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 lda is lda is lithium diisopropyl amide so we have a lithium with a positive charge and we have a nitrogen with two isopropyl groups and two lone pairs this reagent lda 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 dye isopropyl amide with its negative charge is going to take off the hydrogen and so we're going to get a residence stabilized enola 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 can react it with methyl bromide so then this action is going to reform the pi bond 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 two 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 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 reagents 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 at a low temperature and then we can add methyl bromide in 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 thermodynamic 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 hydrogens 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 let's remove the blue hydrogen if we do so we're going to get 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 enolate ions however at higher temperatures the most stable enela ion is the one that's going to form it's going to be the major product to figure out which one is going to be the most stable enola ion we need to look at the carbon-carbon double bond and 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 giving 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 thermodynamic 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 lda would 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 giving 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 enamine so it's going to look like this and now once you have the enamine 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 attacking the methyl group and expel in the 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 h2o 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 monoalkylated products now we can also react the enamine intermediate with other electrophiles not just 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 breaking 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 going to push these electrons to take a hydrogen expelling hydroxide so right now we have the aldehyde again 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 our final answer for this example now we can also react the enamine intermediate with and acid chloride so what do you think the major product for this reaction will be feel free to pause the video and try so we 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 in the final step we need to get rid of the nitrogen group using h2o plus giving back our original ketone and so this is the final answer so what we have here is a diketone so you can react the enamine with an alkyl halide an alpha beta unsaturated aldehyde or ketone you can also react it with an acid chloride to get a diketone you