- 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
How to form different enolates from ketones. Created by Jay.
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- At6:31, why does the kinetic reaction need to take place in a cold environment? I would have assumed the "fast" kinetic product can be formed at room temperature, but the thermodynamic product would need the temperature control.(8 votes)
- The kinetic product benefits from the colder temperature because it has a lower activation energy and the activation energy of the reverse reaction (products reforming the reactants) is to great for the reverse reaction to proceed. Once the kinetic reaction occurs, which is the faster of the two reactions, the kinetic products won't be able to reform the reactants and possibly undergo a reaction forming the more stable thermodynamic product. Consequently, more of the reactants will be locked in to the kinetic product.
At higher temperatures the reaction will favor the more stable thermodynamic product because the reverse reaction can occur for the kinetic products and will have a lower activation energy than the reverse reaction for the thermodynamic products. Thus, kinetic products will be more likely to reform the reactants than thermodynamic products.
Consequently, the higher temperatures allow reversal of the reaction and the higher activation energy of the thermodynamic product will produce an even greater activation energy for the reverse reaction if the initial reaction is exothermic. Higher activation energies indicate less favorable reactions so the products are more locked in to the thermodynamic product.(15 votes)
- What is the reasoning behind the fact that extra akyl substitution leads to more thermodynamically stable? I thought akyl groups are EDG's, therefore, more carbons would de-stabilize the carbon double bond (since carbons are not very electronegative).(12 votes)
- At4:40, what is meant by "more substituted"?(2 votes)
- Why does hydride from NaH deprotonate the enolate, but hydride from LiAlH4 nucleophilically attack the back of the carbonyl to reduce it to the alcohol?(2 votes)
- NaH is a strong base so it will deprotonate the carboxylic acid first. This makes the hydride ion lose/ decrease its nucleophilicity. LAH does not have problem with this so it will strongly reduce carboxylic acid and ester to alcohol. NaBH4 cannot reduce carboxylic acid because it is not reactive enough.(0 votes)
- What is the function of the enolate?(1 vote)
- It's an excellent nucleophile. Look up "alpha carbon chemistry" for examples of the various reactions/roles of such nucleophiles (i.e. alkylation, aldol, etc.).(2 votes)
- Do all bases (not just the big bulky ones like LDA) take the more sterically hindered hydrogen?(1 vote)
- Wait! At3:13instead of kicking the electrons of alpha hydrogen to form double bond, if those electrons are taken by the carbon,then we would have form carbanion,but the carbon left to this(i.e., beta carbon) is electron donating group which destabilizes the carbanion.therefore that shouldn't be a product.where am i wrong??(1 vote)
- Can we use DBN,or DBU instead of LDA to form the kinetic product (in other words least substituted enolate anion)?(1 vote)
- what about the acid catalysed enolate formation? how to predict which enolate will form in this case?(1 vote)
- What's stopping the hydride from forming the kinetic enolate? Probability of this seems higher than forming the thermodynamic one (3/5 vs. 2/5 in the first example)(1 vote)
- Theoretically, nothing is stoping the hydride from forming the kinetic enolate...We're just unable to isolate (or trap it). In fact, we probably would still get a very very small amount of it. However, your probabilities (3/5 vs 2/5) are too simplistic (and somewhat invalid). As mentioned in the video, consider what the most stable double bond would be. Upon deprotonation, the formation of the carbo-anion is much more stable on a more substituted carbon...thus, the thermodynamic product is favored...and more probable.
Also, it is worth noting that NaH is in a solid state. Many Na+ salts (NaH, NaBH4, etc) are insoluble at cold temperatures. This once again favors the thermodynamic product since the reaction would have to be done at higher temperatures (surpassing the Ea for thermo) and for longer periods of time. Higher temp and longer reaction time make the thermodynamic product more probable.(1 vote)
Voiceover: We've already seen how to form enolates from ketones. What happens if you don't start with a symmetrical ketone? Right here, this ketone is not symmetrical. If we look at the right side, we have a methyl group. On the left side of our carbonyl, there is a CH2 and then an R group, an alkyl group. This is different. If we look at our alpha carbons, an alpha carbon is the one next to our carbonyl. It's on the left side. This is an alpha carbon. On the right side, this is an aplha carbon. When we look at alpha protons on the this alpha carbon on the right, there would be three alpha protons. On the alpha carbon on the left, there would be two alpha protons. The question is which one of those alpha protons are we going to take with our base? The answer lies in what kind of base we use and also reaction conditions. If we use a base like LDA, lithium diisopropylamide, we talked about this base in a previous video. It's a very strong base, but it's also very sterically hindered. We have these big isopropyl groups, which means that LDA wants to approach our ketone from the least sterically hindered side. This left side, here, has our alkyl group, which is much bulkier than the hydogren. This could interfere if it approaced from the left side. It's much more likely to approach from the right side and therefore, take one of the alpha protons on the alpha carbon on the right. Let's say that this lone pair of electrons, here, takes this proton, which moves these electrons in here to form a double bond, kicks these electrons off onto our oxygen. We can go ahead and draw the enolate anion that would result. We would have our oxygen up here with three lone pairs of electrons, negative one formal charge, a