English Subtitles for Chem 201. Organic Reaction Mechanisms I. Lecture 01. Arrow Pushing. Part 2

Subtitles / Closed Captions - English

>> All right.

Welcome back to the second part of our first lecture in Chemistry 201. And this is our lecture of arrow pushing. And right now I'm going to give you seven rules for mechanistic arrow pushing. When you draw an arrow pushing mechanism

for an organic reaction, how do you know that you've correctly broken it down into individual mechanistic steps that are elementary and added the correct curved arrows? So let's go ahead and start off with rule number 1. [ Writing on Board ]

This seems like the simples rule. Draw correct Lewis structures. That's one of the first things that you learn when you take general chemistry as an undergraduate. This has several components to it. How do you know that you're drawing correct Lewis structures

and using then correctly? The first part of this rule seems pretty simple, and that is obey the octet rule. So let me just remind you of what the octet rule says. The octet rule says that for second row atoms, not all atoms, just atoms in the second row.

Not hydrogen. Not transition metals. Not sulfur. Not phosphorus. But these atoms right here in my little mini periodic table. Lithium, beryllium we're not going to worry about, boron,

carbon, nitrogen, oxygen and fluorine. For these second row atoms, we cannot have more than eight electrons on that structure. So why is this important? Let me give you an example of how the octet rule comes into play.

The octet rule predicts that you want to have eight electrons on second row atoms. So if you ask yourself why is this methyl cation so reactive? It isn't because of the positive charge. It's because it lacks eight electrons. Boron is about as reactive as a methyl cation to most species

in solution, and it has no charge. But it desperately wants to have an octet of electrons. Likewise, if I draw a carbonium ion, this is another type of carbocation that you won't generally see. There's nothing wrong with having five bonds to this carbon atom in this carbonium ion.

And that may be surprising to you. Let me draw a resonance structure for this carbonium ion that has too many bonds. Here's a resonance structure of this. This carbonium ion is really just methane plus an extra proton attached.

That extra proton is sharing a pair of electrons in one of those four bonds. So there's nothing wrong with five bonds to carbon as long as I don't violate the octet rule. So obey the octet rule and don't violate that rule. Don't exceed eight electrons on carbon.

And if you have less than eight electrons on carbon, that carbon atom's going to be reactive. Okay, so second part of this. Well, let give you another sort of point to note when you obey the octet rule. Quite often in organic chemistry organic structures are drawn

out as molecular formulae. That's because it's easy to type this onto a keyboard. But when you translate this into a correct Lewis structure, if you do so incorrectly -- let's suppose I draw this out as a Lewis structure and I do it like this.

This typewritten version, this molecular formula version for acetamide cannot possibly be this, where I only have four electrons and two bonds to this central carbon atom. That can't be right. That's so -- that is

so desperately far away from the octet rule. So if you start off with this incorrect structure for acetamide, you have no hope of getting the correct mechanism. So start off with correct Lewis structures and obey the octet rule.

That carbon atom is very far from having an octet of electrons. [ Erasing Board ] So the second part of this first rule, draw a correct Lewis structure. [ Writing on Board ]

So part B of this rule is draw every substituent on every atom except for carbon. So if return to this structure for acetamide, the correct structure for acetamide, I could draw like this where I draw every single hydrogen atom. But one thing that happens very quickly is

that organic chemists become very good at adding to the number four. Instantly you should know that there should be four bonds to every carbon atom if you want that carbon atom to be neutral. So organic chemists very quickly stop drawing in the carbon atom symbol, C,

and they stop drawing the substituents hydrogen, because you should be very good at recognizing how many hydrogens are bound to this carbon that only has one bond, and how many hydrogens are bound to this carbon which already has four bonds.

So when you draw correct Lewis structures, you always have to draw the atoms that are attached to every nitrogen, every oxygen. But for carbon, it's assumed that you are instantaneous at adding up to four and filling in mentally the hydrogens that are attached.

The number one problem that people run into in arrow pushing mechanisms, the number one mistake is not seeing the hydrogen atoms that are there. So throughout this course and throughout sophomore organic chemistry, you're going to run

into cases where you need to pull off this hydrogen at this position that is not drawn for you. And so you need to practice with that until you are instantaneous at knowing where the hydrogen atoms are. So if you're not that good already, then practice until you are.

Okay, the second -- the third part of our rule to use correct Lewis structures. That looks good. [ Writing on Board ] So draw every charge. Draw every charge.

So when I take a simple structure like this. This is trimethylamine N-oxide. It's an oxidant in some reactions or used in the Kornblum reaction. There needs to be charges on this structure. And if I don't draw all those charges,

the charges that are necessary for a correct Lewis structure, then I'm going to miss some of the potential reactivity of this molecule. So once I draw the charges on a trimethylamine N-oxide or any N-oxide, I recognize that that oxygen is probably the nucleophilic part

of this molecule. Now, I have to be careful not to avoid the first part of our rule, which is draw correct Lewis structures. Every single part of my body wants to take one of these lone pairs and donate it down to that nitrogen because it's plus, but don't do that.

