Inductive and resonance effects. Created by Jay.
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- What does sigma complex mean, exactly?(7 votes)
- When an electrophile E+ attacks a benzene ring, it forms a resonance stabilized cation with the positive charge distributed largely over the ortho and para positions. This carbocation is often described as a cyclohexadienyl cation or an arenium ion or as a σ complex .(20 votes)
- Do all meta-directing groups tend to be deactivators of the benzene ring?(5 votes)
- Yes, they remove electrons from the ring at all positions, but primarily at the ortho and para positions. This removal of electrons makes the ring less reactive (deactivates the ring), but the meta positions have a higher electron density than the ortho and para positions, so they are meta-directing.(9 votes)
- At2:33and6:46Jay shows resonance forms of benzene with meta directing substituents that have positive charges in the ortho position. The additional resonance forms would have positive charges at the para position as well as the other ortho position.
Would it be correct to say that the partial positive charge on those positions make attaching to an electrophile less likely or is the meta directing effect purely due to destabilization of the intermediate?(4 votes)
- you better refer to @Ernest Zinck's answer for @Hiren Patel's question above. that's directly handling your inquiry
in paraphrase, the regio-selectivity by the substituent for making only ortho and para positions positive plays a role to make meta positions more elctron-dense relatively. this makes them more reactive, meta-directing the structure itself(1 vote)
- At7:35, Sal said that a pi bond and an electronegative element should be present in the substituent to a benzene ring , its a moderate deactivator . Then why NO2 is a strong deactivator?(2 votes)
- I am confused about distinguishing electron donating and electron withdrawing group. Could you please help me with that?(2 votes)
- Electron donating groups usually have lone pairs of electrons that can get mingled with the 6 pi electrons & take part in resonance. Electron withdrawing groups usually have an atom that has a +1 formal charge or is partially positive that is directly bonded to the benzene ring, this withdraws electron density from the ring.(2 votes)
- at4:36could someone explain why the electrons on F are too far away to take part in resonance? I'm just thinking F is the most electronegative atom, so electrons would be close to F and far from C. How are the electrons far?(2 votes)
- The electrons are too far from the benzene ring (close to the F atoms) and are thus not able to participate in resonance.(2 votes)
- What does a deactivator mean? Is it deactivating the electrophillic aromatic substitution, causing the substitution process not occuring?(2 votes)
- Deactivator withdraws electron from something.
Here the deactivating group withdraws electron density from the benzene ring.
When the benzene ring lost electrons to a deactivating group, this benzene ring generates a positive charge.
For an electrophilic aromatic substitution to take place, benzene ring has to act like a nucleophile. Electrophilic group (positive charge, electron deficient) will have to attack nucleophile( electron rich) benzene.
But in the case of deactivator, benzene is not a nucleophile. It actually becomes an electrophile instead by losing electrons to deactivator. Thus electrophile won't be able to attack benzene ring.
Thus electrophilic aromatic substitution won't take place.(2 votes)
- Hello at the end of the video you say a way to find moderate deactivators is to recognize the pie born attached to a electro negative atom what about no2 it has a pie bond but is strong deactivator(2 votes)
- that has a + charge on N. this makes it a strong than moderate deactivator. cause it wants and thus withdraws electrons from benzene ring more eagerly(1 vote)
- Didn't nitrogen have 5 valence electrons? Where's the lone electron?(1 vote)
- N has a positive formal charge, and it is making 4 bonds. it has a full octet but only has 'ownership' of 4 electrons, not 5 as you have stated. There is no lone electron on the N.(3 votes)
- Is it possible to have a meta director that isn't also a deactivator?(1 vote)
- No. Deactivators are meta-directing because of the placement of the + charge in the resonance forms of the sigma complex. The meta position becomes more stable because having two + charges next to each other is particularly unstable, as seen in the resonance forms for the ortho and para positions.(2 votes)
In the last video, we looked at the nitration of the nitrobenzene molecule, and we saw that the nitro group is a meta director because of this plus 1 formal charge on the nitrogen right next to your ring. We also said that the nitro group is a deactivator, which meant that the nitrobenzene molecule reacted more slowly than benzene itself in a nitration reaction. And so let's see if we can figure out why the nitro group is a deactivator in terms of both the inductive and a resonance effect. And we'll start with induction first, which we know is due to differences in electronegativity, and so we could think about this nitrogen here compared to this carbon on the ring. The nitrogen being much more electronegative, and so it can pull electrons closer to itself, and so we can show the movement of electrons towards nitrogen. A positively-charged nitrogen is extremely electron withdrawing, and it could even withdraw electron density from your ring itself. That makes the ring more positive, which would destabilize the sigma complex that results in the mechanism for electrophilic aromatic substitution. Since the sigma complex is less likely to form, that would deactivate the ring towards electrophilic aromatic substitution. Also, you could think about the electron