Predict the mechanism for the following reactions. Provide the...

Question

Predict the mechanism for the following reactions. Provide the full mechanism and draw the final product.

Chemical structures illustrating elimination reactions in organic chemistry, chapter 8.

Accepted Answer

It's flowchart time. So what's the first question? And what's the first answer? Is this negatively charged or neutral? Okay? And it has a weird looking nucleophile. Alright? I don't blame you if you have no clue what that is, but it is negatively charged because there's a k, right? So this would have a negative charge right there. And the k would dissociate, so I'm going to say that it's negative. I'm going to go down the left hand side. Okay? Second question is, is this one of my bulky bases? K? Actually, no. My 3 bulky bases are t butoxide, LDA, lit and p. This is not any of those. So I'm going to say no and I'm going to move down to the 3rd. Okay. The third question is what's my leaving group? Okay. So my leaving group this time it's easy. It's just an alkyl halide. What type of alkyl halide? That would be tertiary. Okay? So there's a tertiary alkyl halide. By the way, this ring is drawn in 3 d, but you don't need to draw your answer in 3 d. That's just a weird way. I'm just trying to throw you guys off. Alright? So tertiary, do I know the mechanism already? No. I need to go to my last step, which is, okay, is this one of my good bases? Now, this is the part that a bunch of you guys probably got messed up on. Okay? And in advance, I'm sorry. This is just a tricky problem, but this is the way you learn. A lot of you guys thought that this is actually going to be OR negative, which is an oxide. Okay? That's actually the first strong base that I list because you see O and you see that this is an R. It actually turns out this is not OR negative. Okay? OR negative would be something like O CH3. See how it's just a single O, it's not 2 oxygens? What this actually is is something that would be o o r negative. Okay? Because notice that there's 2 oxygens, so this is actually a different type of reagent altogether. Instead of being called an oxide, this is actually called an acetate. It's something completely different. And it reacts completely different and guess what? It's not on my strong base list. It's a nucleophile. Okay? So whenever you see 2 oxygens that can resonate with each other, that's not going to be on my strong base category. That's going to be on my nucleophile category. So that means that this is a better nucleophile. And what does that mean for tertiary? That means I'm down here at tertiary. It's not on my list of bases. Okay? And now I have this, this nucleophile. I'm actually going to go back up to the SN1 E1 category. Okay? Because this is not going to be able to do a basic elimination, but it's also not going to be able to do an SN 2 because it's so backed up. So it's going to do SN1 E1. Alright? So now that we know that, let's go ahead and draw all the steps. My first step is that the chlorine leaves on its own. That's going to give me a carbocation that looks like this. Okay? Notice that I drew that carbocation trigonal planar on the plane because it's planar. All right. And then we have 2 different mechanisms that we could do. We have the e one mechanism and we have the sn one mechanism. Okay. Let's do the e one first. Okay. So the way we figure out the e one is to draw the alpha and then figure out how many different betas I have. How many different betas did you guys get? This should be a beta. That should be its own. I'm calling that the green beta. And then these betas are blue because the ones on the ring are both symmetrical. The one on the green is completely different. So that's going to be 2 different molecules, 2 different types of carbons that have that are beta that are beta to the alpha. Now, do all of them have at least 1 hydrogen? Yes. All of them have at least 1, so they can all react. That means I'm going to get 2 different products. I'm going to get the product in the green and I'm going to get the product in the blue. I don't have to draw both mechanisms, but let's at least draw one of them. So one of them would look like this. Let's say that I'm using this hydrogen up here. I would take my the way that it's by the way, it's called in condensed formula would be o o I'm sorry. Yeah. You could just write it this, c h three c o o negative. That's that's the way it's often written. Okay? And I would just do this, Pull off the h and make a double bond. Okay? So that would be one of the compounds that I would get, But I would also get the blue one. The blue one would be if I had eliminated on the ring, and that would look like that. Alright. That's cool. So those would be my 2 e one products. Okay? So as you guys can tell, this one wasn't I wasn't as nice to you guys this time. I gave you multiple products. Okay. I wanted to start off easy and now we're getting a little bit harder. Now, for the sn1, what the SN1 mechanism would look like is that my CH3 COO negative directly attacks. Okay? So then I have to ask, would that make a racemic compound or would that make a chiral compound if I attack there? No. It wouldn't be chiral because both sides of the ring are perfectly symmetrical. So I wouldn't get 4 different groups. So luckily, for this part, I'm only going to get one product. And that product would be a methyl group and then whatever is attached to it, which would just be my O double bond O CH3 or basically the same nucleophile that I had at the top there. You could have also drawn it as 02 c c h three. That would have been fine as well. But you could draw it as the structure. You could draw it as the molecular formula. That's fine. So those would be my 3 structures. I would get my 3 products. I would get 1, 2, 3. I hope that makes sense. I know that was tricky because you had to go you you had to recognize that it wasn't an oxide and go up the other side. But now that you know that, hopefully that will help you get these problems right in the future. 6th problem.

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Predict the mechanism for the following reactions. Provide the full mechanism and draw the final product.

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