Draw the structures of all products you would obtain by saponification of the following lipid with aqueous KOH. What are the names of the products?

Question

Chemical structure of a triacylglycerol showing bonds and functional groups for hydrolysis reactions.

Accepted Answer

Welcome back, everyone draw the structures of all products formed in the saponification reaction between the lipid below and aqueous koh provide their common names. So we're given a Tr Ayl Glycerol and specifically what we want to do in this problem is simply perform a saponification reaction. We're using potassium hydroxide and we have to recall that to produce the products, we simply want to break our carbon oxygen single bonds. Now, first of all, we're going to produce carboxylate salts and the second product will be Glycerol. So first of all, let's draw the structure of Glycerol. We want to draw our we number chain. So we have three carbons and now we simply want to add oh to each part. And this gives us the structure of Glycerol. And this is what we always get in a saponification reaction between Tr Ayl Glycerol and a basic solution such as koh. Now, the other set of products will depend on the nature of each side chain. First of all, we are breaking the carbon oxygen bond and oxygen gets a negative charge for the first fragment. And this gives us carboxylate, right? Those be carboxylate souls, we're going to begin with oxygen with a negative charge and then everything else remains the same. Let's add car bono and then we have ch two, 12 times C three and we simply want to add an iron spectator and that would be Koh Right. So from Koh, we determine that the iron spectator is the potassium cion. Now, we want to understand that this is a carboxylate. So whenever we want to name it, we begin with the name of the cion and that would be potassium. And now for the second structure, well, how do we call that carboxylate? We want to understand that we have a total of 14 carbons and there is non saturation, the chain is linear. So basically the common name for that carboxylate would be laid. Now for the next structure, once again, we are drawing the result on carboxylate with the potassium cion, the ion spectator. So another sold. Now it's different. In this case, we have a straight chain of 16 carbons and that carboxylate will be helmet state. So we are going to name it as potassium imitate for the final product. Once again, let's add potassium as an iron spectator and ensure the result on carboxylate. It has C 27 times followed by ch double bonded to ch, then we have C 25 times. Let's fix our name slightly because we need more space to to this specific structure, right? And now we have C 25 times ch three. So once again, let's begin with potassium and let's observe the nature of the original side chain. It has 16789, 10 carbons. And we can clearly see that starting at carbon number nine, we have a double bond between carbon number nine and sun, right? Carbon number one, then we have seven additional carbon thats eight. And now starting with carbon nine, we have a double bond. So for that specific chain, the common name would be palmetto. So we have potassium Paul May to eight and that's the common name for the final sold. Now, we of course, also want to balance the equation. As we can see there are three iron Spectators. So we're simply adding three K A wage and therefore we have our final answers for this problem. Thank you for watching.

Draw the structures of all products you would obtain by saponification of the following lipid with aqueous KOH. What are the names of the products?

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Key Concepts

Saponification

Lipids

Fatty Acids