|Ch. 1 - A Review of General Chemistry||4hrs & 48mins||0% complete|
|Ch. 2 - Molecular Representations||1hr & 12mins||0% complete|
|Ch. 3 - Acids and Bases||2hrs & 45mins||0% complete|
|Ch. 4 - Alkanes and Cycloalkanes||4hrs & 19mins||0% complete|
|Ch. 5 - Chirality||3hrs & 33mins||0% complete|
|Ch. 6 - Thermodynamics and Kinetics||1hr & 19mins||0% complete|
|Ch. 7 - Substitution Reactions||1hr & 46mins||0% complete|
|Ch. 8 - Elimination Reactions||2hrs & 25mins||0% complete|
|Ch. 9 - Alkenes and Alkynes||2hrs & 10mins||0% complete|
|Ch. 10 - Addition Reactions||3hrs & 32mins||0% complete|
|Ch. 11 - Radical Reactions||1hr & 55mins||0% complete|
|Ch. 12 - Alcohols, Ethers, Epoxides and Thiols||2hrs & 42mins||0% complete|
|Ch. 13 - Alcohols and Carbonyl Compounds||2hrs & 14mins||0% complete|
|Ch. 14 - Synthetic Techniques||1hr & 28mins||0% complete|
|Ch. 15 - Analytical Techniques: IR, NMR, Mass Spect||7hrs & 20mins||0% complete|
|Ch. 16 - Conjugated Systems||5hrs & 49mins||0% complete|
|Ch. 17 - Aromaticity||2hrs & 24mins||0% complete|
|Ch. 18 - Reactions of Aromatics: EAS and Beyond||4hrs & 31mins||0% complete|
|Ch. 19 - Aldehydes and Ketones: Nucleophilic Addition||4hrs & 54mins||0% complete|
|Ch. 20 - Carboxylic Acid Derivatives: NAS||2hrs & 3mins||0% complete|
|Ch. 21 - Enolate Chemistry: Reactions at the Alpha-Carbon||1hr & 56mins||0% complete|
|Ch. 22 - Condensation Chemistry||2hrs & 13mins||0% complete|
|Ch. 23 - Amines||1hr & 43mins||0% complete|
|Ch. 24 - Carbohydrates||5hrs & 56mins||0% complete|
|Ch. 25 - Phenols||15mins||0% complete|
|Ch. 26 - Amino Acids, Peptides, and Proteins||2hrs & 54mins||0% complete|
|Ch. 26 - Transition Metals||5hrs & 33mins||0% complete|
|Amine Alkylation||12 mins||0 completed|
|Gabriel Synthesis||11 mins||0 completed|
|Amines by Reduction||13 mins||0 completed|
|Nitrogenous Nucleophiles||9 mins||0 completed|
|Reductive Amination||10 mins||0 completed|
|Curtius Rearrangement||15 mins||0 completed|
|Hofmann Rearrangement||10 mins||0 completed|
|Hofmann Elimination||11 mins||0 completed|
|Cope Elimination||13 mins||0 completed|
Concept #1: General Reaction
Hey guys, in the next few videos we're going to discuss a really unique type of elimination reaction called the Cope elimination.
It turns out that amines are really easily oxidized even by weak oxidizing agents. In fact, if you just leave some amine in your bench at the laboratory, and it's exposed to open air, even just the O2 gas that's in room air is enough to oxidize amines into hydroxyl amines and even amine oxides.
It turns out that tertiary amine oxides are capable of doing something called a self-elimination and making an elimination product, specifically a Hofmann product. It might have been a long time since we did elimination, but hopefully you guys can remember there was a Zaitsev product and a Hofmann product. The Hofmann product being the one that was the less substituted. We're going to be forming less-substituted elimination products through a Cope elimination.
There's a little bit of theory we want to get into beforehand just in case you read it in your textbook, at least I explained it to you one time. It turns out that first of all, anything that is a tertiary amine oxide can also be referred to as an N oxide. That just makes sense. You're saying that there's an oxygen coming off of an N. That would be basically – let me just circle it for you. That would be this kind of structure right there. So you've got an O directly attached to N.
