Concept: General Reaction3m
In this video, we’re going to break down the mechanism for a reaction called the Hofmann rearrangement. The Hofmann rearrangement is also known as the Hofmann degradation. If you hear that term, just consider it anonymous with Hofmann rearrangement. It's a method to turn amides into primary amines. I do want to let you guys know that this is similar to another mechanism that you may or may not know at this point called the Curtius rearrangement. I’m just going to put here it's similar to Curtius rearrangement. Just letting you know if you have seen my video on that reaction or if you’ve learned about it already, this mechanism is going to remind you a whole lot of that one. They’re kind of similar. If you don't know that reaction yet, it's fine because I’m going to teach you the mechanism anyway. You don't need to know the Curtius to understand this reaction.
But similar to Curtius, there’s going to be two really similar things here which is that this is a reaction that goes through an isocyanate intermediate. I used the term intermediate here very precautiously because I don't like to say intermediate. It actually sounds like it’s got a charge or that it’s like a highly energized species. Isocyanate is pretty stable. I'm just saying it’s like an intermediary structure where we make the isocyanate first and then we add something to it. Let’s just make sure that we're clear on that. Then also similar to Curtius, it’s going to liberate CO2 gas as a byproduct. But other than that, it’s a mechanism all on its own.
Here's the general reaction. We've got an amide and you react it with two different steps. One is you have a base. You have some kind of base that's going to deprotonate the nitrogen, turn it into a nucleophile. We've got an electrophilic Br2. The Br2 is going to be with the nitrogen that attacks. After you're able to add one equivalent of the bromine, what we're going to see is a rearrangement take place and a decarboxylation. That's going to produce R gas or CO2 gas and the part we really care about, the imine. Notice that one thing that happens here, I’m just talking in general terms, is that the R group that was originally on one side of the carbonyl eventually gets attached directly to the N. Notice that before I had them separated by a carbon and now I have them directly attached to each other. That's because I'm able to get rid of the carbonyl in the middle through my decarboxylation that you’re going to see later. In the next video, I’m going to go through the whole mechanism of Hofmann rearrangement so you guys know exactly what to expect.
Alright guys. So, we're going to start off with our amide and H, H and our R group on one side, okay? Well, that's part of the amide and in our first step, we're going to react it with our base. So, usually, we use sodium hydroxide, OH negative. So, my first step, what I'm going to do is I'm going to deprotonate the amide to make it nucleophilic. So, I'm going to take out an H, make a double bond, kick my electrons up to the O, this is now going to give me a nucleophilic structure that looks like this, okay? Now, when this structure interacts with my Br2, remember that guy's Br2 is a pretty good electrophile. Remember, that we have a lot of instances of negative charges attacking Br2. So, we're going to do that, we're going to kick our electrons back down, use this nucleophilic double bond to attack one of the BR's and kick out one, what this is going to give us is a new structure that looks like this, it's a nitrogen with now one hydrogen, actually let me flip that around a little bit, let me put the bromine here, that's the new bromine and the hydrogen there, okay? So, this is what we would call an N-bromo amide, okay? Because, we have a bromine coming straight off of the nitrogen, okay? Well, guys the thing about the N-bromide is that it can react with my OH negative again because it has an H still. So, we're going to do that whole step again, we're going to take our another equivalent over OH minus and you can predict what's going to happen, we're going to form the same type of structure where you get a nucleophilic molecule. So, let's go ahead and bring the structure down, we've got O negative, we've got a double bond to N to bromine and that's pretty much it, and then you know we're always getting water here. So, I mean that you would have water here and you'd have water here, that was, okay? Because your Oh grabbed an H, awesome guys. Well, this is the part that gets a little bit weird, this is the rearrangement step, this is the hard part, this is why this is a tricky reaction. So, I'm going to put here, this is the rearrangement, okay? Because instead of doing what we did before we're going to grab a bromine, we're going to get a strange rearrangement where eventually this R group on this side is going to attach to the N and make our isocyanate, how does that happen? Well, because these electrons are going to make a double bond. Now, notice that these carbon is in a tricky situation right here, this carbon, let me just write that really big, okay? Why? Because if you make a double bond. Now, you've got your four carbons. So, you would assume that you have to make, if you make that bond you have to break a bond and you have to break a bond in order to preserve the octet of the carbon. Now, if were me predicting this mechanism and I was just drawing from my past experience I would think that the next bond that breaks is this one, don't draw it, but that's what I would think, I would think make a bond, break a bond, make a negative charge on that end, great, awesome. But that's not what happens, this is where the rearrangement part comes in, instead of breaking that double bond it's actually going to be more energetically favorable to break this single bond and attach that carbon to the nitrogen. Notice that by doing that I still preserve my octet but now I'm going to get that the nitrogen is attached to R group, okay?
