0:00
Hello everyone.
00:01
Today we're doing chapter 29 problem 40 and this problem asks us to draw the organic product form amino acid lucine is treated with each of these following reagents.
00:10
So if we draw lucine remember lucine one letter code is l you draw lucine 1, 2, 3, 4 bonds or 4 carbons i should say so if we were to take lucine and react it with the reagent in a we see that we have an alcohol group and a pro -native group so we can see that we're gonna have a acid catalyzed esterification reaction occurring.
00:43
So if i use, if i see what we have in lucene, well, what can react with a proton, well, we can see this oxygen can be protonated, and then we can have acid catalyze esterification, replacing this oh group, falling off as water, and being replaced with this methanol group here.
01:02
So the product of a would just be ester formation, and methyl one carbon methyl ester formation for a.
01:21
And if we go to b, we see that we're actually adding an acid chloride with a strong base or a good base.
01:30
So we can see that we need to have a nucleophile to react with this electrophile that we presented here.
01:38
And the base, what's the reason that we have a base here? well, the base is going to make our nucleophile into a better nucleophile.
01:44
So it's going to make it more electron -rich.
01:46
So you can imagine that this periodine is going to rip off a proton from a potential nucleophile, electron -rich nitrogen.
01:53
So it's going to make this into an n -minus, or depending on the ph solution, it's a neutral n, which is going to be more acidic than a positively charged nitrogen.
02:01
And now this strong nucleophile can go through an s -n2 mechanism to attack this acid chloride carbonal carbon, to kick off your good leaving group green chlorine and yield an enacetylated version of leucine.
02:17
So now if i draw the product of b, we see that now we have an enacetylated lucine coming off the nitrogen of the backbone.
02:39
In c, we see that this is the same protecting group, the phenol -protecting group for carboxylic acids in the synthesis of amino acid problems that we've done in the past in this chapter.
02:51
So we see that what we're doing is a carbonylic acid protecting reaction here by addition of this phenol group.
03:10
And in c, and now in d, we are adding an in hydride in the acetylchinhid group.
03:18
And once again, it's going to act as an electrophile.
03:20
Remember, just like in b when we reacted this loosing with electrophile, we added puridine to make the nucal file on losing a stronger nuky.
03:30
So it could attack this electrophile of the reagent d to give us some sort of product.
03:36
So here we see that we're going to have an acetyl nitrogen here, protecting this nitrogen to make sure that this carburexylg acid stays highly reactive while this, while deactivating the nitrogen of the backbone.
04:06
Now, at e, we're adding one equivalent of hcl.
04:10
So what i call it hl is going to protonate ionizable group.
04:14
So everything here is already pronated except this nitrogen.
04:16
So this nitrogen is the only thing that can be protonated here.
04:19
So we're going to protonate nitrogen because it acts as a base.
04:23
It's happy to be protonated and pick up that extra proton to give us product e.
04:38
Now product f, we're adding a strong base.
04:42
So now we're going to deprotonate something that is easily deprotonated.
04:46
So if i were to decrease or decrease amount, of protons in a solution or increase the ph, which one of these groups would lose its proton first? well, the one with the lowest pca.
04:58
So if i were to raise a ph of the solution, anything above two or three, then the first thing that's going to be depotanate is this carboxylic acid.
05:07
So we're going to depotinate this carbacilic acid while leaving this amine alone, like how it is in the starting material in the question stem.
05:16
So now we depotanate this carbonyl this carbonylite as carbonylac acid while not touching this.
05:20
Already deprotonated amine.
05:25
In f, now we have a benzolic acid group, an acid chloride off a benzene ring with the same periodine reagent.
05:36
So we know that this is going to be an electrophile reacting with the nitrogen nucleophile off of lucine, and we know that there's going to be some sort of protection of this backbone nitrogen occurring, just like in a pre -sover to draw the backbone and draw lucian.
05:57
Now if i just imagine nitrogen is going to attack the carbonal carbon and eventually kick off the chlorine leaving group, then we'll be left with this protecting group right here, being this spinoic acid group 4g.
06:17
And now in h, we see that we're adding bach and hydride, so we're adding boch and hydride and the base triathlon amine.
06:31
So once again, this is another aiming -protecting reagent that we use in the synthesis of dipeptides or peptides when you start with just amino acid building blocks.
06:41
So in h, now the nitrogen here, the nitrogen is now going to be bach protected, rendering it inactive and making this carbacilic acid only reactive functional group in this amino acid.
07:03
And now if we see an i, well, i says use a product in d and d.
07:06
Then add this functional group here.
07:11
Let's see what d is.
07:12
So d we deactivated this amine group, so we're adding dcc and this pepta or this amino acid.
07:21
If i ever draw this amino acid out, let's just see how it looks like, because we're using it for inj...