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Current time:0:00Total duration:8:19

Kinetic and thermodynamic enolates

Video transcript

we've already seen how to form enolates from ketones but what happens if you don't start with a symmetrical ketones so 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's a ch2 and then in our group an alkyl group so this is different right if we look at our alpha carbons right an alpha carbon is the one next to our carbonyl so on the left side this is an alpha carbon on the right side this is an alpha carbon and so when we look at alpha protons on this alpha carbon on the right there would be three alpha protons and on the Alpha carbon on the left there would be two alpha protons so the question is which one of those alpha protons are we going to take with our base and the answer lies in what kind of base we use and also reaction conditions so if we use a base like Lda right lithium diisopropylamide we talked about this base in a previous video it's a very strong base but it's also very sterically hindered right 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 but much bulkier than a hydrogen so this could interfere if it approach from the left side so 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 so let's say that the this lone pair of electrons here takes this proton right which moves these electrons in here to form a double bond kicks these electrons off onto our oxygen so we can go ahead and draw the enolate anion that with results all right so 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 hydrogen's attached to that carbon and so this is our enolate anion let's follow those electrons right the electrons in here and magenta moved in to form our double bond and then we could say that these electrons in here moved out onto our oxygen to form our enolate and since lithium is present if you want to do you could put lithium here li+ like that so this is the enolate anion that we would get and we call this the kinetic enolate so let me go ahead and write down here this is the kinetic enolate and it's called the kinetic enolate because this is the one that forms the fastest alright so think about kinetic and speed so this is the one that would form the fastest because we used Lda as our base and because because of the choice of our base and also because of probability alright on the alpha carbon on the right we had three alpha protons right so greater chance of taking one of these as a pro as opposed to these two over here on the left and so the kinetic enolate is the one that forms the fastest let's look at forming another kind of e delayed right so let's think about deprotonating the alpha carbon on the left this time so this is our alpha carbon over here with two alpha protons alright let's use a different base this time let's use sodium hydride so na plus h minus or we could use potassium hydride right these are sources of hydride anions which we know can act as a base so the hydride anion could take this proton right leaving these electrons in here pushing these electrons off onto our oxygen then we can draw a different enol eight alright we can show the double bond now is between those two carbons and then we have our oxygen up here with a negative one formal charge and then we have our methyl group over here and then we left one hydrogen behind and so we have a different enol a once again following some electrons electrons in magenta alright moved in here to form our double bond and then the electrons and blue here moved out on to our oxygen so this is a different enolate so we call this the thermodynamic enolate so let me go ahead and write that so the thermo dynamic enolate let's analyze these these two enol eights that we've formed in terms of stability and to do that we need to look at the substitution at the double bond alright so going back up here to the kinetic enolate I look at my double bond I think about how substituted it is alright we have these two hydrogen's over here on on this side and so that's actually not as substitute as the thermodynamic enolate right if we look at the thermodynamic enolate in the double bond we only have one hydrogen here we have an r group over here right so we also have this alkyl group if you're thinking about it right so this is actually the more stable enolate that forms because we know the more substituted the double bond the more stable the more stable it is so the thermodynamic enolate is the more stable enolate that's more substituted the kinetic enolate is not as stable but it is the one that forms the fastest and so once again you can you can control which one of these analytes you form right depending on the base that you use let's look at a problem all right where we have a ketone alright then we're going to add our two different bases to our ketone so here's our ketone and first let's add some sodium hydride alright we know that sodium hydride puts our ketone under thermodynamic control and so when we identify our alpha carbons right let's go ahead and do that alpha carbons are the ones next to our carbonyl so this would be an alpha carbon over here on the right and then this would be an alpha carbon over here on the Left we think about how many alpha protons we have for the alpha alpha carbon on the right there's only one and for the alpha carbon the left there would be two alpha protons so if I think about thermodynamic control I know that's going to form the more stable enolate right the the thermodynamic enolate and so 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 so let's go ahead and draw this enolate so we would push the electrons in here and then we would have our oxygen up here with three lone pairs of electrons a negative one formal charge and then we still have our methyl group right here so ch3 so showing those electrons in magenta moving in here to form our double bond and so this is the thermodynamic enolate it's more substituted alright so because of this methyl group here right this double bond is more substituted than then the up enough we showed an enolate forming from the other side so this is our thermodynamic enolate what if we use what if you use something like Lda and we make our temperatures very cold so negative 78 degrees Celsius all right so this is going to take a proton 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 so the LDA is going to approach from the left side so let me go ahead and draw that in so here we have here we have our L da right negative one formal charge on the nitrogen so we could show a lone pair of electrons taking this proton right least sterically hindered one these electrons move in here these electrons kick off onto your oxygen alright so let's go ahead and show the product of that we would have our ring alright we would have our methyl group over here we would now have a double bond here and then we would have our oxygen with three lone pairs of electrons and a negative 1 formal charge again if you want to draw your lithium in there this would be the enolate anion that would form alright so showing those electrons let's go ahead and make them in blue this time so these electrons in here moved in to form our double bond and so this is our kinetic enolate alright so once again we have our kinetic vs our thermodynamic the kinetic enolate is not as substituted right because we have a hydrogen here so it's not a substituted it's not as stable but it's favored by low temperatures and strong sterically hindered base like Lda so we're going to form this as our enolate using using this base and these reaction conditions I add a little bit of a higher temperature right and use of a non sterically hindered base like sodium hydride we're going to form the more stable enolate the thermodynamic enolate and so you can control which enolate you form once again based on the type of base that you use and also your reaction conditions