Treatment of p -tert-butylphenol with a strong acid such as H2 SO4 yields phenol and 2 -methylpr

Treatment of p -tert-butylphenol with a strong acid such as...

Key Concepts

Aromaticity and Resonance Stabilization

Aromatic systems are greatly stabilized by delocalized electrons, which distribute electron density evenly across the ring. When a reaction restores or retains aromaticity, the process is thermodynamically favorable, providing a significant driving force for mechanisms that involve aromatic compounds.

Elimination Reactions

Elimination reactions involve the removal of atoms or groups from a molecule, typically resulting in the formation of a multiple bond (such as an alkene). Under acidic conditions, a carbocation intermediate can lose a proton, leading to the generation of an alkene and driving the reaction forward.

Carbocation Formation and Stability

Many acid-catalyzed mechanisms proceed through the generation of a carbocation intermediate. The stability of this intermediate, often enhanced by its substitution pattern or resonance effects, is crucial because a more stable carbocation is easier to form and can further react to yield subsequent products.

Protonation and Leaving Group Activation

Protonation is a key step where a functional group (often a hydroxyl group) is converted into a better leaving group by gaining a proton. This activation makes it easier for the molecule to undergo further transformation such as bond cleavage, as the newly formed leaving group (like water) is much more stable once it leaves.

Acid-Catalyzed Reaction Mechanism

This concept involves the use of strong acids to initiate a reaction by protonating specific functional groups, which in turn makes them more reactive. The acid donates a proton thereby facilitating bond cleavage or rearrangement processes that might otherwise be unfavorable under neutral conditions.

Transcript

0:00 All right.

00:01 So when we treat this terputal phenol with h2s -o -4, apparently we can pop off the turpudal group in the form of two methylopropine and replace that with a hydrogen atom, right? so we're basically doing a reversible electrophilic arometic substitution.

00:20 So a previous question in the chapter talked about the desulfonelination, the desulfonelioration.

00:29 The desulfination mechanism, i think it's 16 .38, and you have to draw that mechanism.

00:36 So this is actually going to be a very similar, a very similar process to that, right? so you're replacing a carbon -carbon bond with a carbon -hydrogen bond.

00:45 It took me a while to think about why this would even happen, but in strong acid, right, you know from previous problems that h -plus can be an electrophile in these electrophilic aromatic substitution reactions.

01:00 And you know that, well, i had to look up why this might be driven forward, right? so two methyl propens actually gas.

01:10 So i think the way that i think about these processes is if you think about delta h minus t delta s, when you make a gas as a product, your entropy increases.

01:22 So if delta s is pretty positive and maybe, even if you increase temperature, this overall term becomes negative to a pretty high degree.

01:34 So the chance that you will make delta g negative by having this super negative term being subtracted from the delta h, there's a chance that at a high enough temperature potentially, you might get a delta g that's below zero, which would make the process spontaneous at some elevated temperature.

01:54 I don't know what kind of stuff your professors will ask you on exams, but that's the kind of stuff that i think about.

02:01 Anyway, so if that's helpful it is, if it's not, just fast forward.

02:05 Right? so the way that you would approach this mechanism is the same way you would approach any electrophilic aromatic substitution.

02:15 So just pretend that your terputal group is a leaving group.

02:19 And you don't have to pop that off in the first step, right? the first step is electrophilic attack of the electrophile by the benzene ring.

02:27 So you're going to take sulfuric acid i think when you're trying to figure these mechanisms out it's helpful on exams to write things out instead of staring at the question until you can do it figure it out in your head i feel like the visual aids are much more helpful and you'll save time just by trying things on scratch paper and stuff that's my advice so you added the h to the turbutal position right there's already an h here now you have that carbocan so how are you going to pop off? how are you going to break that carbon -carbon bonds? right.

03:09 So the reason why i drew these h's is to help you with your arrow pushing.

03:15 Right.

03:15 So all of the carbons on the ring have h's already.

03:18 So if you push electrons, if you try to kick this group out this way, you're going to take the electrons in that.

03:31 Bond with you, right? which means there's not going to be electrons to regenerate aromaticity.

03:37 So it's very likely that we're going to have to take this bond and push it onto the ring, which would mean that we have to give this central carbon on the terbutal group some electrons.

03:51 You have to form the double bonds.

03:52 Also, you know that there's a double bond in the product, which means that right now all of your methyl groups have three hydrogens...

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