Vinyl alcohols (alcohols directly on a double bond) undergo a process called tautomerization. Don't worry too much about it because we will devote an entire chapter to this process next semester, so you aren’t expected to fully understand it yet.
For now, just memorize what the enol and keto forms look like, so you can predict the products that form when you add alcohol to an alkyne.
Concept: Vinyl alcohols yield tautomers.4m
Back when we talk about how we could add groups to double bonds, we discussed how there were three different ways to add water to a double bond to make an alcohol. Well, it turns out that if we add water to a triple bond, we still are going to get that alcohol. The thing is that we're going to get a slightly different product. Instead of just having a single bond with an alcohol, we're now going to have a double bond with an alcohol attached to it. Even though that sounds like a very minor difference, that's actually going to translate into a huge difference in the functional group that we get afterwards. Let's go into this right now. This is going to be the hydration of triple bonds.
It turns out that anytime that you make a vinyl alcohol, that's the name of basically having an alcohol directly on a double bond, that is going to react very uniquely. It's not going to react like the addition reactions that we saw with double bonds. In fact, this is going to do a phenomenon called tautomerization. Now this is a phenomenon that we're not going to fully understand the mechanism for until orgo two, so it's kind of unfortunate that we have to talk about it now, but I'm just going to give you guys like a really quick refresher on what this is, so you guys know what tautomerization is.
Basically, if I were to summarize it, I'm not going to teach you the full mechanism because that would be a whole separate lesson, but all you really need to know is that there you're going to reversibly swap the position of a hydrogen and a pi bond. This is what I'm saying, anytime that you make a vinyl alcohol, this is something special. This is not a regular alcohol. This is an alcohol that is now subject to a phenomenon called tautomerization.
Here, I'll show you. Here would be an alcohol that's directly attached to a double bond. This is vinyl alcohol. And through the tautomerization process, that I'm no going to show you the mechanism for, this is going to turn into a completely different functional group, where basically, my double bond is going to move over here, my H is going to move down here, so these are going to switch places. What's going to wind up happening is that you get a carbonyl formed and instead of this being a CH2, now this is going to turn into a CH3.
What winds up happening is that this turns from a vinyl alcohol to a ketone. How did that happen? Like I said, unfortunately, it would take me 20 minutes to explain this whole thing to you, so instead I'm just going to tell you guys to memorize that a double bond and a hydrogen switch places anytime that you have a vinyl alcohol.
Now we do have some fancy words for this because this is its own thing. Basically, when it's in the vinyl alcohol phase, that's called the enol. That makes sense because ene stands for alkene, -ol stands for alcohol. So it's basically whenever you have an alcohol on that alkene, that would be called an enol. Well, the enol rapidly tautomerizes to a keto form. The keto form is just the ketone or the aldehyde that's produced after tautomerization takes place.
Now, what you notice is that I didn't draw these equilibrium arrows evenly. This is a phenomenon that's constantly in equilibrium, but one of the arrows is much bigger than the other. That's because it turns out that the keto form is going to be favored in almost all cases, highly favored over the enol form. What that means is that immediately upon making any vinyl alcohol or most vinyl alcohols, I can expect it to rapidly transform into the keto phase and the keto side of the equilibrium looks like a ketone or an aldehyde.
Basically, the whole gist of what I'm trying to say is that any time that you hydrate a triple bond, you're actually going to get a ketone or an aldehyde as the product. It's through this process of tautomerization. Now exactly which ones do we get? Let's go ahead and look at each specific reagent.
Both acid-catalyzed hydration and oxymercuration-reduction of any alkyne leads to formation of a ketone. These reactions both yield a Markovnikov vinyl alcohol, which then tautomerizes.
Concept: Markovnikov addition of alcohols yields ketones.3m
There is oxymercuration of alkynes and there's hydroboration of alkynes. When we do an oxymercuration of an alkyne, what we're really doing is we're doing it a Markovnikov addition of alcohol. Remember that oxymercuration is the most popular way, one of the most popular ways to add a Markovnikov alcohol to a double bond. Well, the same thing applies for a triple bond as well.
What that means is that if I have two sites, I have let's say the blue site and the red site and I'm trying to figure out where the alcohol is going to go, it's going to go in the more substituted position, so I would expect that after an oxymercuration, I'm going to get an alcohol right here in the more substituted position.
Now notice that I put the oxymerc reagents down here, but I also included HA over H2O, do you guys remember what that was? That's hydration. This would be an acid catalyzed hydration. Just so you know, both of these create Markovnikov additions. Both of them favor the Markovnikov alcohol, so actually I can use both of them. Even though oxymerc is maybe more commonly used, hydration is still a great choice. Both of these reagents really lead to the same intermediate structure which is going to be this enol.
All right? Are you getting that so far? The reason I'm calling it an enol is because now I have a Markovnikov alcohol on a double bond. But what we know that it's not going to stay like that because enols are not stable. They like to tautomerize. After the tautomerization process, what's the product going to look like?
Well, the product is going to be the same ring, but now instead of having a single bond to O, I'm going to get a double bond to O. Instead of having a double bond to the carbon, I'm now going to have a single bond to the carbon. It turns out that the product of oxymercuration or even hydration is going to be ketones. Anytime that I'm Markovnikov hydrating a triple bond, I'm going to get a ketone as the product.
Now what part of this mechanism should you be able to draw? The first part. The second part, you are fine just to say tautomerization, just label it and then draw the product. Like I said, I'm not going to teach you that full mechanism until we get to orgo two. But for right now, you know at least the general idea of what's going on.
Draw the major organic product(s) of the following reactions including stereochemistry when it is appropriate.
Draw the major product of the following reaction. If two organic products are obtained, draw them both.
Show how you would carry out the following conversion. Specify the reagents you would use to carry it out. If two or more ways of conversion to the same product are possible, show only one of them.
Draw the major organic product(s) of the following reactions including stereochemistry when its appropriate.
What alkyne that does not contain O is best used to make the compound shown?
Draw the major product of the following reaction. If two organic product are obtained, draw them both.
Provide the structure of the major organic product that results when 2-butyne is treated with HgSO4/H2SO4 in water.
Draw the structure of the product that is formed when the compound shown below undergoes a reaction with H2O in the presence of HgSO4 and H2SO4.
Predict the minor and major products for the following reaction. Hint the products are isomers and in equilbrium with each other.
When pent-2-yne reacts with mercuric sulfate in dilute sulfuric acid, the product is a mixture of two ketones. Give the structures of these products, and use mechanisms to show how they are formed.
Reaction of an asymmetrical terminal alkyne with mercury sulfate and sulfuric acid in water would produce which of the following?
Predict the product for the reaction below.
What is the final product of the reaction drawn below?
Which molecule will correctly complete the reaction shown?
Which of the following is the enol intermediate in the acid-catalyzed addition of water to propyne?
Which one of the following alkynes gives a single ketone in the acid-catalyzed hydration of each?
For the transformation shown, select the most appropriate reagent(s) to effect the change.
b) H2, Pd
c) 1. Disiamylborane, 2. HO –, H2O, H2O2
d) K2Cr2O7, H+
e) H2SO4, HgSO4
For the reaction shown, which of the compounds below would be expected major, and final, organic product?