withdrawing effect of this nitro group making the ring less nucleophilic. And so both of those factors mean that the nitro group is a deactivator towards electrophilic aromatic substitution. So induction says that the nitro group is a deactivator. Let's look at resonance next. So we could draw a resonance structure, right, where we take these pi electrons and move them in here, which would push these electrons off onto your oxygen. So let's go ahead and draw that. We have our ring, we have pi electrons in our ring, now, the nitrogen is double bonded to our ring, and we have an oxygen on the right, which has a negative 1 formal charge. The nitrogen is bonded to an oxygen on the left, which now has three lone pairs of electrons and therefore, has a negative 1 formal charge. Let me go ahead and highlight those electrons here. So if these electrons in blue move off onto the oxygen, that gives that oxygen a negative 1 formal charge. This nitrogen has a plus 1 formal charge. And let's go ahead and follow these pi electrons too. So in red, these pi electrons move out here to form this pi bond taking a bond away from that carbon, therefore, giving that carbon a plus 1 formal charge. And you could keep going and draw more resonance structures, and they would have a positive charge in the ring, but I'm going to stop there because the point I'm trying to make is that the nitro group is electron withdrawing, and it makes the ring more positive, which would, of course, deactivate the ring towards electrophilic aromatic substitution. So both the inductive effect and the resonance effect say that the nitro group is a deactivator, and so we can say it's a strong deactivator because of both of those effects. Let's look at another molecule that's considered to be a strong deactivator. So this molecule down here. And we can see that there's a carbon bonded to our benzene ring, and that carbon is bonded to three fluorines. So you could call this trifluoromethyl group, you could call it CF3 group, whatever you want to call it. In terms of electronegativity difference, obviously, fluorine is much more electronegative than carbon so you could think about these fluorine atoms withdrawing a lot of electron density away from that carbon, which would give that carbon a partial positive charge. And so that's an explanation for why the CF3 group is a meta director. So once again, look at the last video for more details about why a positive charge next to your ring favors a meta attack. In terms of the inductive effect on the ring, these are very electronegative atoms and so they can withdraw some electron density away from your ring so you could show the electrons of the rain moving in that direction towards your electronegative fluorines. And so that withdraws electron density from the ring making the ring more positive, which would, of course, deactivate the ring towards electrophilic aromatic substitution. And so this trifluoromethyl group is both a meta director and deactivating, and this inductive effect is so powerful, this CF3 group turns out to be a strong deactivator. We can't have any kind of a resonance effect here because these lone pairs of electrons on one of the fluorine's I'll just choose this one right here-- those are too far away from the ring to participate in resonance. So only in inductive effect, but it turns out to be a very powerful one. Let's get one more example of a meta director, and this one is a moderate deactivator. So first, let's try to figure out why this is a meta director, and so if we look at the atom that is directly bonded to the ring that is double bonded to an oxygen. And we know that oxygen is much more electronegative than this carbon so we could think about this oxygen as being partially negative, pulling some electron density towards it, withdrawing some electron density away from this carbon, giving this carbon a partial positive charge. And so once again, you have a partial positive next to a ring, which makes this group a meta director. In terms of being deactivating, we need to think about resonance structures. So let's think about moving these electrons in this pi bond off on to your oxygen so let's go ahead and draw that one. So we have our benzene ring, and we have, now, our carbon with only a single bond to this oxygen. This oxygen now has a negative 1 formal charge on it like that, and then this carbon is still bonded to this R group here. We took a bond away from that carbon so now this carbon has a full plus 1 formal charge like that. So we can now show a resonance structure for this. We have these pi electrons here can move out into here to form a pi bond between those two carbons. So let's go ahead and show this resonance structure. So we have our ring, these pi electrons. Now, there's a double bond here, and this carbon is still bonded to this oxygen. So this oxygen has a negative 1 formal charge. And there's an R group on it to that carbon. Let's show those pi electrons moving here so these pi electrons moved into here taking a bond away from this carbon so we have a plus 1 formal charge right here. And, of course, once again, you could keep going and show more resonance structures, but I'm only trying to point out that the presence of this group increases the positive charge on the ring, and so the ring is more positive, which means it is deactivating towards electrophilic aromatic substitution. And so this group is both a meta director and a deactivator. Now, let's think about a general way to recognize these moderate deactivators here. So if I go back to this first drawing on the left, we have an atom bonded to a benzene ring, and that atom has a pi bond to an electronegative atom. In this case, it's oxygen, and so that's the pattern that you're looking for. So a pi bond next to an electronegative atom will give you a moderate deactivator.