00:01
Okay, this problem is asking us to predict the product of each of these reactions.
00:03
Okay, so the way that they formatted this, it's basically my compound with a step one and step two of synthesis.
00:09
Okay, so basically my molecule, and that is going to be reaction with step one and step two.
00:14
Okay, so if we're not used to that format, this is the format that we might be more used to.
00:17
Okay, but without mind, let's go ahead and perform these reactions.
00:22
So first up, i have diethyl -ephtane di -oate.
00:24
Okay, because that is the compound name, i'm going to start off with the very back and then work my way from right to left.
00:29
Okay, so dye -o -8, that means i'm going to have two esters on my heptane.
00:33
So that's going to be two esters on my seven -carbon chain.
00:35
So let's go ahead and write that out.
00:37
Seven carbons, just like that.
00:39
And then on the first and last carbon, i'm going to have the connection to my ester.
00:43
Okay, and i know it's going to be on the first and seventh carbon, simply because if they weren't on those first and seventh carbons, then we would have ketones.
00:51
Okay, so those are the structures of my ester so far, but i need to know what is going to be attached to these oxygens.
00:56
Okay, and that is going to be diethyl.
00:58
So that means i'm going to have.
00:59
Two carbons attached to each of those ether -like oxygens.
01:02
So i'm going to have my ethel and my other ethyl.
01:06
Okay, that is the structure of heptane -dioate, and now let's go ahead and perform my reaction.
01:12
Okay, first up, we have sodium ethoxide.
01:15
So sodium ethyxide is going to behave as a base, and that is going to take off the most acidic hydrogen.
01:20
Okay, so if we analyze this compound, again, we have these esters, we have an alpha carbon here, and we also have an alpha carbon there.
01:26
Okay, each of those are going to have acidic protons, and those are the ones that are going to get deprotonated by my sodium methoxide.
01:32
Okay, but because this molecule is symmetrical, we only have to focus on one of those hydrogens.
01:36
So let's just take off this one, for example.
01:39
Okay, so sodium methoxide is going to take off that proton, meaning i'm going to have to move the electrons from this carbon hydrogen bond onto this carbon.
01:46
Okay, so that will result in the following compound.
01:48
I'm going to have my ester on this side, and then my seven carbons, just like that.
01:55
And then over here on this carbon, we deprotonated it, so it has that nucleophilic carbon.
02:00
Okay, so now that nucleophilic carbon can behave as a nucleophile to attack an electrophilic site.
02:05
Okay, this is going to perform a dekmin condensation, otherwise known as an intra -molecular clicing condensation.
02:10
Okay, and that is because we have this nucleophilic carbon, and we already have an ester on the same molecule.
02:16
Okay, intramolecular condensations are going to happen before even inter -molecular condensations.
02:21
Okay, so the carbons or sorry the electrons from this carbon are going to travel all the way to this carbonyl moving the electrons up to the oxygen okay and that is going to make a one two three four five six member drain okay i always label my nucleophile as carbon number one and then my electrophile the site that the nucleophile attacks is carbon number okay or at least the last number okay so first i'm going to do is write out my six six member drain like this in which this is carbon number one, two, three, four, five, six.
02:53
Okay, so attached to carbon number one right here, we have the connection to this ester.
02:59
So i'm going to attach that directly to my carbon number one.
03:03
Okay, just like that.
03:05
Okay, if we analyze carbon number two, we don't have anything of importance aside from carbon number one and carbon number three, but we already represented that on my cyclic structure.
03:13
Okay, so we don't have to write anything else.
03:15
Okay, and then we do have those hydrogens attached to it, but again, in a skyline, delta structures, we don't have to write that out.
03:20
Okay, karma number three, same thing.
03:22
We just, we just have those hydrogens.
03:24
Four, same thing, five, same thing.
03:26
But then six is going to be where it gets a little bit interesting because we have that oxygen with a now negative charge.
03:32
And then i have the presence of this ether like oxygen with its die ethyl group, or that ethel group.
03:37
So i'm going to have the oxygen with its ethyl group.
03:40
Okay, so again, this is a de -cling condensation, otherwise known as an intramolecular condensation in which we move the electrons down from this oxygen to recreate my double bond.
