- 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 enolates from aldehydes using sodium ethoxide and sodium hydride. Created by Jay.
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- How do we know if some specie (such as etoxide) acts as a nucleophile or as a base (when reacting with a carbonile group)?(5 votes)
- If the species is as equally strong of a base as it is a nucleophile (like alkoxides such as ethoxide), then they will attack the less hindered primary (terminal) carbon much more quickly than the secondary carbon, and since the alkoxide is charged (is a strong nucleophile/base) it will react quickly and not wait around for something more thermodynamically favourable to show up. Because there is no leaving group on the α carbon, the alkoxide cannot attack as a nucleophile, therefore it acts as a base and deprotonates the terminal carbon.
If the carbonyl itself is terminal (aldehyde), then deprotonation of that carbon is unlikely because the carbonyl carbon is sp2 hybridized (therefore the C-H bond is stronger), and the more electronegative oxygen would rather grab the electron density anyway. However, this reaction is reversible, and the electrons could easily come back down and expel the newly attached group. This is evidenced by the hydrolysis of esters in aqueous solutions. However, I would expect some equilibria of both reactions to be happening, and isopropanol is itself only marginally more stable than ethanol.
If both the carbonyl and both neighbouring carbons are secondary, the equally strong nucleophile/bases will typically eliminate (deprotonate), again because that can happen more quickly than nucleophilic attack.(11 votes)
- Can someone explain how he derived pKeq = pKa acidL - pKa acidR at3:32? Thanks!(2 votes)
- For simpler typing, let's use 1 for acid L and 2 for acid R. Then
HA₁ ⇌ H⁺ + A₁⁻; Ka₁
H⁺ + A₂⁻ ⇌ HA₂; 1/Ka₂ (when you reverse an equation, you invert its K)
Add the two equations.
HA₁ + A₂⁻ ⇌ HA₂ + A₁⁻
Keq = Ka₁/Ka₂ (when you add two equations, you multiply their K values)
Take the negative logarithm of both sides of the equation.
pKeq = pKa₁ - pKa₂ = pKa(acid L) – pKa(acid R)(6 votes)
- Why does the base attack the alpha-hydrogen and not the hydrogen attached directly to the carbonyl carbon of the aldehyde?(2 votes)
- First of all remember that C-H bonds are not normally considered acidic. Bases attack acidic protons.
If the hydrogen from the alpha carbon is removed, the lone pair on the carbon can be delocalised to the carbonyl through resonance.
If the hydrogen of the aldehyde was removed the situation above cannot happen, the lone pair would be localised on that carbon.
The end result is that the second situation is is not energetically favourable.(4 votes)
- At5:45, it's mentioned that the bubbling out of solution of H2 gas causes the reaction to proceed towards the products. Am I correct in thinking this is due to our understanding of Le Chatelier's principle - as the abundance of one species decreases, the equilibrium will attempt to compensate for this by shifting towards the formation of that species (in our case H2 gas)?(3 votes)
- at5:20why is the carbanion a greater contributor to the enolate anion's resonance structure? I thought the more electronegative atom was most comfortable with the negative charge.(2 votes)
- I think he said that the oxyanion was the greater contributor, but he chose not to draw it.(2 votes)
- Why does equilibrium favour the formation of the weaker acid?
