Kolbe-Schmidt Reaction - Online Tutor, Practice Problems & Exam Prep

Kolbe-Schmidt Reaction Concept 1

Hey, everyone. In this video, we're going to talk about the Kolbe-Schmitt reaction. Now EAS or electrophilic aromatic substitution reaction is used to synthesize ortho-hydroxybenzoic acid, otherwise known as salicylic acid. We're going to say under basic conditions, phenol will be deprotonated to form phenoxide. Now, phenoxide is then carboxylated, electrophile at the ortho position.

Now, here we have our phenol. Here we have carbon dioxide. We're utilizing sodium hydroxide with water and a proton followed by high pressure. This will help us to force the carbon dioxide onto the benzene ring and then having it be protonated into a carboxylic acid at the end. That's essentially the Kolbe-Schmitt reaction.

One key thing to take away from this is we have a phenyl group which is an ortho-para director. Ideally, when we want to add a group to benzene, it would be most advantageous to put it here in the para position so that we can avoid steric clashing. But with the Kolbe-Schmitt reaction, we actually make an ortho product. The reason for this is because of intramolecular hydrogen bonding. So the oxygen of this carbonyl could form an H-bond with the H of the hydroxyl group.

This is intramolecular, meaning within the same molecule. This actually leads to greater stability of the overall molecule. So we have a case where the ortho position is actually better for these two substituents. So, again, remember, OH is an ortho-para director. Typically, we want the next group to go in the para position, but if there's intramolecular hydrogen bonding, that's possible.

It'd rather go ortho. The Kolbe-Schmitt reaction takes advantage of this idea. Right? So just keep that in mind. The whole point is to go from phenol to this salicylic acid.

Kolbe-Schmidt Reaction Example 1

Hey, everyone. So in this question, now we're going to look at the mechanism for the Kolbe-Schmitt reaction. So step 1 involves phenol being deprotonated by sodium hydroxide. Alright. So here, let's draw our phenol.

We're going to deprotonate it with Sodium Hydroxide. So, here this is going to come in and deprotonate, grabbing this H. Oxygen holds on to the electrons. We're going to make our Phenoxide ion, which is resonance stabilized. Because here, once we make it, it could resonate into the Benzene ring.

So this could come here to make a double bond, kicking this bond here. And here, we're going to have another form that it could take. So now this is negatively charged. It could continue to resonate. So this moves here, taking this bond here.

So here goes a negative now. This can resonate here, causing this to resonate here. Okay. So this is negatively now. And then it could technically resonate back, kicking the double bond from the carbonyl back up to the oxygen, recreating benzene.

Here, we're not showing that last one, but it does exist. And in fact, we'll show that. So this could resonate here, keeping this bond up here. And so all the different ways we could show the resonance structures of the phenoxide ion. Now, here, phenoxide ion reacts with carbon dioxide through EAS or electrophilic aromatic substitution.

And we're going to say, here goes our Phenoxide ion here. And in fact, what we're going to do is we're going to just show one of the resonance structures that we had. Let's show this one. And there goes carbon dioxide which is C=O. So this decides to come and attach here, kicking 1 of these pi bonds to the oxygen.

So here goes. The bond connecting to it. Okay. So we've added it. So we've done electrophilic addition.

There's still an H here the whole time. And that's important because we're going to need that H in order to restore aromaticity. So the substitution part is going to come in. As remember, EAS itself is an addition-elimination type of mechanism. And then we're going to have there.

Alright. So then we're going to say here, in order to do this, we're going to use another mole of OH-, which is going to come in, deprotonate, grab this H, fall here to make a double bond, and kick this bond up. And here we're going to write down what that looks like down here. So we've recreated our benzene ring, but now we have this negative oxygen, which needs to be converted into a neutral form. And we also have this negative oxygen that is part of the carboxylic acid that got added.

All we have to do is protonate both of them. Alright. So we're going to protonate both of these with H+. Okay. So here we're going to say this can get protonated by this H+, and then another H+ can be used to protonate this oxygen that's negative.

