How to direct a mixed aldol condensation using a lithium enolate. Created by Jay.
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- Why does LDA form a cyclic ring when the last three times its been used it videos it hasnt?(10 votes)
- I know it looks confusing but it's really not anything different than what he was doing before. In this case, he's just showing it all happening at once instead of separately like he's shown in other videos. I'm only going to cover the first reaction sequence he draws since this is going to be a long answer, but the same methods apply each time.
In past videos he's shown the lithium as just an ionic interaction and not a full bond. You can think of it as the same case here. So, the first step is to protonate the LDA using one of the alpha protons (1st step of the ring cycle). This kicks the lithium off of the LDA (since it's not as ionically attracted anymore because the nitrogen's negative charge is balanced out by the hydrogen's positive charge once it bonds to the hydrogen). Now the lithium is no longer bonded to the nitrogen but, remember, this is all happening very quickly so the lithium doesn't really have a chance to move very far before the end of the reaction. Once the alpha hydrogen is removed, this forms a carbanion at the alpha carbon. The carbanion is less stable than the oxyanion (because the oxygen is more electronegative and can handle that negative charge better), so the alpha-carbon electron pair is going to move in to bond with the carbonyl carbon (forming the double bond; this is shown as the 2nd step of the ring cycle), which pushes one of the carbonyl double bonds off onto the oxygen atom. This gives the oxygen atom a -1 formal charge, which then attracts that positively charged lithium ion that's now floating around but still very close (remember, this happened quickly!). This is shown as the 3rd step of the ring cycle. So, going back through all of that, you can see that it's still the same mechanism that he's been showing in other videos with all of the same steps, he just condensed it into one diagram rather than drawing out each step individually. It was just a way of covering all of the same information a bit more quickly.(9 votes)
- Why did we protonate the hydroxy group again to make water the leaving group at7:00? Based on previous videos in alpha-carbon chemistry, I would assume that we would have a hydroxy group formed from the auto-ionization of water attack the alpha carbon's remaining H group between the carbonyl and the aldol's hydroxy group and kick the electrons up until said hydroxy group left (which would later be protonated by the H+ from the auto-ionization of water)(6 votes)
- as mentioned in the video, an acid workup was done. I think Jay did that to avoid mentioning that a hydroxyl group would be the leaving group. Because in aldol condensations, an E1cb mechanism is happening, which can kick out bad leaving groups (like OH or OR groups) from the compound due to the formation of a double bond(2 votes)
- Aren't you dropping a carbon when you're doing the LDA aldol condensation? The product only has 9 carbons where it should have 10 correct?(2 votes)
- It would have 10 carbons if the starting material reacted with itself, but this is a mixed aldol reaction. The product has 9 carbons because the carbanion that is formed after the addition of LDA has 5 carbons and it attacks the butanal that is added to solution in step 2 which has 4 carbons.(8 votes)
- I'm just wondering at ~7:00, why did the alcohol attack the H+ instead of the Ketone?(3 votes)
- The pKa of protonated ketones is around -7.3, while that of protonated alcohols is around -2.4. This means that the alcohol is about 100,000 times more basic than the ketone!(4 votes)
- Will there be videos on the Claisen condensation, Dieckman condensation, Michael reaction, and Stork reaction?(3 votes)
- Where does the TsOH come in? Is it considered the base in the final step instead of water? TsOH is not mentioned after briefly stating it is step 4 at the beginning.(2 votes)
- TsOH is used in the acid workup step. Remember TsOH is an abbreviation for Toluenesulfonic acid(2 votes)
- How can i identify major product in cross aldol of aldehyde and ketone which gives 4 different products. And which is the rate determining step in aldol condensation?(2 votes)
- at7:40how did that single bond just break onto the oxygen causing water to leave?(2 votes)
- i think the other way around is more reasonable
"water" wants to leave the whole group by stealing pair of electrons for the bond between itself and the carbon it is bound to. because 1) it's a good leaving group and 2) it has +1 formal charge. so it's insane to become a lot more stable with 0 formal charge by doing this
in short, water leaves, breaking the bond of it with the carbon(1 vote)
- What is the point of including these organic chemistry mechanism videos in the MCAT section? Are we really going to be tested on these mechanisms on the exam? They just seem like they've been randomly included without much explanation as to why they're important or how we're going to be tested on them.(0 votes)
- The MCAT could give you the reactants and reagents and expect you to know the products. Or expect you to know if a certain step in the reaction is an SN2 or E1cB, so you'd have to know the mechanism.(4 votes)
- why we canr do cannizzaro reaction in acidic medium ?(1 vote)
- The Cannizzaro reaction involves an intermolecular transfer of a hydride ion (H⁻).
