Lecture 24: Aromatic Substitution with Carbocation as Electrophiles
This is a modal window.
The media could not be loaded, either because the server or network failed or because the format is not supported.
Formal Metadata
| Title |
| |
| Title of Series | ||
| Part Number | 24 | |
| Number of Parts | 26 | |
| Author | ||
| License | CC Attribution - ShareAlike 3.0 USA: You are free to use, adapt and copy, distribute and transmit the work or content in adapted or unchanged form for any legal purpose as long as the work is attributed to the author in the manner specified by the author or licensor and the work or content is shared also in adapted form only under the conditions of this | |
| Identifiers | 10.5446/21619 (DOI) | |
| Publisher | ||
| Release Date | ||
| Language |
Content Metadata
| Subject Area | ||
| Genre | ||
| Abstract |
|
1
12
16
17
18
21
00:00
BenzeneChlorineElectronCarbokationIonenbindungActivation energyResonance (chemistry)Organ donationDoppelbindungAluminiumSubstituentCHARGE syndromeAluminium chlorideFunctional groupChemical reactionReactivity (chemistry)Lone pairHalogenationBiomolecular structureWalkingAromaticityWine tasting descriptorsSunscreenChlorideWursthülleOleumNitrobenzolHalidePhenolReaction mechanismAcidSense DistrictBromideHydrogenPhase (waves)Inclusion (mineral)SubstitutionsreaktionHydrocarboxylierungEthylgruppeThermoformingStickstoffatomBase (chemistry)Acid anhydrideConjugated systemSetzen <Verfahrenstechnik>ElektronentransferCarbon (fiber)ProtonationEisenbromideAcylAlkaneSchwefelblüteAlkylationCobaltoxideSulfonylgruppeGene duplicationAtomAtomic numberWaterElectronic cigaretteSalpetersäureLeft-wing politicsElimination reactionMoleculeConcentrateSulfoneChemical compoundRearrangement reactionNitroverbindungenPropylgruppeChain (unit)SchwefeltrioxidHydrideBurnCarboxylateOctane ratingColourantTidal raceAusgangsgesteinFiningsChemical structureMedical historyVancomycinIodideAlcoholDessertInitiation (chemistry)Active siteFluorideBrown adipose tissueGesundheitsstörungTitrationVerwitterungLamb and muttonSuppositoryNitrogen fixationAreaQuartzRiver sourceResinTrace elementBromoethaneJoint (geology)Pilot experimentFireStorage tankCurryMineralAssetAddition reactionHope, ArkansasFreies ElektronChemical plantKatalaseSteelLeadSolutionOctanePeriodateBiosynthesisCalculus (medicine)Computer animation
Transcript: English(auto-generated)
00:09
All right, we're going to get started. I looked at the schedule, we are a tad bit behind so I'm going to do less breaks today.
00:21
I'm going to be more efficient but not try to go too fast, OK? Questions before we get started, anybody? All right. All right we were talking about bromination, you guys can help by quieting down quickly
00:46
so we can get started. We were talking about bromination last time, electrophilic aromatic substitution with bromine, we got through most of the mechanism and now we want to talk about the very last step which is going to be regenerating the aromatic ring and this is always going to be an E1
01:04
reaction. It will be very similar for every single mechanism in this chapter, the only difference will be what we actually use for the base to do the E1 reaction. So, and really what you do is we're going to pick any one of the resonance structures
01:25
that we just drew for the carbocation intermediate and we're going to eliminate the hydrogen that's bonded to the same carbon that the bromine is bonded to and we're going
01:40
to use Br- to do this deprotonation. And you can see we reestablished the aromatic ring, we're eliminating a hydrogen that's
02:04
beta to the carbocation. Where does the Br- come from? It comes from this right here, when this happens we get FeBr4 and that's in equilibrium
02:23
with FeBr3 plus Br-. So when we do these Lewis acid-base reactions they are reversible just like Bronsted acid-base reactions so that bromide ion can come on and off that bromine, on that iron, so it
02:43
can come on and come off and so we're going to use Br- for that elimination. Questions on that mechanism, anybody? Yes, oh, okay, let's do it, can we add it onto here?
