Understanding Class I Carbonyl Chemistry

Hello organic chemistry students! In this video we're going to introduce carbonyls and what their functional groups are and the classes of carbonyl functional groups. And then we're going to dive into class one carbonyls. Now we've already seen a lot of these functional groups before. We actually have seen them all before, but we're just now going to categorize them in different ways. Now the first ones we want to look at are going to be esters, carboxylic acids, amides, and anhydrides. Now All of these are part of what we call Class I carbonyl chemistry. Why are they grouped together? If we were to look down here at the anhydride, if a nucleophile came into this anhydride, it would attack one of the carbonyls, break open the double bond. Those electrons would donate back down because we can kick out this oxygen and carbonyl. What we would end up forming here is that we have this nucleophile added onto one of the carbonyls of the former anhydride, and we have now just kicked out this oxygen. Now, this oxygen has direct resonance stabilization, which means it is a good leaving group. And that's all what class I carbonyls have in common, or share in common. Good leaving groups. So here, we can get this amine to be a leaving group, this oxygen and carbon chain or this OH right here. And that is why we classify them as class 1 carbonyls. Let's see what happens with some other ones. Down below, we have an aldehyde and a ketone. Now, in each one of these cases, when a nucleophile attacks this carbonyl, we break open the double bond and stop. That's it. Nothing is going to leave. And let's go ahead and look at this and ask ourselves why. So here's the nucleophile, and I'm going to go ahead and show that hydrogen. So before we work up this reaction, we have a negative charge on this oxygen. Now, we had a negative charge on the oxygen right here as well, but it donated down to kick out the leaving group. Here, if this negative charge donated down, we would have to lose this hydrogen and these two electrons as a hydride. Good idea or bad idea? And that's a bad idea. Horrible idea. So we can't kick out a hydride. We're not going to kick out the nucleophile we just added in, unless it's an extremely good leaving group, aka bad nucleophile. And we're not going to kick out this carbon chain, because if we donate these electrons down, and these electrons go under carbon, we've created a carbanion, very unstable. So what's the unique characteristic about class 2 carbonyls, which once again are aldehydes and ketones? They have bad leaving groups. That's the big thing. Class 1 has good leaving groups. Class 2 does not contain a leaving group. Oh, and before I forget, there is one more Class 1 that I forgot about, and those are the acid chlorides. You can imagine a chloride being a wonderful leaving group, as all halogens are. So the big difference between Class 1 and Class 2, leaving groups, and that's it. Now we have a third class, and that is this last one down below. Now, I'm writing Z and R right here because it could be oxygen, nitrogen, oxygen, I said oxygen twice, carbon, hydrogen, anything like that. All we're caring about in this class 3 functional group of carbonyls is that we're looking at this carbon directly attached to the carbonyl. And that's what we call the alpha carbon. So any carbon directly attached to the carbonyl is considered alpha. Any hydrogens on that alpha carbon are called... alpha hydrogens. Now alpha hydrogens are acidic, so if a base came in we could pull this proton, push electrons down between this carbon oxygen bond. Actually I'll go ahead and stop for one second and let's reverse that and I'll just have the electrons go right onto this carbon. And I'm not going to do reversible arrows like that because we're going to use strong bases in this class only. And I have just made a carbon with a full out negative charge. That seems horrible, just like we did up above. But this is unique because the negative charge could donate electrons down and resonate through the carbon-oxygen double bond. And that's what we can't do in class 2. This is unique into class 3. And this is going to allow us to form something called an enolate. This enolate will do different types of chemistry that we will learn about in a future video. But that is class 3 carbonyl chemistry. So while carbonyl chemistry 1 versus 2 is talking about the leaving group ability of one of the groups on the carbon of the carbonyl, class 3 is looking at the hydrogens on the alpha. carbon. So we can have class 3 carbonyl chemistry occurring in class 2 or class 1, because we have alpha carbons in both those cases. But we're not going to be looking at class 1 and 2 and class 3. So even though we have a 1 and a 2, class 3, we just care about this acidic hydrogen right here, and that's it. So now, let's go ahead and talk about class 1 carbonyl chemistry. Alright, so here's our basic reaction for class 1 carbon yields. What we're going to see here is that a nucleophile will come in, attack this carbon of the carbon yield. Let me go ahead and show the arrows. We attack the carbon of the carbon yield, we break open the pi bond. When this is done, we have formed this intermediate right here that we call the tetrahedral transition state. This carbon is now sp3 hybridized. That's very important. Because remember back to our substitution and elimination chemistry, those only happen on sp3 hybridized carbons. Now this is not substitution or elimination chemistry, but the concepts are the same. This is carbonyl chemistry class 1. So the nucleophile attacks the carbon of the carbonyl, that sp2 hybridized carbon, and we form our tetrahedral transition state. But now if you notice, we have equilibrium arrows. This complex could donate down and push out the nucleophile. to reform our starting material. That is 100% possible. Or the negative charge could donate down and kick out the X or the Z, oxygen, the halide, the nitrogen, or whatnot, to form this new carbonyl species and kicking out the X and Z. This is the basic reaction of all class one carbonyls. It's going to be relatively straightforward. But now, how do we know which way this equilibrium goes? If the nucleophile that we have right here is more reactive than the X and Z that are kicked out, we're going to favor pushing this towards the right. Let me say that again. If the nucleophile is more reactive than the X or Z, we push the reaction to the right. Another way of putting it, if the nucleophile is more unstable than X and Z, meaning X and Z is more stable, we want to form that more stable species. But now, if the nucleophile... is more stable than x and z. We're not going to kick out the xz here. When the electrons donate down, we'll kick out the nucleophile and return us the starting material. So that's how we decide which way the equilibrium is being favored. If the nucleophile is strong, we favor the right. If