A hydride ion would not survive in an acidic (H⁺) medium.(2 votes)
Voiceover: Here's another way to do a mixed, or crossed aldol condensation, this time using a lithium enolate. So if we took this ketone and this aldohyde and just mixed them together with some base, we would get a mixture of products. We wouldn't get our desired product, this conjugated enone over here. So if we do this step-wise, we can do a directed aldol condensation. So if we take this ketone and add LDA to it, and then we add our aldohyde, and then in our work up, we can add water and antolymene sulfonic acid as an acid source, we can get our conjugated enone in a decent yield. So let's look at each of these steps one by one. So first we'll start off with the addition of LDA which we know is a strong base. So the strong base is going to take a proton from our ketone, and so let's analyze our ketone here. So the alpha carbons are the ones next to the carbonyl, and so this is an alpha carbon on our ketone, and this is an alpha carbon on our ketone. So which one of those alpha carbons will our strong base deprotonate? So we saw in a previous video, that LDA is going to form the kinetic enolate, so it's going to take a proton from the least sterically hindered side. And so it's going to take a proton from this right side over here to form the kinetic enolate. Right, if I took one from the left, I would be the thermodynamic enolate. So I can go ahead and draw in an alpha proton on our alpha carbon right here. So let's say this is the one that LDA is going to take. So let's go ahead and draw in LDA. So I'm going to go ahead and draw my nitrogen in here, and then I have my isopropyl groups on my nitrogen, and then I'm going to have the nitrogen bonded to lithium here like that. So when this deprotonation occurs, it's actually a cyclic mechanism, so the oxygen is going to be forming a bond with this lithium, the nitrogen is going to be forming a bond with this proton here. So let's go ahead and show the cyclic mechanism. So if these electrons right here on the nitrogen anion, move in and take this proton, we're going to form a bond right here. These electrons will move in here to form a carbon-carbon double bond, and then these electrons would kick off onto your oxygen to form a bond with lithium. So let's go ahead and show the product, and then we'll show the movement of those electrons here. So we have our oxygen, it's going to be bonded to lithium like that, and now we have a carbon-carbon double bond. And then we would have nitrogen bonded to hydrogen now and we have an amine over here on the right. So let's go ahead and follow some of those electrons. So if I'm saying these electrons in blue here on the nitrogen anion are going to pick up this proton to form our amine here, like that, I could say that these electrons in here, once it's deprotonated, they could form our double bonds right in here and then I could also point out that these electrons right here could move out to bond with the lithium to form our lithium enolate. So once again, this is the kinetic enolate, the one that's formed the fastest, And the use of LDA helps us to get our kinetic enolate because these bulky isopropyl groups would prevent deprotonation at the alpha carbon on the left. So now we've formed our lithium enolate, so let's go to the next step. Alright, so in the next step, we're going to add our aldohyde right here So we're going to add butanol, so let's go ahead and look at the formation of our lithium enolase, and then let's add butanol to that. So let's go ahead and draw in butanol. So here we have our carbonyl, and then we would have four carbons, so two, three, four, like that. We know that aldohydes can function as electrophiles, so in the second step we're going to add an electrophilic carbonyl compound. We know this is electrophilic because the oxygen is partially negative and this carbonyl carbon right here is partially positive, like that. And so, in the second step, it's going to be a nucleophile attacking an electrophile, so the enolase is going to function as a nucleophile, so we're going to form a bond between lithium and oxygen over here, and then we're going to form a carbon-carbon bond down here. So let's show the movement of those electrons, the lithium enolate functions as our nucleophile, the aldohyde functions as our electrophile. So if these electrons move into here, then these electrons