00:01
Okay, this problem is asking us to predict the products of these reactions if we were to react each of these with h2s -o -4 and heat so the purpose of specifying that there's heat in these reactions is because in reality if we didn't have heat then we could have a mixture of substitution and elimination products but by specifying heat we know that we're going to prefer to perform the elimination product so that is going to be our major product in each of these okay, so knowing that i'm going to react this alcohol with h2s .04 and that's a site a side note is that this is a tertiary alcohol, and it's a tertiary alcohol because the alcohol is right there.
00:35
Here's my carbon that that alcohol is attached to, and that carbon is connected to one, two, three other carbons.
00:41
Okay, so for each of these, i would recommend identifying what alcohol you have, whether it's primary, secondary, or tertiary, and then proceeding with the reaction, because then you can kind of figure out whether you can do e1 or e2.
00:53
So because i do have a tertiary alcohol on this one, i know i'm gonna prefer to do e1, and that is because otherwise, i would have too much steric hindrance to proceed with an e2 reaction.
01:05
Okay, so knowing that i have h2so4, which is a super strong acid, i'm going to protonate my most basic compound or most basic atom on this molecule, which is my oxygen.
01:13
So it's going to form oh2, just water connected to my carbon.
01:16
And because this is an e1 reaction, that means i'm going to have a unimolecary transition state.
01:20
So that means i'm going to have two sequential steps, not simultaneous, just two different steps.
01:26
So i'm going to result in the expulsion of water.
01:28
I'm going to have this as my intermediate in which i have basically the same exact compound.
01:34
The only difference being that i have the loss of water and a positive charge now on this carbon.
01:41
Okay, so an e2 reaction is where a leaving group leaves, hence this water, and then a base, such as water, because i just had that leave, a base comes in and deprotonates a hydrogen to eventually make an alkyen.
01:53
Okay, so that water has a couple of different options.
01:56
It wants to relieve, the whole purpose of making an alkyne is to relieve this carbon of its positive charge.
02:03
So in order to relieve that carbon of its positive charge, i must deprotonate a hydrogen which is adjacent to this carbon.
02:10
So what i mean by that is this hydrogen, if i are to deprotonate that, it would make an alken right there.
02:14
If i are to deprotonate this hydrogen, it would make an alken right there.
02:17
And if i were to deprotonate this hydrogen, it would make an alken right there, all representing the relief of this positive charge.
02:25
Okay, so i can't deprotonate this hydrogen or this one or this one because they're just too far away.
02:30
It would have nothing to do with relieving that carbon of its positive charge.
02:33
Okay, so knowing all that information, i'm going to go ahead and do this step by step, or very methodically.
02:40
So first things first, i'm going to deprotonate that hydrogen, resulting in the loss of a hydrogen, and i'm going to move the electrons from that carbon hydrogen bond onto the sigma bond to create an alken.
02:51
Okay, and when i do that, it'll make an alkenate right here.
02:54
Okay, so this is considered to be a dye -substitited alkene.
02:58
We know it's a die -substit -elene because if we're looking at what's only connected to my al -keen, we have those two carbons, and we have to identify what is connected to those two carbons.
03:07
So we have connected to this carbon, we have two hydrogens, which aren't considered in the naming of the substituents on my al -keen, but we do consider the carbons.
03:17
So i have this carbon and this carbon immediately connected to that carbon.
03:20
So that means that i have a die -substitedededalcine.
03:23
So die -substited alkenes are relatively stable, but obviously we would want it to be as stable as possible, which would be a tetras -substit al -keen in which we have four connections.
03:32
So i would have two connections to this carbon, and then two connections to this carbon, which i do have.
03:37
Okay, so knowing all that information, i'm going to go on to the next one, because this one isn't particularly favorable, right? i want to be very favorable to make a dye -substited al -keen.
03:47
Okay, so next thing, i'm going to deprotinate the left hydrogen, which is this one.
03:51
If i depronate that hydrogen, i'm going to result in an alken right there.
03:55
Okay, so that is considered to be a tri -substuted alken.
03:59
And it's a tri -substited alken because here's my component of the alkeen, my two carbons, and connected to this carbon, i have one carbon, and then connected to this carbon, i have one, two carbons.
04:09
So that's two plus one equals three.
04:11
That's a tri -substated alkeen.
04:13
Okay, that's pretty stable, but again, we want to make sure that we can, we have, at least we exhaust all our options.
04:20
So knowing that i have a tricepsiated alkene there, i'm going to move on to the next one.
04:24
Okay, and that would be the loss of this proton via h2o.
04:29
Okay, if i lose that proton, i'm going to form the double bond right there.
04:36
So i'll make an alkyne right there, and will this be? this one would be a tetra -substit alkyne because connected to this carbon, i have one, two carbons, and connected to this carbon, i have one -two carbons.