it's very weak, we favor the left right there. Now, like I said, believe it or not, this is the mechanism we're going to see for all class one carbon yields. Very little deviation right here. Let's go ahead and jump right into it. The very first one that we're going to start to talk about is the most favorite. Oops, let's go ahead and change this. And those are the acid chlorides. This also works with acid bromides or acid iodines, but acid chlorides are the classic ones to use. Now, I just want to take this and throw in water or OH minus, either or. When this happens, the oxygen from either species will attack the carbon of the carbonyl, break open the carbon-oxygen double bond. And what do we end up forming here? You got it. Our tetrahedral transition state, where here is that species. And we'll write proton transfer as well in the case of water, because we lose one of these protons when we form this species right here. So now, at this point, this negative charge is going to donate down, and it has a choice to make. Does it kick out OH- or Cl? What's the better leaving group? Oxygen with a negative charge or Cl with a negative charge? And it's Cl because the halogens are far more stable with a negative charge than oxygen is, even though oxygen is an electronegative element. It can't stabilize it that well. So what we've just done here is we've taken this acid chloride, and you might want to write that down. and we've transformed it into a carboxylic acid. Wow! So the reactions that we're going to be covering here in class 1 carbonyls are going to show the interconversion between the different class 1 functional groups. This is how we can take an acid chloride and form a carboxylic acid. Let's move this, oops, let's move this up so we can still see that. Erase that random line. Let's take that acid chloride again. This time I'm going to go ahead and take it and react it with an alcohol, ethanol right here. The lone pair of electrons Electrons on the oxygen can wrap around, attack the carbon of the carbonyl, break open the carbon-oxygen double bond, forming our tetrahedral transition states. And we're going to write proton transfer as well, because we'll lose that proton right here. Now a choice has to be made. Can this negative charge donate down and kick out the Cl- versus if we don't broke electrons from this bond and gave it to oxygen? an O minus. Oh wait, haven't we already answered that question? We did, up above. Cl minus is far more happier with a negative charge than oxygen minus. So we end up kicking out the Cl and we end up forming this group right here. And what functional group do you think this is right here? Starting with an acid chloride, we added an alcohol and we have formed a ester. So now, Acid chlorides with water or OH form carboxylic acids. Acid chlorides with alcohols form esters. Wow, so easy to do. And they work so well because chlorine is a great leaving group. Now, what if I took this acid chloride and reacted it with this amine right here? We have lone pairs of electrons on the amine. They could wrap around, attack the carbon in the carbonyl, break open the carbon-oxygen double bond. And I know right now you're like, this sounds familiar. It is. Same reaction over and over again. We do a proton transfer in here, losing one of the protons on this nitrogen to form this tetrahedral intermediate. The oxygen with the negative charge donates down. We kick out the chlorine because it is the best leaving group in these systems. And we have now just formed an amide. and Cl-. So an acid chloride is, we can take an acid chloride and transform it into a carboxylic acid with water or OH-, into an ester with an alcohol, into an amide with an amine. Wow, we've taken an acid chloride and almost transformed it into each one of the functional groups of class 1 except one of them. Let's take this acid chloride and I'm going to react it now with a... Carboxylic acid. We have a lone pair of electrons on this oxygen that could wrap around, attack this carbon of the carbonyl, break open the double bond to form our tetrahedral transition states. I know, oh my gosh, saying the same thing again. We lost that proton in the carboxylic acid, so there's our proton transfer. We donate electrons down, and chlorine is still the better leaving group. Even though the other compound, when it leaves, we could have resonance, chlorine is a great leaving group. And in this, we have just formed an anhydride. So acid chlorides can be elaborated onto carboxylic acids, esters, amides, and anhydrides. Wonderful. So we see the power of an acid chloride. Do all class I's have this power? And let's go ahead and look at that next. But before I jump into this next slide, I want to ask the question. The nucleophile attacks the carbon of the carbonyl in each case. The negative charge donates down and kicks out the chlorine. Nothing else is required. This is a good reaction. It works very, very smoothly. Let's see if the rest of them do. The next one that we're going to talk about is going to be carboxylic acids. Alright, so with a carboxylic acid, let's go ahead and look at this one right here. I'm going to take a carboxylic acid, and I'm going to see if I can react it with an alcohol. So now, this alcohol is going to wrap around, attack this carbon right here, break open the carbon-oxygen double bond, same thing that we've been seeing before, and now I'm going to show you the tetrahedral transition states. But I'm leaving this proton on right here. But in the acid chlorides, I did not leave it. In the acid chlorides, it does not matter if this oxygen is protonated or not. The chloride is a superb leaving group. Bromine, great leaving group. Iodine, superb leaving group. They all go just fine. Here, this is a problem. So now, in red. If this negative charge donates down, we're going to kick out this alcohol. Not this alcohol, this oh my- right here and what we are going to end up forming is, here's our proton right here, OH-. Hmm, is OH- more stable or less stable than this ethanol? And the answer is less stable, which means this doesn't happen whatsoever. Rather than doing this arrow right here, we will push out the nucleophile that we added in. and just get back our carboxylic acid. Why is that? When we look at the OH-, out of this right here, this OH- or this protonated oxygen, what's the better leaving group? The oxygen that's protonated. Because as the electrons come back down to oxygen, it neutralizes the molecule, neutral carboxylic acid. So we get full reversibility, nothing carrying on. Because... This negatively charged oxygen must react. It's very highly charged. It needs to undergo chemistry. So if only we could add something in that would help stabilize this negative charge to allow for that lower oxygen to lose its proton. So same conditions, nothing different. What are we going to add in now? H plus. That's it. The oxygen that is sp2 hybridized will pick up that