Current time:0:00Total duration:14:15
0 energy points
Video transcript
Voiceover: If we react an aldehyde, or a ketone, with an excess of alcohol, in an acidic environment, we are going to form an acetal. So, over here on the right, is our acetal, and you can see the OR double prime, from our alcohol, and OR double prime, from our alcohol. You're also going to form water in this reaction, and this reaction is at equilibrium, and so there are several things that you can do, to shift the equilibrium to the right, and to make more or your acetal products. One thing would be, to remove the water as it forms, so if you decrease the concentration of this product, your equilibrium is going to shift, to make more of it, and so therefore, you're going to form more acetal. So, another thing you could do, to shift the equilibrium to the right, would be to increase the concentration of one of your reactants. So, you could increase the concentration of an aldehyde, and then that would, once again, shift the equilibrium to the right, and form more of your acetal products. So several things that you can do, in the lab, to increase your yield. This is done in an acidic environment, and so there are a couple different proton sources you can use. You can use something like sulfuric acid, H two SO four, or you could use something like Toluenesulfonic acid, so TsOH R, two of the more common catalysts used to form your acetal. Let's look at a reaction here, and then we're gonna do the mechanism for this reaction. You'll see it's a bit of a long mechanism. I think it's a little bit easier to understand, if you do it for an actual reaction here. So we have cyclohexanone reacting with an excess of ethanol, and using sulfuric acid as our catalyst, and so just looking at this general pattern up here, for predicting the structure of your acetal, We can find this portion of the molecule, and think about adding that to our ring. And so we have our ring here, and then we would have our oxygen, and then our R group, and then our oxygen, and then our R group like that. So that's the product, kind of a funny-looking molecule, but that is the acetal that we would make. So, let's think about a mechanism for this reaction. So, if you have ethanol and sulfuric acid, one of the things that could happen, is protonation of your ethanol. So let's go ahead, and show a protonated ion: So this is one of the possible things that could happen first. So, we have it protonated, like that, and then, we're going to show that functioning as an acid, and reacting with cyclohexanone. So here we have cyclohexanone, and a lone pair of electrons and cyclohexanone are gonna pick up a proton, so a proton from somewhere, and this could be the acid over here, on the left. So these electrons move over here, to form ethanol, and we protonate our carbon EELs. So, let's go ahead and show that. So, step one would be protonation of your carbon EEL, and that is favored, because that makes your carbon, attached to your oxygen, more electrophilic. So let's go ahead and show that. So we would have a proton now, bonded to our oxygen, still one lone pair of electrons on our oxygen, so let's show these electrons in magenta. Took this proton, and that forms this bond, which gives this oxygen a plus one formal charge, like that. And we know that, because of a resin structure we could draw for this, that makes this carbon more electrophilic, so that carbon is going to function as an electrophile, and therefore a nucleophile can react with it. And we have a nucleophile present, of course, that would be ethanol. So, a molecule of ethanol comes along, functions as a nucleophile, a lone pair of electrons attacks our electrophile, kicks these pi electrons off, onto this oxygen: so, that would be the second-step, nucleophilic attack. So, let's go ahead and write these out: so we had step one, protonation of our carbon EEL, so step two, nucleophilic attack. And so when a nucleophile attacks, we would have, this oxygen over here, would now have two lone pairs of electrons around it, so let's show those, so let's make 'em blue here. So these electrons moved out onto our oxygen, like that. And we just formed a bond between the oxygen on our ethanol, and this carbon, so we have a bond here, like that. And this still had a hydrogen attached to it, an ethyl group, and a plus one formal charge, like that. Alright, so next, let's get a little bit of room down here. The third step would be deprotonation, so let me go ahead and write that. So, step three, we deprotonate. So, another molecule of ethanol could come along and function as a base, and a lone pair of electrons on ethanol could take this proton, which leaves these electrons behind on our oxygen. So let's go ahead, and show that. So next, we would have our ring, we would have an OH over here, on the left, let's go ahead and put in those electrons, and then over here, on the right, we would have, this time, two lone pairs of electrons on our oxygen. So let me go ahead, and use green for those. These electrons right in here moved off, onto our oxygen, and so, if you look at that structure closely, that's a hemiacetal. So deprotonation yields our hemiacetal here, which is an intermediate in our reaction. Alright, so next step, next step here is protonations; let me go ahead, and mark this as being step four. We're going to protonate this OH over here, on the left. And so, one of the possibilities would be a protonated ethanol over here, functioning as an acid, so let's go ahead, and draw that. So a plus one formal charge on this oxygen, and a lone pair of electrons picks up a proton, leaving these electrons behind, and so let's go ahead and show that. So we protonate the OH, and the reason why protonating the OH would be good, is that would give us water as a leaving group. So, let's once gain show those electrons; let's use magenta again. So these electrons, right here, picked up a proton, and let's show these electrons as being that bond now. And then over here, on the right, we have, once again, our oxygen, and ethyl, and then we have two lone pairs of electrons, and then, let's keep this lone pair green right here. And then, since we protonated the OH, we get a plus one formal charge on this oxygen here, and, if you look closely, let me use red for this, if you look closely over here, you can kinda see water hiding, right? We know water's an excellent leaving group, so, if these electrons in green moved in here, to reform the double bond, then that would kick these electrons off onto the