Course: Organic chemistry > Unit 11Lesson 4: Nomenclature and reactions of carboxylic acid derivatives
- Nomenclature and properties of acyl (acid) halides and acid anhydrides
- Nomenclature and properties of esters
- Nomenclature and properties of amides
- Reactivity of carboxylic acid derivatives
- Nucleophilic acyl substitution
- Acid-catalyzed ester hydrolysis
- Acid and base-catalyzed hydrolysis of amides
- Beta-lactam antibiotics
Acid and base-catalyzed hydrolysis of amides
The mechanisms for acid and base catalyzed hydrolysis of amides. Created by Jay.
Want to join the conversation?
- I was under the impression that H20 was a better leaving group than NH3. Why is it then, that NH3 is the leaving group rather than H20?(10 votes)
- Yes, it is.
You have a competition between the two leaving groups.
But if H₂O leaves, you get back starting material.
You get the product only from the loss of NH₃.
Every step is an equilibrium, but the loss of NH₃.
The NH₃ is immediately converted irreversibly to NH₄⁺ under the reaction conditions, and this drives the reaction to completion.(18 votes)
- around10:25why does the NH2- takes the H from R-COOH and not the H+ from the solution?(4 votes)
- At this point, the reaction is still strongly basic. There is very little H⁺ in the solution. NH₂⁻ is a powerful base. It will remove protons from any RCOOH or H₂O molecules it happens to hit. There are many more water molecules than carboxylic acid molecules in the solution. By statistics, I would expect NH₂⁻ to react with water and produce OH⁻ ions. Then the OH⁻ ions will neutralize the RCOOH.(7 votes)
- in the base-catalyzed reaction, it would not be favorable for the amino group to be protonated because + charge is not favorable in basic solutions, right? Hence, we only have NH2 as a leaving group?(4 votes)
- In a basic solution, there is so little H⁺ that essentially nothing is protonated.(4 votes)
- why we prefer basic hydrolysis of amides over acidic(3 votes)
- under base catalysed hydrolysis, why does lone pair of oxygen moves to form carbonyl and NH2 leaves? why not OH that leaves instead? isnt NH2- a poor leaving group?(1 vote)
- If the OH leaves, you just get back the starting material.
NH₂⁻ is not a good leaving group, but it does leave, and that gives you a resonance-stabilized carboxylate anion.(3 votes)
- At1:30he mentioned that there is a resonance structure that makes the carbon more electrophilic and more reactive, but I thought that the presence of the NH2 negated this because it can donate its lone pair. Why does resonance destabilize the carbon then?
Also, is it really "acid-catalyzed" since the NH2 gains 2 H and the Carboxylic Acid gains an OH, which makes it seem like an H30+ was consumed.(2 votes)
- The protonation of the oxygen introduces a major resonance contributor that withdraws electrons from the carbon, thus making it more electrophilic and reactive. At this point, the NH2 is not yet attached to the carbon (and thus cannot donate electron density), but does so in the subsequent step because of the above reason. The way to think about the catalytic character of the acid is to think of what you started with and what you ended with. You started with H3O+ and a neutral NH2 on the carbonyl carbon and produced NH4+ and a neutral OH on the carbonyl carbon. So you started with an acid and you generated another equivalent acid. :)(1 vote)
- 8:28Jay says ‘we formed our carboxylic acid’ and then he proceeds to several other steps. Since the carboxylic acid is the intended final product of this mechanism, why not stop right at8:28?(1 vote)
- This reaction is in a strongly basic solution, so the carboxylic acid will instantaneously deprotonate to form the carboxylate anion.(2 votes)
- Which hydrolysis has a higher yield for this reaction?(1 vote)
- Amides are quite stable, so it takes vigorous conditions to hybridize them.
I doubt that there's much difference in yield using either acid- or base-catalyzed hydrolysis.(2 votes)
- at8:29we already formed our carboxylic acid which is exactly same as our final product. However, why we deprotonate again? that means we can deprotonate even final product as well like infinite loop? or the strong base can count like okay I protonate you once so I will not protonate second time?
Please, explain why protonating stops at the final carboxylic acid but not intermediate carboxylic acid.(1 vote)
- At8:29the carboxylic acid product is formed but NH2^- (called an azanide anion) is also formed. The azanide anion is a very strong base which is in the presence of a carboxylic acid which, as the name suggests, is acidic. This leads to an acid/base reaction to deprotonate the acid not necessarily because we want to but simply because it's unavoidable with our reagents.