proton rapidly, and we have now formed what we call an activated carbon yield. That oxygen is a full-out positive charge, meaning we're pulling more electron density up to this oxygen from this carbon, making this carbon partially positive, and that allows the alcohol to attack it rapidly and push electrons to this oxygen. Now we do have this being reversible, just as in all cases, and what we end up forming is ... I know you're like, wait, this is the same exact thing that we just drew up above. And it almost is. What's the difference between this compound and this intermediate here? Neutral, negative. Very reactive, stable. So now, because it's so stable down below, it allows time for that oxygen to lose the hydrogen. This is a reversible process as well. And we've now neutralized that lower oxygen and we have H plus sitting in solution. OH is a bad leaving group. Now if this oxygen donates electrons down or if this oxygen donates electrons down, it doesn't matter which one of those does. The important thing is, this lower oxygen can never donate electrons down because there's no hydrogen on it for it to leave. So what I'm going to do right now is look at this top oxygen. and look at those lone pairs. One of those lone pairs can pick up this proton right here. When we pick up this proton in a reversible process, now once again, please keep in mind that either OH could have picked up this proton. We've now formed what looks like water. The lone pair on the other oxygen will donate down and push this water molecule out. And what we form here is a carbonyl with a hydrogen on it. attached to a carbon chain. We then lose this proton, and when the reaction is all said and done, we have just formed an ester. Wow, so many more steps on the acid chloride. Why? Acid chlorides have superb leaving groups. They're highly activated. We do not need to do this activation of the carbon yield like we're doing here. With a carboxylic acid, we have to add H plus to activate the carbonyl to allow this transition state time to lose this proton, something that we didn't need in the acid chlorides because the chlorine was a superb leaving group. We lose the proton. The proton is picked up by either one of these OHs. Once again, does not matter. Then the other OH will donate its electrons down to push out that water molecule. We lose this proton and we form our ester functional group. Wonderful. So An acid chloride can be converted to a carboxylic acid, an ester, an amide, and an anhydride. A carboxylic acid can be converted to a ester so far. Let's see if we can do anything else. Oops, let me get my pen in here. All right, let's take this carboxylic acid, and I am going to react it with an amine. Now, just like we had up above, the amine... is going to attack the carbon of the carbonyl, but we still are going to form a negatively charged transition state. So we need to add in H+, don't we, to activate the carbonyl. So we would think. In this reaction, nothing happens. We don't form anything. I know. Whoa, wait a second. Why? Nitrogen with its lone pairs is a basic molecule, and it picks up the protons rapidly. While oxygen has lone pairs of electrons, oxygens are not basic. Nitrogens are. So what we form here is an NH3. We're not protonating the carbonyl, and no reaction can take place. So, so far with carboxylic acids, we can form an ester. We cannot form an anhydride. Ugh, problematic for sure. Let me move this up some. Oops, move this up. Let's see if we can form an anhydride. So if I take a carboxylic acid, react it with H+, and I'm going to use the same exact carboxylic acid, we're going to have this oxygen of the carbonyl pick up this proton, of course the other starting material could have picked it up as well, we protonate that carbon yield just like we did before, and now enters in our nucleophile. We have lone pairs of electrons on this oxygen, we can attack the carbon yield, break open the double bond, and we have now formed this OH-OH oxygen carbon yield with a proton on it, and these are reversible. This positively charged oxygen will get its electrons back when that proton dissociates. And we've now formed this OH species and this carbonyl. Proton over in solution. Wow, either one of those oxygens could pick up this proton. And then this oxygen will donate electrons down and push that water molecule out. And following proton transfer, we get out the anhydride. functional group. Wow, that's a lot of arrows. But carboxylic acid reacting with the carboxylic acid under H plus has the same arrows as a carboxylic acid with H plus and an alcohol. This reaction and this reaction are identical. So carboxylic acids can be transformed into esters. Carboxylic acids cannot be converted to amides. but carboxylic acids can also be converted into, you got it, anhydrides. So far, acid chlorides are superior. But notice, I don't talk about how a carboxylic can become an acid chloride. We'll talk about that a little bit later. So now, let's look at the amide functional group. So if we have an amide functional group right here, if I add in H+, and H2O. Could that oxygen of the carbonyl pick up the proton? It should. There's electrons on it, right? It doesn't. Nothing happens here whatsoever. But wait, we had all the conditions to do this, just like we had on the previous page. Why doesn't anything happen here? While this is the way that we draw an amide, the way an amide truly exists in everyday life, is as follows. This is the true form of an amide under normal conditions such as room temperature. These exist in dynamic equilibrium. Equilibrium here. And if I can make that arrow even smaller, I would. 99.99% here, trace amounts here. This right here though is the active form. Or I should say the unstable form. Down below. This is the stable form or the unactive form. So if we're staying in this form 99.99% of the time, amides don't do a lot. So how can we get this stable form into the unstable form? We can do this, transform it up this way, by adding in heats. So in order to do any chemistry on an amide, we must add heats. Now, before I go a step further. If you think about being sick, if you have a bacterial infection or anything like that, you get a fever, don't you? Our body's cranking up the temperature because we're trying to switch this to tomorization event here into here in the bacteria, so our immune system can break down the proteins in the bacteria and kill it. So the heat, the fever is there to shift the population of the amide. That's why we're doing it. Now if the temperature in your body gets too hot, over 105 I believe it is, they pack you in ice. Why? We then start doing a degradation of amides in our brain. We don't want the brain to break down, so that's why we have to stop the reaction. So let's go ahead and take this concept of heat and apply it to our reaction now. So we draw it in this form. This is the way we always draw ammets. Once again, even though it exists in