are going to bond with that carbon, and then these electrons can bond in here, so another cyclic mechanism. So let's go ahead and draw the result of that. We would have these carbons, we would form a carbonyl right here, now we would form a new carbon-carbon bond right here, and then our oxygen would now be bonded to our lithium, and then we'll draw in the rest of our carbons. So we form a lithium alkoxide product. Following those electrons, these electrons in here in red moved in to here to form our carbonyl. The electrons here in magenta, these are the ones that attacked our carbonyl carbon here to form our new carbon-carbon bond. And then, finally, I could say make these electrons in here blue, and then forming a bond between oxygen and lithium to form our lithium alkoxide intermediate here. And so, two cyclic mechanisms form your carbon-carbon bond here. So first, you would deprotonate. We talked about this cyclic mechanism up here. And then once you add your aldohyde, once you've formed your lithium enolate, another cyclic mechanism will give you your lithium alkoxide. So let's look at the next step, the third step, you're going to add water in your workup. And so here, I've drawn the lithium alkoxide a little differently. The electrons in blue that I had over here in the the lithium-oxygen bond, I'm just going to put those on the oxygen this time. We know the oxygen has two other lone pairs of electrons which give it a negative one formal charge, and then lithium will be a plus one charge, and so this is just another way to represent our lithium alkoxide. This is how we've usually done it in our videos here. And so, in the next step, in the work up, if you do aqueous work ups, you add some water here, you would protonate your alkoxide anion, so let's go ahead and show that, so we take a proton here from water, and let's go ahead and draw the products. So, once again, we would have our carbonyl, and then we would protonate the oxygen to form our aldol. So here is our aldol product. Alright, so once you form your aldol, let's go ahead and look at the next step here. We added some tolymine sulfonic acid, so we added a source of protons. So in the final step, to get to our enone, we need to dehydrate our aldol. We've seen how to form our enone using base, and this time we're going to do an acid catalyzed dehydration. So, if you add a source of protons here, we know we still have two lone pairs of electrons on our oxygen. One of those lone pairs could pick up that proton, so we protonate. Let's go ahead and draw what we would make here. We have our carbonyl, and we have all of these carbons, and now we would have this oxygen, still has one lone pair of electrons, so let's go ahead and show those electrons. These electrons in magenta, picked up a proton right here so we have one lone pair of electrons left, which gives that oxygen a plus one formal charge. So we have an excellent leaving group here If we think about these electrons moving off on to the oxygen we have water as a leaving group. So loss of water at this step would yield a cation. So let's go ahead and draw the cation that would form. We have our carbonyl, and we have all of these carbons, and so we lost a bond to this carbon right here, so this is the carbon that's going to form our cations. Let's go ahead and draw a plus one formal charge on our cation like that. And so, in the next step, a base is going to come along, and it's going to take a proton from our alpha carbon. so once again we think about where our carbonyl is, the carbon next to it is our alpha carbon, and so there's a proton on here. So we can deprotonate it with a base. I'm just going to write a generic base, but it could be something like the water that just left in the previous step. So, a base is going to come along, and take this proton, and so these electrons are gonna move in here to form your double bond. That's going to take away your plus one formal charge. So let's go ahead and draw our product. so we would have our carbonyl, and then we would now have a double bond right here, and then we'd draw in the rest of our carbons like that. So if we show those electrons, let's make them blue this time, so these electrons in here, the base takes the protons, and the electrons move in to give us our double bond, and we now have formed our enone as our product. So once again, this is an example of a directed aldol condensation.