double bond over here on the right, and then we still have two hydrogens attached to that carbon. This is our enolate anion. Let's follow those electrons. The electrons here in magenta moved in to form our double bond. Then, we could say that these electrons in here moved out onto our oxygen to form our enolate. Since lithium is present, if you wanted to, you could put lithium here. LI plus, like that. This is the enolate anion that we would get. We call this the kinetic enolate. Let me go ahead and write that down, here. This is the kinetic enolate. It's called the kinetic enolate because this is the one that forms the fastest. Think about kinetic and speed. This is the one that would form the fastest because we used LDA as our base and because of the choice of our base, and also because of probability. On the alpha carbon on the right, we had three alpha protons, greater chance of taking of of these as opposed to these two over here on the left. The kinetic enolate is the one that forms the fastest. Let's look at forming another kind of enolate. Let's think about deprotonating the alpha carbon on the left this time. This is our alpha carbon over here with two alpha protons. Let's use a different base this time. Let's use sodium hydride, so NA plus H minus, or you could use potassium hydride. These are sources of hydride anions, which we know connect as a base. The hydride anion could take this proton, leaving these electrons in here, pushing these electrons off onto our oxygen. Then, we draw a different enolate. We can show the double bond now is between those two carbons. Then we have our oxygen up here with a negative one formal charge. Then we have our methyl group over here. Then we left one hydrogen behind. We have a different enolate. Once again, following some electrons, the electrons in magenta, moved in here to form our double bond and then the electrons in blue, here, moved out onto our oxygen. This is a different enolate and we call this the thermodynamic enolate. Let me go ahead and write that. The thermodynamic enolate. Let's analyze these two enolates that we formed in terms of stability. To do that, we need to look at the substitution at the double bond. Going back up here to the kinetic enolate, I look at my double bond, I think about how substituted it is. We have these two hydrogens over here on this side. That's actually not as substituted as the thermodynamic enolate. If we look at the thermodynamic enolate in the double bond, we'll only have one hydrogen here. We have an R group over here. We also have this alkyl group, if you're thinking about it. This is actually the more stable enolate that forms because we know the more substituted the double bond, the more stable it is. The thermodynamic enolate is the more stable enolate. It's more substituted. The kinetic enolate is not as stable, but it is the one that forms that fastest. Once again, you can control which one of these enolates you form depending on the base that you use. Let's look at a problem where we have a ketone. Then we're going to add our two different bases to our ketones. Here's our ketone. First, let's add some sodium hydride. We know that sodium hydride puts our ketone under thermodynamic control. When we identify our alpha carbons, let's go ahead and do that. Alpha carbons are the ones next to our carbonyls. This would be an alpha carbon over here on the right. Then this would be an alpha carbon over here on the left. We think about how many alpha protons we have. For the alpha carbon on the right, there's only one. For the alpha carbon on the left, there would be two alpha protons. If I think about thermodynamic control, I know that's going to form the more stable enolotes, the thermodynamic enolate. I know that hydride is going to take the proton on the right, leaving these electrons in here and pushing those electrons off onto your oxygen. Let's go ahead and draw this enolate. We would push the electrons in here and then we would have our oxygen up here with three lone pairs of electrons. Negative one formal charge. We still have our methyl group right here, CH three. Showing those electron in magenta, moving in here to form our double bond. This is the thermodynamic enolate. It's more substituted. Because of this methyl group here, this double bond is more substituted than if we showed an enolate forming from the other side. This is our thermodynamic enolate. What if we use something like LDA and we make our temperatures very cold? Negative 78 degrees Celsius. This is going to take protons from the least sterically hindered side, which we know is going to be the left side. Over here on the right side, we have this methyl group. The LDA is going to approach from the left side. Let me go ahead and draw that in. Here we have our LDA. Negative one formal charge on the nitrogen. We could show a lone pair of electrons taking this proton, least sterically hindered one. These electrons move in here, these electrons kick off onto your oxygen. Let's go ahead and show the product of that. We would have our ring. We would have our methyl group over here. We would now have a double bond here. Then we would have our oxygen with three lone pairs of electrons and a negative one formal charge. Again, if you wanted to draw your lithium in there. This would be the enolate anion that would form. Showing those electrons, let's go ahead and make them in blue this time. These electrons in here moved in to form our double bond. This is our kinetic enolate. Once again, we have our kinetic versus our thermodynamic. The kinetic enolate is not as substituted because we have a hydrogen here. It's not as substituted, it's not as stable, but it's favored by low temperatures and strong sterically hindered base like LDA. We're going to form this as our enolate using this base and these reaction conditions. At a little it of a higher temperature, and use of a non-sterically hindered base like sodium hydride, we're going to form the more stable enolate, the thermal dynamic enolate. You can control which enolate you form, once again, based on the type of base that you use and also your reaction conditions.