04:48
So that is a tetras substitute alpine that is considered to be my most stable.
04:52
Okay, so i'm going to perform this as my major product.
04:56
So c, c, h3, and then this is a double bond now.
05:01
And then this carbon is connected to not a hydrogen anymore because i got rid of that one.
05:05
So it's connected to ch3 and ch3.
05:08
Okay, so this is my most stable product, and it's my major product.
05:13
Okay, next step, we'll do this one.
05:16
So this is an alcohol, and the same thing as always, we're going to go ahead and proteinate that oxygen.
05:21
So it's going to result in oh2 with a positive charge, all connected to that carbon.
05:25
Okay, so again, this is either going to behave in an e1 or e2 reaction.
05:29
We're going to do e1 because with secondary, you have a choice between e1 and e2, it's going to preferentially do e1.
05:35
So i'm going to result in the loss of that water.
05:39
I'm going to have this as my product, or at least my intermediate.
05:43
So i have that alkyne unaffected, and i have that positive carbocatin right there.
05:47
Okay, so now i have a choice.
05:48
I have a choice.
05:49
I have have to relieve that carbon of its positive charge, so i have to deprotonate one of these two hydrogens, right? because those are connected to the carbons, which are immediately connected to that carbon.
05:59
So the hydrogen right there, the hydrogen right there, those are going to get deprotonated by water, but only one of them is.
06:05
And the question is, which one? so if i are to deprotonate that one, it would result in an alken right there.
06:10
So this would be the product of that one, so that, and then if i are to deprotonate this one, the one on the bottom, it would result in an alkenet right there.
06:21
So i would get this product.
06:25
Okay, so which one is considered to be the major product? it's going to be this one.
06:30
And that's simply because in this one we form a conjugated system, which it's not necessarily important for this particular problem, but a conjugated system is considered to be more stable than not having a conjugate system.
06:41
Okay, so that's just basically alternating sigma bonds and pi bonds, which is pretty stable.
06:47
Okay, so next up, we have this one.
06:49
So this one, this is, oops, i have a two on that a hundred.
06:52
Okay, so this one, is a secondary alcohol and that's because the carbon in which my alcohol is attached to is connected to one two other carbons.
07:00
Okay so that's a secondary alcohol so i know i'm going to be focusing primarily on an e1 type of reaction.
07:06
Okay so first things first i'm going to go ahead and proteinate that alcohol forming oh2 with a positive charge.
07:13
Okay and again i'm going to make this alcohol slash water leave resulting in the formation of this product or intermediate.
07:21
So duplicate that move it down and i'm going to have a positive on that carbon that my water just left off of.
07:28
Okay, so there's my positive charge.
07:29
And in the previous ones, we had our positive charges, and then we just immediately had our corresponding bases deprotonate the neighboring hydrogen so we can form an alkyne.
07:39
But in this one, we have a secondary carbocadine, and that's not bad, but we have the potential to do even better.
07:46
So similar to some previous problems in which we moved our carbon carbon carbon carbon carbon carbon out around so that we can form a more stable, um, carbonatein intermediate.
07:53
We're going to do the same exact thing.
07:54
Here except we're going to do it in an e1 type of reaction.
07:57
So in the previous ones we did it with s in 1, now we're just doing the same thing except with an e1 reaction.
08:03
Okay, so i have my secondary carbon carbon it has the potential to be even better and i'm going to do that by moving this methyl group onto that carbon.
08:10
So by doing that, i will relieve this carbon of its positive charge and i'll move my positive charge onto this carbon and that carbon because it's connected to one, two, three other carbons, that's going to be a tertiary carbonide on as opposed to a secondary carbonate ion.
08:23
Okay, so let's go ahead and do that.
08:25
I'm going to move this methyl group onto that carbon resulting in this.
08:29
So ch3, ch2, ch, which is now connected to a methyl group.
08:36
And then i have the connection to that carbon, which is connected to ch3 and ch3.
08:42
Okay, so again, this is going to have a positive charge now because i moved electrons away from that carbon and onto this one.
08:50
Okay, so again, this is going to behave in an e1 type of reaction.
08:54
So i need to focus on the hydrogens which are surrounding this carbon.
08:59
Okay.
08:59
And in this case, it's going to be either this one, this one, or this one.
09:04
Okay, and if i were to deprotonate this one, it'll be the same thing as depronate that one.
09:07
So i basically have two options.
09:09
Okay, so if i were to deprotonate this hydrogen, it would result in the formation of an alkeen right here.
09:17
Okay, that would be considered to be a dissubstited alkeen.
09:21
That's because if i'm only focusing on what's my alkeen, i have the connection to one, two other carbons on this carbon and no carbon of that one.
09:28
So that's a die -substited alkene.
09:30
That's not particularly stable.
09:32
We have the potential to be even better than that...