At what position, and on what ring, would you expect the following substances to undergo electro

step 1 Okay, so we have this question. We're given four examples of these molecules that have two aroma

At what position, and on what ring, would you expect the...

Key Concepts

Electrophilic Aromatic Substitution (EAS)

This is a fundamental reaction mechanism in aromatic chemistry where an electrophile replaces a hydrogen atom on an aromatic ring. The process involves the formation of an arenium ion intermediate, followed by loss of a proton to restore aromaticity. Understanding the mechanism is crucial for predicting where an electrophile will attack an aromatic system.

Substituent Effects

Substituents already attached to an aromatic ring can either donate electron density (activating groups) or withdraw electron density (deactivating groups), which profoundly affects both the reactivity of the ring and the orientation (ortho, meta, or para) of further substitutions. Recognizing these effects helps in predicting the most reactive sites.

Directing Effects

Substituents on an aromatic ring are classified as ortho/para directors or meta directors, depending on how they stabilize or destabilize the intermediate arenium ion through resonance or inductive effects. This concept is essential for determining the site of electrophilic attack, especially in molecules with multiple substituents.

Reactivity in Polycyclic Aromatic Systems

In compounds featuring more than one aromatic ring, different rings may exhibit different reactivities based on their substitution pattern and resonance stabilization. Understanding the interplay between rings helps predict which ring is more likely to undergo electrophilic substitution and at which specific position.

Transcript

00:02 Okay, so we have this question.

00:03 We're given four examples of these molecules that have two aromatic rings, and we have to predict which ring and at what position on that ring will the next electrophilic aromatic substitution occur at.

00:16 So with these problems, it's important to take each ring separately and identify what substituents are on each ring.

00:26 And then identify if those substituents are not only ortho -paradirecting, but also activating and deactivating.

00:37 So the more deactivated ring is not going to undergo substitution.

00:43 And by the connection, the more activated ring of the two is going to undergo substitution.

00:49 So the first step is which ring is more activated? and then where are the groups directing the next substituent to go? so if you take a, for example, ring one, ring one on the left, here's ring one, has one and it looks like a or group, r being the other ring.

01:15 So you know that because that auction has one pairs, it's able to electron donate by resonance.

01:21 So this is going to be an electron donating group and a very strong directing one by as it is, right? so you know that methoxy ome is a strong methyl group.

01:32 So o -p -h, ph being this phenol ring, is also a strong electron donating group.

01:39 So that's fun.

01:41 Ring 2 has a methyl group on it and the same group that we were just talking about.

01:48 It still has that o -p -h, that methoxy equivalents, really it's phenoxie.

01:55 But it's a strong electron -donating group, but you also have the methyl group, which is also electron -donating.

02:01 So which ring is more activated, the one that has the more electron -donating groups on there, which is going to be this ring right here.

02:08 So the right ring in example a is going to undergo substitution, and now you're to figure out which positions.

02:14 So when you're fighting between two substituents, you're going to go with the one that's more strongly directing, and that's going to be wherever the alcoxy or wherever the o, the oxygen functional group is directing it to.

02:32 So that is an ortho -paradirector.

02:34 So you have these sites available.

02:36 The parasite is taken up by that methyl group, right? so you're probably going to get substitution at either of.

02:44 Of these orthocytes.

02:45 They are equivalent by symmetry.

02:46 So either one is good to go.

02:49 So then by extension, that's pretty much how we approach these problems.

02:53 Right.

02:53 So look at both rings, which one's more activated? so in the case of b here, ring one has this amino group in the form of nhr, right? so it's a strong autron donating group.

03:10 So that ring is activated.

03:12 Then whichever group is bridging the two rings is going to be present in both rings.

03:19 Right.

03:19 So is the substituent on ring two activating or deactivating.

03:25 Well, bromide is a deactivating group.

03:27 Orthoparad director, yes, but it's a deactivating group.

03:30 So the left ring in this case is going to be the more activated ring and is likely going to go up to undergo substitution.

03:39 The quickest.

03:40 So, right? so the amino group is an ortho power director...