this form nearly all the time, but that's the way we draw it. This time I'm going to write H+, I'm going to write my H2O, and I must always imply heat, so we're in the reactive conformer. Now, This oxygen can pick up this proton because it actually is an sp2 hybridized oxygen. We pick it up. We form the activated carbonyl species. And now water with its lone pair will attack the carbonyl, break open the double bond in a reversible process. And we have now formed... This species here. The nitrogen will pick up one of these protons, giving electrons back to oxygen, and we now have a protonated nitrogen. Wonderful! One of these oxygens with its electrons will donate electrons down and push out our amine. We do a proton transfer, and we have a carboxylic acid. So an amide can be converted into a carboxylic acid with H+, water, and heat. Wonderful. What else can we do with an amide? So an amide, and I'm not going to be pushing the arrows on this one because we now know what those arrows are. They're the same things that we saw in previous slides. So now I'm going to put H plus in and an alcohol and heats. So we know that the alcohol is going to switch places with this amine right here. We know that's the fact. We've seen the mechanism up above. It doesn't change. And we form this functional group here. So an amide can be converted into a ester. You have it. There it is. So we can take an amide, form a carboxylic acid. We can take an amide, form an ester. But we cannot take an amide. and form an anhydride. Sadly, that one doesn't work. So that's our limitations here, right there. So we've talked about the acid chloride. We've talked about the carboxylic acid. We've talked about the amines, the amide systems, right? We haven't talked about esters yet. So in an ester, we have the same problem once again. We have to make sure that we neutralize this tetrahedral transition state. So we're going to add some H plus in. And I'm going to add water in. Now this is not an amide. There is no tautomerization to form this reactive species. It's the ester. It's ready to go. What we are going to form here is a carboxylic acid. So whoa whoa wait a second. I'm noticing a pattern now. No matter what the carbonyl is, the class one carbonyl, when I add water or OH minus, I get a carboxylic acid. Wow. No matter what the class one carbonyl is, when I add in an alcohol, what do I... always get? Oh, I get an ester. Ah, that's interesting. And for the limited ones, when I add in an NH2, we should get an amide out. But here we don't get an amide out either. Why? Because the nitrogens and means will pick up that proton and quench the reaction. So carboxylic acids cannot be converted to amides, nor can esters right here. So We've talked about four of the five class one carbonyl functional groups. What's left? Anhydrides. Now notice I'm not starting a new page. I'm just going to draw the anhydride right here. An anhydride with a nucleophile being present. When it attacks one of those carbonyls, we break open the double bond, donate electrons down, and kick out our super good leaving group to form this resonance stabilized leaving group. Now this is a good leaving group just like chlorine is. So the anhydrides can do the same chemistry that the acid chlorides can do. We can take an anhydride and form a carboxylic acid where this is an OH or we can form an ester where this is an OR group or we can form an amide where this is an NH with R or hydrogen also attached. Same exact chemistry. Wonderful. Taking all of these groups, transform them into carboxylic acids, amides, esters, and anhydrides. What's the one thing we haven't done? Form the acid chloride. How do we form an acid chloride? And we're going to do this right here on this page before we end the video. So how is an acid chloride formed? In one way only. We take a carboxylic acid and we're going to react it with thionyl chloride to form the acid chloride. In this, we've seen this already when we looked at the halide video, the alcohol, not the alcohol, I'm sorry, this oxygen will attack the sulfur, break open the double bond, donate electrons down, and kick out the chlorine. And what we have formed here is this oxygen. Oops, that negative charge should not be there. There's the double bond. Here's this oxygen, hydrogen, attached to sulfur, doubly bound with a Cl-. positive charge here. We can lose this proton right here and I'm going to go ahead and just erase it. Oops. I'm going to erase it. I can't erase it now because I wrote on top of it. So I'm going to have to go ahead and show the proton transfer step. I was just going to erase it and put PT over the arrow. So now we have a chlorine sitting in solution. That chlorine will come in, attack the carbon of the carbonyl, break open electrons, donate electrons down. and do the same exact reaction that we saw when we converted an alcohol into, you got it, a chloride. And that converts them into our acid chlorides as shown. So, with this, in this video, we have very nicely shown how acid chlorides can be converted into the other four class 1 functional groups. Great. We then showed how... Carboxylic acids and esters have limitations, right? They can be converted into alcohols, I'm sorry, carboxylic acids, esters, anhydrides, but not amides. Amides themselves can be transformed into other functional groups of class 1, but we have to add heat, absolutely. There's no ifs, ands, or buts. And then we talked about the anhydrides. So now, we need to come up with an overall order of reactivity of class 1 functional groups. Clearly, the acid chlorides do all the great chemistry. It has the best leaving group on it. The next one that has the best leaving group are the anhydrides. Now, we have to fill in these last three boxes right here. I'll put little lines in. We have the anhydrides, we have the carboxylic acids, and we have the esters. Carboxylic acids and esters are similar to one another, aren't they? They are, so we're going to say they're equal. But do they react better than amides? And the answer is yes. An amide, you have to add heat to do tautomerization and try to get the reaction to go. And that's why they are dead last in terms of their reactivity. So what we're going to do in this class is get rid of this greater than sign. And we are going to say is that the carboxylic acids and the esters have the same functionality reaction rates. So with that, this ends. the video on Class 1 Carbon Neochemistry. Hopefully you've noticed the same mechanism over and over and over again. I know. What I'd recommend doing is watch this video again, take more notes on this, and practice the mechanisms. And come back and watch this video to find out how you did. With that, if you have any questions, please feel free to send me an email. See me at the discussion or office hours. And I'll be happy to help you the best that I can. Hope each of you are doing well. and I look forward to seeing you sometime soon.

