00:01
Okay, this problem is asking us to perform a robinson annulation on two -methyl -one -3 -cyclopentane dione and three -button 2 -oan.
00:06
So this is my two -methyl -1 -3 -cyclopentane dione, in which i have a carbon on the top, carbon on the bottom, and then a methyl group attached to a 5 -carbon ring.
00:15
Okay, and then here's my 3 -bu -2 -own, like that.
00:18
Okay, so robinson -annulation is essentially the base -catalyzed deprotonation of the most acidic hydrogen, and then we're going to perform a michael edition, and then eventually end off with an aldol condensation.
00:30
So what i'm going to do is i'm going to react this, both of these compounds, with sodium hydroxide, or at least have them both in the same environment and then have sodium hydroxide react with them.
00:41
Okay, so sodium hydroxide, we know the super basic base, is going to deprotate the most acidic hydrogen on one of these two molecules.
00:48
So we have a couple options for the most acidic hydrogen.
00:50
We have the hydrogen here.
00:52
The hydrogen is attached to the alpha carbon of this ketone.
00:56
So that is an option.
00:57
We also have one right here, and on the symmetrical carbonyl, we have this one.
01:02
And we also have this one, this hydrogen.
01:05
So that hydrogen, which i just circled, that hydrogen is attached to, one, the alpha carbon, and two, it is sandwiched between these two carbonyles.
01:12
So those two carbonyles are pulling electron density away from that carbon through resonance and the inductive effect.
01:18
Okay, so that's why this hydrogen is considered the most acidic hydrogen on both of these molecules.
01:24
Okay, so that is the proton that's going to get deprotonated.
01:27
In a solution with sodium hydroxide.
01:30
So sodium hydroxide, deprotonate that base.
01:32
So i'll just demonstrate that right here.
01:35
Like that, move the electrons onto that carbon.
01:37
Okay, and then we should end up with the following compound.
01:41
Make sure it has five carbons.
01:45
Okay, and then one pairs there, and then make sure we have our methyl group there, and then finally my carbonyles.
01:53
Okay, now we're going to perform a michael edition.
01:56
So my michael edition attaches my nucleophilic, carbon to the beta position of my alpha beta unsaturated ketone.
02:05
So what i'm going to demonstrate is this nucleophile right here.
02:10
That is a nucleophile because it has low in pairs.
02:12
It is considered a nucleophilic carbon.
02:14
It's going to attach or attack the beta carbon.
02:18
And the reason it attacks this carbon, instead of the base of my carbonyl like we might be used to, is because obviously we're used to attacking this carbonyl right there and usually a ketone or albiide.
02:28
But we have this oxygen, consider an electronegative.
02:33
The electrons are going to move up to the oxygen.
02:35
If i move the electrons up to the auction, the auction will be considered negatively charged, and then we'll have a positively charged carbon right there.
02:41
That positively charged carbon can be, i guess, de -localized by the movement of the electrons from this double bond onto that single bond to create a double bond, therefore moving the positive charge to this primary carbon.
02:52
Okay, so that is why we attack the beta carbon.
02:55
Okay, so next step, we're going to perform that.
02:57
So let's just draw that out.
02:59
Once i move the electrons onto this carbon, i'm going to have to relieve that carbon of its exceeded octet by the movement of electrons from this double bond onto this carbon.
03:11
Okay, so i've stated this in some other videos, but we might be used to seeing, at least in other mechanisms, the electrons moving onto the single bond, and then the eventual movement of electrons into the oxygen.
03:21
What we're going to see is that they eventually cycle back and then move onto the same carbon.
03:26
Okay, so just the demonstration of one arrow, we'll just do, we'll be fine, okay, because we're going to end up with electrons on that carbon no matter what.
03:35
Okay, so i'm going to draw out the product of that, which is as follows.
03:40
I have my carbonyles, which are unaffected, and then on this carbon, i have my methyl group, and then i also have the attachment to my alpha beta unsaturated ketone.
03:51
So this carbon right here, that carbon is this carbon, and then it is attached to this carbon.
03:58
So let's draw that out.
04:00
I have attached to this carbon, which is this one, is my former double bond.
04:05
So my former double bond, which is now a single bond, that single bond has lone pairs, or at least that carbon does.
04:11
Okay, and then is attached to a carbonyl like this, and then attached to that is my methyl group.
04:18
So what we're going to do is protonate that negative charge.
04:23
We're going to protonate that carbon.
04:24
So i'm going to use the water that i had formed by the protonation of sodium hydroxide in the previous couple of steps like that and then we should end up with this in which i have my carbonyles unaffected on the top and the bottom my methyl group unaffected and then we just created a chain like that okay so this is where things get interesting in terms of the whole entire robinson annulation because first off we've just done michael edition but then this of robinson annulation ends off with albal constation although intramolecular aldol constation so what have we just created? we just created a base, right? by the deprotonation water, we just created a base.
05:08
So right now we're in technically basic conditions.
05:11
In basic conditions, we're still going to deprotonate the most acidic hydrogen.
05:16
If we look closely at my molecule, this carbon, which we might have thought would be containing the most acidic hydrogen, has so far four bonds, right? it has the attachment to this methyl, it has the attachment to this carbon chain, et cetera, right? there are no acidic hydrogen there.
05:32
But we also have an acidic hydrogen here.
05:34
That hydrogen is attached to a ketone.
05:38
It's on the alpha carbon.
05:39
So it's considered acidic still.
05:40
So i'm going to deprotonate that and then move the electrons onto that carbon.
05:46
Once i perform that, i should get the same molecule just with deprotonated carbon.
05:52
So let's just highlight that to make things easier and then just draw lone pairs on that essential carbon.
05:58
On that carbon.
05:59
Okay...