Nucleophilic aromatic substitution I
The addition-elimination mechanism. Created by Jay.
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- at 1.08, how does the benzene ring get in the way of the OH- attacking the carbon?(3 votes)
- In an SN2 reaction, the nucleophile has to attack from the back side of the carbon bearing the leaving group.
That means that the OH⁻ would have to get into the very centre of the benzene ring.
There just isn't enough room for even the smallest nucleophile to get in there.(5 votes)
- Is this mechanism also called the ArSN mechanism (Ar for Aryl, SN for nucleophilic substitution?)(2 votes)
- It's usually shortened to SNAr (at least in English)(3 votes)
- Wouldn't which ortho position the leaving group was on relative to the electron withdrawing group change the mech?
Either way it work, but one way your resonance form would be significantly more complicated - the negative charge would have to travel all the way around the ring...(2 votes)
- It doesn't matter, because benzene is symmetric and planar. The ortho positions are equivalent to each other.(3 votes)
- Is the product stabilized by resonance. I wandered when I saw negative and positive charge on the O and N.(1 vote)
- In the the nitro group, the negative charge is delocalized over the two oxygen atoms. The aromatic ring has delocalized pi electrons. More resonance structures can also be drawn that place a positive charge on the ring or the oxygen in the OH group, a positive charge on the nitrogen and a negative charge on each oxygen atom in the nitro group. So yes, the product is highly resonance stabilized. You should try to draw out all the resonance structures; there are at least six.(2 votes)
- Sir said that the substitution cannot take place in an aromatic halogen compound because the carbon atom bonded to the halogen atom is sp2 hybridized(time stamp -0:57to1:06). Is this true for all sp2 hybridized carbon atoms in aliphatic compounds as well?(1 vote)
- Which step in the reaction is the rate determining step(RDS) ? Basically what i want to ask is, provided there had been a bromine and a chlorine atom on both of the ortho position then which group would have gone elimination?(1 vote)
- If there were a Br para to the EWG group as well, to which would the nucleophile add? ortho Br or para Br??(1 vote)
- In first example we have NO2 group that is meta direction but why is Br in ortho position? Br should be in meta position .
- Pretty sure the answer to this is the same as your other question. Just because something is a ortho/para director doesn't mean the meta product cannot ever be formed and vice versa.(1 vote)
- Why cyno
group cannot be directly substituted to benzene ring(1 vote)
- i just wanted to know if Para nitrobromo-benzene with KCN will give 4 nitrobenzonitrile(1 vote)
We've done a lot of electrophilic aromatic substitution reactions. Let's look at the possibility of a nucleophilic aromatic substitution. And so if we start it here with bromobenzene and we add a nucleophile, something like the hydroxide anion right here. So negative 1 formal charge, it 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, it's a concerted mechanism where these electrons should kick off onto the bromine. So if this happens, we would get our benzene ring and we would get now the OH has substituted in for the bromine. And the bromine has left in the form of an anion. 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 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 an SN1 type mechanism, we know the first step in that is dissociation. So these electrons in here are 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, that would get a plus 1 formal charge like that. And so we form a carbocation. The problem is this is a very unstable carbocation. We can't really draw any resonance structures for it. And so since it's an unstable carbocation 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 do a 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 in this case, those two groups are both ortho to each other. 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, attacking the carbon, once again, that's bonded to our halogen. This time 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. 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 OH attached to our ring, too, 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 the formal charge. And now the nitrogen is bonded to an oxygen on the left, also with a negative 1 formal charge like that. The nitrogen itself has a plus 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 on our hydroxide anion, you could think about them as being this bond now. And these pi electrons in here like that, 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 moved out onto the oxygen on the left to give it negative 1 formal charge. And so the first step in this mechanism is the addition of the nucleophile. 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 that negative charge that you're adding actually ends up on this oxygen here, 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 withdraws 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, the one showing the negative charge on the oxygen. So 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 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 onto some of 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 would kick off onto 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 OH attached to our ring. And now we have our nitro group back the way it looked before with a negative 1 formal charge on this oxygen. And this oxygen over here double bond to the nitrogen, the nitrogen having a plus 1 formal charge. So the electrons in green over here on this structure, so these electrons moved back into here to form this bond. And the electrons in blue moved back into here to re-form your ring this way. And we're done. We've seen that the OH 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. So that was an example of a situation where the two groups are ortho to each other. 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 so therefore we can perform nucleophilic aromatic substitution. And so I'm going to use the same nucleophile as before. I'm going to use hydroxide so we have a negative 1 formal charge. 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 here. And then that pushes these electrons off onto our oxygen once again. So let's go ahead and show again all those electrons after they have moved here. So we have our bromine still attached to our ring. Our OH 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 nitrogen here being positively charged. So two negatively charged oxygens 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, when we added some electrons to our ring, that negative charge 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 that would push these electrons back into here. Which pushes these electrons over to here. And then these electrons come off onto our leaving group. And so the bromide anion leaves, and we now have our OH has substituted in for it. So we have our ring. We now have our OH on there. And we have, of course, our nitro group. So over here, double bonded to this oxygen, this nitrogen has a single bond to this oxygen and negative 1 formal charge and a plus 1 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 tried 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 onto 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.