03:13
Let's add it onto here. I recommend drawing the first one over again but you could draw it from this one and what
03:21
it would look like, a lot fancier and just continue with the green arrow, yeah? Kind of cool, huh? But it looks more complicated that way. A little easier from the last resonance structure, a little harder from that resonance structure, not as obvious, not as intuitive.
03:41
Okay let's look at the energy diagram, it looks a lot like the energy diagram for an SN1 reaction, an E1 reaction, we have a large energy of activation, this is the rate determining step because we are disrupting aromaticity.
04:06
We are disrupting aromaticity but we are not doing it for long because in the second step of the reaction we're going to, we get this resonance stabilized carbocation in the second step we reestablish aromaticity and that's a very very fast reaction here.
04:21
Very fast second step. And that leads us right into nitration and like I said, that mechanism that we did for bromination, this is going to be very similar, the only thing that's going to be different is how
04:42
we make the electrophile in the very beginning. The rest of it's going to look very much the same. So make electrophile, let's draw that. We have nitric acid and sulfuric acid. Concentrated nitric acid, concentrated sulfuric acid is what we use for this reaction.
05:02
So this is what nitric acid looks like and the reason we're adding the sulfuric acid is to protonate the oxygen because we're going to lose, oh let me fix that, we're
05:24
going to protonate the oxygen that's on the right hand side.
05:43
I like the fact that I hear a lot of clicking because that means you're changing colors which I should be doing right now but I'm trying to speed up a little bit here. Not the obvious oxygen to protonate, you might think oh well the oxygen, the other oxygen on the left is the one that should be protonated but it actually is the one
06:04
on the right because we're going to be losing that group as water. And so what's going to happen here is that we're going to have the electrons on oxygen here form a new double bond between the oxygen and the nitrogen and then we're going to kick off this leaving group as water and that's why we add the sulfuric acid.
06:23
So this is our souped up electrophile. It's called a nitronium ion and that's going to be our electrophile. In the second phase of this reaction the benzene is going to attack the electrophile.
06:45
So here we have benzene attacking nitronium ion and so any one of those pi bonds of the benzene can attack, I'll just choose this one here, we're going to attack the
07:04
nitrogen and then we're going to kick electrons up onto oxygen. And I'll put the nitro group right here, I'm going to draw it out the first time
07:22
so you see what it looks like and then I'm just going to abbreviate it as NO2. I always put the hydrogen on the same atom that the new electrophile has come in on. That's the hydrogen we're going to eliminate. On the other side of that double bond that we attacked with that will get the positive
07:42
charge. And we definitely have resonance stabilization, you should always be able to draw three resonance structures moving a positive charge around three different spots in the benzene ring.
08:02
So like I said now I'm going to abbreviate this NO2 here and then we can go here. And you want to make sure that you check to make sure that you've drawn three resonance structures, there will be, you will be getting points for every resonance structure that
08:21
you show on the final exam. But be careful, don't draw too many, if you draw extra resonance structures that are wrong you get those cancel out ones that are right. So I've had people draw twelve resonance structures. So we don't, you know, we can't pick out the right answer in a sea of answers.
08:40
So you want to make sure that you're really certain and sometimes people will duplicate resonance structures so just make sure that you have three resonance structures when you draw your carbocation intermediate. And then we regenerate the aromatic ring, I'm going to redraw the first one.
09:13
We've got some possible different groups, bases to use here, we could use water, we
09:22
could use the conjugate base of sulfuric acid, those would be the two likely possibilities here. I'll just use the conjugate base of sulfuric acid and this is our E1 elimination, we're going to grab this hydrogen and we're going to make a new double bond here to reestablish aromaticity.
09:51
So as you can see step two and step three these are very similar except all we're doing is changing the electrophile, changing the base that we use to regenerate the aromatic
10:00
ring so really the first phase of this reaction is the only thing that's different. Alright so that's nitration, that's a great way to put a nitro group on a benzene ring, we also have sulfonation and I'm going to abbreviate this one even more I'm just going to actually show you how to make the electrophile.
