00:02
Okay, this problem is having us do the third and fourth steps of the hageman's ester synthesis.
00:06
So in the previous problem, number 53, we did the first and second step, so if you haven't seen that yet, i would recommend going back and making sure that you know exactly how we got to this compound, which i'm about to draw.
00:17
So we have my di -carponeal, and then that has my ester attached to it, all connected to basically the same molecule.
00:26
Okay, so ketone.
00:28
Okay, so we have to recognize that this is going to undergo an aldol condensation reaction, and we know that this is going to undergo an aldol condensation reaction, and we know that we know that.
00:32
That because it stayed in the problem and also because if we look at my hagerens ester, we have a cyclic molecule, we have a six -member drain.
00:40
And the only way to accomplish that is if we perform something like an alvelle condensation.
00:44
Okay, so what i'm going to do is recognize the aldol condensation reagent, which is a base, such as sodium ethoxide.
00:51
So my sodium ethyde is going to accomplish two things.
00:53
It's going to one deprotonate in acidic hydrogen, and two, leave a carbane ion that is going to attack a carbonyl to eventually form my cyclic compound.
01:01
So let's go ahead and analyze the potential hydrogens that are going to be deprotinated.
01:06
So before i do that, i look closely at this molecule and i see that i have this structure and this structure.
01:13
So if i look closely, those are the same thing, only separated by this single carbon.
01:17
So as far as determining which hydrogen i need to deprotate, i know that i only have to analyze one because it's considered basically symmetrical.
01:23
Okay, so let's just analyze the bottom one for now.
01:26
Okay, so analyzing the bottom one, i have an alpha carbon here and i have an alpha carbon there.
01:30
Okay, i don't have any other alpha carbon, so i know that those are the only two hydrogens that i have to consider.
01:36
Okay, so on paper, this hydrogen would probably be considered more acidic.
01:41
And the reason behind that is because it's sandwiched between this carbonyl and that carbonyl.
01:45
So that makes electron density being pulled away from that carbon, resulting in an acidic proton.
01:50
Okay, but if i remember back to aldo condensations from prior problems and previous knowledge, then i would know that in order for an aldoacol conundation to occur, the carbon in which i am deprotonating in my first step must have at least two hydrogens.
02:04
And the reason behind that is, let's just say i deprotonate this hydrogen, or deprotonate the carbonate the carbon, resulting in a carbonyon.
02:09
We're going to result in the nucleophilicity of that carbon, and then that can go ahead and attack a carbonyl.
02:15
But in the condensation component of the reaction, we have to deprotonate that, we have to deprotonate that carbon again so that we can form an alkyne.
02:23
Okay, so as apparent by this carbon, it only has the one hydrogen.
02:26
That means that we cannot proceed with, the further reaction of an outlaw condensation, okay, because we need at least two.
02:32
So if we look at this one, on the left, we have three hydrogens.
02:36
So that is perfect.
02:37
We can actually go ahead and deprotonate one of those right now, okay, because that will result in an alveconensation in the future.
02:42
Okay, so i'm going to go ahead and deprotate a hydrogen, resulting in the nucleophilicity of that carbon, and that results in this product.
02:51
So basically the same thing, the only difference being that instead of my hydrogen there, i have now low impairs.
03:05
Okay, so that's where i'm at right now.
03:07
And of course, i need to get to my six -membered ring, according to my product.
03:12
This carbon nucleophile has a couple options.
03:15
It can, one, attack this carbon, it can attack this carbon, or it can attack that carbon.
03:19
We have to see what the resulting product is going to be in order to decide.
03:24
So if i were to attack this carbon, it would result in a one, two, three, four -membered rain.
03:30
Four membered rings are not ideal.
03:31
One because my hedrimand ester is a six member drain and two because this is this has 90 degree angles that has too much bond angle strain which we know is not ideal so let's not attack that one instead let's see if what would happen if we attack this one so if i attack that carbon it would result in a one two three four five six member drain okay six member drain we're on track right because we have a six member drain in my product but if i attack an ester the carbonyl associated with an ester that will result in a chlycing condensation reaction which we know no is not the same as an aldo condensation product.
04:04
So that would not be ideal, in fact.
04:06
Okay, next up, we have the carboneneal on the left, the top left, which is this one.
04:11
So if my nucleophile attacks that carbon, it would result in a one, two, three, four, five, six member rain, which is ideal.
04:18
One, because it's a six member rain, and two, because it would result in an aldol conundation product, because it's a ketone.
04:23
For aldeacconensation, we would want to have a ketone or aldehyde, so it's perfect.
04:28
Okay, so let's go ahead and draw that out.
04:30
So when i attack that carbon, i'm going to have to move the electrons up to that oxygen.
04:34
Okay, so the resulting product should be as such.
04:37
So i know that i'm making a six remember doing, so i'll go ahead and draw that right now.
04:41
My one, two, three, four, five, six.
04:44
My one, two, three, four, five, six corresponds to the numbers that i wrote on that molecule.
04:48
Okay, so attached to one, i have number two.
04:51
Okay, and then attached to two, i have a carbonyl.
04:53
Three, i have my ester.
04:57
Four, nothing, just hydrogens.
04:59
Five, i have the connection to this.
05:02
Esther, and then six, i have the connection to this auction and then this methyl group.
05:11
Okay, so that is where i'm at right now, and i need to decide what to do next.
05:15
So i see this auction with a negative charge.
05:17
Obviously in organic chemistry, if we don't need to have a negative charge, we shouldn't have that.
05:21
So what i'm going to do is i'm just going to go ahead and deprotonate it using ethanol.
05:25
Or sorry, protonated...