Treatment of p -bromotoluene with NaOH at 300^∘ C yields a mixture of two products, but treatmen

step 1 All right. So this is talking about benzene. It's kind of weird putting a triple bond in a six -

Treatment of p -bromotoluene with NaOH at 300^∘ C yields a...

Key Concepts

Nucleophilic Aromatic Substitution

This reaction class involves the replacement of a leaving group on an aromatic ring by a nucleophile. The substitution does not proceed by the typical electrophilic aromatic mechanism but instead often involves alternative pathways when the aromatic ring is not activated by strong electron-withdrawing groups.

Benzyne Mechanism

Under very harsh conditions, such as high temperatures, aromatic nucleophilic substitution can occur via an elimination?addition process where a leaving group and an ortho hydrogen are eliminated to create a reactive intermediate known as benzyne. Nucleophilic attack on the benzyne intermediate can occur at either of the two positions adjacent to the triple bond, leading to the formation of different regioisomers.

Substituent Effects on Regioselectivity

The presence of electron-donating groups like the methyl group affects the regioselectivity of nucleophilic attack on the benzyne intermediate. Although such groups are typically ortho/para directing in electrophilic aromatic substitution, in the context of the benzyne mechanism they influence the stability and formation of the intermediate and determine which positions are more susceptible to nucleophilic attack. This leads to differences in the number of products formed, as seen in the case with para- versus meta-substituted bromo compounds.

Transcript

0:00 All right.

00:01 So this is talking about benzene.

00:04 It's kind of weird putting a triple bond in a six -membered ring, but they're intermediates.

00:10 They don't last very long.

00:11 Benzines are super reactive.

00:12 So we're told in this reaction in this question that we're treating p.

00:18 Bromotoliumium -talumin with sodium hydroxide and heating the crap out of it.

00:22 And we only get two products.

00:24 So why is that? and then on the next panel, i have the other scheme where we get, three products.

00:32 And the reason for that is essentially symmetry.

00:38 So you'll notice that when we have the base here, so i drew the resident structures because you need to push your arrows onto an existing double bond and then kick the leaving groove off the other side of that double bond.

00:54 So for example, if i'm going to go ahead and use hydroxide here and do this, i need to push onto a double bond and then kick off my bromide.

01:04 So then that is going to give me a benzine that looks like, let me make sure i draw this right, that's going to give me a benzine that looks like this, right? so you know from the chapter that when you use sodium hydroxide, now you have water around, water is going to very quickly attack either of those carbons, right? so it's either going to attack that one at that three position, or it's going to attack at the four position.

01:37 So whichever way it goes, so i'll do the three position first.

01:43 Whichever way it goes, you're going to get a mixture of products here.

01:47 One's going to be the one three phenol, and the other one is going to be the one four phenol.

01:59 Right.

02:00 So there's two products right there.

02:02 So the other place, so if you'll indulge me, let me erase what i drew up here.

02:11 So you can also have it pick off the proton that is at the four, at the five position.

02:17 Right.

02:17 So we just picked the one off the three position.

02:20 Now we're going to deprotonate off the five position with respect to that methyl group.

02:25 Right...

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