Hydrolysis of Thioesters - Online Tutor, Practice Problems & Exam Prep

Hydrolysis of Thioesters Concept 1

Hey everyone. So in this video, we're going to take a look at the hydrolysis of thioesters. Now, recall that thioesters are sulfur analogs of just regular esters. And we're going to say here that acid-catalyzed hydrolysis of a thioester produces a carboxylic acid and a thiol. And we're going to say here that the reaction takes place via a nucleophilic acyl substitution mechanism.

If we take a look at the basic premise of a NAS or nucleophilic acyl substitution mechanism, we have step 1 which is protonation, step 2 which is nucleophilic attack, step 3 is yet again a proton transfer, but here we're saying proton transfer because we've already done protonation in step 1. We're going to say step 4 is a leaving group, and then we're going to say step 5 is deprotonation. Steps 1, 3, and 5, although their names are different, they're basically talking about the movement of a proton, an H⁺ from one structure to another. Right? So they have that in common with each other even though the names are slightly different from one another.

The premise still remains. Right? So when we're talking about acid-catalyzed hydrolysis of thioesters, this is going to be the basic setup for a majority of the mechanisms that you're going to see.

Hydrolysis of Thioesters Example 1

Alright. So here goes our thioester and we're using H3O+. So step 1, we're going to say that the carbonyl oxygen is protonated by the hydronium ion, the H3O+ ion. The way this works is that oxygen is going to use its lone pair to grab an H+ from H3O+. The bond breaks and goes to oxygen.

Doing this creates initially this structure. Oxygen makes 3 bonds so it's positively charged. And water is created because we lost an H+ from H3O+. Step 2, we're going to say the water molecule attacks the carbonyl carbon. So we're going to say now the water comes in. It's going to attack here, breaking this bond up to oxygen.

So now water is neutral. And here goes the water molecule that is attached to the carbonyl carbon. Next, we're going to say, a proton is transferred from the positively charged oxygen to the sulfur. So our proton transfer, so basically, one of these hydrogens on this water molecule would will transfer itself over to the sulfur. Doing that gives us this.

It's going to be an OH2 now, so now it's no longer a good leaving group. Sulfur is making 3 bonds. It's positively charged. Ideally, it wants to make only 2 bonds. It's making one extra, so that makes it a good leaving group.

So now we're going to say, an oxygen atom pushes its lone pairs to kick out the thiol. So this is going to come down here to make a double bond, kicking this whole thing out. So what we're going to get now is oxygen again making 3 bonds, so it's positively charged. And then we're going to have this OH2, which is basically this OH2 that I moved over. And remember, we've kicked out this thiol right here.

Right here that we can see. And so, lastly, step 5, the newly reformed carbonyl oxygen is deprotonated regenerating the hydronium ion. So water comes in and it's going to grab this H+. The bond breaks and oxygen holds on to the electrons. So now we have our carboxylic acid as a product, but then also remember, we made this thiol as well.

Right? So this would be the basic mechanism when it comes to, well, a simple mechanism of acid-catalyzed hydrolysis of a thioester. Alright. So keep in mind the various steps that we saw and the movement of your arrows. So when you're not new to organic chemistry, we know that each step is important and vital because it leads to the next one.

Alright. So, again, our final products here should be a carboxylic acid and a thiol.

Hydrolysis of Thioesters Concept 2

Hey, everyone. So in our continued discussion of hydrolysis of thioesters, we're going to take a look at it from a basic angle. When we're talking about basic hydrolysis of a thioester, we use the term saponification. We're going to have the saponification of thioesters. The saponification of thioesters produces a carboxylate anion and a thiol.

Remember, the medium, the environment is too basic for carboxylic acid to remain. It would be deprotonated, and we're going to see a carboxylate anion form instead. Nevertheless, we're still going to also get a thiol within this reaction. Because it's undergoing a base hydrolysis, the general steps are step 1, nucleophilic attack, step 2, loss of our leaving group, and step 3, a proton transfer. Alright.