10:23
The second part and the third part I'm going to leave for you to try on your own. So here's sulfonation, we have two different conditions, concentrated sulfuric acid or fuming sulfuric acid which is sulfuric acid that contains sulfur trioxide also in there.
10:41
So depending on which one you're starting with, let's see what that looks like. If you're starting with concentrated sulfuric acid this is what it would look like.
11:08
We're going to get a proton transfer here from one sulfuric acid molecule to the other,
11:25
okay just like that, and then we're going to lose water. So very similar to the previous one, we're going to lose water here.
11:45
Water's going to leave and that gives us, there's our souped up electrophile.
12:01
If you use fuming sulfuric acid, which is what I usually use when I'm doing this reaction so let's put a little note there, I use this one. You start off with the sulfur trioxide and then you just protonate it.
12:22
So if you use this, if you start with the first one, this is what it would look like. And we're just going to protonate here, going to grab that proton, break the hydrogen oxygen bond, let's throw some lone pairs on that oxygen and that
12:45
gives you the same thing, just like that.
13:00
Okay so that's how you make the electrophile. And then step two and three, benzene attacks, and then you regenerate the aromatic ring, it's going to look just like the previous one. So that's really good practice for you. Alright so an interesting thing about aromatic sulfonic acids, this is a reversible reaction and can be driven in either direction.
13:22
Okay so here would be the equilibrium. And you can imagine that since water is a product here, if you have dilute, if you add water so you dilute the acid, it's going to go to the left. Okay so increase the concentration of water, it drives the equilibrium to the left.
13:43
If you use concentrated sulfuric acid, then you have little water, it's going to drive the reaction to the right. So this is actually very useful in synthesis, I'll show you an example when we get towards the end of the chapter, that forward and reverse is actually very useful. And so for the forward direction, concentrated fuming sulfuric.
14:15
That's the way to do it.
14:22
For the reverse direction, so you're going to heat in water or dilute sulfuric acid.
14:46
Very useful and what we're going to be doing now is putting a sulfonyl group onto direct incoming substituents or to block a position where normally substituents would go and then once we have our groups on there then we can take it right back off again. So actually very, very useful.
15:05
Right, questions so far, anybody? We're going to talk about Friedel-Crafts now. Friedel-Crafts, there's alkylation and there's Friedel-Crafts acylation. We'll talk about alkylation first. And in this reaction we have RCL, this can be primary, secondary, or tertiary.
15:31
ALCL3, that is a Lewis acid catalyst.
15:42
For bromination we used FEBR3, iron tribromide. That's also a Lewis acid catalyst. So that's a common thing to do in electrophilic aromatic substitution. Now what you do is you incorporate an R group
16:01
onto the benzene ring. So the big deal here is going to be making the electrophile, we'll do the making the electrophile and then I'm going to do an abbreviated step two and three. Okay, so and this is going to be kind of remind you
16:20
of chapter nine where we had a different mechanism whether it was primary, secondary, or tertiary. So if it is secondary or tertiary alkyl halide you're going to form a carbocation. So that hopefully is going to seem, wow, just deja vu from chapter nine.
16:42
And then you can imagine if it's primary, primary we're not going to form a primary carbocation. So let me give you an example of what this would look like.
17:01
Let's use tert-butyl chloride. It doesn't have to be chloride by the way, it is a ALCL3 but it can be a bromide or an iodide. So they don't have to match, the halogens do not have to match. So this is ALCL3 and this is going to do what Lewis acids like to do
17:22
which is accept a pair of electrons from something, especially something with a lone pair, very favorable.
17:53
So this is going to be a Cl plus and an ALCL3 minus. We're changing charges after the aluminum accepts a pair of electrons.
18:04
Okay, and then what's going to happen is that we're going to break the bond to the carbon and the chlorine to make up tertiary carbocation.
