Triacylglycerol Reactions: Hydrolysis - Online Tutor, Practice Problems & Exam Prep

Saponification is a base-catalyzed hydrolysis of triglycerides, producing glycerol and carboxylate salts of fatty acids, essential for soap production. The mechanism involves nucleophilic acyl substitution, comprising nucleophilic attack, leaving group loss, and proton transfer. Acid-catalyzed hydrolysis, using strong acids like HCl or H₂SO₄, also yields glycerol and fatty acids through a stepwise process. Both methods highlight the importance of ester bond cleavage in lipid chemistry, emphasizing the role of hydroxide ions and hydronium ions in forming the final products.

Triacylglycerol Reactions: Hydrolysis Concept 1

Now, saponification represents base-catalyzed hydrolysis. Because of this, we're going to say under this reaction, the hydroxide ion leaves the ester bond to create salts of the fatty acids and glycerol. If we take a look here, we have our ester linkages these red bonds here, they're going to be severed when we do saponification. We're using our Sodium Hydroxide and some heat in this case. We're going to add Hydrogens to the Oxygens that help to make the Glycerol molecule and we're going to say here we're making salt of the fatty acids. That means we're not going to make Carboxylic Acid Fatty Acids, we're going to make Carboxylate Anions. So, basically, we're going to have negatively charged Oxygens and because the metal that's formed with the base is Na, we have Na positive near it. So these will represent our 3 salts of the fatty acids. Now, here we're going to say the salts of the fatty acids are used in the creation of soaps. So this is one important application of the Saponification of a triglyceride molecule. It helps in the creation of soaps that we use. We're going to say here, when Sodium Hydroxide is used as our base, this helps to make a solid soap. And then if we use Potassium Hydroxide, KOH, that helps us to make liquid soap. So again, saponification represents base catalyzed hydrolysis in this instance of a Triglyceride or Triacylglycerol molecule and we're going to say that it's important in the creation of soaps. Based on the type of base that you're using, you can either make hard soap, a solid soap, or you can make a liquid soap.

Triacylglycerol Reactions: Hydrolysis Example 1

This example question says, "Draw the starting Triacylglycerol used when its complete basic hydrolysis created 2 Laurate Salts, 1 Palmitate Salt, and a Glycerol Molecule." Alright. So, here we're working backwards. And what we're going to say first is that these 2 salts came from fatty acids. Here, this laurate salt comes from lauric acid, and this palmitate salt came from palmitic acid. Now, both of these fatty acids represent saturated fatty acids. Lauric acid has 12 carbons and no pi bonds, and then palmitic acid has 16 carbons and no pi bonds.

Now, here, we're working backwards. We're going to start out with our glycerol base. So we have CH₂CHCH₂, and we're going to say they're connected to oxygens. To create our starting material, we're going to make ester linkages, so these oxygens are connected to the carbonyls of these fatty acid chains. So, lauric acid or laurate salts are going to be the first two, that's 12 carbons, so 2, 4, 6, 8, 10, 12. And we do it again. 2, 4, 6, 8, 10, 12. And then finally, the last one is a 16-carbon chain. So 2, 4, 6, 8, 10, 12, 14, 16. This would represent our starting material, our triglyceride molecule or our triacylglycerol molecule.

So again, we're just working backwards knowing what the salt was in terms of its fatty acid form is essential to determine how many carbons it possesses, and how many pi bonds it possesses. Then from there, we just have to make ester linkages between the fatty acids and the glycerol base. Alright. So the glycerol part or that connects the fatty acids to it. And so just keep that in mind when asked to figure out what our starting material would be.

The salt of a fatty acid has its ending of “ic acid” changed to “ate”. Based on this information provide the name of the salt created from the following saponification reaction. 

• A. Potassium Laurate
• B. Sodium Oleate
• C. Potassium Myristate
• D. Potassium Lactate

Triacylglycerol Reactions: Hydrolysis Concept 2

In this video, we're going to take a look at the Base Catalyzed hydrolysis mechanism of Tri-Acyl Molecules. Now, here we're going to say it follows an SNAC type of mechanism which is Nucleophilic Acyl Substitution. Now, remember we've talked about this type of mechanism when we covered carboxylic acid derivatives. Here we are going to say that it is composed of three primary steps. Step 1 involves a nucleophilic attack. Step 2, we have our loss of leaving group. And step 3, we typically have some type of proton transfer.

