Hofmann Elimination

Now we're gonna talk about a specific reaction called the Hoffman Elimination. So, guys, the name Hoffman elimination could be a little confusing because we've used both of these terms before. We've talked plenty about elimination reactions in organic chemistry one, and we learned that they could either follows it saves rule or Hoffman's rule. Okay, so But when I mentioned Hoffman elimination here, I'm not referring to that. I was referring to a specific reaction called Hoffman elimination. Okay, not Hoffman's rule regarding elimination. So this is also referred to as exhaustive methylation or Hoffman degradation. So if you hear any of those words, or if your professors call it any of those other terms, just know this is the same reaction. Okay, Now the name Hoffman elimination actually is helpful for us because it's gonna serve to remind us that we're gonna produce ah Hoffman elimination product, meaning that it is gonna follow Hoffman's rule. But specifically, it's gonna be within a meat. OK, Now, guys in organic chemistry one, we learn that alcohol's can be eliminated, dehydrated to form double bonds. Okay, that was actually a reaction called dehydration. Well, in the same way alcohol's can. I'm sorry. Amines can, too. But they need to be turned into a good leaving group. Okay, Because a means as they are, if you're just to kick off NH two negative, Terrible leaving group, that's, like the strongest base ever. So the point of the first step of this reaction is gonna be to try to make the nitrogen a good leaving group. Then we can kick it out and do an elimination reaction with the base. Okay, so that's exactly what the first step is. So, guys, the first step is going to be some kind of alcohol. Hey, lied usually an alcohol iodide, and it's you may see it written as in excess, or you may not. Okay, In the absence of data telling you excess or a number of equivalents, always assume that there's excess. Okay, so in this case, I don't have to write excess, but your professor may be nice and right excess. Okay, Now, what that's gonna do is it's gonna react with my nitrogen multiple times, okay? It's gonna make it a great leaving group. Then we're going to do in. My second step is after the nitrogen is a great leaving group It's a quaternary ammonium. Then what we're gonna do is we're gonna react it with silver oxide, which is going to serve as my elimination region. It's gonna be a base. It's gonna turn. It's going to generate a base which can then eliminate. Now, notice that my nitrogen had two options of where to eliminate. I could have either eliminated along the blue line here or along the red line here. One of them was more substitute. One of them was less substituted. I went with the least substituted product. Okay, It's not gonna be a major minor product scenario here, pretty much almost gonna get 100% of the less substituted product. And we'll talk more about that when we get into the mechanism. But let's substituted. So just for right now I just want you guys to memorize these regents. We've got some kind of a mean reacting with an excess of alcohol. Hey, lied. And some mixture of silver oxide and base or water. Okay, so that's going to generate the least of studio Joaquin. Now, in the next video, I'm gonna show you guys the full mechanism for this reaction