oxygen, and then we would have water. So, this is the dehydration portion, so we're gonna form water. So let me go ahead, and mark this as being the next step, right? So, in the next step, when those electrons kick in there, so this would be step five, we're going to lose H two O, so the dehydration step. And we would be left with, once again, our ring, and, this time, a double bond to this oxygen, with an ethyl coming off of that oxygen like this. So let's go ahead and make sure we still have a lone pair of electrons on this oxygen, and a plus one formal charge, and the electrons in green, so these electrons in here, moved in here to give us our double bond once again. And, once again, we have a plus one formal charge on the oxygen, so if you drew a resonance structure for this, you would actually have this carbon as being very electrophilic. So, once again, we're going to get a nucleophile attacking our electrophile in the next step, so this would be step six. So, step six would be a nucleophilic attack. So, in step six, a nucleophile comes along, once again, ethanol is our nucleophile, so here is ethanol, so let's go ahead and show ethanol right here, with lone pairs of electrons. And one of these lone pairs of electrons, of course, would attack our electrophile, so nucleophile attacks electrophile, and that would push these electrons in here off onto this oxygen. So let's go ahead, and draw what we have next. Alright, so we now have an oxygen, with still a hydrogen on it, and ethyl right here, a lone pair of electrons, a plus one formal charge on this oxygen. So, let's highlight those electrons: so, in magenta here, these electrons formed a bond, so that oxygen is now bonded to that carbon. And then over here on the right, we have an oxygen, with an ethyl group, and now there are two lone pairs of electrons on this oxygen. So, we are almost there, right, last step. So, step seven would be a deprotonation step. So in step seven here, all we have to do is take that proton off, and we would form our acetal product. So, once again, we could have a molecule of ethanol come along, and function as a base, and so, a lone pair of electrons take this proton, leaving these electrons behind, on the oxygen, and then finally we are able to draw our acetal products. So we would have, let's go ahead and make this a little bit more angled, so on the left, we would have our oxygen, with an ethyl, and then this carbon is also bonded to another oxygen, with an ethyl coming off of it like that. And so, let's go ahead and show those final electrons here, on our oxygen like this, and, once again, highlight these electrons, came off onto our oxygen. So, we've formed our acetal product. So, a very long mechanism, and it's a little bit challenging; I think it was just easier to go through an actual reaction for this, so, that's a long one. Let's do two quick problems, to think about the acetal product here. So, let's look at this next reaction. So here, we have acetaldehyde, and then here we have butanol. This time, we're gonna use Toluenesulfonic acid, as our acid catalyst, and one of the things you could do is increase the concentration of one of your reactants, and if you increase the concentration of acetaldehyde, you can actually drive this reaction to completion. If you think about the structure of the product, we know that we're going to be adding on this portion of our alcohol, to this carbon, and that's going to happen twice. So therefore, we need to make sure we have two carbons, and those are our two carbons, and then we have that carbon bonded to an oxygen. We need to have four carbons in our product: So, one, two, three four. And then we know that it's gonna be bonded to another oxygen, and so one, two, three four. So, that would be our acetal product. So, let's highlight some carbons here, so we can follow along. So, this carbon right here, would be this carbon on the right. This carbon, that used to be our carbon EEL carbon, is going to be right here, and then, let's switch colors for the butanol molecule. So, oxygen right here, would be this one, and this one, and then we have one, two, three four; so we have one, two, three, four; one two, three, and four. So counting your carbons is one of the techniques you can use to figure out your final acetal product. Let's do one more reaction here. So, this would be a ketone, so we have a four-carbon ketone, so butanone; reacting it with ethylene glycol, and, once again, we use Toluenesulfonic acid, as our catalyst. And this one's a little bit different, because we can see we have a diol, as one of our reactants; up here, we just had butanol, only one OH, but this one has two on it. So, trying to figure out the product here, sometimes it helps just to run through the mechanism really quickly, and so the Toluenesulfonic acid is going to help us to protonate our carbon EEL, and then we have our nucleophile attack, so one of these OHs is going to attack here. And so, without going through all the steps in the mechanism again, that was obviously a pretty complicated mechanism, I'll jump to one of the later steps of the mechanism, where we have already lost water, so minus H two O, so we've already gotten past the dehydration step. And then that would give us this as our intermediate, so there is actually gonna be a plus one formal charge on this oxygen. And then we have these two carbons over here, and then our other OH on this side, so let's go ahead, and color-coordinate some of our atoms once again. So, this oxygen has already bonded, we've already lost water, so that oxygen is this oxygen, right here. And then, we still have another OH on this molecule, and that's this one over here, like that. So when we get to this step, we're actually gonna get an intra-molecular, nucleophilic attack. So, these electrons are going to attack this carbon, and kick these electrons off, onto this oxygen. And so, when you think about the final product, you're actually gonna get a cyclic product here, a cyclic acetone. So, we would have our four carbons, and then we would have this oxygen, and then two carbons, and then this oxygen, and they're both bonded to this carbon right here. And so, once again, let's highlight some of those carbons: so this carbon right here, and this carbon right here, or this carbon, and this carbon, and, in our final product, like that. And so, this is a cyclic acetal that we have formed, so a little bit trickier than the previous reactions. So in the next video, we'll see a use of cyclic acetals as a protecting group.