You need to add the source of protons for the second half of the reaction at10:40to make the solution essentially acidic enough for the carboxylic acid to exist.
Hope that helps.(2 votes)
- At8:13, you made a correction stating the amino nitrogen is usually protonated first. So does the amide anion ever form in this case?(1 vote)
- The NH₂⁻ ion would not form in water because it is a stronger base than OH⁻.
It would immediately react with water to form OH⁻.(2 votes)
Voiceover: We've already see that amides are generally unreactive due to resonance stabilization, however you can get them to hydrolyze if you use harsh reaction conditions. If you use strong acid or a strong base and you heat things up for several hours, you can hydrolyze amides. If we take a look at this amide right here, we can break this bond using acid and heat and form a carboxylic acid. Let's look at the mechanism for acid-catalyzed hydrolysis of amides. The first step is protonation of the carbonyl oxygen. Lone pair of electrons on this oxygen picks up a proton from hydronium, leaving these electrons behind. We've seen in previous videos that protonating your carbonyl oxygen makes your carbonyl carbon more electrophilic. Let me draw in the NH2 here, we can follow those electrons along. The lone pair of electrons right here on the oxygen is going to pick up a proton from hydronium, forming this bond right here and giving the oxygen a plus one formal charge. To think about a resonance structure for this, it's going to withdraw some electron density from our carbonyl carbon here, making that carbon more electrophilic. Therefore it's going to react with a nucleophile. The nucleophile that's present would be water. Let me go ahead and draw water in here. If you deprotonate hydronium you would form H2O. Water is going to function as a nucleophile, attack our electrophile, which is our carbon, which would push these electrons off on to the oxygen. If we draw that we would now have on the left side an OH with two lone pairs of electrons. Let's follow those electrons in here. These electrons in green move off onto the oxygen. We're going to form a bond between the carbon and this oxygen here. This oxygen has also two hydrogens on it, lone pair of electrons, plus one formal charge. Let's say the electrons in blue. These electrons right here in blue form a bond between the carbon and that oxygen. That gives the oxygen a plus one formal charge. We also have our NH2 down here. In the next step we have a plus one formal charge so we're going to deprotonate to get rid of that plus one charge. Water's going to come along again, this time function as a base. We just saw water act as a nucleophile, and now water's going to act as a base. It's going to take a proton, leaving those electrons behind on the oxygen, and that gets rid of our formal charge. On the left side we would have an OH and we would also have an OH now on the right side. Let me go ahead and show those electrons. If we're going to deprotonate, take that proton, leave these electrons behind on the oxygen, those are those electrons right there. I could draw these electrons in blue to remind us of which ones they are. And then we still have an NH2. This time when I draw the NH2 I'm going to draw it out. I'm going to show lone pair of electrons on the nitrogen, I'm going to draw in those hydrogens. In the next step this amine is going to function as a base. We have hydronium present which we know is an acid. H3O plus. I'll draw in everything here. The amine is going to act as a base and hydronium is going to act as an acid. The amine is going to take a proton from H3O plus, leaving these electrons behind on the oxygen. Let's get some more room here. We have another acid base reaction. We're going to protonate the amine because that's going to give us a better leaving group. Let me go ahead and draw in everything else. Once again we have our OH on both sides. I'm going to draw those in. We still have our R group. Then over here our nitrogen now has four bonds. That gives nitrogen a plus one formal charge. Let's follow those electrons, let's make them magenta. These electrons right here on the nitrogen. I'm going to take this proton. We could say it's this one right here, giving the nitrogen a plus one formal charge. This gives us a good leaving group. If you look over here you can see ammonia is hiding as our leaving group. In our next step we're going to reform our carbonyl. These electrons right here are going to move in to reform our carbonyl. Once again, too many bonds to this carbon, so these electrons have to come off on to the nitrogen. Let's go ahead and draw what we would form. We would have an R group, we would have our carbon now double bonded to this oxygen, a plus one formal charge on this oxygen. Let's show those electrons. The electrons in here, in red, move in to reform our carbonyl. We still have an OH bonded to our carbon, but we pushed the electrons in here in blue off onto the nitrogen, so we form ammonia. Let me go ahead and draw in ammonia over here. We have ammonia now, NH3 with a lone pair of electrons on that nitrogen. I'll say that the electrons in blue were the ones