Predict the major monoalkylation products you would expect to obtain from reaction of the follow

step 1 All right. So we are going to be methylating eight species here. And they're asked to predict th

Predict the major monoalkylation products you would expect...

Key Concepts

Friedel–Crafts Alkylation

This is a reaction where an alkyl halide reacts with an aromatic compound in the presence of a Lewis acid catalyst (such as AlCl?) to form an alkylated aromatic product. The Lewis acid activates the alkyl halide by forming a complex that generates a potent electrophile, typically a carbocation or a related species, which then attacks the aromatic ring.

Electrophilic Aromatic Substitution (EAS)

EAS is a fundamental reaction mechanism for aromatic compounds whereby an electrophile replaces a hydrogen atom on the ring. The reaction involves the formation of a non-aromatic sigma-complex (arenium ion) intermediate followed by deprotonation to restore aromaticity, allowing the introduction of various substituents into the aromatic ring.

Activating and Deactivating Substituents

Aromatic rings often contain substituents that can either donate or withdraw electron density. Activating groups (typically electron-donating groups) increase the electron density of the ring and enhance its reactivity toward electrophiles, whereas deactivating groups (typically electron-withdrawing groups) reduce the ring's reactivity. The nature of these substituents influences both the rate of the electrophilic aromatic substitution reaction and the stability of the intermediate sigma-complex.

Directive Effects (Ortho/Para vs Meta Directors)

Substituents on aromatic rings also affect the position at which new substituents are introduced in electrophilic aromatic substitution. Electron-donating groups generally direct incoming electrophiles to the ortho and para positions, while electron-withdrawing groups typically direct to the meta position. This directing effect is a result of the substituents’ resonance and inductive properties, which stabilize or destabilize the intermediates formed during the reaction.

Effect of Lewis Acid Coordination on Substituent Behavior

In reactions involving Lewis acids like AlCl?, certain substituents on the aromatic ring can interact directly with the Lewis acid. This interaction can alter the electronic properties of the substituent and the overall ring, sometimes leading to changes in reactivity or altering the typical directing effects. Such complexation can in some cases deactivate or modify the expected orientation for electrophilic aromatic substitution.

Transcript

00:01 All right.

00:01 So we are going to be methylating eight species here.

00:07 And they're asked to predict the major monoalkalation products you would expect to obtain from the reaction with chloromethine and aluminum chloride.

00:17 So we're going to go in order here.

00:18 I'm going to be giving some commentary as we go because i don't necessarily think that this question is very well worded, only because of some exceptions you know about friedel crafts that i think are being ignored here.

00:31 So we're going to assume that all of these reactions happen, and we're just going to go solely based off directing effects.

00:38 Right.

00:39 So they want the major products.

00:40 We're going to give them the major product.

00:42 So let's start with the bromination products, right? so bromine is an orthoparadirector, although deactivating.

00:48 And you may remember that deactivated rings are not going to undergo frugal crafts, but these slightly, deactivated rings, the halogenated rings are okay, but anything less deactivated than that tends to not undergo friedel crafts.

01:04 Also remember that friedel crafts has a tendency to poly substitute.

01:09 So because the question said, we're going to have a monoalkalated products, we're going to ignore that as well.

01:15 So we're going to probably get a mixture of ortho and para bromotoluin here, probably favoring the para substituent.

01:28 The para substitution just from the bromine being a pretty large atom.

01:32 It's going to disfavor the orthopositions due to some steric interaction there.

01:38 So i know from the textbook said that anilins tends to not react by fredo crafts because there's an amino group that is able to undergo protonation.

01:50 And that's true, but there's also another reason, which is that the nitrogen tends to bind to metals, right? so atoms with loan pairs will form coordination.

02:01 You don't know about this yet, maybe, but they'll form coordination, they'll coordinate to the metal, which will inhibit the aluminum to be inactivated, to be available to activate the chloro, the alkyl chloride for freeofrafts to make the carbocation.

02:19 So phenols are also less active in that regard, but the amino groups being stronger bases are going to inhibit the aluminum.

02:28 So that's why the textbook says, you can't deactivated rings and amino compounds do not undergo fridyl crafts.

02:35 But we're going to pretend that this is fine.

02:38 Phenols tend to be okay, but i think the yields are just pretty low of these compounds, but we're going to ignore all of that.

02:44 Right.

02:44 So anyway, the, so you have two ortho and para directing groups.

02:51 So the effects of the methoxy, sorry, the effects of the phenol are probably going to override the effects of the bromine because it's a stronger activator.

03:07 So you have a tendency to want to put that at either of these ortho positions to the methoxy or pari or parat, sorry, the methoxy, the phenol, or para to the phenol.

03:17 And being that bromine is an ortho parodirector, i would imagine you're going to get some contribution.

03:22 So i'm going to say there are going to be two major products here.

03:27 Probably the, you probably get some mixture.

03:33 So you'll get a methyl group there, but you'll also form this compound.

03:39 The toxi, sorry, metabromine and that ortho.

03:49 I'm going to make some room here.

03:53 I'm going to move e over and i'll move f over because you're probably also going to get this one.

04:06 Right so there's two orthopositions probably disfavors that one because you're shoving a methyl group that's relatively sterically encumbering to the orthopositions right between those two substituents so you're probably not going to you might not want to draw this one but it's still going to be it's still directed there by both of the groups on the ring um so you can't you can't discount it um this is i would say this generally doesn't go fridalcraft's alcalation because it is an amino group, which the textbook did say tends to have a hard time undergoing fidel crafts, but we're going to listen to the question and say, what products would i expect here? i wouldn't expect any products, but if it were going to be to react, it's going to be ortho -parad directed by the amino group, but the parat positions are already taken up by the chloride, so i'm going to draw two products here.

05:03 Really, it's one product because of symmetry, but it's probably going to go ortho to the amino group.

05:11 And then you could put it on the other side, but it's the same compound.

05:13 If you just rotate, you'll have that position.

05:16 You'll have that in the other position as well.

05:19 I'm not going to put, i'm not going to say the methyl, the orthoposition to the chlorines are going to be favored at all because the amino group is so much more activating than the chlorine is deactivating.

05:33 It's a much stronger director.

05:34 So i'll say that's the major product.

05:37 We have a tri -substuted ring here, right? so you have the nitro group that wants to put this meta, but you have both of the chlorines that want to put it ortho to the chlorines and maybe para, which is the same position here.

05:53 So because the nitro group, both of the chlorines are directing it to this position, i'll say the major product is probably going to be between the chlorines.

06:03 Even though that might be relatively sterically encumbering.

06:07 Why can't...

06:09 There we go...

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