03:48
And in the process of doing that, i'm going to have to get rid of a leaving group.
03:51
And that leaving group is going to be this ethyxide.
03:54
So the product of that is going to be as follows.
03:57
I'm going to have my six -membered ring like that.
04:01
On this first carbon, i have my ester.
04:04
And then on the carbon right next to it, i'm going to have the presence of my carbonyl.
04:08
Okay, and that carbonyl originated from the movement of electrons down from the oxygen to recreate my double bond.
04:14
Okay, and then we did have to get rid of ethoxide.
04:16
So that is going to be present in the solution as well.
04:19
Okay, so ethoxide, very similar to how it behaved as a base in the first reaction in the first step is going to behave a base as a base in this component of the reaction as well because, again, this is a base and we still have an acidic proton.
04:32
That acidic proton is right here.
04:35
Okay, so using this base, i'm going to take off that proton, simply because that proton is acidic, and that is because it's sandwiched between these two carbonyles, making it particularly more acidic.
04:45
Okay, so after i take off that proton, i'm going to have to have to move the electrons between that carbon hydrogen onto this carbon right there.
04:51
Okay, so that is going to yield this product in which i have my six member drain.
04:56
And it's basically going to be the same thing as before, except for the fact that i have those lone pairs present on this center carbon.
05:04
Okay, so that is the final reaction after step one.
05:07
That's the final product.
05:08
Okay, but again, we don't have just step one.
05:10
We have step two.
05:11
That is going to be hcl.
05:13
So hcl, all it's going to do is simply proteinate this basic carbon.
05:18
So i have h .l.
05:20
I'm going to use these electrons to attack that h, the hydrogen, and protonate that site.
05:26
So my final product is going to be this.
05:33
Okay, where i simply have the hydrogen there, but we don't have to represent that, simply because in skeletal structures, we don't have to show that extra hydrogen.
05:40
Okay, so that will be my final product.
05:42
And now let's move on to this last, or this second one.
05:46
Okay, so for the second one, we have pentanoic acid, reaction with pbr3, br2, and that's going to be followed up with hydrolysis.
05:53
Okay, so let's go ahead.
05:53
And write out our pentanoic acid.
05:56
That's going to be five carbons, and then oic acid corresponds to a carboxylac acid.
06:01
So similar to the ester problem, the diethylophthypth, and diolate, we know that that carboxylac acid is going to be present on the either first or last carbon, simply because if it wasn't, at that point, it wouldn't be a carboxylac acid anymore.
06:13
Okay, so i have my one, two, three, four, five carbons.
06:17
Let's just put it on this last carbon right here.
06:19
So there is my carboxylac acid.
06:22
Okay, that is pentane, or pentanoic acid.
06:24
Okay, so we're going to rack this with pbr3, br2, and that is going to undergo hydrolysis.
06:34
Okay, so these reagents correspond to the hvz mechanism.
06:39
That is h.
06:40
Volhardt zelensky.
06:41
Okay, and what that does is essentially retain my carbonsolk acid, but it places a bromine on this alpha carbon.
06:50
Okay, so the product of that is going to be this.
06:52
Five carbons, carboxylac acid, and then i have my bromine.
06:57
On that position.
06:58
Okay, if you're curious about the mechanism, it's in the textbook, but basically what we do is we create an, um, a carbox, sorry, an asyl bromide.
07:07
And then from that acolyde, we go ahead and brominate this alpha carbon.
07:11
And then after that, the purpose of hydrolysis is to, is to reform my carbacylacilic acid from that, um, acolyde.
07:18
Okay, so that is going to be the final product.
07:21
Okay, next up we have acetone reacting with lda slash thf in which thf is simply just a solvent, and then we're going to have a step one of slow addition of ethyl and a step two of hcl.
07:32
Okay, so let's go ahead and try this out.
07:33
So acetone, i'm going to have this compound, basically just a ketone.
07:38
Okay, and that is going to react with lda.
07:40
So the purpose of having lda as opposed to something like sodium hydroxide, for example, is that lda is a super strong non -nuclilic base.
07:49
Okay, what it's going to do is it's going to deprotonate every single one of these acetones.
07:54
Okay, so again, i have my very strong base of lda.
07:58
That's going to take off this acidic proton, moving the electrons onto that carbon.