18:32
And we have ALCL4 and you can imagine that this ALCL4 is going to do the same thing that the iron tribromide did. Those chlorines can come on and off the aluminum.
18:43
So we're going to be using chloride ion in the very last step of this reaction. Okay, so that's what happens if it's secondary or tertiary you're going to make a carbocation. If it is primary, don't form a primary carbocation.
19:09
So what we're going to do is we're going to keep this attached to ALCL3.
19:21
All right, so let me give you an example here. We'll use ethyl chloride. No, you know what, we'll use ethyl bromide so you can see that it doesn't matter if we change the halogen.
19:42
So the lone pair on bromine going to attack the aluminum. Aluminum's happy because it has fulfilled its octet rule. It's got eight electrons now.
20:02
And so rather than having that group leave, that's going to be our electrophile. So the electrophile is going to change. So that's our electrophile. So E plus for abbreviation for electrophile. And then on this one, this is going to be our electrophile.
20:24
So I'll show you the arrow pushing for the second part of this reaction. So you can see how it's different between the two. So if secondary or tertiary,
20:42
it's going to look like this. Benzene's going to attack the tertiary carbocation.
21:06
Don't forget our positive charge there. We're going to form a carbocation, resonance stabilized carbocation. I always put the hydrogen on the same carbon where the incoming group just came.
21:21
I'm leaving the other hydrogens off, but this one I like to include here, because that's the hydrogen I'm going to be eliminating when I do the final phase of this reaction. So there's that, plus two more resonance structures.
21:44
If it's primary, let's see what that looks like.
22:05
And again, this could be bromide, this could be iodide, this could be chloride. And then we have a positive charge on bromine. And so what we're going to have
22:21
is the arrow's going to come from the pi bond, just like in the previous example. We're going to attack carbon, and then this whole group is going to leave. And see how we bypassed forming a primary carbocation.
22:50
Again, I'm putting the hydrogen on that same carbon. There's going to be plus two more.
23:01
On the test, you will need to draw those right rather than write plus two more, okay? So I'm just trying to save a little time and space here. And then two ways to show the final step. We're going to do it from here. You can use a Cl minus.
23:21
We've also got bromide ion. You can use bromide ion to do this final elimination. Like that.
23:40
Or, you can do it how the book does it, which is this way. I'll show you how the book does it. AlCl3 minus, and we need to change colors, don't we? Let's do this color right here.
24:03
In the book, they have the arrow, and this is important, not from the lone pair on chlorine. It comes from this aluminum chlorine bond. So you can have it do that also. The second arrow would look exactly the same. So we'll just put a green arrow there.
24:21
Second arrow would look exactly the same. Really important that you don't have this come from the lone pair on chlorine. If you're going to have it come from the lone pair on chloride ion, that's okay, or bromide ion. But if you're doing it this way, then the arrow has to come from this aluminum chlorine bond.
24:40
So all of those are fine choices for you. Questions. Now we've learned a lot of different ways to make carbocations. You don't have to use this way to make carbocations. Pretty much any of the ways we've talked about making a carbocation can work for this reaction.
25:04
So let's take a look at that on the next page. So how did we learn to, okay, so you'll recognize both of these ways. By protonating an alkene, we can make a carbocation.
25:23
What chapter was this from? Yeah, that was from chapter 10. Okay, so that can give you, that gives you a carbocation like that. This carbocation does not care where it came from.
25:42
It will work equally well as the AlCl3 example. So that's another way to do it. We can also protonate an alcohol. That was what chapter? That was chapter nine. So we're going back a ways.
26:04
So the carbocation doesn't care how it was formed. It's going, the benzene ring is, the benzene doesn't care how the carbocations form. It's going to still do the same thing no matter how the carbocation is formed. And it also doesn't matter how the carbocation is formed
26:20
with respect to what do we know that carbocations like to do? They like to rearrange. So when they rearrange, so this is going to go to this. This is another way to make that carbocation. We got to watch out for rearrangements.