- The weaker acid will be a relatively stronger base than the stronger acid. Hence the equilibrium will favour the weaker acid being protonated and the stronger one being deprotonated.(1 vote)
- What benefit do you get from using sodium ethoxide over a strong base such as LDA?(1 vote)
- LDA amount will determine which position the enolate will be formed if your molecule is unsymmetrical. This is explained in the Kinetic and Thermodynamic elongates video.(1 vote)
- The anion formed during base catalysed keto-enol tautomerization is called enolate anion. What is the name given to the intermediate cation formed during acid catalysed keto-enol tautomerization?(1 vote)
- We learned in school that keto-enol tautomerization is only possible in an basic milieu. In acid the ketones remain ketones and vice versa.(1 vote)
- Why does equilibrium favour the formation of the weaker acid?(1 vote)
- At4:54, why doesn't the electrons of the alpha hydrogen directly form a double bond with the carbon and kicking the electrons of double bonds in carbonyl group to oxygen to form directly enolate anion (just like E2 reaction,since :H- is a strong base)?(1 vote)
Voiceover: In the last video we saw a formation of an enolate anion in one of our mechanisms. In this video and the next one we're going to go in much more detail about how to form enolate anions and what base to use. For example, this base right here is the ethoxide anion which you could get from sodium ethoxide, so Na plus OEt minus, and this could act as a base and take a proton from an aldehyde or a ketone. In this case we have an aldehyde, so this is acetaldehyde here, and we need to find the alpha carbon, so the alpha carbon is the one next to the carbonyl carbon, so this is the alpha carbon right here on acetaldehydes. There are three hydrogens attached to that alpha carbon, we have three alpha protons, so our base could take any one of those three alpha protons. I'm just going to draw one in here to simplify things. And so we could show our base taking this proton and leaving these electrons behind on our carbon, so we can draw the enolate anion that would form. So we have our cabonyl here and then we could show these electrons on this carbon now, so let me go ahead and follow those electrons, so in magenta these electrons move off on to this carbon to form a carbanion. Remember there's also two other hydrogens attached to that carbon there. And we could draw a resonance structure for this, so we could show these electrons in magenta moving in here to form a double bond, push these electrons off on to the oxygen, and if we do that we'll have a resonance structure showing the negative charge this time on the oxygen. So the oxygen gets a negative one formal charge now, and then we have a double bond over here on the right, and so this would be our other resonance structure. So the electrons in magenta moved in here to form our double bond, and then let's make these electrons in here blue, move off on to the oxygen to make an oxyanion. So this is our enolate anion here, so we have the carbanion form and then the oxyanion form of our enolate anion. So your enolate anion here, which is once again extremely important for reactions we'll talk about later. So our base has formed our enolate anion. If you think about what happens to the base, if you protonate ethoxide you're going to form ethanol. So we'll go ahead and show ethanol also formed here in this reaction. So we have an equilibrium here between our aldehydes and our enolate anion, and to figure out which direction is favored we need to know some pKa values. So the pKa value for this aldehyde is approximately 17, and the pKa value for ethanol is approximately 16. And so one way to figure out which direction is favored is to use these equations down here. We could first find the pKeq by taking the pKa of the acid on the left, so the acid on the left is our aldehyde, so the pKa is 17, and from that number we subtract the pKa of the acid on the right. The pKa of the acid on the right is 16, which is ethanol. So 17 minus 16 gives us one, and then to find the Keq we can take 10 to the negative pKeq, so 10 to the negative one is equal to point one, which is obviously less than one, and so we know that the equilibrium favors the reactants, the equilibrium is back in this direction so the equilibrium favors formation of the aldehyde, so at equilibrium we're going to have some aldehyde present, we're also going to have some enolate anion present here. So if you choose sodium ethoxide as your base you're going to have both the aldehyde and your enolate anion present. Another way to figure out which direction the equilibrium goes is to think about which one is the weaker acid. So we have these two acids here. Let me go ahead and change colors. We have a pKa of 17 and a pKa of 16. The lower the pKa the more acidic something is, so ethanol is more acidic than R aldehyde and the equilibrium favors formation of the weaker acid, and so since the aldehyde is the weaker acid the equilibrium favors formation of this weaker acid. So that's another way to think about which direction for the equilibrium. What if you wanted to completely make your enolate anion? So one thing you could do is add a base like a hydride, so here we have once again acetaldehyde here. So we have acetaldehyde, and this time our base is the hydride anion, so we could get that from something like sodium hydride, Na plus H minus, or potassium hydride, K plus H minus. And so once again we identify our alpha carbon, which is this one, with three alpha protons. So we can just go ahead and draw one of them in there. And we could show our hydride anion functioning as a base, taking this proton, leaving these electrons behind on that carbon. So let's go ahead and show the enolate anion that results. So we have our carbonyl here, and then we have our electrons on this carbon, giving it a negative one formal charge, so electrons in magenta move out on to this carbon, forming the carbanion. And just to save time I won't draw any oxyanion, but that's the one that's actually a greater contributor to the resonance hybrid here. So we have our carbanion and then we would also form hydrogen gas, so if this hydride anion picks up a proton we would form H2. And so let's show those electrons, so the electrons in red here on our hydride anion pick up this proton and forming this bond and so we get hydrogen gas, which would bubble out of solution, and since this is going to bubble out of solution we're going to drive the reaction to completion, so we're going to push the reaction to completion and so we're going to get our enolate anion so we're pretty much going to get complete formation of our enolate anion here. So in the next video we're going to talk about another base called LDA which can also give you complete formation of your enolate anion.