Let me just say right off the bat that you can see that's going to violate the octet rule. So don't do it like this. Just leave it with these charges, the plus and the minus, the separated structure. That's called an ylide, because it gives you an indication

of the kind of reactivity you should expect from that molecule. The last part of our rule has to do with drawing lone pairs of electrons. Draw every lone pair of electrons. 95 percent of every organic reaction mechanism

out there starts off with an arrow coming from a lone pair. In almost all the cases. And if you don't draw them, you may forget that those things are important. So if I don't remember to draw the lone pairs, this is a molecule called t-butyl isonitrile.

I'll draw the charges in. If I don't remember to draw that lone pair on carbon, I may miss that that's the nucleophilic end of the molecule. Likewise, if I take a typical mechanism. Here's an example of a reaction between an enol

and a molecular bromine. The reaction involves carbon-bromine bond formation and starts off with -- or it looks like it starts off with forming a carbon-bromine bond. But the reason why this double bond is so reactive is because there are lone pairs on this oxygen atom

that are donating into this pi star making it nucleophilic. So, the reaction starts off with a lone pair on that oxygen. And if you didn't draw it, you might forget that that's why that carbon-carbon double bond is so reactive. Okay, so draw correct Lewis structures. There's a flip side to this

about drawing correct Lewis structures. And that is what is a good Lewis structure? Frequently we're going to draw structures that aren't Lewis structures. And I'll show you when you'll confront those. [ Erasing Board ]

[ Writing on Board ] So there's two types of bonds that you draw in chemical structures that you have to watch out for, that is hydrogen bonds and dative bonds. I'll show you examples of both. Do not use curved arrows with these types of bonds.

So, the main rule is for those two types of bonds, when you draw chemical reactions, omit curved arrows as a way to represent the making or breaking of hydrogen bonds or dative bonds. So let me give you an example. I'll start off by showing a regular arrow pushing mechanism.

So here's a carbon-iodine bond. And when correctly drawn, if you ionize that iodine atom off, you should get two different ions out of this. If I really am taking electrons away from carbon, I should end up with a carbocation that has only six electrons. And if I really am getting this pair of electrons solely

to iodide, that iodide should pick up a negative charge. Initially in this bond, there were two electrons, one belonged to iodine and one belonged to carbon. But if I take carbon's electron away, that's why I end up with a minus on iodide. And I end up with a plus charge on the carbon.

Okay, so what would happen if I tried to do this kind of arrow pushing with a hydrogen bond? Let me just start off by drawing a hydrogen bond. And typically that's represented with dash lines in biology and chemistry. So here's a hydrogen bond between some sort of an alcohol

or water and a carbonyl. This could be the backbone of a protein. If I try to use this kind of arrow pushing to represent the breaking of this hydrogen bond, I'm going to be forced to draw a structure that leaves me with a positive charge on this proton

and a negative charge on this oxygen. And that can't possibly be correct. This started off, this oxygen atom here started off with two electrons, or two lone pairs. And if it still has two lone pairs, it's going to be a neutral oxygen, not this minus charge

that I'm forced to end up with. So, what's the correct way to represent the cleavage of this hydrogen bond? The correct way is to simply omit the curved arrow. Don't draw this curved arrow. I think that the reader or the viewer can tell

that you've just broken the hydrogen bond. You don't need to draw the curved arrow to show people that. So just draw the step and leave out the curved arrows that will lead you to the wrong conclusions. Let's take an example of another type of bond,

and that's dative bonds in organometallic complexes, typically. Also in some organic structures. Here's an example of dative bonds. So here's a nickel, tricarbonyl triphenylphosphine complexes. And it has four bonds here.

This is not a Lewis structure. If it were a Lewis structure, I'd have to have charges on for example oxygen or on nickel. These are examples of dative bonds, dative meaning donor. If I were to show that the phosphine can dissociate like this and draw out the products,

it's going to be the same situation. I'll be forced, if nickel were neutral before, it now has to be a nickel cation. And that's not true. Dissociation of a phosphine does not generate a nickel cation. And if that were a regular bond and I cleaved it,

it would force me to end up with a negative charge on phosphorus. And that's not right. So what's the correct way to deal with this cleavage of a dative bond? Don't show the curved arrow. Just show the reaction.