Now we've already seen a lot of these functional groups before. We actually have seen them all before, but we're just now going to categorize them in different ways. Now the first ones we want to look at are going to be esters, carboxylic acids, amides, and anhydrides.

Now All of these are part of what we call Class I carbonyl chemistry. Why are they grouped together? If we were to look down here at the anhydride, if a nucleophile came into this anhydride, it would attack one of the carbonyls, break open the double bond. Those electrons would donate back down because we can kick out this oxygen and carbonyl. What we would end up forming here is that we have this nucleophile added onto one of the carbonyls of the former anhydride, and we have now just kicked out this oxygen.

Now, this oxygen has direct resonance stabilization, which means it is a good leaving group. And that's all what class I carbonyls have in common, or share in common. Good leaving groups. So here, we can get this amine to be a leaving group, this oxygen and carbon chain or this OH right here. And that is why we classify them as class 1 carbonyls.

Let's see what happens with some other ones. Down below, we have an aldehyde and a ketone. Now, in each one of these cases, when a nucleophile attacks this carbonyl, we break open the double bond and stop.

That's it. Nothing is going to leave. And let's go ahead and look at this and ask ourselves why. So here's the nucleophile, and I'm going to go ahead and show that hydrogen.

So before we work up this reaction, we have a negative charge on this oxygen. Now, we had a negative charge on the oxygen right here as well, but it donated down to kick out the leaving group. Here, if this negative charge donated down, we would have to lose this hydrogen and these two electrons as a hydride. Good idea or bad idea?

And that's a bad idea. Horrible idea. So we can't kick out a hydride. We're not going to kick out the nucleophile we just added in, unless it's an extremely good leaving group, aka bad nucleophile. And we're not going to kick out this carbon chain, because if we donate these electrons down, and these electrons go under carbon, we've created a carbanion, very unstable.

So what's the unique characteristic about class 2 carbonyls, which once again are aldehydes and ketones? They have bad leaving groups. That's the big thing. Class 1 has good leaving groups.

Class 2 does not contain a leaving group. Oh, and before I forget, there is one more Class 1 that I forgot about, and those are the acid chlorides. You can imagine a chloride being a wonderful leaving group, as all halogens are. So the big difference between Class 1 and Class 2, leaving groups, and that's it. Now we have a third class, and that is this last one down below.

Now, I'm writing Z and R right here because it could be oxygen, nitrogen, oxygen, I said oxygen twice, carbon, hydrogen, anything like that. All we're caring about in this class 3 functional group of carbonyls is that we're looking at this carbon directly attached to the carbonyl. And that's what we call the alpha carbon.

So any carbon directly attached to the carbonyl is considered alpha. Any hydrogens on that alpha carbon are called... alpha hydrogens.

Now alpha hydrogens are acidic, so if a base came in we could pull this proton, push electrons down between this carbon oxygen bond. Actually I'll go ahead and stop for one second and let's reverse that and I'll just have the electrons go right onto this carbon. And I'm not going to do reversible arrows like that because we're going to use strong bases in this class only.

And I have just made a carbon with a full out negative charge. That seems horrible, just like we did up above. But this is unique because the negative charge could donate electrons down and resonate through the carbon-oxygen double bond. And that's what we can't do in class 2. This is unique into class 3. And this is going to allow us to form something called an enolate.

This enolate will do different types of chemistry that we will learn about in a future video. But that is class 3 carbonyl chemistry. So while carbonyl chemistry 1 versus 2 is talking about the leaving group ability of one of the groups on the carbon of the carbonyl, class 3 is looking at the hydrogens on the alpha.

carbon. So we can have class 3 carbonyl chemistry occurring in class 2 or class 1, because we have alpha carbons in both those cases. But we're not going to be looking at class 1 and 2 and class 3. So even though we have a 1 and a 2, class 3, we just care about this acidic hydrogen right here, and that's it.

So now, let's go ahead and talk about class 1 carbonyl chemistry. Alright, so here's our basic reaction for class 1 carbon yields. What we're going to see here is that a nucleophile will come in, attack this carbon of the carbon yield. Let me go ahead and show the arrows. We attack the carbon of the carbon yield, we break open the pi bond.

When this is done, we have formed this intermediate right here that we call the tetrahedral transition state. This carbon is now sp3 hybridized. That's very important. Because remember back to our substitution and elimination chemistry, those only happen on sp3 hybridized carbons. Now this is not substitution or elimination chemistry, but the concepts are the same.

This is carbonyl chemistry class 1. So the nucleophile attacks the carbon of the carbonyl, that sp2 hybridized carbon, and we form our tetrahedral transition state. But now if you notice, we have equilibrium arrows. This complex could donate down and push out the nucleophile.

to reform our starting material. That is 100% possible. Or the negative charge could donate down and kick out the X or the Z, oxygen, the halide, the nitrogen, or whatnot, to form this new carbonyl species and kicking out the X and Z.

This is the basic reaction of all class one carbonyls. It's going to be relatively straightforward. But now, how do we know which way this equilibrium goes?

If the nucleophile that we have right here is more reactive than the X and Z that are kicked out, we're going to favor pushing this towards the right. Let me say that again. If the nucleophile is more reactive than the X or Z, we push the reaction to the right.

Another way of putting it, if the nucleophile is more unstable than X and Z, meaning X and Z is more stable, we want to form that more stable species. But now, if the nucleophile... is more stable than x and z. We're not going to kick out the xz here. When the electrons donate down, we'll kick out the nucleophile and return us the starting material.

So that's how we decide which way the equilibrium is being favored. If the nucleophile is strong, we favor the right. If it's very weak, we favor the left right there. Now, like I said, believe it or not, this is the mechanism we're going to see for all class one carbon yields.

Very little deviation right here. Let's go ahead and jump right into it. The very first one that we're going to start to talk about is the most favorite.

