 This problem asks What would be the carbocation? Intermediate if this substrate was treated with sulfuric acid So do you guys remember what happens when you treat an alcohol with an acid? So it turns that bad leaving group remember the hydroxyl group is a bad leaving group because remember the hydroxide is a good Nucleophile, so if you remember us talking about that so you have a good nucleophile Previous it's going to be a bad leaving group if it's trying to leap Okay, so we have to make it into a good leaving group if we want to do one of these Sn1 or e1 reactions. Okay, do you guys recall that? So if we treat this with sulfuric acid, we're going to make the E1 intermediate which is the carbocation So what is the carbocation? We're going to make would you guys? Do you guys have an idea or should I draw a mechanism? Do you want me to draw it or do you want to do you want to guess? But a positive chart right so let's draw the just let's draw the carbocation in our media and Then we'll go back and Do the mechanism? So that'd be the carbocation intermediate and is this intermediate stable with this rearrange would you expect it to rearrange? No, why not? Okay, so it's not going to rearrange so that's the intermediate Let's draw the mechanism really quick to get to that intermediate and drawing mechanisms write your lone pairs draw your bonds on Really helps you're having a hard time remembering the Lewis structure of sulfuric acid memorize Okay, it's a good one to memorize So acid base reactions remember they happen fastest, right? So that's going to happen first. We've got the Bronson base here Bronson acid a bath mechanism So now we've got this intermediate here, right? Which is and we've also of course got the conjugate base of sulfuric acid So now this is a good leading group, right? Why? Because that can make a small molecule Okay, not to mention. We're also making a tertiary carbocation. That's the second step giving us our desired Carbocation Well Because Remember, what's the next step of the there's another step, right? We're just not showing it. This is the e1 reaction So the next step would be This is the acid base reaction. We're about to show the e1 here, but this problem doesn't ask for So this is the intermediate of an e1 reaction and we do want to keep going with the e1 Would that help? Well, so we would get Cut two different products from this reaction. Okay, one of them would be more stable than the other one So the first product would be if you can imagine water here Deprotonating that proton giving this the double bond there. That's the more stable product The Zaytzo product So if you recall the more substituted the alkene is the more stable it is so that's going to be the higher in the product mixture Concentration, this is going to be the higher concentration in the product And then you're going to have another product and that same intermediate Which is going to be lower in concentration some other water molecule Like that Yeah, of course at the end you're going to have the Hadronium ion and that's really the catalyst of this reaction Because this will go back and start the reaction with one of the starting materials again So honestly you could have thought That The alcohol itself doesn't necessarily need to deprotonate sulfuric acid here It could be you know water deprotonating sulfuric acid because remember what we said is Sulfuric acid in these reactions is always mixed with at least some small amount of water okay, and of course down here you have the Hydronium ion molecule. The thing is I want you to remember that you can freely rotate around these Sigma bonds, right? So if that's the case then if you wrote this product as a different product That's wrong. Okay. That's the same product as that other product. Do you understand that? So that was the same product as that other Just rotate around that city. Any questions on that? Okay, you sure okay