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Current time:0:00Total duration:9:47

Video transcript

we've done a lot of electrophilic aromatic substitution reactions let's let's look at the possibility of a nucleophilic aromatic substitution and so if we started here with bromobenzene and we add a nucleophile something like the hydroxide anion right here so negative one formal charge could function as a nucleophile and it would of course attack the carbon that is bonded to our halogen here and so when the nucleophile attacks here if we're thinking about an sn2 type mechanism that's a concerted mechanism where these electrons should kick off onto the bromine so if this happened right we would get our benzene ring and we would get now the O H has substituted in for the bromine and the bromine has left in the form of an anion right so the bromide anion here which has a negative 1 formal charge and is relatively stable on its own so it's a decent leaving group so the problem with this is that when our nucleophile is attacking this carbon right here this is an sp2 hybridized carbon which is part of this benzene ring of course and the benzene ring is going to get in the way of the nucleophile attack via an sn2 type mechanism and so because of that ring because you're working with an sp2 hybridized carbon the nucleophile can't attack in the proper orientation and so the an sn2 mechanism is not possible so sn2 does not occur at an sp2 hybridized carbon and so this reaction doesn't proceed this way what about an sn1 type mechanism so if we thought about thought about sn1 type mechanism we know the first step in that is dissociation so these electrons in here coming off onto the bromine and so we go ahead and draw our benzene ring and since we took a bond away from the carbon that's bonded to the bromine all right that would get a +1 formal charge like that and so we form a carbo cation the problem is that this is a very unstable carbo cation we can't really draw any resonance structures for it and so since it is unstable carbo cation it's not likely to form and so an sn1 type mechanism is highly unlikely it is actually possible if you have an incredible leaving group but for our purposes we'll say it's extremely unlikely so sn2 is out sn1 is out and so you might think that you can't nucleophilic aromatic substitution but as a matter of fact you can and let's take a look at the criteria in order to do so so your ring must have an electron withdrawing group so withdrawing some electron density from the ring we have that here of course this nitro group is electron withdrawing the ring must have a leaving group and we just saw how our halogen here can act as a leaving group the leaving group is ortho or para to the electron withdrawing group well and in this case those two groups are both ortho to each other so so this molecule could undergo nucleophilic aromatic substitution and so let's think about once again our hydroxide functioning as the nucleophile so we have our negatively charged hydroxide anion functioning as a nucleophile right attacking the carbon once again that's bonded to our halogen this time we're going to we're going to move some electrons around so these PI electrons are going to move into here to form a double bond and these electrons are going to move out onto the oxygen so let's go ahead and show the result of all those electrons moving so we have our ring all right we have our PI electrons here we have our bromine still attached to the ring so let's go ahead and put those lone pairs of electrons on it and now we have an OHA tattoo our ring - like that so let's put those lone pairs of electrons on now our nitrogen is double bonded to our ring like that the nitrogen is still bonded to an oxygen on the right the oxygen on the right still has a negative 1 formal charge and now the nitrogen is bonded to an oxygen on the left with also with a negative 1 formal charge like that the nitrogen itself has a +1 formal charge so let's see if we can show the movement of all of those electrons here and so if I show these electrons in magenta right on our hydroxide anion those you could think about them as being this bond now and these PI electrons in here like that right these you could think about moving out to here to form a PI bond with the nitrogen and then finally these electrons right in here right moved out onto this oxygen on the left to give it a negative 1 formal charge and so the first step in this mechanism is the addition of the nucleophile all right so let me go ahead and write that so it's the addition step where we add our nucleophile to the ring and when you do that you're of course adding electrons to the ring and those and that negative charge that you're adding actually ends up on this oxygen here right so this oxygen has a negative 1 formal charge it's able to stabilize that since it's very electronegative and so the presence of your electron withdrawing group right withdraw some electron density from the ring it allows some of that electron density to be temporarily stored on your electron withdrawing group and that stabilizes this intermediate here you can actually draw some other resonance structures but this is the one that we're most concerned with one showing the negative charge on the oxygen so this once again this electron withdrawing group is able to stabilize that negative charge that was added to the ring and it's only able to stabilize it because it is ortho to the leaving group and so in our next step alright so in our next step we're going to show the leaving group leaving so this is the elimination step so the elimination of the leaving group and again that nitro group was temporarily holding on to some of that that negative charge and those electrons and they're going to move back into here which would push these electrons back into here and then these electrons will kick off on to your bromine and so we can go ahead and show the final product now where we have our benzene ring and the bromine has left in the form of the bromide anion so we have an OHA attached to our ring and now we have our nitro group back right the way it looked before with a negative 1 formal charge on this oxygen and this this oxygen over here double bonded to the nitrogen the nitrogen having a +1 formal charge so the electrons in green over here on this structure so these electrons moved back into here right to form this bond and the electrons in blue right move back into here to reform your ring this way and and we're done we've seeing that the O H has substituted in for the halogen here so this is called the addition elimination mechanism and you can see why first you add your nucleophile and then that electron density is temporarily stored in the electron withdrawing group and it comes off again to eliminate your halogen like that alright so that was an example of a situation where the two groups are ortho to each other let's do one let's do one where the two groups are para to each other so let's look at this next example here and so once again we have an electron withdrawing group our Nitro group we have a leaving group our halogen and and so therefore we can perform nucleophilic aromatic substitution and so I'm going to use the same nucleophiles before I'm going to use hydroxide so we have a negative 1 formal charge right and our nucleophile is going to attack of course the carbon that is bonded to our leaving group our halogen which would push these electrons over to here which would push these electrons over to here and then that pushes these electrons off onto our oxygen once again so let's go ahead and show again all of those electrons after they have moved here so we have our our bromine is still attached to our ring all right ROH is now attached to our ring like that so it would have two lone pairs of electrons on it and we have some PI electrons here PI electrons here nitrogen is once again double bonded to our ring now it's bonded to this oxygen on the left which has a negative 1 formal charge and the oxygen on the right now has a negative 1 formal charge like that with this sense with this nitrogen here being a positively charged so to negatively charged oxygen x' and one positively charged nitrogen and so once again this is the addition step the addition of the nucleophile to the ring and now that negative charge right when we added some electrons to our ring that negative charge is is partially residing on this oxygen here and so our electron withdrawing group stabilizes this intermediate the next step of course is elimination and so if these electrons move back into here right that would push these electrons back into here which pushes these electrons over to here and then these electrons come off on to our on to our leaving group and so the bromide anion leaves and we now have roh has substituted in for it so we have our ring we now have our Oh H on there and we have of course our nitro group so over here double bonded to this oxygen this nitrogen is a single bond to this oxygen and negative one formal charge and a plus one formal charge like that and so this is an example of the addition elimination mechanism where your electron withdrawing group and your leaving group our para to each other if you try to do this where the two groups are meta to each other you'll see that it doesn't work you can't show that electron density out on to this oxygen you can't show your electrons moving around to do that so it's only when those groups are either ortho or para to each other