The mechanism of the Fischer esterification was controversial ...

Question

The mechanism of the Fischer esterification was controversial until 1938, when Irving Roberts and Harold Urey of Columbia University used isotopic labeling to follow the alcohol oxygen atom through the reaction. A catalytic amount of sulfuric acid was added to a mixture of 1 mole of acetic acid and 1 mole of special methanol containing the heavy ¹⁸O isotope of oxygen. After a short period, the acid was neutralized to stop the reaction, and the components of the mixture were separated.

(a) Propose a mechanism for this reaction.
(b) Follow the labeled ¹⁸O atom through your mechanism, and show where it is found in the products.

(a) Propose a mechanism for this reaction.

(b) Follow the labeled ¹⁸O atom through your mechanism, and show where it is found in the products.

Accepted Answer

Welcome back everyone. People had different ideas about the mechanism of Fisher Esterification, and it wasn't until 1938 that Irving Roberts and Harold Urey of Columbia University used isotopic labeling to track how the oxygen in the alcohol moved through the reaction. Using the same method, propose a mechanism for the reaction of propanoic acid and methanol that has the isotope o 18 in it with acid as a catalyst. So lets consider our starting materials, and as we understand, propanoic acid is the starting material, so we're going to draw carboxyl and the parent chain has a total of 3 carbons because that's propanoic acid. We are going to begin with the first step of the reaction, and that's the protonation step. We're using an acid as a catalyst. We're going to represent it as hydronium h three o plus. And essentially in the first step of this mechanism, we are performing the protonation step at oxygen to form a better electrophilic center. What we want to do is bond hydrogen to oxygen, which tells us that oxygen gets a formal positive charge. From here, we can essentially understand that this given intermediate is resonance stabilized, so we can essentially move our pi electrons onto oxygen. There is no necessity to show all of the intermediates. Specifically, we are primarily interested in the one that has carbon with a positive charge. This essentially explains an increased electrophilicity of carbon. Right? And from here, this is where we can introduce the molecule of an alcohol. We are using methanol, so we can represent it as c h three o h. And we are going to label that specific oxygen with a star. That would be oxygen 18. And we want to perform the nucleophilic attack step. That's our next step in the mechanism Add the electrophilic carbon. Let's go ahead and draw the resultant intermediate. Now we have c Oh groups bonded to carbon, and we are going to bond that oxygen with a methyl group. We're going to add a star to make sure that we understand which oxygen we are considering, And we also have hydrogen still bonded to oxygen. Right? So we're going to expand it because this is going to be important in the mechanism. From here, our next goal is to deprotonate the resultant oxonium ion and we can use another structure of methanol. So we're going to take c h three o h We're going to add a star, and now methanol is going to act as a proton acceptor. This results in a neutral intermediate. So we're simply removing that specific hydrogen. And in the next step, we are going to consider a protonation step. Now we can either use an acid or we can use the protonated alcohol for the next step, but since it is an acid catalyzed reaction, let's just move on with the acid. So from here we simply want to repeat the protonation step, and now we are going to protonate the oh group. So we're simply drawing another formal unit of hydronium, and we want to illustrate the protonation step at oxygen. The proton capture followed by the formation of water. Now what we're going to do from here is essentially form another structure of oxonium, and we're going to add that hydrogen to oxygen, which essentially creates a formal positive charge on oxygen. Now there is one more thing to consider. Since it is an acid catalyzed hydration, instead of using methanol as a base for the previous step, let's actually use water, right? Because if we want everything to be consistent, notice that in the first step we have formed water. Right? So now water will be best to accept that proton because we are essentially forming hydronium that we are using later. Right? So that should make sense, and therefore instead of using methanol as a base, we're going to use water to make everything consistent. So we're just going to introduce h c o. Now everything is consistent. We have a protonated intermediate, which is oxonium, and from here we can simply observe the loss of the leaving group. However, for the simplicity of this mechanism we can simply show that we can form a pi bond, followed by the loss of the leaving group. Right? And this essentially allows us to form a protonated carbonyl, so oxygen gets a formal positive charge. Now notice that we still have 1, 2, 3 carbons, so we're going to introduce 3 carbons. 1, 2, 1, 2, 3, right? And now on the other side, we have oxygen bonded to the methyl group. We want to label that specific oxygen and we also have to understand that we have removed water from the previous structure. And of course, the final step is the deprotonation step. Right? So in the deprotonation step, we simply want to use the previously formed water molecule that now acts as a base or basically a proton acceptor, and we are forming the ester. Plus h three o plus. Now notice that the alcohol's oxygen essentially appears at the alkoxy part of the ester, right? Since we have labeled that specific oxygen, we can easily track where it comes from and where it ends up. That's it. We have our mechanism. Thank you for watching.

Key Concepts

Fischer Esterification

Isotopic Labeling

Reaction Mechanism

Watch next