And at the end, doing that gives us our phenol back and then our carboxylic acid that is ortho to it. Alright. So that's how we get our salicylic acid at the very end. So this is the general mechanism when it comes to the Kolbe-Schmitt reaction.

Provide the product created when phenol reacts with the following reagents. 

Hey, everyone. So let's take a look at the following practice question. It says, provide the product created when phenol reacts with the following reagents. So here we have our phenyl molecule, and we know that the first two lines where we have carbon dioxide, NaOH, water, and pressure followed by H+ catalyst, well, this represents the Kolbe-Schmidt reaction. We know in that reaction, we're going to get the addition of a carboxylic acid that is ortho to the hydroxyl group here. So we're going to put a carboxylic acid here. So that's what we have initially with steps 1 and 2. Now for steps 3 and 4, what's occurring here? Well, we have an anhydride followed by again H+ catalyst. Remember our reactions dealing with carboxylic acid derivatives. We know that when we react an anhydride with an alcohol, we're going to make an ester. So what's going to happen next is we're going to bring in this anhydride. And since it's a symmetrical anhydride, we can just say a portion of it will take the place of the hydrogen of the hydroxyl of this salicylic acid. So now we're going to have our oxygen here. Here's still our carboxylic acid. And again, this portion here is getting added. So in turn, we just created an ester group for our final product. So just remember, steps 1 and 2 were just the Kolbe-Schmidt reaction when we have our phenyl group reacting with reagents from steps 1 and 2 to create a carboxylic acid that is ortho to our hydroxyl group. And then we just had to remember our reactions when it comes to carboxylic acid derivatives where we have an anhydride reacting with an alcohol in order to produce an ester at the end.

Provide the product created when benzene reacts with the following reagents. 

Hey everyone. So here in this practice question, it says, compounds provide compounds A through D when benzene reacts with the following reagents. Alright, we're starting out with Benzene, and in the first step, we're going to have nitration, where we have our nitric acid reacting with sulfuric acid. We know that's going to add an NO2 group or nitrile group to benzene. But then remember, on the bottom we have reduction occurring, where we have tin(II) chloride with H3O+. That's going to reduce my NO2, my nitro group, to NH2. So here, our compound A is Aniline, this molecule here. Next, we have NaN02 over HCl. Well, that would transform my NH2 into a diazonium salt. So we'd have N triple bonded to N, and then we'd say here that heat will overhead. Well, that would change it into an OH group. So, in fact, we're going to remove the diazonium salt, and we're going to change it into an OH group. This would represent compound B. Next, we are going to react it with carbon dioxide pressure, NaOH, and water over H+ catalyst. These are the reagents that are involved in the Kolbe-Schmitt reaction. So basically, it's going to add a carboxylic acid that is ortho to the OH group. So we're going to make this compound here, which remember is a product of the Kolbe-Schmitt reaction. Now, finally, we're reacting what we have here as compound C with an alcohol and an H+ catalyst. Remember, when reacting alcohol with a carboxylic acid, we're trying to create an ester. So this will be an esterification or Fischer esterification. So we know that this portion of the alcohol would replace this OH group of the carboxylic acid. So what we'll have at the end is this. Here's our OH. And again, this portion of the alcohol comes in and replaces the OH of the carboxylic acid. So this would represent our compound D. Right, these are each of the compounds we make A, B, C, and D, following these list of reagents given in the following practice question.

Supply the reagents necessary to accomplish the following transformation.

• A. 1. CH₃CH₂NHCH₃ xs2. NaBH₄/H₃O⁺
• B. 1. SOCl₂2. CH₃CH₂NHCH₃ xs3. LiAlH₄ xs/H₃O⁺  
• C. 1. LiAlH₄ xs/H₃O⁺ 2. PCC/CH₂Cl₂ 3. CH₃CH₂NHCH₃ xs
• D. 1. CH₃OH/H₃O⁺ 2. CH₃CH₂NHCH₃ xs3. DIBAH/H₂O