08:02
Okay, so sodium hydroxide, it would also deprotonate that alpha carbon, but we're going to have a mixture of products because we're going to have a mixture of deprotonated acetones and a mixture of non -depotanated acetones.
08:14
Okay, and then that would be prone to a potential self -al condensation.
08:18
Okay, and we want to avoid that because we have the addition of these two other compounds.
08:22
So, that base.
08:23
So, that base.
08:23
Is going to deprotonate every single one of these molecules, and that is going to essentially avoid the self -aldo condensation.
08:30
Okay, so i'm going to get this as my product of that, in which i have my deprotonated acetone.
08:37
Okay, so after that, i'm going to have a slow addition of aphal acetate.
08:41
Okay, so ethyl acetate is going to look like this.
08:43
I'm going to have my, i recognize eight as corresponding to an ester.
08:48
So i'm going to write down the basic ester compound.
08:51
Okay, and then acetate, that corresponds to just, just a ch3 on this side, and then the ethyl corresponds to whatever is attached to my oxygen.
08:59
So that's going to be two carbons attached to the oxygen.
09:02
Okay, so that is ethyl acetate.
09:04
And the reason that we add it slowly is because we want to avoid the deprotonation of this carbon.
09:09
Instead, we want to just perform my chlycin condensation in which this nucleophil carbon from my deprotonated acetone attacks directly this carbonyl.
09:17
Okay, and again, we want to avoid the deprotonation of that alpha carbon.
09:21
Okay, so after i attack that electrophilic site of the carbonyl of my ethelacetate, i'm going to have to move the electrons up to this oxygen.
09:28
And that is going to result in this compound, in which i have my acetone.
09:34
And connected to this carbon, which was my nucleophile in the previous step, that is going to be connected to this new carbon, which was the carbeneal in my previously formed ethyl acetate.
09:43
And then connected to that carbon, we have the presence of my oxygen with a negative charge, my ethel group, my ethoxy group, and then my methyl group.
09:52
Okay, so again, this is a clicing condensation in which we're going to have to move the electrons down from the oxygen onto the single bond to make a double bond.
09:58
In the process of doing that, i'm going to have to make my ethoxide into a leaving group.
10:02
Okay, so the product of that will be as follows.
10:04
I'll have my acetone structure, and then on the beta carbon, i have my presence of my other ketone.
10:12
Okay, so that is the structure so far.
10:14
But again, because we had to get rid of ethoxide in that step as a leaving group, that ethoxide is going to behave as a base.
10:19
And that base is going to deprotonate this alpha hydrogen.
10:22
Okay, and it's going to have to move the electrons up to that carbon to make a nucleophilic slash basic site.
10:29
Okay, just like this, lone pairs on that central carbon.
10:34
Okay, so the purpose of hcl is to basically just protonate that site.
10:38
Okay, so we're already done with step one, and we can just go ahead and react it with hcl to protonate that basic carbon.
10:45
Okay, so using hcl, a very strong acid, we're going to protonate that basic carbon.
10:50
And that's going to form my final product of this compound.
10:56
Okay, so that's number three.
10:59
Now let's move on to number four, or part d.
11:02
Okay, so this is diethyl, two ethyl, hexane, dioate, reaction with sodium, ethoxide, and h .c .l.
11:07
Okay, so right here, i recognize that i have dioate.
11:10
That means i'm going to have two esters, and then the hexane corresponds to a six carbon compound.
11:14
The two ethel corresponds to an ethel group on the second carbon, and then the diethel corresponds to the presence of two ethel groups on my oxygens of my etherlyxions of that ester group.
11:24
Okay, so breaking it down again, diowate on my six carbon compound.
11:28
So let's draw out that six carbon compound first.
11:31
One, two, three, four, five, six.
11:32
Here i have my esters on my first and last carbons.
11:39
Okay, that is the basic structure of my hexane dye oate so far, but i still have to add this diethyl group, and i still have to add this two ethyl.
11:45
So that two ethyl is going to correspond to the presence of an ethyl group on the second carbon of this six carbon chain.
11:51
So i have one, two, three, four, five, six, carbons on that second carbon right there, i'm going to have the presence of an ethyl group.
11:59
So one, two carbons...