Transcript:
So let’s begin our discussion with this reaction. Let’s say we have cyclopentanone and in the first step, Let’s react it with LDA. And in the second step, we’re going to react it with methyl bromide. What’s the major product of this reaction? Well, first we need to know what. Ld a is LD. A is lithium diisopropylamide. So we have a lithium with a positive charge and we have a nitrogen with two isopropyl groups and two lone pairs. This reagent. Ld a is a strong base and the purpose of it is to remove the alpha hydrogen. So in the first step, we’re going to deprotonate the acidic hydrogen, so the diisopropylamide with its negative charge, it’s going to take off the hydrogen, and so we’re gonna get a resident stabilized inally ion, and so here’s the other resonance form that we could draw. Now in the next step, we can take one of the two resonance forms. It really doesn’t matter which one and then we could react it with methyl bromide. So then this auction’s gonna reform the Python, causing this Pi Bond to break attacking the methyl group, expelling the bromide ion and so this is how we can alkylate a ketone, so that’s how we can add a methyl group to the alpha carbon of a ketone. Now granted, we could use the other resonance structure, so the carbon with the negative charge is the nucleophilic carbon and it could attack the methyl group to giving us the same product, so let’s work on some practice problems, so let’s say we have this ketone here. And in the first step, we’re going to react in with LDA, and in the second step, we’re going to use Ethyl Bromide, draw the major product of this reaction, so all we need to do is replace the alpha hydrogen with the R group, so this is going to be the major product and that’s it now. Let’s say if we have an unsymmetrical ketone such as this one. What the agents do we need in order to put the methyl group on? Let’s say the right side of the ketone. And what reagents should we use if we want to add a methyl group to the left side of the ketone? Now, if we want to add it to the left side, this is associated with the kinetic product we need to use. Lda, add a little temperature and then we can add methyl bromide and the second step now to get the other product, we need to use a base that is not sterically hindered sodium hydride at room temperature followed by methyl bromide. So this product is the thermal dynamic product, whereas this one is the kinetic product. Now let’s go over the first reaction. So there’s three alpha hydrogen atoms that can be removed. The blue alpha hydrogen or one of the Red Alpha hydrogen’s sodium hydride is a strong base and it can remove the blue hydrogen or the green hydrogen. Now it’s a small base so it can remove both. It’s not sterically hindered, so lets. Remove the blue hydrogen. If we do, so we’re going to get and this intermediate. So the double bond will be on the right side. If the hydride ion removes the green hydrogen, the double bond will be on the left side and so we can get both of these in LA. Ions, however, at higher temperatures, The most stable inally ion is the one that’s gonna form. It’s gonna be the major product to figure out which one is going to be the most stable in LA. Ion, we need to look at the carbon-carbon double bond see which one is more substituted. This carbon-carbon double bond has three. R groups, whereas this one only has two. Therefore, this is more stable, enolate ion and so it’s going to be used to form the major product. So then we can react this with methyl bromide, given us this product as our major product. Now, if we use a sterically hindered base like LDA and at low temperature, this will favor the kinetic product over the thermal dynamic product, and so LDA is a bulky base so as a result, it prefers to go for the hydrogen atom that is more accessible, so these hydrogen atoms. The secondary hydrogen atoms are more accessible than the tertiary blue hydrogen atom, so LDN prefer to grab this hydrogen and so this is going to lead to the kinetic product and then in the final step, we can react this with methyl bromide, given us this product with the methyl group on the left side. Now there are other ways by which we can alkylate a ketone and that’s by using an enamine intermediate, so let’s react it with a secondary amine in order to get the ENA mean, so it’s gonna look like this, and now once you have the enemy, you can react it with an alkyl halide, so let’s use methyl bromide as an example, so the lone pair from the nitrogen will be used to form a double bond, causing this pi bond to break attack in the methyl group and expel in a bromine atom. So now the nitrogen has a positive charge and we’ve added the methyl group to the alpha carbon, so at this point to get rid of the nitrogen group and convert it back to the ketone form. All I need to do is add h3o, plus, which could be a mixture of water and HCL, and so that’s another way in which you can alkylate a ketone. The advantage of going through this route is you don’t have to use a strong base like Lda, and also it’s a very good way to get mono alkylated products. Now we can also react the enemy in intermediate with other electrophiles not just Alkyl halides so another electrophile that we could react it with is the alpha beta unsaturated aldehyde. So in this example, the lone pair from the nitrogen will be used to form a PI bond, causing this double bond to attack the beta carbon, causing these electrons to move over here break in this Pi Bond. So now we have a double bond in this region and we have an oxygen with a negative charge and right now. The nitrogen atom has a positive formal charge. Now we’re going to react this with water, so we’re going to take a lone pair from the oxygen atom form a double bond and it’s gonna push these electrons to take a hydrogen expelling hydroxide. So right, now we have the aldehyde again and our next step is to get rid of the nitrogen atom and to do that. We’re going to use h3o plus, and so this is a final answer For this example. Now we can also react the enamine intermediate with an acid chloride. So what do you think the major product for? This reaction will be feel free to pause the video and try it. So you know what’s going to happen. The enamine will attack the acid chloride and so now the nitrogen atom has a positive charge and the oxygen atom has a negative charge in the next step. We need to get rid of the chlorine atom. It’s a good leaving group, so right, now we have a ketone now on the final step. We need to get rid of the nitrogen group using h3o plus giving back our original ketone. And so this is the final answer. So what we have here is a die key tone, so you can react the enamine with an alkyl halide and alpha beta unsaturated aldehyde or ketone. You can also react it with an acid chloride to get a die ketone.