00:01
Okay, this problem is asking us to predict the product of each of these reactions.
00:03
So in this problem, we have a situation in which we're given the names of the compounds, and then we're just asked to give the reactions with these other names.
00:10
So we're not given any structures.
00:12
We just have to come up with those structures ourselves.
00:14
So first up, i have this one three cyclohexid dion that is going to be a cyclohexane that has two ketones on it, in which those ketones slash carboneals are going to be located on carbons number one and carbon number three.
00:26
So i'm going to have my cyclohexane like this, and then i have my two ketones on which.
00:30
Ketones slash carbonyles present in carbon number one, and then two carbons away from that, three.
00:35
So one, two, and three, four, five, six.
00:38
Okay, so that is my one three cycle hexane diode.
00:41
And that is going to undergo a reaction with lda and thf.
00:45
So in this case, lda is a strong base.
00:47
Thf is just the solvent that it's in.
00:49
Okay, so lda is going to depotinate my most acidic hydrogen, which is the hydrogen associated with this alpha carbon.
00:54
Okay, so i'm going to take off that proton using lda, which is a very strong base.
00:58
I'm going to go ahead and take off the hydrogen, moving the electron on this carbon.
01:01
Okay, so after that, i'll get this as my result of that, in which i have my cyclo -hexane, my carbonyl there, carbonyl here, and then my loan pairs on that alpha carbon.
01:11
Okay, and then that is going to be followed up by an addition of an allial bromide.
01:16
So an allel bromide is going to look like this.
01:18
I'm going to have my alkyne, and then over here i'm going to have my carbon, and then my bromine is going to be present, allelic to this alkyne.
01:27
So again, this is an allelic bromide.
01:29
And then this carbon, nucleophile, is going to attack on this position.
01:33
Okay, it's going to attack that position because it is right next to my bromine, which can behave as a leading group, just like that.
01:39
Okay, so the result of that is going to be this, in which i have my cyclohexane, my two carbonyles on carbonyl number one and carbon number three.
01:47
And then connected to this carbon, i have a connection to my former connection to my bromine, which we got rid of that.
01:53
So now we just have a connection to my other carbon, which is connected to my alken.
01:58
Okay, so now that is my structure.
02:00
That is the final product.
02:03
Okay, now we're going to react that with gamma, butyrol actone, which is going to look like this.
02:08
I'm going to have a four carbon compound.
02:10
I'm going to have an alcohol, but the alcohol is going to be attached to my ch3.
02:15
So basically, it's going to look like this in which we have my carbonyl, in which i have that ketone present.
02:23
And then again, we're going to have my alcohol, but it's not going to be an alcohol because it's going to be connected to my gamma carbon.
02:30
So the gamma carbon is going to be the carbon that is three carbons away from a carbonyl.
02:33
So here's alpha.
02:35
Here's beta.
02:36
Here's gamma.
02:38
Right there.
02:40
Gamma.
02:41
And then we have the connection to that auction.
02:43
So just like that.
02:43
That is gamma butyrol actone.
02:46
Okay.
02:46
So now let's go ahead and react that with lda and thf.
02:49
So again, thf is just a solvent for my super strong base of lda.
02:53
Lda is going to take off this acidic proton, the alpha hydrogen.
02:57
Okay, i'm going to accomplish that with my base.
03:01
Okay, and then that base is going to take off that proton, moving the electrons onto this alpha carbon.
03:06
Okay, so the product of that is going to be, as such.
03:09
I'm going to have my ester, my carbon compound, just like this, in which this is alpha, beta, and gamma still.
03:19
Okay, and then on that alpha carbon, we're going to have the presence of those lone pairs.
03:24
Okay, now we're going to have those lone pairs attack.
03:26
Something and that something is going to be my methyl iodide.
03:30
So that is going to look like this.
03:31
We're going to have an iodine attached to a methyl group.
03:34
So just ch3 connection to iodine.
03:37
Okay, so now this carbon nucleophile can go ahead and attack this ch3 and i'm going to have to release this iodine in the process, behaving that iodine as a leaving group.
03:46
Okay, so this is going to be my product in which i have my compound like this and then attached to this carbon, we just added a new carbon because we had a niphylite, attack, an electrophile, and now we're just connected to that ch3.
03:59
Okay, so that is going to be my final product for that one.
