At what position would you expect electrophilic substitution to occur in each of the following s

At what position would you expect electrophilic substitution...

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 temporary formation of a non-aromatic carbocation intermediate, often called the arenium ion or sigma complex, which is later deprotonated to restore aromaticity. The nature of the intermediate and the reaction conditions largely determine the reactivity and selectivity of the substitution.

Directing Effects of Substituents

Substituents already present on the aromatic ring influence the position at which electrophilic substitution occurs. Activating groups, which typically donate electrons, direct incoming electrophiles to the ortho and para positions due to enhanced electron density in these regions, while deactivating groups, which withdraw electrons, often direct substitution to the meta position because the formation of the arenium ion is less destabilized at these positions.

Activating and Deactivating Effects

A substituent's ability to activate or deactivate an aromatic ring stems from its electronic effects, such as donation or withdrawal of electron density via resonance and inductive effects. Activating groups increase the ring’s reactivity towards electrophiles, whereas deactivating groups reduce the overall reactivity. The interplay of these effects determines both the rate of reaction and the regioselectivity during electrophilic aromatic substitution.

Transcript

00:01 Right.

00:02 So in this question, we're given some dye -substituted benzene structures, and we're told to predict where the next substitution is going to go.

00:14 So really with these groups, you're going to have, in many cases, groups that aren't as, they don't have similar directing strength, whether it's deactivating or activating.

00:29 They're not going to be similar.

00:30 So the one, the key thing here is the one that is more strongly directing, whether that group is activating or deactivating is going to control the major product in the group of products that you're probably going to yield from the next substitution reaction.

00:49 Sorry, but the next, yeah, the next substitution reaction is fine.

00:52 Right.

00:52 So in this case, you have a methoxy group and then meta to that, you're, you know, you have meta to that.

01:00 Have a bromine, right? so the bromine would dictate.

01:04 So i'm going to use red to dictate my, stop it.

01:10 So bromine is an ortho -powered director, right? so according to bromine, you would want to add at either of those three positions.

01:19 But methoxy is also an orthoparad director.

01:23 So, right, so ortho -to -those, you get the same location for all three.

01:31 Essentially.

01:32 So there really isn't a question here as to where the substitution is going to go.

01:36 But the methoxy is much more strongly directing.

01:41 It's one of the stronger activating orthopar directing groups.

01:45 Bromene, remember, is actually deactivating even though it's still an ortho -power director.

01:50 So in this case, i would actually say that you probably get a mixture of those three products, but you probably get an excess of the para product, the one that's para to the methoxy and ortho to the less sterically hindered side of that bromine, only because of steric.

02:11 So i'm thinking if you rotate that methoxy, you're probably going to have some steric effects happening.

02:17 So you block the ortho positions.

02:20 You still see substitution there.

02:22 You'd still see those products, but you'd probably favor the paraposition in that case.

02:27 So in the second one, you have a slightly different situation, right? so you have an amino group and then ortho to that you have a bromine...

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