- Alpha-carbon chemistry questions
- Aldol reactions in metabolism
- Keto-enol tautomerization (by Jay)
- Enolate formation from aldehydes
- Enolate formation from ketones
- Kinetic and thermodynamic enolates
- Aldol condensation
- Mixed (crossed) aldol condensation
- Mixed (crossed) aldol condensation using a lithium enolate
- Retro-aldol and retrosynthesis
- Intramolecular aldol condensation
Learn about mechanisms for acid and base catalyzed keto-enol tautomerization. Uncover how aldehydes or ketones, with a dash of acid or base, can transform into an enol. Explore the role of alpha protons and double bonds in this transformation, and understand the significance of chiral centers in the process. Created by Jay.
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- 3:17Why does the H2O act as a base, rather than a nucleophile?(8 votes)
- It just so happens that all Lewis Bases are Nucleophiles. A lewis base is any molecule that donates electrons, normally from a lone pair, to another molecules empty orbital. They are "searching for a nucleus for the electrons to go so are nucleophilic. This other molecule is called an electrophile and is a Lewis Acid. They are "searching" for electrons so are electrophilic. So to cut a long story short, a base is a nucleophile and a acid is a electrophile.(9 votes)
- This may be a silly question but does the "keto" term apply to both ketones and aldehydes or only ketones?(5 votes)
- It's not a silly question and, yes, the C=O group is often called a keto group whether it is part of an aldehyde or ketone.(19 votes)
- What reaction takes place faster or is "easier" to do, acid-catalyzed or base-catalyzed tautomerization? For example, would a chemist in a lab choose an acidic solution or a basic solution if they wanted to make the enol form in greater concentration than it normally is for a given molecule?(7 votes)
- Depends on the purpose of the reaction. Generally speaking (from my own experience in a synthesis lab) the base catalyzed tautomerization is more practical. The enolate anion (a strong nucleophile) allows us to insert various R groups at the alpha position. Furthermore, a very common reaction in organic synthesis is the formation of a silyl enol ether. Silyl enol ethers are formed using a strong base and TMSCl. The silyl enol ethers are intermediates to many reactions in organic synthesis (Mukaiyama-Michael addition, Saegusa–Ito oxidation, haloketone formation, etc). Hope that helps.(6 votes)
- is it always necessary that in enol form the double bond has to be always closer to the alcohol group(3 votes)
- Yes, definitely....then only the oxygen atom can donate electrons and form /share bonds with the other hydrogen bonds...(5 votes)
- In the base catalyzed reaction, why doesn't -OH do a nucleophilic attack the carbonyl carbon instead of the alpha hydrogen?(4 votes)
- The hydrogens on the alpha carbon are more acidic, so OH is more likely to pull off that proton rather than go for a nucleophilic attack on the carbonyl carbon.(2 votes)
- How do I identify the alpha carbon? For example in halyde compounds which contain more than one halogen atom?(3 votes)
- You look for the functional group of the molecule (eg. aldehyde, ketone group etc.) and then find the carbon right next to that (excluding the carbon of the functional group)(1 vote)
- how to number the alpha carbon in a ketone, since it has two R group? If i change a Keto to an Enol, while the pi bond be in the longer strand or the shorter strand?(2 votes)
- alpha carbons simply means the adjacent or first carbon next to a functional group. (side note: there's also beta, gamma, delta.. etc carbons which correspond to the 2nd, 3rd, 4th, carbon away from the functional group).
In the case of ketone this just means it has 2 alpha carbons. Not sure what you mean by 'shorter or longer strand', but you're alpha carbon needs to have the hydrogen to undergo ->enol tautomer.