26:40
And so that's going to also go to the same thing. Okay, so it doesn't really matter how you make it. But we do always, anytime we form a carbocation by any means, we have to watch out for rearrangements. So for example, if you do this reaction with the straight chain propyl chloride, for example,
27:07
you actually get two products. You get straight chain is actually the minor product, believe it or not. We showed how to form the straight chain. That is 35%. And then we get rearranged.
27:22
So that propyl rearranges to an isopropyl. And that's actually the major product.
27:41
All right, so how are we rearranging that product if we aren't making a primary carbocation? That's the burning question. So let's see how this looks. Mechanism for carbocation rearrangement, well, is that for secondary and tertiary, really straightforward. Let's draw that for secondary and tertiary.
28:01
This is just a review of chapter nine, when we first started talking about rearrangements. So this will be a good review for you.
28:21
All right, so the chloride ion attacks the AlCl3. That's certainly one way to do this. So we form a Lewis acid base adduct, and then the chloride is going to leave.
28:51
So negative charge on the aluminum, positive charge on the chlorine, and so that's going to leave. We're going to form a secondary carbocation
29:00
and flashing lights should be going off in your head, light bulbs, something, telling you that, wow, I'm forming a secondary carbocation, I got to watch out for rearrangements. And this one has a really nice one, two hydride shift that will convert that secondary carbocation
29:21
into a tertiary. So that's what it would look like if you were starting with a secondary alkyl halide,
29:41
that's what the rearrangement would look like. If instead you're starting with a primary and we're going to rearrange but we don't want to make a primary carbocation, what's going to happen is we're going to do a one, two hydride shift at the same time as the AlCl3 is leaving.
30:02
So these two things happen simultaneously. So we get a rearranged product without actually ever forming a Lewis acid base adduct, I mean without ever forming a primary carbocation. So this here, we're going to attack the aluminum, just like that.
30:22
We form our Lewis acid base adduct, let me fix that, I'm going to have to draw out one of those hydrogens. So we have that, we have CH2ClAlCl3,
30:48
negative charge on aluminum, positive charge on chlorine. And what can happen is that this hydrogen can migrate at the same time that group is leaving. So here's what it would look like. Arrow comes from the carbon hydrogen bond,
31:03
moves over here, simultaneously this group is leaving. So we've just formed a more stable secondary carbocation without actually making a primary carbocation. And as you can see from the results of the experiment
31:23
that we get, this is our major product. So more stable secondary carbocation.
31:52
All right, that's Friedel-Crafts alkylation. Let's talk about Friedel-Crafts acylation.
32:11
This will work with acid chlorides, it will work with acid anhydrides, and there's a mechanism for both. I'm only going to use this type.
32:27
I will only test on this type. What is this type? Type where you use an acid chloride as your acyl group. I mean as your group, not an anhydride.
32:42
All right, so what is an acylation? Well, it means that we're going to incorporate, we're going to add a carbonyl onto the benzene ring. This whole entire group right here that's been bonded to the benzene ring is called an acyl group.
33:04
Much more about that in 51C. All right, so since I'm not going to use anhydride, I'm not going to show you, I'm going to show you the mechanism for making the anhydride, the activated electrophile, the souped up electrophile.
33:23
So I'm just going to use the acid chloride. All right, so what's going to happen is the chloride's going to attack the AlCl3.
33:47
This is not strictly correct, but this is the way we're going to learn this. If I was teaching this to you in 51C, I would teach you a different way, but for now this is good enough. So it looks like that, okay?
34:06
If you're curious about the exactly 100% correct way to do this, you can come ask me after class, but I will not be testing you on that. All right, so you can imagine what's going to happen. This whole group is going to leave just like we've been doing. We're going to break the carbon chlorine bond
34:33
to form a very strange looking compound that doesn't look very stable, and that's exactly, by design, it isn't very stable, which is why benzene is happy to attack it, okay?
34:44
Very, very electrophilic. You can draw it like that. It is resonance stabilized, so it has a resonance structure that we can draw here
35:00
where we have the positive charge on oxygen. So the positive charge can be on carbon, the positive charge can be on oxygen, and this is called an acylium ion. So you can see that the root comes from the acyl.
