Hemiacetal

In this next page, we’re going to discuss one of the products that happens when a neutral alcohol attacks a carbonyl and that's called hemiacetals. Let's just start off with one big disclaimer. That disclaimer is that technically the word acetal is used to describe the product of an alcohol and an aldehyde while the word ketal is used to describe an alcohol and a ketone. However, it turns out that professors are lazy, even textbook are lazy. They prefer, since aldehydes and ketones are essentially the same molecule in terms of their reactivity in nucleophilic addition is identical. Instead of using the distinction of acetal ketal, hemiacetal hemiacetal, instead of saying all that, we're just going to use the acetal version.
Whenever you see one of these gem-diether products, because notice that the N part of an acetal reaction is they have two ether groups. They're in a germinal position. They’re germinal. Whenever you have this, we're not going to worry about the R group so much. We’re not going to worry was it originally an aldehyde or a ketone. I don't really care. I’m just going to call it an acetal even though technically it might be a ketal. But it's really an industry standard thing where professors are not specific about the difference between an acetal and a ketal. If your professor specifically always makes that distinction, then by all means go with what they’re saying. But I’m just letting you know that even like online home works and a lot of textbooks don't really care about the difference between an acetal and a ketal. Awesome. Disclaimer over.
You should know this by now. Hemiacetals are only stable when they are cyclic or when they’re a ring form. Here I have another picture of a cyclic hemiacetal. Notice that I have a central carbon that has the four groups that I’m looking for. What is a hemiacetal? A hemiacetal is a functional group with either two Rs or two Hs or a mix. It doesn't matter. And an OH and an OR in a germinal position, so an alcohol and an ether in a germinal position. That’s a hemiacetal. Notice that this molecule is also a hemiacetal because I’ve got my H, I’ve got my R, I’ve got my OH, my alcohol and my ether, my OR.
When it’s a cyclic hemiacetal, you're stable. But if it's not cyclic, then you're not going to be able to end up at the hemiacetal. let me show you guys the general overview of this reaction. It turns out that when you react a carbonyl with one equivalent of alcohol, you're going to get what we call a hemiacetal. When you react it with the second equivalent of alcohol, it's going to be called an acetal and it’s going to make that germinal diether. The mechanism for the first step and for the second step is almost identical. The only way to really get it to stop at the hemiacetal is to make that cyclic version because if it’s not cyclic, it’s just going to pass straight through hemiacetal stage and go straight to acetal.
I hope that’s making sense so far. Now what we’re going to do in this next video is I’m going to show you the exact acid-catalyzed mechanism to create a hemiacetal. 

Another disclaimer before we begin, the mechanism that I'm about to show you contains a resonant structure as you can see here. We're going to fill this one out. I like to add resonant structures into my mechanisms because I think it makes it more clear where the arrows are being moved to. The only thing is that your professor may not use a resonant structure, your professor just might decide to push the arrow without it. If that happens it's okay, you can draw it his way, her way, you can draw it my way, it doesn't matter because in the end of the day as long as the arrows are going in the same places it doesn't matter if you add a resonant structure or not. This is going to be true with a lot of the mechanisms in clutch prep, I go out of my way to try to make the mechanisms extra clear and your professor might not really be explaining every arrow so feel free to draw it my way even if it's slightly different than the way your professor drew it just know that the arrows are being pushed the same way in the end so it's okay that you draw my mechanism which it should be equivalent to the one your professor is drawing. That being said, why don't you help me out with what the first step of this mechanism is?

Since it's acid-catalysed what's the very first thing we should be doing? Protonation. So the very first thing we're going to do in an acid-catalysed mechanism is protonate. Now I noticed that my acid in this case is a protonated version of alcohol, what I'm essentially using is R O H 2 positive. You could've used any acid source guys it doesn't have to be that, you could have used H plus, you could have used H 3 O plus, I'm just doing it like this because then the conjugate is going to make more make for you guys, the conjugate base but you could use any other acid source. So what that's going to make is a resonant structure because I'm going to get a positively charged oxygen but we know that this double bond could join the oxygen to make a lone pair and then I would get a formal charge on the carbon. So I like drawing this resonant structure because it makes it clear to me that that carbon is now very electrophilic, even more electrophilic than it was when it was unreacted. So notice that the whole point of the acid catalyst is to make this carbon even more reactive than it was before. So reactive in fact that alcohol is going to want to attack it so that's the next step guys. So the next step is what we call a nucleophilic attack. I'm just going to put N A, nucleophilic attack. So nucleophilic attack is going to attach the O, make a new single bond and we're going to get a protonated version of an ether attached to that central carbon. What do you think the next step is guys? Deprotonation, so we have to regenerate that catalyst. I'm going to take my alcohol, my neutral alcohol, and since I start off with an alcohol acid catalyst I need to regenerate it. So then I would just grab the H and lo and behold what do I have at the end? Now I have my O H, my O R, my H and my H. On top of that I have my catalyst still there.

So awesome guys, so that is a hemiacetal and I always like to draw it in this cross structure because I like to always keep it consistent in terms of what I'm looking at so that when I try to recognize the functional group later I always just put it into that cross and I'm like do I have all the groups that I need? Just a little peculiarity of mine. So anyway that was the acid catalyst mechanism to get to a hemiacetal. Not bad at all, right? Now are we going to stay there? No because this hemiacetal is not cyclical the way I drew it so this hemiacetal is either going to go back to the original carbonyl or it's going to keep reacting with alcohol to get to an acetal. More on that later. Now in this next video I want to show you guys the base catalyst version of the same reaction.

As an overarching principle of carbonyl chemistry, the based-catalysed mechanisms for reactions are almost always going to be easier than the acid-catalysed ones. The reason is because in acid-catalyst we're trying to protonate, deprotonate, to make things reactive. For base-catalysed mechanisms, the reagent is already going to start off reactive because you're making it a strong nucleophile and specifically for ketones and aldehydes we know that strong nucleophiles can do nucleophilic addition. So that just means that this reaction here that I'm going to show you it's just a nucleophilic addition reaction. So what happens? My nucleophile is O R negative, why? Because alcohol in the presence of a base, remember this is base-catalysed, is going to react with the base to give me an oxide O R negative.

So that oxide since it's a negatively charged nucleophile can do nucleophilic petition on its own just like any other negatively charged nucleophile we've worked with. So I would go ahead attack the carbonyl carbon, make my tetrahedral intermediate and I've got that O negative I have to take care of but that O negative can protonate with the conjugate acid of my base or of my nucleophile so then I could just grab one of the H's to regenerate the base that I would've lost in the other prior reactions. So then what I would wind up getting is I would get my hemiacetal and I have my O H on one side, my O R on another, my H, my H, and I've got some O R left over that can react with another carbonyl. So guys, hope that made sense. This one's a whole lot easier than acid-catalysed so let's move on to what happens in the second step which would be acetals.