If you're trying to get crazier than that like "what if both alpha carbons have hydrogens, which one will form the pi bond?" - I dunno, need more study lol. but I would guess from that point you would need to factor in which alpha carbon would be possible to react with. So steric hindrance? also resonance, stability, etc..(3 votes)
- Great video! One question: at10:31, he points out the alpha carbon in the ring structure. This carbon would have two H atoms bonded to it, which would make it achiral, is that correct? I think my misunderstanding may be based on whether chirality is a prerequisite for the reaction here...Thanks for any help!(1 vote)
- This may be a silly question but why in10:31in first case keto form is more favoured but in the second one enol is more ?(2 votes)
- Why attacking the alpha's hydrogen ?(1 vote)
Voiceover: If you start with an aldehyde or a ketone and add a catalytic amount of acid or base, you'll find the aldehyde or ketone is going to be in equilibrium with this product over here on the right which we call an enol. The name enol comes from the fact that we have a double bond in the molecule, so that's where the EN part comes in, and we also have an alcohol. You can see the OH over here so that's where the OL comes in. This is the enol form and then over here this is the keto form. The keto form and the enol form, and these are different molecules. They're isomers of each other so we call them tautomers and they're in equilibrium with each other. They're not different resonance structures. Let's see if we can analyze our aldehyde or ketone to see how to form our enol. If we look at the carbon that's next to the carbonyl carbon we call this the alpha carbon, and there are two hydrogens attached to the alpha carbon in this case. Let me go ahead and draw those in. Those are called the alpha protons. If we think about transferring one of those alpha protons from the alpha carbon to the oxygen, even though it's most likely not the same proton, it just helps to think about doing that. We can also think about moving the double bond. Over here on the left, the double bond is between the carbon and the oxygen and we're moving that double bond over here between the two carbons. Transferring one alpha proton and shifting your double bond converts the keto form into the enol form. Then, we also have a hydrogen, right? Over here, we still have a hydrogen left on this carbon, so let me go ahead and draw in that hydrogen. That's this hydrogen in blue here. That's how to think about converting a keto tautomer into an enol one. Let's look at the acid-catalyzed mechanism for this. If we start with our aldehyde or ketone and add H three O plus, the first thing that's gonna happen is protonation of our carbonyl and so a lone pair of electrons picks up this proton like that. We can go ahead and draw that. We would protonate our carbonyl so now our oxygen would have a plus one formal charge. Let me just go ahead and draw in those electrons here. Let's say we started with an aldehyde. We'll make this an H. The lone pair of electrons on our oxygen picked up a proton, like that. We can draw a resonance structure for this. We can move these electrons off onto our oxygen so let's go ahead and show a resonance structure. We would have our R group, all right, and now we would have our oxygen with two lone pairs of electrons. Let me go ahead and draw in those two lone pairs of electrons on our oxygen. Then we took a bond away from carbon, right? We took a bond away from this carbon so this carbon right here. Plus one formal charge on that carbon. Then we could show the movement of those electrons. These electrons right here I'm saying moving out onto the oxygen, like that. This is our intermediate here. All right. We know that our alpha carbon has two protons on it. Once again, let's find our alpha carbon. Here it is right here. We know we have two protons attached to it, two alpha protons, if you will. In the next step of our mechanism we're gonna get a molecule of water acting as a base. Let me go ahead and show a molecule of water here. The water's gonna take one of those alpha protons. Let's say once again, it takes this alpha proton and leave these electrons behind. They're gonna move in here to form our double bond. Let's go ahead and draw our product. We would have our R group here and now we would have a double bond formed between our two carbons and then we would have our oxygen, and then we would have two lone pairs of electrons on our oxygen. We would have our hydrogen, and then we would have another hydrogen right here. Let's go ahead and follow some of those electrons. Let's go ahead and make these electrons in here blue. These electrons are gonna move in, in here. It doesn't really matter which one you say it is, let's just say it's that one to form our double bond, and then, the electrons in red moved off onto this oxygen, and then we said that these electrons were in magenta. You can see that we have formed our enol here. This is our enol and then we started with our keto form like that. Keto-enol tautomerization. Let's look at the base catalyzed version. Once again, we start with our aldehyde or ketone but this time we're going to add a base. Something like hydroxide. We find our alpha carbon. Here's our alpha carbon. Once again, with two alpha protons. I'm gonna go ahead and draw in those two protons here. The base is gonna take one of those protons. Let's say it takes this one over here on the right. That leaves these electrons behind on this carbon. Let's go ahead and draw the resulting anion here. We would have our carbonyl like that. Once again let's say we started with an aldehyde and then we would have a lone pair of electrons on this