35:21
This is an acyl group. And now you can imagine that the next thing that's going to happen is the benzene ring is going to attack that, and you can have it attack either one of those resonance structures, your choice. I'm going to use the first resonance structure because that's the way students usually show it,
35:41
and that's perfectly fine here. So benzene is going to attack this very, very, very electrophilic acylium ion.
36:10
What did I do? I will put the acyl group right here.
36:21
I will put the hydrogen on that same atom, and I will put the positive charge on the other side of the double bond that I just attacked with. You can use the second resonance structure. This is what your arrow pushing would look like.
36:47
This is actually a better resonance structure because all atoms have an octet, but again, I don't really care which one you use. If you use that one, you're going to attack carbon, and then you're going to push electrons up onto oxygen. That will lead you to the same exact thing
37:00
that you got the other arrow pushing way. So you're going to get that plus two resonance structures, and then we're going to have chloride ion come and do an elimination.
37:21
This is our final phase, the E1. We're going to grab that hydrogen. We're going to form a new carbon-carbon bond, and it will look like this, acyl group incorporated onto the benzene ring. So we've just done bromination. We've done nitration.
37:41
We've done sulfonation. We've done Friedel-Crafts alkylation. We've done Friedel-Crafts acylation. One of those is going to be on the final, guaranteed. And of course, on the final, you will draw these resonance structures because you'll be getting points for them. Questions?
38:01
So, so far, pretty straightforward. Some new mechanisms to learn. Now it's going to get a little more complicated because what happens if we're not starting with benzene? What if we're starting with a benzene ring that already has a group on it? Where's the new group going to go? We need to be able to predict that
38:20
and plan that and know why. Okay, so here, let me give you an example here. You will be impressed with. This is phenol, and it's so reactive that you don't even need concentrated nitric acid
38:42
and concentrated sulfuric acid, like the example we did. This is rapid at room temperature. And you form two products. I'll give you the amount on each.
39:02
The nitro group can go on in the para position and the nitro group goes on in the ortho position. So we get ortho para only.
39:26
Ortho para only, no meta. That's example number one. If we compare this, if we have a nitro group on the benzene ring already,
39:43
we have to use fuming nitric acid, which is more powerful. We also have to use concentrated sulfuric acid and 100 degrees centigrade.
40:02
So we have to heat that to get that to go on. And you'll be very surprised if you haven't read the chapter yet, that in this case, we get meta only. No ortho, no para, meta only.
40:28
So you can see that a group that's already on the ring may make it easier or harder to put on a new substituent. And it can also direct where the new substituents are going to go.
40:41
So phenol is more reactive and nitrobenzene is less reactive. So that's not something you're gonna actually have to memorize because I'm gonna explain all of it
41:03
and you will understand why that is the case. So it will be one last thing that you need to memorize because you're gonna understand why that's the case because it's all gonna make sense when we get through this section. All right, and so what we're gonna be doing is we're gonna be dividing the groups that are already on the benzene ring into ring activating, ring deactivating,
41:23
and then a third category, halogens. And each of those has their own, each of those has their own reactivity and their own directing effects. So here's ring activating. A group that makes it easier to introduce new substituents is ring activating. Ring activating substituents are electron donating.
41:43
This is the order of ring activation. These guys, they make benzene electron rich. And in this reaction,
42:01
in electrophilic aromatic substitution, the benzene is the nucleophile. So anything you can do to make it more electron rich is gonna make it more reactive. And so these guys here, I would say, these all the way up to OR are strongly activating.
42:23
And R and AR, AR meaning just another benzene ring attached, these would be weakly activating. And these are all going to be better than if it's just straight benzene, no substitution.
42:54
Do all the strongly activate, well, what do all of these have in common except for the weakly activating, what would you say?
43:03
I heard it, lone pairs. Okay, let's throw some lone pairs on here. And the fact that R and AR don't have lone pairs is that's one of the factors that's making them weakly activating. To be strongly activating, you have to have lone pairs.