So once again, leave out this curved arrow. People can see that you broke the nickel-phosphorus bond. People can see that. So if you leave out the curved arrow, you're not forced to draw those incorrect charges there. Okay, so rule number one in general,

draw correct Lewis structures. Not hydrogen bonds, not dative bonds. Draw the lone pairs. Draw every hydrogen atom on everything except a carbon. And put your charges into those Lewis structures. [ Erasing Board ]

[ Writing on Board ] So rule number two. Every curved arrow should start with either a bond or lone pair. Every curved arrow should start from one of those two places. And the reason why they start from one of those two places is that's the two different ways

that we represent filled orbitals. Or more specifically, filled frontier orbitals. Every bond and every lone pair in an organic molecule is a representation of a filled canonical orbital. So let me show you examples of how to use that correctly.

So here's a hydroxide anion. And there's three lone pairs on there. And if you show that attacking acid aldehyde, then the correct way to depict that is start with a lone pair and then have that attack a carbon atom. Now break this carbon-oxygen bond, start your next arrow

from the pi bond between carbon and oxygen. So each arrow here starts with either a bond or a lone pair. That's a correct use of lone pairs. So here's another example of correct arrow pushing. If I want to protonate this carbon-carbon double bond in a hydration reaction, I would start off with hydronium ion.

And then start my first arrow from this carbon-carbon pi bond. And I'm very picky about this. Don't start with an arrow that touches the bond. Make sure your arrow isn't touching the pi bond. And start from the middle of the pi bond. Meaning you can't actually tell

which of these two carbon atoms I'm going to protonate. That's a real source of frustration for many students. They feel like they want to start the arrow from one of these two carbon atoms. That's not what arrow pushing is meant to represent. It's meant to represent

that this carbon-carbon pi bond is attacking that proton. Here's another example of correct usage of arrow pushing. If I draw a 1, 2 rearrangement here, an alkyl shift, I would start this first arrow from the middle of this carbon-carbon bond and show that carbon-carbon bond moving over to this atom.

And I'll be very careful to avoid that charge there and show my arrow ending on the carbon atom. Okay. So these are examples of how to push arrows. Let me give you an example of how not to do it. Then I'll give you a second part to this rule. [ Writing on Board ]

And the rule is the arrows do not start on charges. Do not allow your arrows to start on charges. So here's an example. Suppose you have a hydroxide, which is a nucleophile, and a carbonyl group. If you push your arrows starting with the charge,

you'll predict the correct intermediate as the product for that elementary reaction step, but you shouldn't do this. This is the wrong way to push arrows. Charges don't form bonds. Electrons form bonds. Specifically, the interaction of filled orbitals

with unfilled orbitals creates new bonds. So electrons don't form bonds -- sorry, charges don't form bonds. Electrons form bonds. Just to emphasize why this is so bad, you got lucky here. In 50 percent of cases, if you're using charge as a way to do your arrow pushing, starting from or ending

on charges, you'll predict the incorrect result in 50 percent of cases. So let's take an example. Let's suppose I have borohydride. This is a reaction that's covered in every organic chemistry course

at the undergraduate level. And that's borohydride reduction of carbonyls. If I start my arrow from this boron in order to attack this carbonyl, are you really going to make a fifth bond to boron and violate the octet rule? That's what that arrow pushing says.

It worked for you over here with hydroxide. It will utterly fail you over here. So, remember, never start arrows with charge. Start them with electrons. Start with lone pairs or start with bonds. Those are the two choices.

The second sort of caveat that we have -- [ Writing on Board ] -- is that arrows should not start on atoms. And the problem is, I'm going to mislead you by saying things like oh that bromide pops off. Listen to what I'm saying.

I'm saying that bromide pops off. Oh, bromide pops off. If you draw it like that, you're not representing anything that's related to filled orbitals and unfilled orbitals. You've decided to use your arrow

to represent the trajectory of that bromine atom. The correctly drawn mechanism should have the electrons starting from the carbon-bromine bond. And you've been misled by the fact that I said the bromine pops off. Another example.

I'm going to say protonate that double bond. Oh, let's stick a proton on that double bond. My description suggests that you should try to depict that. That maybe you might want to try to depict that hydrogen, that proton, popping onto one of those two carbon atoms. And that's a misuse of arrows.

That's incorrect. The correct arrow pushing mechanism should have the C-C double bond attacking, not this. It's not meant to -- arrow pushing does not represent -- not meant to represent the trajectories of hydrogen atoms. It should be the C-C pi bond acting as the base

and abstracting that proton. So that's the correct arrow pushing for than mechanism. Okay, so make every arrow start on a bond or a lone pair because those are representations of filled orbitals. [ Erasing Board ]

Okay rule number three. [ Writing on Board ] So rule number three is to make every curved arrow end on either an atom or a bond. And the reason is, these are representations, our best representations.