that came off right here onto the nitrogen. We're almost to our final product of a carboxylic acid, all we have to do is deprotonate. Ammonia is going to function as a base. It's going to take this proton leaving these electrons behind on the oxygen. Of course that's going to give us our carboxylic acid. Let me go ahead and draw in the carboxylic acid here. Let's make these electrons, let's say they're green here. These electrons here in green come off onto the oxygen and we now have our carboxylic acid. Let me draw in the OH here. If I take ammonia and I add a proton to it that would form NH4 plus, ammonium. Let me see if I can squeeze that in here. We would have the ammonium ion, NH4 plus. The electrons in blue, these electrons in blue take this proton so we can form, let's say that bond, and we have our products. We formed a carboxylic acid and we formed ammonium. This last step here favors the formation of the products. This is what helps to drive this reaction to completion to make our carboxylic acid. Let's look at another way to hydrolyze an amide. This time we're going to use a base. We just did it using acid, so this time we're going to look at the mechanism using base. Once again we have our amide, we're going to add sodium hydroxide this time. The hydroxide anion is a better nucleophile than water so for this mechanism right away our hydroxide anion is going to function as a nucleophile, it's going to attack our carbonyl carbon here, pushing these electrons off onto the oxygen. Let's go ahead and show what we would make. We would now have on the left side an oxygen with three lone pairs of electrons, negative one formal charge. These electrons in here come off onto the oxygen. What else do we have bonded to this carbon? We have our R group, we have our NH2, and we have our OH. Let me go ahead and draw in the OH here. Let's highlight that pair of electrons on our sodium hydroxide. Let's say these electrons right here on hydroxide are the ones that form this bond between the carbon and the oxygen. Now we can reform our carbonyl. We're going to reform our carbonyl here. I should put in the lone pair of electrons on this nitrogen here, because when we reform our carbonyl, let's say these electrons move in here to form our carbonyl, then these electrons would come off onto the nitrogen. Let's go ahead and draw what we would have now. We have an R group, we would have our carbonyl has been reformed, we would have our oxygen and our hydrogen. We formed our carboxylic acid. Let me put these electrons here in blue so we can see which ones they are, like that. Let's think about what we lost here. We lost NH2. Let me go ahead and change colors. Let me make this yellow again. We have NH2 but let's think about our electrons. We already have a lone pair of electrons on the nitrogen, and we got another lone pair. Let me show where that came from. These electrons right here came off onto the nitrogen too, let's make them these right here. That gives the nitrogen a negative one formal charge. This is the amide anion which is a pretty strong base. We have a strong base and we have an acid. We're going to have an acid-base reaction. The base is going to take a proton from the acid. Let's say these electrons in red take a proton from the acid, leave these electrons behind. Let's go ahead and draw the product. We would have the conjugate base to a carboxylic acid, which is a carboxylate anion. I'm going to draw in the lone pairs on the oxygen here which gives the oxygen a negative one formal charge. Let's use green here. These electrons in here come off onto our oxygen so we form our carboxylate anion. Let's think about the other product. If we have NH2 minus and we're adding a proton to that then we would form NH3. We form ammonia. Let me go ahead and draw in ammonia here. Put my lone pair of electrons on the nitrogen. These electrons here in red pick up this proton. Let's say it forms this bond right here, so we make ammonia. Once again this acid-base equilibrium favors the formation of our product, so this helps to drive the reaction forward. We drive the reaction forward. Finally let's think about how we would get from a carboxylate anion to our carboxylic acid product. In the first step we added our sodium hydroxide, that gives us this. If we wanted to go from our carboxylate anion to a carboxylic acid we need a source of protons. The carboxylate anion picks up a proton and then forms our carboxylic acid. That's two ways to hydrolyze an amide. Let's take a look at our reaction here. Let's say our goal was to go from benzamide to benzoic acid. We just learned two ways to do that. One way to do that would be to add water and an acid. We get H3O plus and if we heat things up we know that we can hydrolyze our amide that way and give us our benzoic acid, so that's one possibility. That's acid-catalyzed amide hydrolysis. Or we could use base. Or, if we wanted to form a carboxylic acid, first thing we would do is sodium hydroxide and heat things up, Na plus OH minus, once again heat things up. That of course would give us a carboxylate anion, so we need to protonate in the second step to form benzoic acid as our product.