Oops, let's go ahead and change this. And those are the acid chlorides. This also works with acid bromides or acid iodines, but acid chlorides are the classic ones to use. Now, I just want to take this and throw in water or OH minus, either or.

When this happens, the oxygen from either species will attack the carbon of the carbonyl, break open the carbon-oxygen double bond. And what do we end up forming here? You got it. Our tetrahedral transition state, where here is that species. And we'll write proton transfer as well in the case of water, because we lose one of these protons when we form this species right here.

So now, at this point, this negative charge is going to donate down, and it has a choice to make. Does it kick out OH- or Cl? What's the better leaving group? Oxygen with a negative charge or Cl with a negative charge? And it's Cl because the halogens are far more stable with a negative charge than oxygen is, even though oxygen is an electronegative element.

It can't stabilize it that well. So what we've just done here is we've taken this acid chloride, and you might want to write that down. and we've transformed it into a carboxylic acid. Wow! So the reactions that we're going to be covering here in class 1 carbonyls are going to show the interconversion between the different class 1 functional groups.

This is how we can take an acid chloride and form a carboxylic acid. Let's move this, oops, let's move this up so we can still see that. Erase that random line.

Let's take that acid chloride again. This time I'm going to go ahead and take it and react it with an alcohol, ethanol right here. The lone pair of electrons Electrons on the oxygen can wrap around, attack the carbon of the carbonyl, break open the carbon-oxygen double bond, forming our tetrahedral transition states. And we're going to write proton transfer as well, because we'll lose that proton right here. Now a choice has to be made.

Can this negative charge donate down and kick out the Cl- versus if we don't broke electrons from this bond and gave it to oxygen? an O minus. Oh wait, haven't we already answered that question?

We did, up above. Cl minus is far more happier with a negative charge than oxygen minus. So we end up kicking out the Cl and we end up forming this group right here. And what functional group do you think this is right here?

Starting with an acid chloride, we added an alcohol and we have formed a ester. So now, Acid chlorides with water or OH form carboxylic acids. Acid chlorides with alcohols form esters.

Wow, so easy to do. And they work so well because chlorine is a great leaving group. Now, what if I took this acid chloride and reacted it with this amine right here?

We have lone pairs of electrons on the amine. They could wrap around, attack the carbon in the carbonyl, break open the carbon-oxygen double bond. And I know right now you're like, this sounds familiar. It is. Same reaction over and over again.

We do a proton transfer in here, losing one of the protons on this nitrogen to form this tetrahedral intermediate. The oxygen with the negative charge donates down. We kick out the chlorine because it is the best leaving group in these systems. And we have now just formed an amide. and Cl-.

So an acid chloride is, we can take an acid chloride and transform it into a carboxylic acid with water or OH-, into an ester with an alcohol, into an amide with an amine. Wow, we've taken an acid chloride and almost transformed it into each one of the functional groups of class 1 except one of them. Let's take this acid chloride and I'm going to react it now with a...

Carboxylic acid. We have a lone pair of electrons on this oxygen that could wrap around, attack this carbon of the carbonyl, break open the double bond to form our tetrahedral transition states. I know, oh my gosh, saying the same thing again. We lost that proton in the carboxylic acid, so there's our proton transfer. We donate electrons down, and chlorine is still the better leaving group.

Even though the other compound, when it leaves, we could have resonance, chlorine is a great leaving group. And in this, we have just formed an anhydride. So acid chlorides can be elaborated onto carboxylic acids, esters, amides, and anhydrides.

Wonderful. So we see the power of an acid chloride. Do all class I's have this power?

And let's go ahead and look at that next. But before I jump into this next slide, I want to ask the question. The nucleophile attacks the carbon of the carbonyl in each case. The negative charge donates down and kicks out the chlorine.

Nothing else is required. This is a good reaction. It works very, very smoothly. Let's see if the rest of them do. The next one that we're going to talk about is going to be carboxylic acids.

Alright, so with a carboxylic acid, let's go ahead and look at this one right here. I'm going to take a carboxylic acid, and I'm going to see if I can react it with an alcohol. So now, this alcohol is going to wrap around, attack this carbon right here, break open the carbon-oxygen double bond, same thing that we've been seeing before, and now I'm going to show you the tetrahedral transition states. But I'm leaving this proton on right here. But in the acid chlorides, I did not leave it.

In the acid chlorides, it does not matter if this oxygen is protonated or not. The chloride is a superb leaving group. Bromine, great leaving group.

Iodine, superb leaving group. They all go just fine. Here, this is a problem. So now, in red. If this negative charge donates down, we're going to kick out this alcohol.

Not this alcohol, this oh my- right here and what we are going to end up forming is, here's our proton right here, OH-. Hmm, is OH- more stable or less stable than this ethanol? And the answer is less stable, which means this doesn't happen whatsoever. Rather than doing this arrow right here, we will push out the nucleophile that we added in. and just get back our carboxylic acid.

Why is that? When we look at the OH-, out of this right here, this OH- or this protonated oxygen, what's the better leaving group? The oxygen that's protonated. Because as the electrons come back down to oxygen, it neutralizes the molecule, neutral carboxylic acid.

So we get full reversibility, nothing carrying on. Because... This negatively charged oxygen must react.

It's very highly charged. It needs to undergo chemistry. So if only we could add something in that would help stabilize this negative charge to allow for that lower oxygen to lose its proton.

So same conditions, nothing different. What are we going to add in now? H plus. That's it. The oxygen that is sp2 hybridized will pick up that proton rapidly, and we have now formed what we call an activated carbon yield.

