Current time:0:00Total duration:13:43
0 energy points
Video transcript
Let's look at the reaction for Friedel-Crafts alkylation. So we start with our benzene ring, and to benzene we're going to add an alkyl chloride, and our catalyst is aluminum chloride. And the end result is to substitute an R group, the R group that was on the alkyl chloride, for a proton on the aromatic ring. Let's look at the mechanism for Friedel-Crafts alkylation. We start with our alkyl chloride and we add our aluminum chloride, which we've already seen can function as a Lewis acid. So it could be a Lewis acid because it can accept a pair of electrons. And so our Lewis base over here is going to be the alkyl chloride. It's going to donate an electron pair. So I'm going to say this electron pair is going to be donated to that aluminum. So we'd go ahead and draw the result of R-Lewis acid base reaction here. So now the chlorine is bonded to the aluminum, and the aluminum is bonded to these other chlorines here, and I'm going to leave off the lone pairs of electrons on those just to save time. In terms of formal charges, aluminum now has a negative 1 formal charge, and this chlorine now has a plus 1 formal charge like that. So we know that halogens are relatively electronegative, and so this chlorine here could take these electrons, we could show these electrons in here moving off onto that chlorine. And when we do that, we're taking a bond away from R alkyl group and so therefore, R alkyl group is left with a positive 1 formal charge. And this is a carbocation. And this is our electrophile in our mechanism for electrophilic aromatic substitution. And since we actually form a carbocation in this mechanism, a carbocation intermediate, rearrangement is actually possible so we have to be careful when we're looking at a Friedel-Crafts alkylation reaction in terms of predicting the products. Our other product here, right? Well, we would have our aluminum still bonded to these chlorines, and of course, it's bonded to this other chlorine too. And this chlorine now has three lone pairs of electrons around it so these electrons in here, ones that were in the bond between our R group and our chlorine left, and they are now on our chlorine like that. And so our aluminum still has a negative 1 formal charge in there like that. All right, so now we have generated our electrophile, which is our carbocation. And we know, of course, that our benzene ring is going to react with our electrophile. So here is our benzene ring, and here is our carbocation so positively charged. So we know that the pi electrons in our benzene ring are going to function as a nucleophile and attack the electrophiles so negative charges are attracted to positive charges like that. And so we can go ahead and draw the results of our nucleophilic attacks so we now have our hydrogen still bonded here. And once again, I could add the alkyl group to either of those two carbons. Let me just highlight those again. So either one of these two, but just to be consistent with how I've been doing it in the previous videos, right? I'm going to add the R group to the top carbon here, which means that I took a bond away from the bottom carbons. Let me go ahead and highlight that one. So this, of course, lost a bond, which means that is where our positive 1, a formal charge, would go so. Positive 1, formal charge, a carbocation. And just to save time, I'm not going to draw the other resonance structures for this ion. OK? So remember, this will represent our sigma complex, and you can watch the previous videos for how to draw resonance structures for this carbocation. And so in the last step of our mechanism here, we're going to reform our aromatic ring, and so we need to deprotonate our sigma complex in order to do that so we could think about these electrons in here moving out to take that proton, and these electrons moving in here to take away the plus 1 formal charge and reform our aromatic ring. So now we have our aromatic ring reformed to R group is on our ring like that. Our other products would be HCL and also we would regenerate our catalyst, aluminum chloride like that. So let's go ahead and follow some electrons. All right? So these electrons in magenta right here. When the proton leaves, those electrons in magenta reform this right here. And I can say that these electrons in here are forming a bond between the chlorine and the proton like that to form HCL. So that's our mechanism for Friedel-Crafts alkylation. Let's look at a few examples of this reaction. So we'll start with benzene, and to benzene we are going to add two chloropropane. So go ahead and draw two chloropropane. That's going to be our alkyl chloride. And once again, we need our aluminum chloride as our catalyst. And so the way I do Friedel-Crafts alkyl reactions is I think about what sort of a carbocation am I going to get. And so without drawing the entire mechanism, I know that this chlorine is going to leave during the mechanism, and if the chlorine leaves during the mechanism, well, that means that we took a bond away from this carbon right here, and so that's where our plus 1 formal charge is going to be. So we have a plus 1 formal charge on that top carbon right there. If we think about a possible rearrangement, this is a secondary carbocation and there's no possible way that this carbocation could rearrange to form a tertiary carbocation. And so secondary carbocations are relatively stable, and so this will be the carbocation that reacts with our benzene ring. And so you could think about once again, these pi electrons forming a bond with that carbon so let's go ahead and draw the results of our nucleophilic attacks. So we have our ring, we have our other pi electrons, we have our hydrogen that's already on our ring, and now we formed a bond to three carbons so let's go ahead and once again highlight some electrons here. So these electrons in magenta have now formed a bond between those two carbons right there. And always double check yourself and count your carbons so we have one, two, three carbons on that carbocation. So we have to make sure we have the same number over here-- one, two, and three. So in our last step, we know that deprotonation of our sigma complex, right? So this is a positive 1 formal charge here. Deprotonation