If we take a look here at this example, it says, provide the mechanism for the base catalyzed hydrolysis for one of the fatty acid chains in the following triglyceride. So here we have our triglyceride, the three fatty acid chains are the same, and we're using our Sodium Hydroxide with some heat. Now, because they're all the same, we can just focus on any of the chains, it really doesn't matter because they're all the same. So step 1 here, we're going to use the hydroxide ion as a nucleophile to attack the carbonyl carbon. So here, remember this oxygen will be partially negative and this carbon connected to it is partially positive. This is going to come in here and kick this bond up to the oxygen. Doing that now produces this structure here. We've attached our OH group to that formerly carbonyl carbon. We've broken that pi bond, so now it sits as a lone pair on this oxygen. Oxygen now has a negative charge.

Step 2 it says, recreate the pi bond of the carbonyl group to kick out the Carboxyl group of the ester. So basically, what happens here is this is going to come back down here to remake our double bond, and this group will serve as a leaving group. So it just gets jettisoned out. When we do that, that gives us now here's the group that's been jettisoned out and it's a negative oxygen. And here, we've created a carboxylic acid.

Now here we're going to use that newly created or newly formed carboxylic acid to protonate the alkoxide ion. Because remember, in base catalyzed hydrolysis, we create our carboxylate anion. We don't create a carboxylic acid at the end. So here, we're going to say this alkoxide ion is going to deprotonate this carboxylic acid removing this H. Oxygen holds on to the electrons since it's more electronegative. So we'll have this OH has been created plus we have our carboxylate anion. Okay. So 1, 2, 3, 4, 5. 1, 2, 3, 4, 5 as our answer here. OK. So this would be our carboxylate anion that's formed. And remember, this OH is just a portion of the glycerol backbone that we have. We would do this 2 more times if we wanted to basically sever the other 2 ester bonds that are still present within our original triglyceride molecule. Right. So just remember, these are the steps that we'd take in order to do our base catalyzed mechanism for a triacylglycerol molecule or triglyceride molecule.

Triacylglycerol Reactions: Hydrolysis Concept 3

In this video, we're going to take a look at a triacylglycerol reaction in the form of hydrolysis. Now, hydrolysis can be either acidic or basic. In this first video, we're going to look at acid-catalyzed hydrolysis. Now, under this type of reaction, an ester bond is hydrolyzed to create a glycerol molecule and three fatty acids. Now, here we're going to say it occurs stepwise in the presence of a strong acid. The acids are typically hydrochloric acid or sulfuric acid. When we say stepwise, that means that all three fatty acids don't come off all at once. First, we do hydrolysis of the first ester linkage to break off the first fatty acid, then the second one occurs, and then the third one occurs. In this illustration here, we're going to show our final products as having three fatty acids. We're going to know that they all didn't come off all at once. This is the end result of going through three successions of acid catalyzed hydrolysis to get rid of all three fatty acids. Now, here if we take a look, we have our ester linkages here in red, we're using water and H⁺. Remember, hydrochloric acid and sulfuric acid are strong acids. In an aqueous environment, you typically see them represented as H⁺. You could still see H₂O over HCl or H₂O over H₂SO₄, but typically again, we see these two strong acids in this form here. Now, what happens is our ester linkage is cut. So we cut these linkages. When we cut these linkages, the oxygen that is part of my glycerol molecule in blue, that oxygen gains a hydrogen. And in that way, we've created our OH groups for glycerol. So we've just created our glycerol molecule here. And then our fatty acid. Remember, a fatty acid has a hydrocarbon tail and then it has a carboxylic acid head. To make this into a Carboxylic Acid, we add OH to my carbonyl group. This is hydrolysis. We're using water to cut the ester linkage. We're basically adding water back. An H on the glycerol oxygens, OH on the carbonyl carbons. Now, here besides acid catalyzed hydrolysis, we have enzymatic hydrolysis. Now, this is a similar reaction done under milder conditions that instead uses the digestive enzyme lipase. So if we were to use the enzymatic approach, we'd use lipase instead of an acidic environment. This would still create our glycerol molecule and three fatty acids. We put it here because we're essentially making the same products. The pathway to get to them is different, but we're still making the same products at the end. So just remember, whether you're using acid catalyzed hydrolysis or enzymatic hydrolysis, your products will be a glycerol molecule and three fatty acids.

