|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|
|Addition Reaction||7 mins||0 completed|
|Markovnikov||5 mins||0 completed|
|Hydrohalogenation||7 mins||0 completed|
|Acid-Catalyzed Hydration||17 mins||0 completed|
|Oxymercuration||20 mins||0 completed|
|Hydroboration||27 mins||0 completed|
|Hydrogenation||7 mins||0 completed|
|Halogenation||6 mins||0 completed|
|Halohydrin||17 mins||0 completed|
|Carbene||13 mins||0 completed|
|Epoxidation||8 mins||0 completed|
|Epoxide Reactions||9 mins||0 completed|
|Dihydroxylation||9 mins||0 completed|
|Ozonolysis||7 mins||0 completed|
|Ozonolysis Full Mechanism||25 mins||0 completed|
|Oxidative Cleavage||8 mins||0 completed|
|Alkyne Oxidative Cleavage||6 mins||0 completed|
|Alkyne Hydrohalogenation||3 mins||0 completed|
|Alkyne Halogenation||2 mins||0 completed|
|Alkyne Hydration||6 mins||0 completed|
|Alkyne Hydroboration||3 mins||0 completed|
Ethers are generally unreactive. However, epoxides (3-membered cyclic ethers) are highly strained, so they are able to react with nucleophiles in ways a typical ether could not. Let’s take a look.
Concept #1: Acid-Catalyzed Epoxide Ring-Opening
Now that we have epoxides and know how to make them, it's important to know what can we do with them. It turns out that ethers are highly unreactive. Ethers barely react with anything. But because of the fact that epoxides are three-membered rings, they're very, very highly strained. What that means is that even though ethers as a function group are not very reactive, epoxides are because they have so much potential energy locked up in those bonds. They're going to want to break open first chance that they get.
So it turns out that we have two different ways that we can open the epoxide ring and they have to do with different reagents that we use. These are called the acid-catalyzed ring opening reaction and the base-catalyzed ring opening reaction. Let's start off with acid and then we'll work on to base.
So in an acid-catalyzed reaction – notice first of all that I'm using the words acid-catalyzed. That actually already gives us a hint about the mechanism. Remember when I said whenever you have an acid catalyzed mechanism, what's the first step going to be? Protonation. You always need to protonate first. If it says acid-catalyzed, you always need to protonate first. So what that means is that the very first step is if I'm using an acid with an epoxide, the O, the nucleophilic O, is going to go ahead and grab the H from the acid because that H has a positive charge.
So now what I'm going to get is a molecule that looks like this, where everything is still in the same exact place, except that I've got a formal charge on this O because the O has too many bonds.
So now what happens? I've got a Cl- and the Cl- wants to break open the ring, but it's trying to decide which side is it going to break. Is it going to break the most substituted, the tertiary, or the least substituted, the secondary? That's supposed to be a three by the way. Let me draw that again. The tertiary or the secondary? The answer to this is that the chlorine or whatever anion we're using, even if it was just a neutral nucleophile, is going to be the most attracted to the side of the ring with the most positive character.
So what we're looking at is the side of the ring that could stabilize that carbocation the best because that – I'm sorry, I just said carbocation. It's not a carbocation. It's just a cation. It's just a positive charge. But that positive charge can delocalize a little bit into those two different atoms. So the question is which side is going to be the one that has the most positive character, the secondary or the tertiary. The answer is the tertiary.
So I'm going to go ahead and attack here. If I make that bond, I have to break a bond. So I'm going to break the bond to the O. What this is going to give me is a new compound that looks like this. Let's say that this bond right here is the red bond that I'm drawing here. Oops. That red is the same as this. What is it going to have on it? Well, let's say that this chlorine attached from the front, right there. What else would it have? Well, it would have two methyl groups, so I'm going to draw methyl, methyl.
And then on the other side of the red bond what am I going to have? Well, if the chlorine attached from the back, what that means is that the alcohol is going to go – I mean from the front, the alcohol is going to go to the back. By the way, this alcohol that I'm drawing here is just that right there. Notice that it's still attached in this position. So it's still going to be attached there.