But what we notice is that there's a really weird type of bond there. Notice that I have an arrow pointing to the O, which is something that we won't really talk about a whole lot in organic chemistry at the college level. That contains what's called a dative or a dipolar covalent bond. What does that exactly mean? Let me just define that in case you forgot or in case you never heard of this before.
It's a covalent bond. That means there is equal sharing of electrons. It's a strong bond. It's covalent just like any other covalent bond you learned about. But we call it dipolar or dative because you've got two charges that are separated on that bond. You've got a positive and a negative. Notice that they never go away. You've got your positive. You've got your negative.
And it's known as dipolar because one of the species is the one that's giving its electrons away to the other. In this case, we know that the nitrogen has a very nucleophilic/basic lone pair, so it's going to go ahead and donate that lone pair to the oxygen forming that dative bond.
Now, it could very well be denoted just as a straight line because as you guys know covalent bonds are always just drawn as a stick. You could just draw it as a stick that would be fine, but in some texts, it's denoted as an arrow, so it would be nitrogen, arrow, oxygen, which for the purposes of this class, I just want you to recognize it as a bond. This is going to be a covalent bond just like any other bond. Awesome.
Now I just cleared that out of the way so we can talk about the reaction and the mechanism. First of all, before we go into the mechanism, let's just look at the general reaction. What we've got is an amine that we're first going to oxidize. This is hydrogen peroxide. Hydrogen peroxide, not the strongest oxidizing agent, but I just told you guys, even O2 gas, just room air is enough to oxidize an amine. So definitely hydrogen peroxide can.
What you're going to form is that dative tertiary amine oxide or the N oxide. Now what's special about these guys is that N oxides can self-eliminate. What we're going to wind up getting is in heat, heat is going to wind up separating these and you're going to wind up getting a Hofmann alkene. So you can see that basically we had two options. We could have either gone in the red direction or in the blue direction here. We went with the less substituted version so that's Hofmann and you wind up getting a hydroxyl amine as a byproduct. I'm just going to write here this is a byproduct. We don't really care about it.
Now in the next video, what I'm going to do is I'm just going to break down the Cope elimination mechanism and then we'll do a practice problem.
Concept #2: Mechanism
Alright guys, so the Cope elimination is a concerted mechanism and you guys might recall concerted means it's all one step so that means it's going to have a transition state or an intermediate, what do you think? transition state, okay? It's going to have a transition state that all forms and collapses in one step. Remember, that we also denote transition states with a little double dagger, okay? So, let's go ahead and just bring down the structure from the top. So, you can just see how exactly how the structure would react. So, I'm going to draw like this, nitrogen. Now, what I am going to do for the sake of the mechanism guys is I'm going to move the methyl groups up. So, I'm going to move this methyl here, this methyl here and I'll move the lone pair here, okay? You'll see why in a second. So, I'm just rotating it. So, when I oxidize I'm just going to right here, O, in general that stands for any type of oxidation, what I'm going to get as a product for my first step is I'm going to get my N methyl, methyl, O, negative charge, positive charge. Now, you guys might remember this as that dipolar or dative bond, I didn't draw it as an arrow because you don't have to, that's just some older text, draws an arrow, but I'm just going to draw it as a line because the same thing, it's a covalent bond, okay? Wonderful, okay. So now, how does this work? Well, in this step what's going to happen is that that, and by the way, I put a positive, there needs to be a negative there. So, I'm sorry, I don't know why I put a positive there, should be a negative on the O, a positive on the N like I have above. Now, we're going to get is a concerted elimination mechanism on the least substituted R group so that means that the least substituted would be this one right here, this hydrogen is the most accessible to the O, the other one would be a little bit more difficult. So, we're going to get some arrows forming here, we're going to get that the O attacks the H, okay? Then, we're going to get that the H form, does an illumination reaction, makes a double-bond and releases the N, okay? So, we're going to wind up getting is a transition state that looks like the following it's going to have now a single bond there but we're now going to have a dotted line, partial bond to N, those bonds are still there.