Triacylglycerol Reactions: Hydrolysis Example 2

In this example question, it says, "Draw the fatty acid products for the following reaction." So, here we have our triacylglycerol molecule or our triglyceride. We're using H₂ over our strong acid, hydrochloric acid. We know that essentially what happens here is that we're going to sever our ester linkages. The oxygens that are part of our glycerol molecule, they gain an H. Then, these carbonyl groups here are going to gain OH groups to make our three fatty acids. So, what we're going to do here is attach OHs to these carbonyl groups.

So there goes one fatty acid there. Let's see, here we have one chain, and then we have this pi bond. There goes my other fatty acid. And finally, here we count another chain. So, those are my three fatty acids that have been made. From this, we make the glycerol molecule and then these three individual fatty acids. They are all different from one another, so we don’t have to draw them the same way. We have three distinct types of fatty acids within this product.

Provide the common name of the fatty acids produced from the acid-catalyzed hydrolysis of the following triacylglycerol.

• A. Palmitoleic acid
• B. Lauric acid
• C. Palmitic acid
• D. Myristic acid

Triacylglycerol Reactions: Hydrolysis Concept 4

Hey everyone. So in this video, we're going to take a look at our acid-catalyzed hydrolysis mechanism for Triacylglycerol Molecules or Triglyceride Molecules. Now, here when it comes to this, it follows an NAS type of mechanism. Remember, this represents Nucleophilic Acyl Substitution, a mechanism that you saw when it dealt with carboxylic acid derivatives. Because it's acid-catalyzed in nature, there are going to be more steps than you typically would see if it were just base catalyzed. Here, our steps are proton transfer, so steps 1, 3, and 5. We have a nucleophilic attack for step 2, and then we have a loss of leaving group for step 4. Alright. So here if we take a look, we're going to provide the mechanism for the acid-catalyzed hydrolysis for the top fatty acid chain of the following triglyceride. So, we're talking about this one right here. Notice when we're dealing with this acid-catalyzed hydrolysis, we have water with some type of strong acid. Remember, this strong acid would protonate the water molecule to create our Hydronium ion. So, this feeds into step 1. In step 1, we are going to protonate the carbonyl oxygen with our Hydronium ion. So what's going to happen here is we're going to have this oxygen coming in and it's going to grab an H+ from the Hydronium ion. When it grabs that H+, it would create what we have down here in step 2. Our Carbonyl oxygen has gained an H+, so now it only has 1 pi bond and is positively charged because it's making 3 bonds. We're going to say here, we're going to use the newly created water molecule to attack the carbonyl carbon. So, this water molecule will come in and it's going to attack this carbonyl carbon. Carbon can't make more than 4 bonds, so this bond breaks and goes to the oxygen. That gives us step 3. In step 3, our water molecule is now attached to this carbonyl carbon. That oxygen is making 3 bonds, so it's positively charged. Here, we're going to perform a proton transfer between the attached water molecule and the Alcoxy oxygen. Here, to do that, we need to use a water molecule that's hanging around. Remember, this is occurring within an aqueous environment, so there are tons of water molecules present. What happens here is this water molecule will grab an H+. And what we're going to have here initially, and we gotta draw this out. Okay. So now this is just a regular O, not because it lost one of its H+. That water molecule that gained an H+ is now a Hydronium ion. And now this oxygen can get protonated by it. So we grab that H. Grabbing that H creates what we have here in step 4. That oxygen now has an H on it, so now it's making 3 bonds, so it's positively charged. We're going to say we're going to create a double bond between the hydroxyl oxygen group and attach carbon, and then kick out the alkoxy leaving group. So basically, what happens here is this comes down to make a double bond which causes this to get kicked out. Doing that now gives us this portion right here. So, we've kicked out this portion that's still part of the glycerol backbone. We have an R-O, which is this O, and the carbonyl group has been reformed. That oxygen is making 3 bonds, so it's positively charged. Now, step 5, we deprotonate the double-bonded hydroxyl group, Oxygen with a water molecule. So this comes and it's going to remove this H+, and oxygen holds on to the electrons since it's more electronegative. So then, we're going to have Carbonyl's been recreated. We have our carboxylic acid plus our Hydronium ion. Now, technically, we also have the portion of the glycerol molecule that still has its OH still. So these will be the things that we have. Remember, when we're doing acid-catalyzed hydrolysis, we will create a carboxylic acid as one of our products. Here, we showed the mechanistic approach to get to that carboxylic acid. And remember, we started out with our Hydronium ion which acts as an acid catalyst so we had to recreate it at the end. So that's why we have Hydronium ion at the very end. Here, the squiggly line with the OH is just a portion of the glycerol backbone that we're showing that's still around in terms of all of this. If we wanted to continue with this, we have 2 more fatty acid chains that could also undergo acid-catalyzed hydrolysis as well. But here, we're only asked to do it for 1 of them, and that's all we needed to show. Alright. So just keep notes in terms of this mechanistic process when it's under acidic conditions.