carbon, the carbon in red here. Let me go ahead and identify those electrons so these electrons in here in magenta have moved off onto this carbon like that, which gives that carbon a negative one formal charge. It's a carbanion. There's still a hydrogen attached to that carbon in red. This hydrogen right here is still attached to it, I'm just not drawing it in so we can see a little bit better. All right. This is one form of the anion that we could have. We could draw a resonance structure to show the other form, so if we moved these electrons in magenta into here and pushed these electrons off onto this oxygen. Let's draw the resonance structure. We would have our R group here, we would have a double bond, and then our oxygen would have three lone pairs of electrons giving it a negative one formal charge, and then we would have our hydrogen over here. The electrons in magenta moved in here to form our pi bond and then we can say that these electrons in here moved off onto our oxygen. We could go ahead and show that. Let me just go ahead and put the other bracket on here. We have two forms of this anion. This is called the enolate anion. This is the enolate anion. This is going to be extremely important in future reactions. You can see the enolate anion has two resonance structures. One where we're showing the negative charge on the carbon. That would be this one right over here. The negative charge on the carbons. This is our carbanion form, so carbanion. Then we also have a resonance structure where the negative charge is on the oxygen so we could call this oxyanion. If you think about which one contributes more to the overall hybrid, oxygen is more electronegative than carbon and so, it's better able to have a negative one formal charge on it. The oxyanion contributes more to the resonance hybrids. All right. Let's think about the last step in our mechanism to form our enol. If we think about our oxyanion, all we'd have to do is protonate that oxygen here. We could just go ahead and draw a water molecule. We have a water molecule. This time water's going to function as an acid it's going to donate a proton. Let say these electrons in blue take this proton, leave these electrons behind, and so from our oxyanion, we can go ahead and draw our enol product. We have our R group here. We would have our double bond, we would have our oxygen, all right. Now protonated like this to form our enol product. Let me just go ahead and show those electrons in blue. Picked up a proton here to form our enol. That's how to get there using base-catalyze. Once again, we will talk much more about the enolate anion in future videos here. Let's look at a situation where the alpha carbon is a chiral center. Let's look at this right here. Here's our alpha carbon. Let's just say it's a chiral center. If R and R double prime are different from each other we would have four different things attached to this carbon. The alpha carbon here is SP three hybridized with tetrahedral geometry. Let's say it's either the R or the S enantiomer. It doesn't really matter which one. You can see now we have only one alpha proton. Only one alpha proton but because there is an alpha proton we can form an enol. In either an acid or base-catalyzed mechanism we could think about the proton here in red, you could think about transferring one to this oxygen and moving your double bond, and then we form our enol. Here is our enol. Now let's look and see what happened to the carbon in red right here. On the left, the alpha carbon was SP three hybridized with tetrahedral geometry. Now, this carbon is SP two hybridized with trigonal planar geometry. Whatever stereochemical information we had over here on the left, whether it was the R or the S enantiomer, it's been lost now that we've formed the enol. The enol is achiral, it's flat, it's planar. When we reform the keto form, so one of the possibilities is to form the enantiomer that we started with but the other possibility is to form the other enantiomer. You can see that's what I've shown here. I've shown the hydrogen now going away from us and our R double prime group coming out at us. This is the enantiomer. Because we formed the enol we can get a mixture of enantiomers. Enolization can lead to racemization. We can get a mixture of enantiomers and if we wait long enough, we can get an equal mixture of these guys. This one and this one would be in equilibrium with our enol form. That's something to think about if you have a chiral center at your alpha carbon. Let's look at two quick examples of keto and enol forms. Over here on the left we have cyclohexanone and on the right would be the enol version of it. You could think about one of these as being your alpha carbon, right, and you could move these electrons in here and push those electrons off. You could see that would give you this enol form. It turns out that the keto form is favored. The equilibrium is actually far to the left favoring formation of the keto form. Even under just normal conditions, so not acid or base-catalyzed. There's only a trace amount of the enol presence however, there are some cases where the enol is extra-stabilized and that's the case for this example down here. We have the keto form and we have the enol form. Once again, you could think about these electrons moving in here, pushing those electrons off giving you your enol form. This is a specially-stabilized enol, right? This is phenol right here. We know that phenol has an aromatic ring. The formation of the enol form is extra-stabilize because of this aromatic ring. This time the equilibrium is actually to the right and much more of it is in the enol form than in the keto form. In this case, we have some special stabilization.