00:01
In this problem, we want to draw out the structures of glycerol -tripalmatate as well as glyceroliate, and then we want to determine which one of these structures should have a higher melting point.
00:15
If we consider the root words in each one of these final structures, the glycerol part should indicate that we are starting with glycerol, which is shown on the top, and for tripalmitate, we should recognize that that is derived, from palmitic acid, which is shown below.
00:35
And then for trioliate, we should know that that corresponds to the 18 carbon structure of oleic acid shown here.
00:43
And for each case, the tri means that we have three of either palmitic acid or oleic acid reacting at each one of the oh groups in glycerol, since there are three of them.
00:58
And for each one of these reactions, the three, the molecules of either palmitic acid or oleic acid are going to form condensation reactions and therefore esters with each one of those oh groups from glycerol at their respective oh groups from the carboxylic acid.
01:19
Again the condensation reaction means that water is going to leave as an ester forms at each one of those sites.
01:29
So we all have a total of three esters at each one of those.
01:33
So we can start with the structure for glyceryl tripalotate.
01:56
We can draw out the structure of glycerol using these carbons.
02:03
These two are attached to ch2.
02:06
This one is attached to ch, and they are all directly bonded to an o.
02:12
After the ester forms, all of these hydrogens will leave to form water.
02:18
And then the bond will form with this end of each carboxylic acid.
02:29
And so we have to just draw that out and then draw it three times at each one of those sites for the oh groups from glycerol, and then draw the condensed structure of the remaining carbon chain.
02:44
So again, considering that we can draw glycerol, in that backbone structure, we can have ch2 bonded to a ch, bonded to a ch, bonded to a ch2, and each one of these are bonded to an o as part of that ester, and then the c -double bond o, and then since this is palmitic acid that was used to form these three esters, we can count out to the left of this carbon from the carboxylic acid, how many ch2 groups? 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, and then attached to this terminal methyl group.
03:54
So for each case, it's just ch2 14 times and a methyl group.
04:05
And it's tri -palmitate, so we just do that three times.
04:20
And that is the final structure for glycerol -tri -palmatate.
04:27
And now for glyceroliate, we do the same thing, but with the slate nuances to the structure of oleic acid compared to palmitic acid...