Propose products (if any) and mechanisms for the following AlC...

Question

Propose products (if any) and mechanisms for the following AlCl₃-catalyzed reactions:

c. 3-chloro-2,2-dimethylbutane with isopropylbenzene

Accepted Answer

All right. Hi everyone. So for this question you must predict the product and draw mechanism for the reaction of todd align with neo pento chloride. So let's begin and let's begin by first noticing the fact that we have to Halloween which is aromatic and an okay, Oh, hey, light. So when we have an aromatic system with an oak Yohei light in the presence of a catalyst such as a L. C. L. Three. Recall that this is going to mean that we're talking about a Friedel crafts um calculation reaction right. Articulation. Because what we're doing is we are calculating told we are adding an alcohol group to taliban, which in this case is this neo pencil group that we have associated to our chlorate. So before we begin or before we go ahead and we draw the final product first, recall that in this instance, as soon as our chlorine is going to go ahead and leave forming a cat eye on, we end up with a primary carbon cat eye on. Now, that's not ideal. Right? But the neo pencil group is capable of doing an all kill shift. Right? Because what can happen is that one of the method groups um adjacent to this primary carbon here, it can go ahead and it can shift towards the primary carbon cat ion to form a tertiary carbon cat island, which naturally is going to be more favorable. Right? So before we begin, I'm gonna redraw Halloween. And we have to also remember the fact that substitue ints or rather existing substitue ints on an aromatic system are going to direct the placement of the new substitute wint. So with that being said, we have uh an aromatic benzene ring with a method group attached to it, right? That's our Halloween. Now. Methyl groups are also called a kill groups and all kill groups. Recall that there are an example of what we refer to as activating groups, right? They are activating groups because they donate electron density towards the aromatic system, therefore activating it. So, another name that we have for activating groups is Ortho slash para directors. Because the activating group is going to direct the new substitue in 22 places. Either the Ortho or the pair of positions. So if our method group is on carbon one of the benzene ring, the Ortho position is carbon two and the para position is carbon four. So we have two possibilities, but in this case we have to predict the main product, right? There's one answer in this case. Generally speaking, when we're talking about either an Ortho or a pair of product, it's gonna be the pair of products that is more favored because regardless of your substitue in, if they're on opposite sides of your ring, right, there's gonna be a lot less Starik hindrance than if they were adjacent to each other on the ring, in the Ortho position. So that is why um the para position is going to be favored and therefore I'm going to draw our rearranged Neil pento group on the para position. So it's going to look like that because remember that um these three methyl groups on carbon two of the neo pencil um group one of them is going to shift towards carbon one which was the primary carbon. So to illustrate that point, let's go ahead and let's redraw um our neo pencil chloride and let's get started with our mechanism right? Because recalling what we're talking about free to craft calculation or electra filic aromatic substitution in general. Our first step is going to be formation of the electro file, right? It's gonna be formation of the carbon. So what's going to happen first like we discussed before is um that the electro file is going to form first and that's going to happen because this chloride group of our neo pencil chloride is going to go ahead and attack aluminum. It's gonna go ahead and attack aluminum. And so we are going to end up with our primary carbon cast iron but not for long. Right? So this is gonna be now a L. C. L. Four minus That's our 4th ceo and now aluminum has a negative charge. So now um recall that we talked about before that the primary carbon cat ion is not favorable. Right? So this is where are all kill shift is going to come in Our methyl shift I should say. And so one of these methods groups here is going to jump towards carbon one which is our car, broke down and we are going to end up with a more stable tertiary carbon copy on instead. So this is gonna be our new carbon cat. I'll and now we can go ahead and introduce our ptolemy if I draw told me like that. Now what's going to happen is this is going to be our attack, right? Because our carbon cat ion is going to be our electro file and now one of the pi bonds of our talking weed is going to go ahead and attack the car, broke a tie on. And because we mentioned that the pair of product is favored, it's going to be the pi electrons on the para position that goes ahead and attacks are carbo katia. So we're going to end up with our di substituted benzene. This is our Halloween. But now notice how the pi bond is no longer present, that third pi bond because those electrons were lost, so to speak, while forming the bond with this tertiary um carbo cat eye on. So we're going to end up actually with a cat eye on, on the ring, right where that pi bon used to be. But this can be fixed when we remember that we have hydrogen associated to um or ring here. So let hydrogen if I scroll down is going to come and save the day because this is where our A. L C 04 minus is going to come in. So if I go ahead and I redraw A. L. C. L. Four. See this is our minus church. So now what's going to happen is that one of these chloride ions is going to detach itself from R A L. C. L. For minus and take um that hydrogen. So this bond is going to dissociate from aluminum and it's gonna go towards um this hydrogen on our Halloween instead. So in our next step, I should actually add that the pair of electrons that was initially tethering our hydrogen to the carbon of the benzene ring. It's going to jump towards where the carbon cat eye on on the ring is So with that in mind, aromatic city is now restored. We have our three pi bonds. Once again, our method group stays the same and we have our local group over here and now A. L. C. L three has been formed once again. Right, because our fourth chloride has dissociated and because our fourth chloride went ahead and attacked the hydrogen of the ring, we're going to end up with hcl as another side product. So with that being said, we went through the entire mechanism to justify the formation of our final product. So with that being said, I'm just gonna do one quick scroll through, write your first step is always going to be formation of the car woke a dile. Sorry, not your carbon cat eye on your electoral file, which in this case was a primary carbon copy, so that's not favorable. Which is why we had a methyl shift into a tertiary carbon cat ion, which arch arguing went ahead and attacked. However, during the attack, aromatic city was lost, but aromatic city was once again restored after r a L c 04 minus regenerated into A. L. C. L three. Alright, so if you stuck it out to the end, thank you so much for watching and I hope you found this helpful.

Propose products (if any) and mechanisms for the following AlCl₃-catalyzed reactions:
c. 3-chloro-2,2-dimethylbutane with isopropylbenzene

Key Concepts

Friedel-Crafts Alkylation

Carbocation Stability

Regioselectivity in Electrophilic Aromatic Substitution

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