That oxygen is a full-out positive charge, meaning we're pulling more electron density up to this oxygen from this carbon, making this carbon partially positive, and that allows the alcohol to attack it rapidly and push electrons to this oxygen. Now we do have this being reversible, just as in all cases, and what we end up forming is ... I know you're like, wait, this is the same exact thing that we just drew up above.

And it almost is. What's the difference between this compound and this intermediate here? Neutral, negative.

Very reactive, stable. So now, because it's so stable down below, it allows time for that oxygen to lose the hydrogen. This is a reversible process as well. And we've now neutralized that lower oxygen and we have H plus sitting in solution.

OH is a bad leaving group. Now if this oxygen donates electrons down or if this oxygen donates electrons down, it doesn't matter which one of those does. The important thing is, this lower oxygen can never donate electrons down because there's no hydrogen on it for it to leave. So what I'm going to do right now is look at this top oxygen.

and look at those lone pairs. One of those lone pairs can pick up this proton right here. When we pick up this proton in a reversible process, now once again, please keep in mind that either OH could have picked up this proton. We've now formed what looks like water.

The lone pair on the other oxygen will donate down and push this water molecule out. And what we form here is a carbonyl with a hydrogen on it. attached to a carbon chain.

We then lose this proton, and when the reaction is all said and done, we have just formed an ester. Wow, so many more steps on the acid chloride. Why?

Acid chlorides have superb leaving groups. They're highly activated. We do not need to do this activation of the carbon yield like we're doing here. With a carboxylic acid, we have to add H plus to activate the carbonyl to allow this transition state time to lose this proton, something that we didn't need in the acid chlorides because the chlorine was a superb leaving group. We lose the proton.

The proton is picked up by either one of these OHs. Once again, does not matter. Then the other OH will donate its electrons down to push out that water molecule.

We lose this proton and we form our ester functional group. Wonderful. So An acid chloride can be converted to a carboxylic acid, an ester, an amide, and an anhydride.

A carboxylic acid can be converted to a ester so far. Let's see if we can do anything else. Oops, let me get my pen in here.

All right, let's take this carboxylic acid, and I am going to react it with an amine. Now, just like we had up above, the amine... is going to attack the carbon of the carbonyl, but we still are going to form a negatively charged transition state.

So we need to add in H+, don't we, to activate the carbonyl. So we would think. In this reaction, nothing happens. We don't form anything. I know.

Whoa, wait a second. Why? Nitrogen with its lone pairs is a basic molecule, and it picks up the protons rapidly.

While oxygen has lone pairs of electrons, oxygens are not basic. Nitrogens are. So what we form here is an NH3.

We're not protonating the carbonyl, and no reaction can take place. So, so far with carboxylic acids, we can form an ester. We cannot form an anhydride.

Ugh, problematic for sure. Let me move this up some. Oops, move this up.

Let's see if we can form an anhydride. So if I take a carboxylic acid, react it with H+, and I'm going to use the same exact carboxylic acid, we're going to have this oxygen of the carbonyl pick up this proton, of course the other starting material could have picked it up as well, we protonate that carbon yield just like we did before, and now enters in our nucleophile. We have lone pairs of electrons on this oxygen, we can attack the carbon yield, break open the double bond, and we have now formed this OH-OH oxygen carbon yield with a proton on it, and these are reversible. This positively charged oxygen will get its electrons back when that proton dissociates. And we've now formed this OH species and this carbonyl.

Proton over in solution. Wow, either one of those oxygens could pick up this proton. And then this oxygen will donate electrons down and push that water molecule out.

And following proton transfer, we get out the anhydride. functional group. Wow, that's a lot of arrows. But carboxylic acid reacting with the carboxylic acid under H plus has the same arrows as a carboxylic acid with H plus and an alcohol.

This reaction and this reaction are identical. So carboxylic acids can be transformed into esters. Carboxylic acids cannot be converted to amides. but carboxylic acids can also be converted into, you got it, anhydrides.

So far, acid chlorides are superior. But notice, I don't talk about how a carboxylic can become an acid chloride. We'll talk about that a little bit later.

So now, let's look at the amide functional group. So if we have an amide functional group right here, if I add in H+, and H2O. Could that oxygen of the carbonyl pick up the proton? It should.

There's electrons on it, right? It doesn't. Nothing happens here whatsoever.

But wait, we had all the conditions to do this, just like we had on the previous page. Why doesn't anything happen here? While this is the way that we draw an amide, the way an amide truly exists in everyday life, is as follows.

This is the true form of an amide under normal conditions such as room temperature. These exist in dynamic equilibrium. Equilibrium here. And if I can make that arrow even smaller, I would.

99.99% here, trace amounts here. This right here though is the active form. Or I should say the unstable form. Down below.

This is the stable form or the unactive form. So if we're staying in this form 99.99% of the time, amides don't do a lot. So how can we get this stable form into the unstable form? We can do this, transform it up this way, by adding in heats.

So in order to do any chemistry on an amide, we must add heats. Now, before I go a step further. If you think about being sick, if you have a bacterial infection or anything like that, you get a fever, don't you?

Our body's cranking up the temperature because we're trying to switch this to tomorization event here into here in the bacteria, so our immune system can break down the proteins in the bacteria and kill it. So the heat, the fever is there to shift the population of the amide. That's why we're doing it. Now if the temperature in your body gets too hot, over 105 I believe it is, they pack you in ice. Why?

We then start doing a degradation of amides in our brain. We don't want the brain to break down, so that's why we have to stop the reaction. So let's go ahead and take this concept of heat and apply it to our reaction now. So we draw it in this form. This is the way we always draw ammets.

Once again, even though it exists in this form nearly all the time, but that's the way we draw it. This time I'm going to write H+, I'm going to write my H2O, and I must always imply heat, so we're in the reactive conformer. Now, This oxygen can pick up this proton because it actually is an sp2 hybridized oxygen.

