positively charged here. The A- that left comes back to deprotonate that oxygen to make it neutral. So this comes in here, grabs this H here. So what we get at the end is this: Here goes our neutral oxygen. We have our 5-membered ring that we created as a result, and the methyl group that's on it.

This is what we get in the end through our specific catalysis. Now, if we're going to do general catalysis, remember this happens by concerted steps, so it's all at one time. So let's look at general now. Here we're going to draw our structure again. And so then we're going to have this. And then we’re going to have our reaction all in one step. We can have this hydroxyl group hitting here at the same time this pi bond is breaking to grab this, and this bond is breaking here. So we have quite a few arrows moving here.

So then we're going to get this structure; again, here goes our oxygen making three bonds and being positively charged. And then A- comes back deprotonating that oxygen. So now we're going to have our neutral structure at the end, and we have lone pairs here. I didn't show them in the top one, but they're there. Right. So both processes happen by nonconcerted and concerted effort. But, at the end, we're still going to get our same product.

This is what happens if we're following our specific catalysis versus our general catalysis.

Provide a mechanism and appropriate bases for both, specific and general base catalysis.

Hey everyone. So in this practice question it says, Provide a mechanism and appropriate basis for both the specific and general base catalysis. In this reaction, we have base catalyzed alpha chlorination. Here, our alpha carbon possesses 2 hydrogens on it. And through the mechanism, we would replace both of those alpha hydrogens with chlorines. So let's look at the specific catalysis first. So if we're looking at specific catalysis first, remember here we'd use a strong base in order to do this. The base of choice would be LDA. Now we're gonna say the way this would work is here is our structure, we're going to use LDA, which would remove one alpha hydrogen right from the beginning to create our enolate. The alpha carbon keeps the electrons. Here is our enolate that's been formed. Remember, this is resonance stabilized, so I could make a pi bond and kick this pi bond between carbon and the oxygen up to the oxygen. Here, I'm not gonna show that. I'm just gonna go straight into this reacting with chlorine. So this would come in, get this, pick this out, and we'd have that chlorine attached now to that alpha carbon. This would happen twice, so we'd get another time. So, again, there's another alpha hydrogen here. So the process would begin again where this removes a hydrogen. The alpha carbon holds on to the electrons. Here goes our enolate again. We'd react with another mole of Cl2. And now we have attached 2 chlorines to my alpha carbon by specific base catalysis. Now if we're looking at general, remember this one here happens in a concerted effort or considered concerted step, so all at once. And we use a weaker base, so we're gonna use aqueous hydroxide. In this one, we'd have still our structure. We're gonna use our aqueous base here. It would come here, grab this H here, this bond would break, and we'd use it to grab onto this chlorine, and this one would go away. So we're gonna make here. We would have the bond between OH- and the H, this bond would be forming. And then we'd say that this bond here is going to be and, actually, just to make it easier to track, let's just put this hydrogen over here. It's forming a connection with this OH-. So, the bond between them is forming. The bond between the alpha carbon and the hydrogen is breaking. The bond between the chlorine and the alpha carbon is forming. The bond between the chlorine and the other chlorine is breaking. So this will be our transition state. And at the end, once everything's finished forming and breaking, we'd have that chlorine on the alpha carbon. Now we'd start this process again, so we bring this down here. So there goes my chlorine is here. And then we still have one more alpha hydrogen here. Again, our aqueous hydroxide, our weaker base form, would come in, grab this H here. The bond would break. We use that bond to grab onto this chlorine and this would leave. So, again, we make another transition state. So we have this chlorine over here is is still attached. We're gonna say the bond between the alpha carbon and the hydrogen is breaking because it's forming a connection with this hydroxide. At the same time, this alpha carbon wants to form a bond with another chlorine, and the bond between the chlorines is dissolving. And at the end what we have now, once all the bonds have finished forming and breaking, we'd have 2 chlorines attached to that alpha carbon. So this would be our general catalysis mechanism. Here again, we're using a concerted step. So everything's happening at once to add the first chlorine and then again to add the second. Here, we're using a weaker base, so we're using aqueous hydroxide instead of the stronger LDA like we would in our specific catalysis. Right? So this is the approach we take for both types of catalysis. In terms of this base catalyzed alpha chlorination reaction.