Let's talk about a solvent that likes to react with carbonyls, and that's water. Water loves reacting with carbonyls to make a molecule called a hydrate. But remember that a hydrate is just a gem-diol. Germinal diol means that they're both attached to the same carbon.
The mechanism is pretty straightforward. What winds up happening is that the lone pairs on the oxygen are attracted to that electrophilic carbon and you get the formation of a tetrahedral intermediate, TI for short. What that's going to give us is a negatively charged oxygen and a positively charged water. Notice that the reason that this O is positively charged is because this is the water that came from here and now it has one extra bond. Then what you get is a proton transfer. This step is called a proton transfer where the O literally just plucks an H off of another part of the tetrahedral intermediate.
If this looks unfamiliar to you, you haven’t done a lot of proton transfers yet, get used to it. A lot of these solvents that attack carbonyls, some of them are going to have proton transfers. It’s something that you should be aware of. This is pretty interesting so as I mentioned before, this means that if you're in a lab and you mix, you have a 50% solution of 2-butanone and that 50%. That means that if the other 50% is water, then it's not just going to be that you have 50% water and 50% 2-butanone. It’s actually going to be that you’re going to have some percentage in there is going to be a hydrate where the water is interacting with the ketone to me a gem-diol. You actually already might have experienced this because in your biology lab, if you guys have taken Bio 1 or 2, and if you’ve ever smelled those nasty animals that they bring out for you to cut open and look at, sometimes you’ll have to maybe cut open an earthworm or a bunny. I don’t know, depending on animal cruelty. Regardless, they're always soaked in what we usually call formaldehyde. Formaldehyde we think has that nasty smell of a dead thing that they're preserving. But actually when you're in lab cutting open that animal, it’s actually not the formaldehyde that you’re smelling. It’s formalin. Formalin is the specific hydrate that’s made from formaldehyde. When it reacts with water, it makes formalin. Formalin is what gives off that smell.
It turns out that you’ve actually already experienced the hydrate in your life possibly or you will. If you take Bio 1 or 2, you’re going to smell these dead animals that are being preserved. That is the smell of formalin which is a hydrate, not the smell of formaldehyde by itself. It turns out that this reaction is not really synthetically useful because the larger the R groups get, the more bulky that tetrahedral intermediate is going to be and the less favored it is. The equilibrium is going to be greatly shifted to the left, the greater, the bigger that the R groups are. As your R groups get bigger and bigger, you’re going to have more and more original carbonyl and less and less hydrate. You can imagine that if you have a ten-carbon chain on both sides, it’s going to be very difficult in terms of sterics to form a hydrate and it’s going to be much easier to keep it as a carbonyl.
The only time that you would actually get a predominant of the hydrate is with an extremely small carbonyl like formaldehyde like we have here, where formaldehyde and water actually gives a majority of formalin but it's because you have the smallest R groups possible which is just Hs. In that case, it’s favored. But if you have larger R groups, then you're going to usually shifts towards the carbonyl in terms of your equilibrium.
Let's do a mechanism. Show the whole mechanism and then predict the equilibrium for the product and then I'll show you the answer.
Example: Show the mechanism, predict the equilibrium2m
So the mechanism was pretty straightforward. You would take your water, your oxygen, you would attack the carbonyl carbon, you're going to get at tetrahedral intermediate that has an O negative and an O H 2 positive with an isopropyl and a phenol on either side then we know we're going to do a proton transfer and that's going to give us our hydrate. O H, O H, benzene and isopropyl.
So not that hard. Now it also asks for equilibrium. Now I noticed that I kind of messed up because I drew a forward arrow, it's not forwards arrow, it's an equilibrium arrow. So let's draw those in. So the equilibrium arrows for both of these steps would be shifted towards you think the right or to the left? What do you think is more favoured the hydrate or the original ketone? So you guys, these R groups are definitely bigger than hydrogen. They're pretty bulky so the equilibrium is going to be greatly shifted to the left and only a tiny bit is going to go forward. In fact it might be on the order of less than one percent hydrate.
So that's why hydrates they're interesting to understand in terms of the theory of solvent attacking carbonyls but synthetically we don't really use these because they're so in favour to form that really you can't really get a stable gem-diol out of it. The gem-diol is going to eventually go back towards being a carbonyl. So anyway that's the end of this reaction. Let's move on to the next topic.
Give the product for the following reaction.
Propose a mechanism for conversion of the dianion to the ketone under mildly acidic conditions.
Rank the following compounds in order of increasing amount of hydrate present at equilibrium.
What is the product of the following reaction?
Draw the expected products for the following reactions in the boxes provided.
Show the mechanism for the formation of the following hydrate in acidic and basic conditions.
The reaction shown is classified as
A) a nucleophilic substitution.
B) an electrophilic substitution.
C) a nucleophilic addition.
D) an electrophilic addition.
The hydration reaction of acetaldehyde is shown. Rank the hydration reactivity of the four carbonyl compounds shown below from the most reactive (1) to least reactive (4).
Which carbonyl compound most prefers to be in the hydrate form?