04:03
Okay, and next up we have this 2 -7 octane dion reacting with sodium hydroxide.
04:07
So 2 -7 octane dion is going to have a ketone, in fact, two of them.
04:11
That's going to be on my 8 -carbone compound, and then those carbonyles are going to be present on carbon number 2 and carbon number 7.
04:17
So 8 -carbon compound, and then on carbon number 2 and carbon number 7, we're going to have my carbonyles representing in a diome, two ketones.
04:27
Okay, and this is going to react with sodium hydroxide.
04:30
So sodium hydroxide is going to take off an acidic proton.
04:33
Okay, the question is which one? because we have multiple alpha carbons.
04:36
We have one right here, we have one right here, we have one right here, and we have one right here.
04:41
So because these two are equivalent to these two, we only have to really focus on these two, for example.
04:48
Okay, so my sodium hydroxide is going to take off the proton of one of these two carbons.
04:51
The question is which one? so the best way to analyze this is to determine what would be the corresponding products be if are to perform my intra molecular reaction.
05:01
So again, if i take off this proton, for example, i'd have a set of long pairs on this carbon, and that carbon is going to behave as a nucleophile, and we'd attack this carbonyl.
05:10
That would form a one, two, three, four, five memberdroid.
05:15
Okay, pretty good, five memberdroid.
05:16
What about if we're to attack on the other carbon? so if we were to attack this carbon instead, we'd have a set of long pairs on this carbon, and it would look something like this.
05:26
Which would have those lump hairs on that carbon, which can go ahead and attack this carbonyl.
05:30
So that's going to look something like this.
05:31
It's going to be a one, two, three, four, five, six, seven membered ring.
05:36
Okay, so we're comparing the formation of a five member drain to a seven member drain.
05:40
Okay, the one that is more favorable is a five member drain.
05:42
Okay, so this is just a little side note.
05:45
In terms of rein favorability, it goes six is greater than five, which is greater than seven.
05:50
Okay, so every time we want to form a six member drain, we want to, if we have, if we don't have the ability to make a five member green then we would form a seven member green okay so those are the preference levels and now let's go ahead and accomplish that so again my sodium hydroxide is going to take off that most acidic hydrogen which is going to be out for debate but it's going to form the most stable compound which is going to be if i deep protein this hydrogen okay that's going to move the electrons onto that carbon resulting in the formation of this intermediate so eight carbon compound ketone there ketone here and a set of lone pairs on that carbon.
06:25
And again, these lone pairs are going to behave as a nucleophile to attack this carbonyl intra -molecularly.
06:31
Okay, and it's going to form my five carbon compound because we have one, two, three, four, five.
06:36
Okay, so attached to carbon number one in my five member, we're going to have the attachment to this ketone.
06:46
So again, carbon number one is going to have that ketone on it.
06:49
Okay, carbon number two right here is not going to have anything of importance aside from carbon number one in carbon number three, but we already represented that in my five member ring.
06:58
Okay, three, same thing, four same thing, but five is where we have the presence of my auction with a negative charge and also my methyl group.
07:06
Okay, so this right here corresponds to this right here, all attached to carbon number five.
07:10
Okay, so that's where we're at, and we have to get rid of this auction with a negative charge.
07:14
Okay, but we can't just get rid of that auction with a negative charge by moving the electrons down because i would have to have a good leaving group.
07:20
So that would be corresponding to a chlycin condensation.
07:22
Instead, we're going to have the protonation of the with a water.
07:27
Okay, so we formed water by the protonation of my sodium hydroxide.
07:31
So we should be making it an alcohol from that.
07:34
So again, we have a five member drain like that.
07:37
We have my ketone up there, my methyl group, and my alcohol over here.
07:43
Okay, so the right next step is determining is that my final product.
07:46
So this is the aldol addition product, but generally aldol addition products are not the most favorable thing.
07:51
Okay, the equilibrium would actually shift back towards my reactants.
07:55
But in terms of continuing this reaction, if i are to go even further than this and go to my condensation reaction, it's going to want to prefer that condensation reaction because we would have the resulting alpha beta and saturated ketone, which is particularly favorable.
08:10
Okay, so we're going to have the presence of my hydroxide due to the deprotonation of water in the previous step.
08:16
So that sodium hydroxide is going to take off the most aesthetic proton...