Now, I'm going to get a single bond to O why is it a straight line because remember that nothing really happened with that bond, the end to the O is the covalent bond that stays intact but now I've also got a dotted line down here to H, that's supposed to be an H and another dotted line to this carbon and then finally. Notice that we're making a double bond. So, there should be a double line here, a dotted line here, showing that there's a partial double bond being created, okay? So, now what I'm going to do, I'm trying to kind of duck out of the way here, is also guys. So, perfect. Now, we're going to do is we're going to just draw this in brackets and with our transition state, okay? So, there you have it, that's all going to kind of form in and decompose at the same time and what that's going to do after the transition state is complete is you're going to get your, I'm sorry, your Hofmann product. So, there's my double bond plus, we're going to get our hydroxyl amine, which is N, methyl, methyl with an OH, this is what we call a hydroxyl amine and again guys this is a byproduct, I don't care too much about it, what I care more about is this product here, the Hofmann elimination product, okay? So, that's your mechanism. Now, what we're going to do is going to do an example, that's drawn a little bit differently and I love the fact that's kind of it's drawn differently it's to throw you off a little bit. So, I want you to think about this mechanism try to draw it off the best you can but most importantly try to get the right product here and then I will go ahead and give you the answer. So, go for it.
Example #1: Predict the Product
Alright, so hopefully you are able to recognize that this is first of all a cope elimination, why? Because I've got a tertiary amine, I've got an oxidizing agent, which is usually hydrogen peroxide and then I'm adding heat to favor elimination. Now, remember, I didn't really get to mention this in the mechanism above but why does heat favor elimination? You guys remember that? Guys, it had to do with entropic effects, had to do with the fact that when you eliminate you're creating more particles and you're increasing entropy. So heat is, you guys remember, from Gibbs free energy, heat is going to favor on high entropy, so that's why we always go towards an elimination reaction, the more you jack up the heat, okay? So, just a little bit of theory there. So, you've got our oxidation, we've got our heat, what does this look like? Well, the first step is going to be to add the O, okay? Now, just you guys know, I also didn't get to mention this in the prior video, that the oxidation step is a mechanism that is not well described because it turns out that there's a lot of different oxidation mechanisms that could create this amine oxide, meaning that you're not responsible for the first step, so I'm going to just say here you don't need mechanism. In fact, if you look in your book, your book doesn't show you the mechanism for this because professors are the debate is still raging on, in your chemistry department, on what exactly medic mechanism is, which one is the major one that creates that oxidation step, okay? There's a lot of different ways to oxidize those amines. So, now we've got our N-Oxide and we've got a few, we have to figure out what's the next step, well, it's going to eliminate with something. Now, I did forget a methyl here, so I add that methyl, perfect. Okay guys, so what's going to be the elimination step? Well, we look at the carbon that's attached to the N, so we're going to look at this carbon, I'm going to say out of those, out of the carbons extending from that one, which one is the least substituted? Well, we're going to notice that we have really three different choices, we could either make a double bond here, on the ring there, I'm going to make that blue or we could make it down the double bond- the ring there and that would be green or we could make it down the methyl group here, which I'll make black, okay? Which of these looks like the least substituted double bond to you? Well, it turns out guys that blue and green are the same thing, because this is a symmetrical cyclopentane, so really, I'm just going to delete green because is the same thing as blue. So, now between blue and black, which one's better? The one that's least substituted is going to be the substituent coming off the chains, that's black, so that means that the mechanism is actually going to take off the hydrogen on this carbon, okay? So, we can go ahead and draw that really quick, it's going to be a little weird looking but it's going to look at this like negative to the H, make a double bond kick off the end, okay? Now, you're going to get your transition state, which we don't need to draw, that's you know, that's if it was a full mechanism question we would draw that but, all I really want is a product here. So, now we know that the product is going to be this, it's going to be a double bond sticking off of the cyclopentane plus your N CH32 and that has an OH sticking off it, that's your hydroxyl amine, why? Because at first you had just an O but now you grab the H, so it's OH. You might be wondering, Johnny, why is that a neutral? Because remember, we gave it electrons, we broke the bonds of the R group, so now the nitrogen only has 3 groups around it, so it can be neutral, okay? So guys, this is your final product, the one we really care about is the substitution products, we made a Hofmann elimination and we are done with this topic, okay? Hofmann Elimination Product. Alright, so I hope that made sense guys, let's move on to the next video.
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