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Current time:0:00Total duration:9:37

Video transcript

one way to make an ester is to use a Fischer esterification reaction so if you start with a carboxylic acid and you add an alcohol and a source of protons right you're going to form your ester and you're also going to make water in this process it's important to note that the oxygen and the r prime group come from your alcohol and so we'll see that in the mechanism also this reaction is at equilibrium so if you want to make more of your ester product right you have to shift the equilibrium to the right and so there are a couple ways to do that one thing you could do would be to decrease the concentration of water right which would shift your equilibrium to the right to make more of your ester you could also do something like increase the concentration of the alcohol and now would shift the equilibrium to the right as well let's take a look at the mechanism to form esters so we start with our carboxylic acid and we're going to protonate the oxygen right so a lone pair of electrons and the oxygen can pick up a proton right leave these electrons behind and so we can show that we have now protonated the oxygen right here so it gets a +1 formal charge all right protonation of your carbonyl activates your carbonyl it makes your carbon more electrophilic so this carbon is now more electrophilic so we saw that in some of the previous videos and so if the carbon is more electrophilic right our next step it makes sense is going to be a nucleophilic attack so and a molecule of our alcohol comes along and you can think about a lone pair of electrons on the oxygen attacking this carbon right here all right pushing these electrons off onto your oxygen so let's go ahead and show the results of our nucleophilic attack so we have now have an oxygen up here it had one lone pair on the left now it has two lone pairs right bonded to a hydrogen and then over here on the right is the bond that we just formed the bond between carbon and oxygen so let's show those electrons in magenta right so this lone pair on the oxygen right forms the bond between carbon and oxygen so there it is alright also on this oxygen there's still a hydrogen they're still in our prime group right and one lone pair of electrons giving this oxygen a plus 1 formal charge and there's still an O H bonded to our carbon like that the next step is to get rid of this plus one formal charge on our oxygen so another molecule of alcohol comes along and this time acts as a base so in the previous step alcohol acted as a nucleophile and this step is going to act as a base it's going to take this proton right and leave these electrons behind on the oxygen so let's go ahead and get some more room down here all right so we're going to deprotonate so let's go ahead and show what we would make alright so we have our carbon bonded to this top oxygen right two lone pairs of electrons on it a hydrogen and our group off to the left I'm going to draw this oxygen down here with lone pairs of electrons so we can show the next step alright and then this oxygen right here now has two lone pairs of electrons on it and it's still bonded to our our prime group so let's show those electrons the electrons in blue here alright move off on to the oxygen so that's the deprotonation step next step is to protonate the o h at the bottom so let's get some more even more room to show that okay so we're going to show this o H down at the bottom here being protonated so I'm going to draw in I'm going to draw in a source of protons right down here right so our protonated alcohol plus one formal charge on our oxygen R prime right here so lone pair of electrons on this oxygen can pick up a proton leave these electrons behind right so we have a protonation step and the reason why this this protonation is favored is because this is going to create an excellent leaving group right so if you look closely you're going to see water there as our leaving group so let's go ahead and show that so now this oxygen down here right has been protonated so it has a plus one formal charge on this oxygen so let's show those electrons so these electrons right here on this oxygen right pick up this proton so forming this bond right to that proton so let's go ahead and draw in the rest of what we have we have our oxygen with two lone pairs of electrons and our R prime group like that alright so in the next step right we just formed water as our leaving group in here right if you can see it right if these electrons in here were to come off on the oxygen you can see that's water so when these electrons on this top oxygen moving here to reform our double bond right that's when that's when these electrons in here in blue are going to come off onto our oxygen and water is an excellent leaving group so let's draw what we have now so once again everything is at equilibrium so we're going to reform our double bond right and this top oxygen now has a +1 formal charge so let's show those electrons let's make them red here so these electrons in red right are going to move in here to reform our double bond this carbon is still bonded to an R group on the left side right this carbon is bonded to an oxygen and our our prime group over here the oxygen has two lone pairs of electrons all right and we just lost water so let me go ahead and draw water down here alright so loss of h2o at this step all right we're almost to our final product because all we have to do is deprotonate right here and we'll form our ester so we could show another molecule of alcohol coming along so our prime two lone pairs of electrons right taking taking this proton right here right leaving these electrons behind on our oxygen so let's go ahead and draw the final structure of our ester right so we would have our C double bond o with two lone pairs of electrons all right so let's show let's show these electrons in here right move off onto our oxygen like that and then our carbon is still bonded to this oxygen and we have our R prime group and so our end results right is to form our ester and water okay so that's a little bit of a long mechanism let's take a look at some reactions to form esters using the fischer esterification reaction so let's start with this molecule over here on the left so this is salicylic acid and if we add methanol and we use sulfuric acid as our source of protons we're going to form an ester and this is one of those those famous labs that's always done an undergraduate organic chemistry so we think about the mechanism remember remember that this oxygen on our alcohol and in this case this methyl are going to add right so we're going to lose this o H on our carboxylic acid right and we're going to put this oxygen in this methyl group on in place so let's go ahead and and draw the product all right so we would have our benzene ring right here and we would have our carbon double bonded to our oxygen and we would have the oxygen from the alcohol from methanol and then our methyl group like that and then we still have ROH right here the reason why this is one of those classic undergraduate labs is this is wintergreen right so this is an incredible smell it's always a lot of fun to do this in an undergraduate lab because the lab smells great when you're done so the synthesis of wintergreen all right let's look at another Fischer esterification this one is a little bit different this one is an intramolecular Fischer esterification so if we look at our starting molecule on the left alright this time we have our carboxylic acid and our alcohol and the exact same molecule and we have all these all these single bonds in here right which we know we can have some free rotation so if we draw the molecule a different conformation so let's go ahead and do that so we have our carboxylic acid up here all right and let's count how many carbons we have so let me use let me use red for that so we have one we have carbon one two three four and five so we have five carbons so let's go ahead and draw them in so there's carbon one two three four five and then we have our Oh H so let me go ahead and number those carbons alright so this is carbon one carbon two carbon three carbon four and carbon five and so in a different conformation alright we can think about we could think about this oxygen attacking this carbonyl and the mechanism alright so we know we know that we're going to lose right we know that we're going to lose this o H right we know we're going to lose this hydrogen and so we can stick those together and think about our final product all right so we're going to form right and Esther but it's a different Esther than what we've seen before right so we have our carbonyl right here and then we have this oxygen alright so this oxygen is a member of our ring now and that came from this oxygen so we can see our five carbons one two three four five right we can think about losing water here and so our product is called a lactone okay so this is an a Lac tone here so it's an ester that's in a ring here we have a six membered ring where oxygen is one of the members of the Ring alright so a pretty cool intramolecular fischer esterification reaction to form a lactone