So again, remember when we're talking about saponification of a thioester, we're talking about base hydrolysis, basic hydrolysis of a thioester.

Hydrolysis of Thioesters Example 2

Ethyl propanethioate. So here we have our thioester, and it's reacting with aqueous hydroxide ion. Right. So this is a basic solution. Now, in step 1, we're going to have the hydroxide ion attack the carbonyl carbon forming an alkoxide.

So here, the oxygen with its lone pair comes in, hits the carbonyl carbon breaking the bond, and kicking it up here. What we're going to make now is a negatively charged oxygen since it's making one bond and has 3 lone pairs. And here goes the hydroxide we just added. This is an alkoxide ion because we have a negative oxygen connected to this carbon chain. Now, in step 2, we're going to say the alkoxide ion pushes its electrons to kick out the thiol as a thiolate anion.

Alright. So this is going to come down to make a double bond, and it's going to kick this out. So what we're going to make is a carboxylic acid plus this thiolate anion. But here's the thing, we're not allowed to have this type of environment. Steps 3 a and 3 b are connected to one another.

They're both part of proton transfer. Here in step 3 a, we're going to say the thiolate anion deprotonates the carboxylic acid forming a carboxylate anion. Carboxylic acid is very acidic, so it's naturally going to become deprotonated by the thiolate anion. So here, we're going to say it comes in, removes this H⁺, oxygen holds on to the electrons. As a result of this, we make our carboxylate anion, and we have a thiol.

And then for step 3 b, we're going to say a thiol, it has a pK_a of roughly around 10, and is sufficiently acidic to remain deprotonated in a solution of hydroxide ion, which typically has a pK_a of around 15.7. So what's going to happen here is this O_H is strong enough that it's going to deprotonate this. The thiol here gives up its hydrogen, the electrons go to the sulfur. So here we have R, negatively charged sulfur plus water.

So these were the steps, all the steps when it comes to the basic hydrolysis of a thioester. Right? So just keep in mind, we're including a basic environment, so it's hard for these various acidic groups to stay in their protonated form. It's very easy for them to become deprotonated.

Draw the structures of products A and B in the following reaction sequence.

• A.
• B.
• C.
• D.

Thioesters can react with nucleophiles to yield substitution products. Write a mechanism for the reaction of S-methyl ethanethioate with dimethylamine.

Everyone. So in this practice question, it says thioesters can react with nucleophiles to yield substitution products. Write a mechanism for the reaction of S Methylethane Thioate with Dimethylamine. Alright. So here, nitrogen is basic in nature. This one means basic in nature. So it's going to act as a nucleophile and attack the carbonyl carbon of this thioester. So, what we're going to get initially is this structure. As nitrogen is making four bonds, it's going to be positively charged. So the next step is going to be a proton transfer. Here, the hydrogen off of nitrogen is going to move towards the oxygen to create an OH group. So now, nitrogen is making three bonds, its ideal number of bonds, so it is no longer positively charged. Next, we're going to have the loss of our leaving group. In this case, we're going to say that oxygen decides it wants to make a double bond, and when it does, it's going to kick off this sulfur. So what we're going to get now is this oxygen making three bonds, which is going to make it positively charged. And we have this structure here, plus this negatively charged sulfur. Alright. Next, we're going to have yet another proton transfer where this sulfur is going to deprotonate this OH. Oxygen holds on to the electrons, so then we're going to get this neutral amide plus this thiol. Now remember, because of the acidity of this thiol, it could exist in equilibrium with more amine that's still swimming around because remember we're dealing with excess amine. So we can have this thiol here plus another mole of this amine, And in equilibrium, we can have the sulfur is going to get deprotonated to become S-negative again, plus this nitrogen which is making four bonds and becomes positively charged. Alright? So, again, we make these structures once we've gone through the mechanism. But because of the acidic nature of a thiol, it could exist in equilibrium with the excess amine that's still floating around within the solution. So you'd also have to show this equilibrium equation at the end. Alright. So that would be our final mechanism.