of our sigma complex, we get rid of that positive 1 formal charge and reform our aromatic ring so we've seen that these electrons would move in here, and so we have our product. So we have our aromatic ring like that. And then we have this isopropyl group coming off of our ring so we form isopropyl benzene as our final product. Let's do one more Friedel-Crafts alkylation. This one will have a possible rearrangement so let's go ahead and start with benzene again. So we have our benzene ring, and to benzene we are going to add butyl chloride here so one, two, three, four carbons. And then a chlorine like that, and then aluminum chloride, once again, as our catalyst. And so if you're approaching this problem, once again, I like to think about what sort of a carbocation is going to form in the mechanism. So I know that this chlorine would kick off during the mechanism, and so that would take away a bond from this carbon right here so we're going to get a plus 1 formal charge on that carbon. So if I go ahead and draw out my four carbons here, and I know there's going be a plus 1 formal charge on that carbon on the end, which we, of course, know this is a primary carbocation, and primary carbocations are not very stable so there could be some rearrangement here. And if you think about what's the best way to rearrange our carbocation to see if we can form a more stable one, you know that there are hydrogens attached to the carbon next door to that positive 1 formal charge. And so we get a hydride shift here. So we've seen how to do that in earlier videos. So you could think about the hydrogen and these two electrons, all of them are going to move over here. So if we go ahead and show the result of that hydride shift so that hydrogen and those two electrons are going to move over there, that takes away the plus 1 formal charge of the carbon on the end. And we took a bond away from this carbon now so that's this carbon over here so that is where the plus 1 formal charge is going to go. And so now we have a secondary carbocation. So just to refresh your memory, this is a secondary carbocation because the carbon in magenta is bonded to two other carbons. So we have a secondary carbocation, which we know is more stable than a primary carbocation. And so if we can think about two possible products here. So our benzene ring could react it with our primary carbocation or more likely it's going to react with our secondary carbocation over so let's go ahead and draw the results if it reacts with a secondary carbocation. So I could think about my electrons in here and my pi bonds going all the way over here to react with that carbocation, that secondary carbocation, and so we can go ahead and show the result of our nucleophilic attack. So once again, we have our pi electrons. We had a hydrogen already bonded to that ring there, and we're going to form a bond to what used to be our carbocation so we have four carbons, and one of them is going to branching like that. So let's go ahead and analyze what we've done so far. So plus 1 formal charge on our sigma complex. So our electrons, our pi electrons in here, are going to be these electrons right here so forming our carbon-carbon bond like that. And so when we look at what's attached to that-- let me go ahead and highlight some of these carbons here. So I'll start in red. So this carbon right here I'm going to say is this carbon like that. So in magenta this carbon down here in magenta is, of course, this carbon right here in magenta, the one that's bonded to our ring. And then, of course, we could follow a few more carbons here. So I'm going to say that this carbon in green that's, of course, this one right here. And then finally, this carbon right here in blue is this carbon. So always count your carbons and make sure that you have the right branching for something like this. So in the last step, of course, deprotonation of our sigma complex, these electrons would move into here like that. And we have one of our final products here. So let's go ahead and draw this one so we would have that as our major product for this reaction. It is possible for the primary carbocation to react as well so let's go ahead and draw the result of that. So if these pi electrons instead went for the primary carbocation, they'd form a bond with that carbon right there. And so we could go ahead and draw the result of that so we would have our ring once again. And once again, we have our hydrogen, and this time we would have four carbons coming off of our ring. So one, two, three, four like that. And once again, positive 1 formal charge on our sigma complex. So let's go ahead and once again the highlight those electrons in magenta. So the electrons in magenta right here they formed this carbon-carbon bond. And they're bonding to this carbon this time. So the one in the end, so I'm going to say it's that one right there. So next we have the carbon in magenta right here, and so that would be this one. And just to be consistent with our colors, I'll go with green next so this carbon is green and so that one is green right here. And then finally, this carbon in blue is this carbon in blue. The reason I'm taking the time to show all of these carbons is I find that this is one of the areas where students will make the most mistakes so you have to be careful about following your electrons and counting your carbons when you're drawing your final products here. So deprotonation of our sigma complex, these electrons move into there and we would form butylbenzene as the minor product in this reaction so let's go ahead and draw that out. So we have four carbons coming off of our benzene ring, and this would be the minor product since it comes from the less stable carbocation. The primary carbocation is not as stable as the secondary carbocation. And so this shows one of the limitations to the Friedel-Crafts alkylation reaction. You can't always control the alkyl group that you're putting on to your ring because of the possible rearrangement because there's a carbocation present in the mechanism. So if your goal was to make butylbenzene, you wouldn't be able to make it in extremely high yield using a Friedel-Crafts alkylation, and so we'll see a better way to do that in the next video, which is on Friedel-Crafts acylation.