Propose a possible mechanism for the following reaction. 

Here it says, propose a possible mechanism for the following reaction. So here we have our Triacylglycerol molecule, or triglyceride. It's undergoing acid-catalyzed hydrolysis. A consequence of this is that we're going to recreate our glycerol molecule plus three fatty acids. Right. So, since this is acid-catalyzed hydrolysis, let's work this out. And so we have this portion here. Okay. So there goes our initial structure, and it's going to undergo acid-catalyzed hydrolysis. So what happens first is we're going to protonate our carbonyl oxygen. So this is going to come in and grab an H+. So we'll get here our protonated oxygen. So this is going to be a long process in terms of writing this out. Alright. So we still have our other two fatty acid chains that are still involved. Alright. So we have this portion here. Alright. So what's going to happen next is we're going to have our water molecule that's been formed come in and attack this carbonyl carbon, kicking this bond up. So let's keep going. So we have our OH here. Okay. The water molecule is attaching. Okay. So here I'm drawing them a little bit more spaced out. Okay. So now, we need to do a proton transfer. So yet another water molecule comes in. It's going to deprotonate this. Oxygen holds on to the electrons. And then we have, this is an OH now. Almost done, kind of. So then we have this structure here. We have the other fatty acid chains that are still unaffected so far. So we've recreated our Hydronium Ion. So, this will deprotonate, and it does this so that it can become a good leaving group. Let's keep going. Okay. So now it's gained a hydrogen. So it'll be positively charged since it's making three bonds. We have this OH here. There we go. And then these portions are still the same. Since it's a good leaving group, we're going to have a loss of the leaving group. This oxygen decides to remake its double bond, and it's going to jettison this portion out. So what we're going to have now is there goes our OH. We still have our other two fatty acid chains attached. And then we have this fatty acid over here, positive. One lone pair. So we have that so far. Now, this can be deprotonated to give us a carboxylic acid at the end by a water molecule. Comes in and deprotonates it. So we have a carboxylic acid at the end. Alright. So then Alright. And it's going to have those carbons there. So we've created this carboxylic acid, and then we still have this portion which needs to undergo acid-catalyzed hydrolysis two more times because we have two more chains there. But for simplicity's sake, we're just going to say this. We're going to say that this acid-catalyzed hydrolysis process happens two more times. This line represents it happening to this chain here, and then this other line represents it happening to this chain here. So, at the end, we'd have our glycerol molecule, plus our three carboxylic acids, plus just hydronium ions floating around. Okay. So this would be the mechanism that we talk about in terms of this particular practice question.

An optically inactive triacylglycerol molecule undergoes base-catalyzed hydrolysis to produce 1 glycerol molecule, 1 mole of laurate and 2 moles of myristate. Determine the structure of the triacylglycerol molecule.

