What I want to do for this question is I just want to show you guys the full mechanism. I'm going to draw the full mechanism step by step and you guys are just going to follow me as we go along. So in my first step, I've got my double bond and my double bond is a nucleophile. I'm looking for an electrophile, something that I can grab a proton from. That's going to be my sulfuric acid. So I go over here and now notice that I drew the sulfuric acid a little bit weird. This actually just has to do with the fact that the way that sulfuric acid is drawn right here, it's very difficult to take a proton off of it while it still makes sense chemically. It's better if you draw it in this notation, so that you can actually see where the H is attached to. Oh, it's attached to an O. It's not attached to an S. So I know this is a little bit tricky, but I am going to ask you guys to memorize how to draw sulfuric acid when you're trying to deprotonate it. I know it sucks, but this is going to make your life easier when it comes to actually drawing the product. So here we go. We've got our double bond and I'm looking for my electrophile. That's going to be this H because my H is acidic. So I take my double bond. I grab an H. Give the electrons to the O. What I wind up getting is a carbocation right here. Cool? All right. So I've got my carbocation. What's the next step? I've got that plus I've also got ${\text{OSO}_{3}^{-}}$. So is that going to attack? Well, not yet because on top of that, this carbocation is not very stable. It's secondary and if it moves over, it could become tertiary. So what we're going to do here is a shift. What kind of shift would it be? Well, in this case, this is a methyl group. I only have methyl groups surrounding this carbon. So what that means is that I can't do a hydride shift. Remember that hydride shifts are always going to be preferred. But I don't have any hydrogens on that carbon, so that means I'm forced to do a methyl shift. So I'm going to grab any of these methyls, it doesn't matter which one, and I'm going to move it over to this carbon. What this is going to give me is a constitutional isomer at the end that looks like this, 1, 2, CH3, shift. And now what I'm going to get is a new carbocation that looks like this, where now my positive charge is down there. Why is that? Because I used to have 4 bonds right there and now I only have 3. So now I've got my carbocation and now we're faced with the dilemma of what attacks the carbocation. And I've basically got 2 different nucleophiles. I've got the anion from my acid and I've also got water. One of these nucleophiles is negatively charged and then one of them is neutral. So which of these do you think is the stronger nucleophile? It's usually going to be the negative charge. So I would say that actually the conjugate base of my acid, the ${\text{OSO}_{3}^{-}}$, would be the stronger nucleophile than I would expect it to attack. But we've got another variable in the mix. This is actually going to be a repeating theme throughout this section. It turns out that even though the ${\text{OSO}_{3}^{-}}$ is the stronger nucleophile because it's negatively charged, there's just going to be a whole lot more water. So imagine that I've got one of these guys because I just donated the acid, whatever. And then I've got like a billion of these guys. Imagine that I've got a billion water molecules and I've got only 1 of the ${\text{OSO}_{3}^{-}}$. Even though the sulfuric acid anion is going to be stronger nucleophile, there's not much of it to go around. So what that means is that this carbocation is very easy to attack. It's very easy for water to attack as well. So it's just going to attack with the first thing that collides with it. And because of the way that I have this solution mixed up, I'm always going to have way more water than I am going to have anion. So what that means is that even though water is weaker, it's going to be the one that attacks. Okay? That all has to do with the ratio of water I have to acid. And just so you know, for a hydration, that's usually going to be a lot of water and only a little bit of acid just to catalyze the reaction. So I hope that makes sense why the water attacks and why the sulfate doesn't. So now what we're going to do is we're going to draw the final, or not the final, but the new product, which is that I have an H2O+ because the O still has those 2 H's on it. Now that O has a formal charge because it doesn't want to have 3 bonds. So can you tell me what the last step is going to be in this reaction? The last step is always going to be in any acid catalyzed mechanism, it's always going can be deprotonation. So what that means is that the last step I need to take away a
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- 1. A Review of General Chemistry5h 5m
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- Reactions of Amino Acids: Esterification7m
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- 34. Nucleic Acids1h 32m
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- 36. Synthetic Polymers1h 49m
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10. Addition Reactions
Acid-Catalyzed Hydration
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