00:01
Okay, this problem deals with the reformatky reaction in which we have a zinc bromide attached to this alpha carbon, and then the alpha carbon is going to behave as a nucleophile to attack either an aldehyde or a ketone or a ketone or a ketone or a ketone or a ketone or a ketone or a bit.
00:12
It can attack virtually any carbonyl.
00:15
But with zinc bromide, it's going to make this carbonyl, or specifically this alpha chlorides, it's going to make this carbonyl, or specifically this alpha -chlorids, esters, ketones, alvehydes, it can attack virtually any carbonyl.
00:26
But with zinc bromide, it's going to make this carbonyl, or specifically this alpha carbon, less reactive.
00:32
It's going to make it less nucleophilic.
00:33
So this is not going to perform a self -closin connoissation, for example.
00:37
Okay, and that is simply because this carbon is not as nucleophilic as it would be to have a self -closin connoissellation.
00:44
Okay, so with that in mind, let's go ahead and proceed with these reactions in terms of breaking them down.
00:48
So for this one, i have my ester, and then i have this connection to my alcohol.
00:54
Okay, so with reformatucky reactions, we have to start off with my ester.
00:59
So i'm going to have my och3 on this side and then my carbonyl on this side.
01:02
So that is what i'm going to start off with.
01:04
Okay, but the question is, what am i going to be attached to? so as i know, attached to the alpha carbon, i'm going to have the connection to my zinc bromide.
01:11
Okay, the zinc bromide allows my carbon, my alpha carbon to be nucleophilic.
01:14
Okay, that nucleophilic carbon can attack virtually a, well, at least a reactive ketone or alvehyde.
01:21
So it's going to either attack a ketone or an alvehyde in this case.
01:23
And in this case, we have the connection to one, two, three, four carbons attached to this alpha carbon.
01:29
So that's four carbons, meaning that i'm going to have a one, two, three, four carbon compound, and that is going to be an aldehyde.
01:36
Okay, so basically this carbon nucleophile attacks this carbonyl, moving the electrons up to the oxygen, and then that oxygen is going to have a negative charge, but it's eventually going to get protonated by water in this solution.
01:47
Okay, so those are the components of this reaction.
01:50
Okay, what about this one? so this one, as we can see, we have a carboxylic acid.
01:55
Okay, with reformatomy reactions, we have to have an ester.
01:58
Okay, and that is depicted in the first reaction and the example given above.
02:02
Okay, so we have to have that carboxylic acid originate as an ester.
02:07
Okay, so i'm just going to transform this into, let's just say, anethoxide attached to that carbonyl.
02:12
And then again, we're going to have the carbon.
02:14
That alpha carbon is going to be connected to zinc bromide.
02:17
So zinc bromide.
02:18
Okay, in addition to that zinc bromide, we're also going to have the connection to my ch2, ch3.
02:23
And we also have a connection to these three carbons over here, but these three carbons are connected to my alcohol.
02:28
So i don't have to worry about that yet in terms of this singular molecule.
02:32
Okay, so again, we have the connection to my ch2, ch3 in addition to that zinc bromide on that alpha carbon.
02:38
Okay, so that is that molecule, and we're going to rack that with a one, two, three carbon compound in which i have an albehyde.
02:44
So one, two, three carbons, aldehyde.
02:47
Okay, this nuclear fluid carbon can go ahead and attack that carbonyl, moving the electrons up to the, the oxygen, which is going to make an oxygen with a negative charge, but that is eventually going to get protonated by water.
02:57
Okay, and that is going to form almost my compound, but again, we want to hydrolyze that ester into forming my carboxic acid.
03:04
So it's going to undergo this reaction right here between these two compounds, and then it's going to react with acid water and heat in order to form my desired compound.
03:13
Okay, next step, we have this one.
03:14
So this one is going to be a little bit different than the ones that we've previously analyzed in the fact that i have my carbocosok acid, and i also have this alpha beta unsaturated system.
03:23
So that is characterized by the presence of my alken between my alpha and beta carbon...