Now we’re going to talk about a specific reaction called the Hofmann elimination. The name Hofmann elimination can be a little confusing because we use both of these terms before. We've talked plenty about elimination reactions in Organic Chemistry 1. We learned that they could either follow Zaitsev’s rule or Hofmann’s rule. But when I mention Hofmann elimination here, I'm not referring to that. I’m just referring to a specific reaction called Hofmann elimination, not Hofmann's rule regarding elimination. This is also referred you as exhaustive methylation or Hofmann degradation. If you hear of those words or your professors call it any of those other terms, just know that this is the same reaction. The name Hofmann elimination actually is helpful for us because it’s going to serve to remind us that we're going to produce a Hofmann elimination product, meaning that it is going to follow Hofmann’s rule but specifically it’s going to be with an amine. In Organic Chemistry 1, we learned that alcohols can be eliminated, dehydrated to form double bonds. That was actually a reaction called dehydration. In the same way, amines can’t do but they need to be turned into a good leaving group because amines as they are, if you were just to kick off an NH2 negative, terrible leaving group. That’s like the strongest base ever. The point of the first step of this reaction is going to be to try and make the nitrogen a good leaving group so then we can kick it out and do an elimination reaction with a base. That's exactly what the first step is. The first step is going to be some kind of alkyl halide, usually an alkyl iodide. You may see it written as in excess, or you may not. In the absence data telling you excess or number of equivalence, always assume that there’s excess. In this case, I don’t have to write excess but your professor may be nice and write excess.
What that's going to do it’s going to react to my nitrogen multiple times. It’s going to make it a great leaving group. Then what 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 going to do is we're going to react it with silver oxide, which is going to serve as my elimination reagent. It's going to be a base. It’s going to generate a base which can then eliminate. 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 substituted. One of them was less substituted. I went with the least substituted product. It’s not going to be a major or minor products in there. You’re pretty much always going to get 100% of the less substituted product. We'll talk more about that when we get into the mechanism.
Just for right now, I just want you guys to memorize these reagents. We’ve got some kind of amine reacting with an excess of alkyl halide and some mixture of silver oxide and base or water. That's going to generate the less substituted alkene. In next video, I’m going to show you guys the full mechanism for this reaction.
Alright guys. So, let's looking for 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 going to be to make the amine a 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 amine alkylation, okay? And, this is one of the few times that amine alkylation is actually going to be helpful because remember that amine alkylation is notorious for going all the way to the quaternary amine and poly alkylating, that's exactly what's going to happen here, except that's a good thing for us because I want my amine have a positive charge, it's going to turn into a good leaving group that way. So, remember recall that from that mechanism I do an SN2, it's an SN2 mechanism, and you wind up getting a molecule that now looks like this where I've got N with still two H's. So, I'm going to put your NH2 but now it's got a methyl group and a positive, okay? Fast forward, two more reactions, imagine doing this two more times. So, imagine I've got my CH3-I times two, this is the part that's called exhaustive methylation because I'm literally methylating this thing until it's done reacting, until I exhaust the amount of reactions that can take place. So, imagine that I do this two more times, which as I stated earlier guys by the way I missed that methyl, is a complication of alkylation that it's going to alkylate multiple times, what we're finally going to get is a nitrogen with three R 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 leaver group sucking, okay? So, now the next part is using Ag2O, right? So, we've got your silver oxide, okay? And for my silver oxide what we're going to find is that the silver oxide isn't really a strong base in regards to ways you use it before but in the presence of iodine, because we're going to have a lot of iodine floating around, we've got a bunch of I negatives, right? What's going to wind up happening is that the Ag2O and the iodine are going to combine to form inorganic, this is an inorganic reaction AGI and OH negative, you don't need to know the mechanism for this part because this is an inorganic reaction, this is like just if you mix two different inorganic reagents together and you get some kind of net ionic equation bla, bla, bla, the interesting part about this for us is the OH negative because now that I've got that OH 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 R groups and a positive, okay? And you guys might recall that I said that you're always going to get the Hofmann product not the Zaitsev. So, I've got two beta hydrogen's 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 is called beta elimination because this is my alpha carbon and these are both beta to it, so this is what we call a beta elimination, okay? With OH negative and usually guys OH negative would yield a more substituted product, we'd usually get a Zaitsev product. So, I would expect that I would actually make my double bond towards the top but I'm going to go with the red one, why? let's just draw this mechanism really quick, why is it going to be favored for my OH negative to attack the bottom H, make a double bond and kick out my nitrogen, giving me a molecule that looks now like this, with my N CH3,3 as a leaving group. Notice, that's neutral, but the question is why, and guys the answer is it's complicated, you can go to your textbook and in textbooks probably have a much better more thorough explanation that I'm going to 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 an E2 reaction to take place for a beta elimination E2 reaction you need anti-coplanar geometry or what's also called anti-periplanar, okay? And the anti-coplanar geometry is going to be the most energetically favorable, when you're going on to the least substituted side rather than the more substituted side, the answer for this can really only be described using a Newman projection.
So, in your textbook there will probably be a Newman projection. Remember, Newman projections are ways to visualize single bonds and basically the Newman projection is more stable, the Newman projection that is required for the anti-coplanar 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 products with the total bond going up because if you were to draw a Newman projection of the more substituted side, we find that there's going to be a really ugly Gauche interaction with the R group and it's going to wind up hitting or basically interacting with the nitrogen leaving group, it's better for us to have the least substituted side attack because you're not going to get that Gauche interaction, you're going to get a much further distance between the R group and between the nitrogen leaving group that's really bulky, okay? So, just letting us know probably more theory than you need but if you want the really true detailed explanation you can definitely read up your textbook, will have some nice new project for you to look at, okay? All I care about is that you know that it's the Hofmann product least substituted double bond, okay? So, let's move on to the next problem, see if you guys can get the whole thing right and then I'll solve it.
So guys, the short answer for this is literally just draw a Hofmann 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 nitrogen is going to become a leaving group and you're going to be the less substituted side. Now, just to kind of reinforce what we learned. Remember, that the first part is my amine alkylation, right? My amine alkylation is going to react three times. So, really, this is times three, it may not say times three, be astute to assume that that's going to happen. So, after my first step what I wind up getting is something that 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 OH negative, which is going to do an E2 attack on the less substituted side and give you your Hofmann double bond, okay? Awesome guys, that's it for Hofmann degradation exhaustive methylation, Hofmann elimination, let me know if you guys have any questions, let's move on to the next video.