Here in this question, it says an optically inactive triacylglycerol molecule undergoes base-catalyzed hydrolysis to produce 1 glycerol molecule, 1 mole of Laurate, and 2 moles of Myristate, and it says to determine the structure of the triacylglycerol molecule. Alright. So here, they're telling us we're making a glycerol. Okay? And if you look, Laurate, Myristate, these represent the carboxylate anion forms of their carboxylic acids. The carboxylic acid here would be Lauric acid, and here this would be Myristic acid. Lauric acid has 12 carbons and no pi bonds. Myristic acid has 14 carbons, no pi bonds. Another important piece of information they tell us here is that it's optically inactive, which means it's not chiral. This would mean that there's some level of symmetry involved with this initial triacylglycerol molecule. To determine if it's chiral or not, we look at carbon number 2. So here goes carbon number 2, it would be connected to carbon number 1, and carbon number 3. To have some level of symmetry, the 2 myristic acids would have to be connected to carbons 1 and 3. So counting the carbons: 2, 4, 6, 8, 10, 12, 14. The same thing with this one: 2, 4, 6, 8, 10, 12, 14. Then, Lauric acid, which is here on carbon number 2, counts as 2, 4, 6, 8, 10, 12. This would have to represent our initial triacylglycerol molecule before it underwent basic hydrolysis. Alright. So, this will be our starting material and it's not chiral because we would see for carbon number 2, this group and this group are the same. And because of that, it's not connected to 4 different groups and can't be chiral making it optically inactive.

Here’s what students ask on this topic:

What is saponification and how does it relate to soap production?

Saponification is the base-catalyzed hydrolysis of triglycerides, typically using sodium hydroxide (NaOH) or potassium hydroxide (KOH). This reaction breaks the ester bonds in triglycerides, producing glycerol and carboxylate salts of fatty acids. These carboxylate salts are the key components in soap. When NaOH is used, it produces solid soap, while KOH results in liquid soap. The process involves nucleophilic acyl substitution, where the hydroxide ion attacks the carbonyl carbon, leading to the formation of glycerol and fatty acid salts.

What is the mechanism of base-catalyzed hydrolysis of triacylglycerols?

The base-catalyzed hydrolysis of triacylglycerols follows a nucleophilic acyl substitution (NAS) mechanism. It involves three main steps: nucleophilic attack, loss of the leaving group, and proton transfer. First, the hydroxide ion (OH^-) attacks the carbonyl carbon of the ester bond, forming a tetrahedral intermediate. Next, the intermediate collapses, expelling the alkoxide ion as a leaving group. Finally, a proton transfer occurs, resulting in the formation of glycerol and carboxylate anions (fatty acid salts).

How does acid-catalyzed hydrolysis of triacylglycerols differ from base-catalyzed hydrolysis?

Acid-catalyzed hydrolysis of triacylglycerols uses strong acids like hydrochloric acid (HCl) or sulfuric acid (H₂SO₄) and proceeds through a stepwise mechanism. It involves proton transfer, nucleophilic attack, and loss of the leaving group, similar to base-catalyzed hydrolysis. However, the products are glycerol and free fatty acids, not carboxylate salts. The process is slower and requires more steps due to the need for protonation and deprotonation at various stages.

What are the products of enzymatic hydrolysis of triacylglycerols?

Enzymatic hydrolysis of triacylglycerols, typically catalyzed by the enzyme lipase, produces glycerol and three fatty acids. This reaction occurs under milder conditions compared to acid or base-catalyzed hydrolysis. The enzyme specifically targets the ester bonds in the triglyceride, facilitating their cleavage and resulting in the same products as acid-catalyzed hydrolysis: glycerol and free fatty acids.

What role do hydroxide ions play in the saponification process?

In the saponification process, hydroxide ions (OH^-) act as nucleophiles that attack the carbonyl carbon of the ester bonds in triglycerides. This nucleophilic attack initiates the hydrolysis reaction, leading to the formation of a tetrahedral intermediate. The intermediate then collapses, expelling the alkoxide ion and forming carboxylate anions (fatty acid salts) and glycerol. The hydroxide ions are crucial for breaking the ester bonds and driving the reaction to completion.