So, once I do that what's going to happen I get my isocyanate, okay? So, by the way, we've still got a problem, if you make of new bonds in the N now that N is going on a formal charge if you don't kick something out, we're going to kick out the Br, okay? So, now we get our isocyanate, which is oxygen double bond carbon, okay? That's from the electrons coming down with double bond. Now, double bond nitrogen and then the nitrogen is attached to an R group, cool, right? And, this is our isocyanate. Now, guys isocyanate is an important molecule on its own and there's a ton of different things that isocyanate can react with, it can react with amines, it can react with alcohol, it can react with water to make all kinds of structures but in this case we're only going to react it with one thing, which is the base that we use for deprotonation, we're going to react with base. So, when we react with base, where do you think a nucleophilic attack would add to my isocyanate, what do you think is the most electrophilic atom, or the most positively charged atom on the isocyanate? good job, you guys got this one, it's the carbon, right? So, we've got that positively charged carbon because of our strong dipoles pulling away from it, we're going to make a bond and break a bond, okay? What, that's going to give us is now a double bond OC with an OH on one side, right? And then on this side, we're going to have a nitrogen with an R group. Now, I'm going to skip a step if you guys don't mind, because notice that I would have gotten a negative charge on that N, right? I would have gotten a negative but I'm just going to protonate it because eventually that gets protonated. So, I'm going to say this plus H, right? Would give me a proton there, awesome guys. So, now we've basically got what we want, okay? Because look. Now, what I have is a nitrogen that's directly attached to an R group and remember that, what I'm trying to make is a primary amine. So, now I'm just thinking is there any way it can get rid of this carboxylic acid as long as I could get rid of the carboxylic acid I could keep the amine and I would get my product, okay? What reaction can we use to get rid of it decarboxylation, okay? So, I can decarboxylate and I can get rid of this thing, okay? So, what this is going to look like is we're going to take our OH minus, we're going to grab an H, make a double bond, break the bond to the N, okay? This is going to make our CO2. So, what you see is that we're going to get now as our products, we're gonna get N, H, H, R, this is our primary amine, okay? Plus, we're going to get carbon, double bond O, double bond O, this is your CO2 gas, okay? Plus you're going to get water, which we made plenty of, okay? Now, the one I really care about is the primary amine, okay? Always I care about the organic product, that's the one your professor cares about, that's the one that's important to draw as a product, CO2 gas is a byproduct, water reactions take water, nothing special, okay? So, guys, I hope that made sense also, if you are familiar with the Curtius rearrangement, hopefully you were able to pick up on some of those similarities because the more similarities and patterns you guys can recognize the more of an organic chem beast you're going to be, okay? So, that's it for this video, let's move on to the next.
Complete the following reaction by drawing the structure of the principal organic product in the box. Draw the correct stereochemistry in the product where necessary. Be sure your product possesses the correct number of carbons and other atoms.
The final product of the following series of reactions is:
What is the reactant for the following chemical reaction?
What is the product of this reaction?
What intermediate, leading to the major product, is formed in this reaction?
Which reaction could be used to produce compound A?