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Describe the chemical reaction that takes place in thesaponification process. What are the reactants and products?...

Question

Describe the chemical reaction that takes place in thesaponification process. What are the reactants and products?

Describe the chemical reaction that takes place in the saponification process. What are the reactants and products?



Answers

What is saponification? What are the products formed in th reaction?

Your soul, CEPAL Unification is the process is the Kruse's oh formation off soap by the acting A strong these, like in your which leaned shy Chris Reid. So this is the reaction that occurs. Um, so the common reaction that are closes you have Ah, chai. Let's ride. So can you. Having any? Which. And this gives you any clothes. So negative. Plus it lists roll. Okay, on since it's try, which means it has three are groups, so this would be three like the simplest role is already there. So the common reaction is ch teau so well, first right, the structure off Bliss Roy, Let's have any are and then it has C h or C who are dash C H 20 C. The born who are double dash plus any which this gifts C h tool. Oh, and h c it poet ch teau poet. Plus you get oh, c o n e our Nash si o n e our number last CEO and e. Okay, so this is this role and this is the These are the structures off soup. And this would also have a partial positive and a partial negative charge. OK, so this is how you get it. So this would be tree? No. Okay,

In this problem, we need to write out the reaction of the pontification of a zero die, only eight mono commentate, and we need to react it with an a o. H. We can start by drawing out the structure of glass. Zero Dy Oli eight Mono pal Meditate using the structures of visceral Dominic acid and Oleic acid we know from the name that die corresponds to two equivalents of Oleic acid. Reacting at two of the O. H sites from glycerol in the remaining O. H from glycerol is where we formed a bond with mono, one equivalent of paramedic acid, and that would correspond to imitate and only eight in each case. Because after we form those bonds with the O. H groups from the carb oxalic acid portion of on the nick and Oleic acid, the condensation reaction removes water and that ball that bond forms and therefore makes this the structures of each one of these acids, an anti on which corresponds to only eight and pal imitate in each case. So when we begin by forming those bonds at those three sites on cholesterol, we can draw out visceral in the way in the back bone structure of having ch two group bonded to a CH group on it to another CH two group. And then all of those air bonded to and Oh, and the H has been removed as part of water and condensation reaction. And it's joined with the other each that that was removed, that the remaining oxygen is where that bond forms. So we can begin by drawing out the structure of the Ciro Di Oli eight mono pal imitate in that way, starting with the backbone structure of visceral, we have ch two ch ch two and now each one of these are bonded to. And oh, and then a carbon Neil see double bond. Oh, and then two of these are going to have the only eight, and I on and the other one will have the commentate and ion attached to the rest of that portion of the glycerol backbone. After that, condensation reaction occurs to form this molecule that we're going to react with nao each in the PSA pontification reaction. So starting with two oly aids moving to the left of where this bond forms from this carbon, Neil, see that we have 12 34 567 ch. Two groups. So we can begin by drawing that condensed structure of ch 27 And then we have a C H double bonded to another ch. So we can include that ch double bond ch. And then we have after that end of that double bond, we have 1234567 more ch two groups and then a terminal methyl group. So ch two seven ch three and remember that it's die only eight. So we have to of those structures on the ends of this molecule. So again, we draw this seem condensed formula or to only eighths. And then we have on the remaining site a mono So one pal imitate ion. So going to the left of where that bond form from the curb. Neil, we have one, 23456789 10 11 12 13 14 ch. Two groups attached to a terminal methyl group. So the mono pal imitate portion, of course, zero DiIulio mono pal imitate can be drawn a ch two 14 ch three And now we react this with three equivalents of any O h in the electoral negative oxygen is going to attack that Kurban Neil Carbon and the electrons will go up to this oxygen which will then kick out the leaving group, which are the and ions in each case. So when that happens, we form. We reform that alcohol, you regenerate glycerol, which is what we started with. And then we will also have to of the only eight and ions and one of the palm it eight and ions that air booth in a combined with the positive charge from the n A plus cat ion. So we can write that all out because this so each group we know we'll attack that carbon Neil to form that bond and so that will produce cholesterol. So that's one of the products so glycerol. And we can draw that structure as ch two o h ch. I wanted to know each and another ch two wanted to an o. H. So that's one of the products foreign from this reaction. And then after this leaving group of this only eight and I on leaves with a minus one charge, it will pair up with the positive one charge from the sodium ion to form that compound twice since we have two of these old yate and ions present in the starting structure. So we have to sodium only eight. And so we can draw out what one of those would look like with a coefficient of two. We could say we have two of these on pounds where we have ch three c h 27 c h double bond, ch C H 27 and then we were going to have a c 02 because it is in an ion, in this case, over on this. And now, since this is a leaving group and that hydrogen is gone, So this is a deep resonated and I, on form of only eight, with a minus charge combined with sodium with a positive charge to make this too sodium only eight. And so those are three of the products that are formed, and the last one will be that one, that one equivalent of sodium imitate so we can draw that out. And again, we just have one of these kids mono pal imitate in the original structure, and you see it's ch three ch 2 14 And then again, C o to minus is again. That is the anti on that's pal imitate. We dio from the rate to the left we we end up with to Oxygen's and that one carbon corresponding to that co two minus and I on or PAL imitate. And then again, that's combined with an A plus to make this a sodium pal imitate compound, and so those are the products of this reaction.

During the chemical reaction. Bonds are broken, new bonds are created and energy is exchanged. The new bonds that created that result in the formation of a new compound.

All right. So this question is asking us to draw the reaction between glycerol and steer acid. Now, cholesterol is a three carbon compound with 308 trips attached to it and steer Kassid is a saturated fatty acid. Now again, a saturated fatty acid has our carb oxalic acid on sugar. But the end so the O. H connects to the sea Double Barnet. Oh, and then it has repeating C H two units, followed by a C H three. Now, in the cases to uric acid, it has an 18 carbon chain tail. So this end first year acid is going to be 17. And then, if you include the seizure ondas your 18th carbon tail. Now, if we want to react arse steering acid with our cholesterol to give bliss a rel, try steering. In order to do that, one of these fatty acids chains is going to attached to one of these apartments. So we want to make a try. Siri, we need to have three fatty acid chains. No, What happens is this hydroxy over break here of charges with one of these hydrogen ins to produce water. And when that happens, a bond forms between this extra oxygen right here in this carbon rate here. So you go ahead and react. Are clitoral in our Starik asset? We're going to get something that looks like this with our oxygen are free oxygen from our saturated fatty acid, which is still attached to that. See double bonnet. Oh, and then has those repeating units to make our steering acid. Now, that was just one reaction. So we need to make sure we do it two more times to get was there. I'll try Steer E And for each one of our body acid chains that we attached Work, literal. We're going to produce one molecule of water. So this is what our tryst Eerie. Closer. I'll try. Steer e product looks like right here. And then we have to add our three water molecules. So again, here's our cholesterol. We're reacting with our steer. IC acid to produce Was there all try steer e plus water


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