43:25
Okay, so that's what they all have in common. So note, all have lone pairs
43:43
except the weakly activating. So that's what you're gonna look for for activated rings are rings that have substituents that have lone pairs. There's ring deactivating. The group that makes it harder
44:01
to introduce a second substituent is ring deactivating. Ring deactivating substituents are electron withdrawing. And I would say that all of these guys here are strongly deactivating.
44:24
And then over here, ketones, aldehydes, amides, esters, carboxylic acids. Okay, so you see the difference here between this, there's a carbonyl right here and there's a carbonyl right here, what's the difference? Yeah, this is the one, this is the group
44:41
that's actually bonded to the benzene ring. Whereas here, the carbonyl itself is actually bonded to the benzene ring. What do these all have in common? So we'll know what to look for. Hmm? No?
45:04
Well let me just draw this nitro group out. Good idea to know what nitro group looks like. So that may give you a hint. Look at the charge on the nitrogen. Look at the charge on the nitrogen. So the ones that are the most deactivated
45:22
have a full positive charge on the atom bonded directly to the aromatic ring. The rest of these guys have partial positive. Partial positive, partial positive, partial positive, partial positive. So all, all have partial positive or full positive
45:54
on the atom that bonded directly to the aromatic ring.
46:12
That's what's making them deactivated. And you're right about the electron withdrawing. They are electron withdrawing. That positive charge is electron,
46:21
that's an electron deficient atom, it pulls electrons towards itself. All right, so we're gonna talk about, there's rules for ring activating substituents and there's rules for ring deactivating substituents. We will start with rule one for ring activating. Ring activating substituents direct incoming substituents
46:42
into the ortho-para positions. They are ortho-para directing. I've repeated the order here of activation. The reasoning is an electron donating group stabilizes the carbocationic intermediate only when it is ortho-para to the site of substitution. So what we're gonna do is we're gonna show ortho attack,
47:04
we're gonna show meta attack, and we're gonna show para. And when we do this, we're going to be able to show why these substituents are gonna be going into the ortho and para position. So what I'm gonna use for this one is just the nitronium ion.
47:23
If I had this on the test, you would certainly want to, you would certainly want to show how you formed that. But what I'm gonna have here is this double bond coming in attacking nitrogen, kick electrons up onto oxygen. If I have meta attack, let's show what that looks like.
47:45
Same double bond's going to attack, except I'm gonna put the nitro group in a different position. If I have the para attack, then the nitro group is gonna go here.
48:03
Now I have to use this double bond. Okay, same idea though. So now let's put this nitro group in the ortho position. Positive charge goes on the other side,
48:21
and then let's draw in our double bonds. And again, that hydrogen there is the one we're gonna eliminate to reestablish aromaticity, and it's there so I don't accidentally put a double bond there when I'm drawing resonance structures. If it's gonna go into the meta position, this is what it looks like.
48:41
Now the positive charge is here, and then these other double bonds stay exactly where they were. If it's going into the para position, here's my NO2, here's my hydrogen, here's my positive charge right here, just like that. And now what we wanna do is we wanna draw
49:00
resonance structures for all of these. And you'll notice I have the skeletons drawn here to save me some time, and what you'll notice is in the ortho attack and in the para attack we actually get four resonance structures. When we do meta, we don't get four. All right, so let's go ahead and move electrons around here.
49:22
So the positive charge is gonna go here. The positive charge can also go here. Nope, I didn't draw that right.
49:44
So we have double bond here, double bond here, positive charge here, and you can see that I gotta be careful not to put a double bond here where this nitro group is. Here's what that lone pair does for me. It allows me to kick in electrons right here
50:01
to make an extra resonance structure. Remember, that's a high energy intermediate. We've interfered with aromaticity. That's a high energy intermediate. We've made a new resonance structure, and that one is actually a specially stable resonance structure. We'll talk about why that is,
50:21
but you could probably figure it out by looking at it right now, why that's the best one. We will continue this on Wednesday.