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 - Video Tutorials & Practice Problems
Triacylglycerol Reactions: Hydrolysis Concept 1
Video transcript
Triacylglycerol Reactions: Hydrolysis Example 1
Video transcript
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.
Potassium Laurate
Sodium Oleate
Potassium Myristate
Potassium Lactate
Triacylglycerol Reactions: Hydrolysis Concept 2
Video transcript
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
Video transcript
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
Video transcript
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.
Palmitoleic acid
Lauric acid
Palmitic acid
Myristic acid
Triacylglycerol Reactions: Hydrolysis Concept 4
Video transcript
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.
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.