Alright, guys, let's look into this mechanism a little further. I just brought the top molecule down so we can work with it. So, guys, as we mentioned, the very first step of this is gonna be to make the Amina good leaving group. But what I failed to mention is that this is a reaction we've already learned or that I have another video for If you haven't learned it yet, you can look it up called a mean calculation. Okay. And this is one of the few times that a mean calculation is actually gonna be helpful. Because remember that Amina Al Collection is notorious for going all the way to the Quaternary, Amine and Paulie calculating That's exactly what's gonna happen here. Except that's a good thing for us because I want my A mean to have a positive charge. It's gonna turn into a good leaving group that way. So remember, recall that from that mechanism I do an s N two. It's an s and two mechanism, and you wind up getting a molecule that now looks like this where I got N with still two h is gonna put your NH two. But now It's got a methyl group and a positive. Okay, fast forward to more reactions. Imagine doing this two more times. So imagine I've got my ch three. I times two. This is the part that's called exhaustive methylation Because I'm literally methylated this thing until it's done reacting until I exhaust the amount of reactions that could take place. So imagine that I do this two more times, which, as I stated earlier, guys, by the way, miss that metal is a complication of calculation that it's gonna calculate multiple times. What we're finally going to get is a nitrogen with three are groups on it and a positive. Okay, now, what's great about that, guys, is that now, is that positive? Is solidly on there. It's a great leaving group. Okay, So now I could go ahead and I could do an elimination reaction, kick out the nitrogen and not have to worry about the leaving groups sucking. Okay, so now the next part is using a G 20 right. So you got your silver oxide. Okay. And for my silver oxide, what we're gonna find is that the silver oxide isn't really a strong base in regards to ways use it before, but in the presence of iodine, because we're gonna have a lot of iodine floating around. We got a bunch of my negatives, right? What's gonna wind up happening is that the G 20 and the eyes and they're going to combine to form in organically. There's an inorganic reaction a g I and Ohh, negative. You don't need to know the mechanism for this part because this is an inorganic reaction that is just a few mixed two different inorganic regions together and you get some kind of net ionic equation blah, blah. The interesting part about this for us is the O. H. Negative. Because now that I've got that O. H negative, I can run an elimination reaction. So, as I stated earlier, let's just go ahead and bring down the final structure that I was working with. Once again, I've got nitrogen with three are groups and a positive. Okay. And you guys, um, might recall that I said that you're always gonna get the Hoffman products, not the sites. If so, I've got to beta hydrogen is that I could use. I've got let's say, the blue beta hydrogen here or I've got the red beta hydrogen. Remember, guys, how you this called beta elimination? Because this is my Alfa Carbon, and these were both beta to it. So this is what we call a beta elimination, okay? With O h. Negative and usually guys O H negative would yield, um, or substituted product would usually get a sites of products. I would expect that I would actually make my double bond towards the top, but I'm gonna go with the red one. Why? Let's just draw this mechanism really quick. Why is it gonna be favored for my O h negative to attack the bottom h, make it double bond and kick out my nitrogen giving me a molecule that looks now like this with my end c h 33 as a leaving group. Notice that it's neutral. But the question is, why? And guys the complex the the answer is it's complicated. You can go to your textbook in your textbooks. Probably gonna have a much better, more thorough explanation that I'm gonna give you now. But basically, guys, it has to do with the fact that the nitrogen is a very bulky leaving group. So that means that remember that for e two reaction to take place for a beta elimination e to reaction, you need antico plainer geometry or what's also called anti Perry Plainer. Okay. And the antico plainer geometry is gonna be the most energetically favorable when you're going into the least substituted, um, side rather than the more substituted side. The answer for this can really only be, um, described using a Newman projection. So in your textbook, there will probably be a Newman projection. Remember, Newman Projections Airways to visualize single bonds. And basically the Newman projection is more stable. The new one projection that is required for the Antico planner is more stable, along least substituted side least substituted side. Okay, so that's the reason that we're getting this product and we're not going to get the product with the Taliban going up. Because if you were to draw a Newman projection off the Mawr substituted side, what we find is that there's gonna be a really ugly gosh interaction with the our group, and it's gonna wind up hitting or basically interacting with the nitrogen leaving group. It's better for us. Toe have the least substituted side attack because you're not gonna get that. Gosh, interaction. You're gonna get much further distance between the our group and between the nitrogen leaving group. That's really bulky. Okay, so just let you guys know probably more theory than you need. But if you want the really true detailed explanation, you could definitely beat up, your textbook will have some nice new and projections for you to look at. Okay. All I care about is that you know that it's the Hoffman product least subsided. Taleban. Okay, so let's move on to the next problem. See if you guys could get the whole thing right, and then we'll solve it.

So, guys, the short answer for this is literally just draw. Ah Hoffman product. Okay, there you go. That's the answer. So you could be done with this in three seconds, as long as you know what's going on, because you know that that nitrogen is going to become a leaving group. I'm gonna go to the less substituted side now, just to kind of reinforce what we learned. Remember that the first part is Miami inoculation, right? Miami in calculation is going to react three times. So, really, this is times three. It may not say times three be stopped to assume that that's gonna happen. So after my first step, what I wind up getting is something. It looks like this nitrogen with three methyl groups on a positive. Okay, now, that's the first step. The second step is just generating your O. H negative, which is going to do and e to attack on the less substituted side and give you your Hoffman double bond. Okay, awesome guys. That's it for Hoffman degradation. Exact exhaustive methylation. Hoffman elimination. Uh, let me know if you have any questions. Let's move on to the next video