We pick it up. We form the activated carbonyl species. And now water with its lone pair will attack the carbonyl, break open the double bond in a reversible process. And we have now formed...

This species here. The nitrogen will pick up one of these protons, giving electrons back to oxygen, and we now have a protonated nitrogen. Wonderful!

One of these oxygens with its electrons will donate electrons down and push out our amine. We do a proton transfer, and we have a carboxylic acid. So an amide can be converted into a carboxylic acid with H+, water, and heat. Wonderful. What else can we do with an amide?

So an amide, and I'm not going to be pushing the arrows on this one because we now know what those arrows are. They're the same things that we saw in previous slides. So now I'm going to put H plus in and an alcohol and heats. So we know that the alcohol is going to switch places with this amine right here. We know that's the fact.

We've seen the mechanism up above. It doesn't change. And we form this functional group here.

So an amide can be converted into a ester. You have it. There it is.

So we can take an amide, form a carboxylic acid. We can take an amide, form an ester. But we cannot take an amide. and form an anhydride. Sadly, that one doesn't work.

So that's our limitations here, right there. So we've talked about the acid chloride. We've talked about the carboxylic acid. We've talked about the amines, the amide systems, right? We haven't talked about esters yet.

So in an ester, we have the same problem once again. We have to make sure that we neutralize this tetrahedral transition state. So we're going to add some H plus in. And I'm going to add water in. Now this is not an amide.

There is no tautomerization to form this reactive species. It's the ester. It's ready to go.

What we are going to form here is a carboxylic acid. So whoa whoa wait a second. I'm noticing a pattern now. No matter what the carbonyl is, the class one carbonyl, when I add water or OH minus, I get a carboxylic acid.

Wow. No matter what the class one carbonyl is, when I add in an alcohol, what do I... always get? Oh, I get an ester. Ah, that's interesting.

And for the limited ones, when I add in an NH2, we should get an amide out. But here we don't get an amide out either. Why?

Because the nitrogens and means will pick up that proton and quench the reaction. So carboxylic acids cannot be converted to amides, nor can esters right here. So We've talked about four of the five class one carbonyl functional groups.

What's left? Anhydrides. Now notice I'm not starting a new page.

I'm just going to draw the anhydride right here. An anhydride with a nucleophile being present. When it attacks one of those carbonyls, we break open the double bond, donate electrons down, and kick out our super good leaving group to form this resonance stabilized leaving group.

Now this is a good leaving group just like chlorine is. So the anhydrides can do the same chemistry that the acid chlorides can do. We can take an anhydride and form a carboxylic acid where this is an OH or we can form an ester where this is an OR group or we can form an amide where this is an NH with R or hydrogen also attached. Same exact chemistry. Wonderful.

Taking all of these groups, transform them into carboxylic acids, amides, esters, and anhydrides. What's the one thing we haven't done? Form the acid chloride.

How do we form an acid chloride? And we're going to do this right here on this page before we end the video. So how is an acid chloride formed? In one way only. We take a carboxylic acid and we're going to react it with thionyl chloride to form the acid chloride.

In this, we've seen this already when we looked at the halide video, the alcohol, not the alcohol, I'm sorry, this oxygen will attack the sulfur, break open the double bond, donate electrons down, and kick out the chlorine. And what we have formed here is this oxygen. Oops, that negative charge should not be there. There's the double bond. Here's this oxygen, hydrogen, attached to sulfur, doubly bound with a Cl-.

positive charge here. We can lose this proton right here and I'm going to go ahead and just erase it. Oops.

I'm going to erase it. I can't erase it now because I wrote on top of it. So I'm going to have to go ahead and show the proton transfer step. I was just going to erase it and put PT over the arrow.

So now we have a chlorine sitting in solution. That chlorine will come in, attack the carbon of the carbonyl, break open electrons, donate electrons down. and do the same exact reaction that we saw when we converted an alcohol into, you got it, a chloride.

And that converts them into our acid chlorides as shown. So, with this, in this video, we have very nicely shown how acid chlorides can be converted into the other four class 1 functional groups. Great. We then showed how...

Carboxylic acids and esters have limitations, right? They can be converted into alcohols, I'm sorry, carboxylic acids, esters, anhydrides, but not amides. Amides themselves can be transformed into other functional groups of class 1, but we have to add heat, absolutely.

There's no ifs, ands, or buts. And then we talked about the anhydrides. So now, we need to come up with an overall order of reactivity of class 1 functional groups.

Clearly, the acid chlorides do all the great chemistry. It has the best leaving group on it. The next one that has the best leaving group are the anhydrides. Now, we have to fill in these last three boxes right here.

I'll put little lines in. We have the anhydrides, we have the carboxylic acids, and we have the esters. Carboxylic acids and esters are similar to one another, aren't they? They are, so we're going to say they're equal.

But do they react better than amides? And the answer is yes. An amide, you have to add heat to do tautomerization and try to get the reaction to go.

And that's why they are dead last in terms of their reactivity. So what we're going to do in this class is get rid of this greater than sign. And we are going to say is that the carboxylic acids and the esters have the same functionality reaction rates.

So with that, this ends. the video on Class 1 Carbon Neochemistry. Hopefully you've noticed the same mechanism over and over and over again. I know.

What I'd recommend doing is watch this video again, take more notes on this, and practice the mechanisms. And come back and watch this video to find out how you did. With that, if you have any questions, please feel free to send me an email.

See me at the discussion or office hours. And I'll be happy to help you the best that I can. Hope each of you are doing well. and I look forward to seeing you sometime soon.

Transcript for:Understanding Class I Carbonyl Chemistry

Transcript for:
Understanding Class I Carbonyl Chemistry