On top of that, I'm still going to have two other groups. I'm going to have a propyl and I'm going to have an H. So that H would just stay where it is in the front. Obviously, you don't need to draw H's. At the end of the day, I could have just drawn this without the H. I could just erase that if I wanted to.
Does that make sense? So notice what my product is here. Well, my product is that I'm going to get – what my product is is that I'm going to get a nucleophile attaching to the more substituted position and I'm going to get an alcohol at the least substituted position. Just so you guys know, this is unique to acid-catalyzed ring opening. Base-catalyze ring opening is actually going to be different.
Acid-catalyzed ring openings favor addition to the MOST substituted carbon (thermodynamic control)
Concept #2: Base-Catalyzed Epoxide Ring-Opening
So let's go ahead and check out base catalyze now let me know if you have any questions about acid let's go on to base, OK? So in the base catalyze ring opening what I have is that I'm starting off with a nucleophile right away and there's no protonation step because obviously I don't have an acid, OK? So what that means is that all I have is an OH negative or some kind of nucleophile looking for a way that it can break open this ring, OK? Because the ring is highly strained, well there's always negative since there's no positive charge it's not going to be attracted to the side that's more substituted it's actually going to be attracted to the side the easiest to access so for this one I'm actually going to go ahead and look for the least substituted position and that's the one that's going to favored because that's the easiest one to form so in this case if my choice is between secondary and tertiary I'm actually going to go for secondary so I'm going to go ahead and attack that carbon which isn't going to break that bond and what that's going to give me is a new molecule that looks like this, where let's say this is my red bond again and this is the red bond here what I would have is let's say that the OH attached from the front, let's say that my nucleophile attached from the front that means that yep I would get my purple group and I would get my hydrogen which I don't have to draw. On the other side, what I would now get is an O- on the back side, OK? Now notice that once again these products are going to be anti to each other because every time you break open a three-membered ring they're going to be anti because they're going to snap out of place and they're going to just go as far from each other as possible, OK? And then I have 2 methyl groups so I'm just going to write that there, 2 methyls, OK? So that's the end of that step now typically we're going to assume that the O negative will protonate, OK? So typically there will be a protonation step sometimes your professor will add over water or something like that and if there was water usually we just assume that the end product is protonated so if it was protonated what would it give us, OK? What it would give us is actually going to be what we call a diol, OK? So check this out look what I'm gonna get at the end, what I'm going to get is I'm going to get an OH going towards the back and in an OH going towards the front, what's the relationship between these alcohols? Well first of all we call this a diol because there's two alcohols that's easy just diol there's two of them, that's actually a special type of molecule that we'll study more in depth, OK? So we have a diol but also the relationship will be vicinal remember that this vicinal is the word that we used to say that they're next to each other that's the same thing as saying 1-2, OK? So we have vicinal diol but on top of that these are anti because they are trans to each other which means that it was an anti-mechanism so the entire idea behind this is that this is what we would call an anti-vicinal Dihydroxylation that just means we're adding 2 alcohols, OK? Why is this important? Well this isn't the only way this isn't the only type of base catalyze opening you could do in fact you could have used any nucleophile I could have used a nitrogen or another type of oxide or whatever but this specific one where I use NaOH is very important because it's really one of the only ways to make these ant vicinal diols that means I have two alcohols facing opposite directions next to each other, this is important because there's another reaction that you're supposed to know called Synviscinal diols and Synviscinal diols and dihydroxylation is a completely different mechanism so these are going to go hand in hand and you're supposed to recognize both of them, OK? If you haven't learned syn vicinal diols already you will learn it soon because I'll teach it to you, OK? But anyway I just want us to know that it's supposed to be......You're supposed to be able to compare these two reactions to each other and you're supposed to be able to know when do you use an apoxide with NaOH and when do you do the syndiol reaction, alright? So I hope that these two mechanisms made sense if I were to sum it up once again it would be acid is attracted to the most substitute, base catalyze is attracted to the least substituted you always get an alcohol at the end regardless and that's really all you need to know and then you're always going to get anti products because you're breaking up into 3 membered ring, alright? So I hope that made sense guys let's go and move on.
Base-catalyzed ring openings favor addition to the LEAST substituted carbon (kinetic control).
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