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Current time:0:00Total duration:13:43

Video transcript

let's look at the reaction for friedel-crafts alkylations 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 can 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 pairs I'm going to say this electron pair is going to be donated to that aluminum so we go ahead and draw the result of our 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 +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 our alkyl group and so therefore our alkyl group is left with a positive 1 formal charge and this is a carbo cation alright and this is our electrophile in our mechanism for electrophilic aromatic substitution and since we actually form a carbo cation in this mechanism a carbo cation intermediate rearrangement is actually possible so we have to be careful when we're looking at in a friedel-crafts alkylation reaction in terms of predicting the products our other product here all 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 art 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 carbo cation and we know of course that our benzene ring is going to react with our electrophile so here's our benzene ring and here is our carbo cation so positively charged so we know that the pi electrons in our benzene ring are going to function as a nucleophile and attack the electrophile 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 hydrogen still bonded here and once again I could add the 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 in the previous videos right I'm going to add the the R group to the top carbon here which means that I took a bond away from the bottom carbon so let me go ahead and highlight that one right 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 carbo cation and just to save time I'm not going to draw the other resonance structures for this ion okay so remember this this will represent our sigma complex and you can watch the previous videos for how to draw resonance structures for this carbo cation 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 +1 formal charge and reform our aromatic ring so now we have our aromatic ring reform to our group is on our ring like that our other products would be HCl and also we would regenerate our catalyst aluminum chlorides like that so let's go ahead and follow some electrons all right so these electrons in magenta right here right when the proton leaves those electrons in magenta reform this right here and I can say that these electrons in here right are forming a bond between the chlorine and the proton like that form HCL so that's 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 two benzene we are going to add two chloro propane so go ahead and draw two chloro propane 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 alkylation reactions is i think about what sort of a carbo-cation am i going to get and so without drawing the entire mechanism i know that this chlorine is going to leave right during the mechanism and if the chlorine leaves during the mechanism it will 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 with a plus 1 formal charge on that top carbon right there if we think about a possible rearrangement this is a secondary carbo cation and there's no possible way that this carbo cation could rearrange to form a tertiary carbo cation and so secondary carbo cations are relatively stable and so this will be the carbo cation that reacts with our benzene ring and so you can think about once again these pi electrons alright forming a bond right with that carbon so let's go ahead and draw the results of our nucleophilic attack so we have our ring we have our other PI electrons we have our hydrogen that's already on our ring and now we form the bond right 2 3 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 carbon so we have one two three carbons on that carbo cation so to make sure we have the same number over here one two and three so in our last step we know that deprotonate 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 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 alright let's do 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 two benzene we are going to add 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 carbo-cation 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 +1 formal charge on that carbon so if I go ahead and draw out my four carbons here and I know there's going to be a plus one formal charge on that carbon on the end which we of course know this is a primary carbo cation and primary carbo cations 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 and see if we can form a more stable one you know that there are hydrogen's attached to the carbon next door to that positive one formal charge and so we get a hydride shift here so we've seen how to do that in earlier videos all right so you can think about the hydrogen and these two electrons right all of them are going to move over here all right so if we go ahead and show the result of that hydride shift right so that hydrogen and those two electrons are going to move over there that takes away the +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 one formal charge is going to go and so now we have a secondary carbo cation so just to refresh your memory this is a secondary carbo cation because the carbon in magenta is bonded to two other carbons so we have a secondary carbo cation which we know is very is more stable than a primary carbo cation and so if we can think about two possible products here so our benzene ring could react with our primary carbo cation or more likely it's going to react with our secondary carbo cation over here so let's go ahead and draw the result if it reacts with the secondary carbo cation so I could think about you think about my electrons and here my PI bonds alright going all the way over here to react with that carbo cation that secondary carbo cation and so we can go ahead and show the result of our nucleophilic attack all right 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 our two what used to be our carbo cation and so we have four carbons and one of them is going to be branching like that so let's go ahead and analyze what we've done so far so plus one formal charge on our sigma complex so our electrons are pi electrons in here alright are going to be these electrons right here so forming our carbon-carbon bond like that and when we look at what's attached to that let me go ahead and highlight some of these carbons here so I'll start I'll start in red so this carbon right here I'm going to say is this carbon like that so in magenta right this carbon down here in magenta right 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 right that's of course this one right here and then finally this carbon right here in blue right 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 protonation 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 carbo cation 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 carbo cation right they would 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 we have four carbons coming off of our ring so one two three four like that and once again positive one formal charge on our sigma complex so let's go ahead and once again highlight those electrons in magenta so the electrons in magenta right here right they formed this carbon-carbon bond and they're bonding to this carbon this time so the one on the end so I'm going to say it's that one right there okay 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 these carbons is I find that this is a one of the areas where students will make the most mistakes all right so we 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 butyl benzene 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 carbo cation the primary carbo cation is not as stable as the secondary carbo cation and so this shows one of the limitations to the friedel-crafts alkylation reaction you can't always control the alkyl group that putting on to your ring because of the possible rearrangement because there's a carbo-cation present in the mechanism so if your goal was to make butyl benzene 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 isolation