They're not good, but they're the best that we've got. These are representations of unfilled orbitals. And this is one of the hardest parts of mechanistic arrow pushing. Every single curved arrow that you draw is meant to represent the interaction of a filled orbital

with unfilled orbitals. But unfilled orbitals are not part of Lewis structures. So how do you know where there's an unfilled orbital? Let's take some examples of arrow pushing mechanisms. Where are there unfilled orbitals in here? So one hint is every single place that you have a bond,

there's also an anti-bonding orbital. If have a sigma bond here, then there's a sigma star antibonding orbital on the back side of that. So when I draw this arrow pushing mechanism and I break this H-Cl bond,

there must be some antibonding orbital that I'm donating into. So when I use a lone pair here to attack that proton that's attached to chlorine, if I'm breaking the bond, then I'm donating into a sigma star orbital between the H and the Cl. Everywhere there's a bond,

there's also an antibonding canonical orbital. So by ending my arrow on an atom, I'm showing that I'm breaking a bond and adding to an antibonding orbital. So here's the next type of place where you don't actually have a bond that's breaking.

If you're not breaking a bond -- so suppose I add to this carbocation, this methyl carbocation, and I use a lone pair. I can also have atoms not end on atoms because they're breaking a bond. Here I'm not breaking any bonds.

But I'm adding to an empty p orbital. There's an empty p orbital here that's not part of a correctly drawn Lewis structure. And that's an empty orbital and it's fair to add to that. And sometimes we draw those in. Sometimes I'll draw, sketch out some sort of an orbital

that might look like this. Sometimes with some phasing in there. But whether I draw that empty p orbital or not, there is an empty p orbital situated on that carbon atom. Here's another example of ending my arrows on atoms and bonds. Here's an elimination reaction.

This is part of an E1 elimination reaction. There's a carbocation at one of those two carbon atoms in the middle of this structure. So if I correctly draw that arrow pushing mechanism, what I'm going to show is that I'm adding to this hydrogen atom and breaking this sigma bond between carbon and hydrogen.

If there's a bond, that means there's an antibonding orbital. The antibonding orbital that's not drawn is sigma star H carbon. So it's on the backside of this. And now these electrons will now add to an empty orbital. And that's adding to an empty p orbital situated on carbon,

which I haven't drawn. I didn't draw the up and down lobe of that empty p orbital. So these are examples of correct arrow pushing where you can see that all of these arrows end on either atoms, or in this case, they can end on bonds. And I can use that as a second way to represent the fact

that this ends up forming a pi bond. So there's two different ways to represent this. You can make this double bond swing down and form a second bond to carbon, a pi bond. Or you can make these electrons add to that empty p orbital. And both of those are acceptable

and commonly used in organic chemistry. Okay, so once again, the important caveat is that arrows do not end on charges. Charges don't form bonds. Electrons form bonds. And you'll desperately want

to make your negative charges attack positive charges. This is a major driving force in organic chemistry. I've tried to dissuade you. Don't use arrow pushing to represent the interaction of charges. And that's not to say that charges aren't important

in bond-forming events. But we're not using our arrows to represent the interaction of charges. Okay. Rule number four is not really driven by some fundamentals of quantum mechanics. Or any fundamental rule or law of nature.

I'm going to give you a rule that's a safety device to prevent you from going out of control. [ Writing on Board ] We have something in this class called the three arrow rule. And it relates to curved arrows. So as long as you're taking Chem 201, I expect you to abide

by this three arrow rule. So the idea is you're not allowed to draw any more than three curved arrows in an elementary reaction step. [ Writing on Board ] So let me give you an example of how to abuse this rule. [ Writing on Board ]

So here's an elimination reaction. We take this alkyl chloride beta-chlorocyclohexanone. We treat this with a strong base like sodium methoxide. And this ends up doing an elimination reaction. And there's two possible products that we can draw for this elimination reaction.

One would be the conjugated enone. And the other would be the beta-gamma unsaturated non-conjugated enone. This is the major product. You see none of this non-conjugated enone here. Why is that?

It must have something to do with the carbonyl. There's clearly a proton that's being removed. And so if I were to draw this mechanism, I might draw something like this. I might start off over here with this sodium oxygen bond. That' ends up being broken in the end.

I make sodium chloride. And now I have to do something with these electrons and the H-C bond when I break that bond. Now, if I swing them down and pop off the chloride, well, then I'm not explaining why I didn't get elimination here. There's also protons down there.

The reason why this proton gets abstracted is because those electrons can donate into the carbonyl. Now, when I do that, I have to break this carbon-oxygen bond. And then in the second step, lone pairs on this oxygen swing back down and then create some sort of an enolate

that pops out the chloride ion. And how do I represent this? There's no enolate drawn on here. Maybe I can just draw a second bond here and then use those to swing over. Ultimately I end up with something that is wrong

in every single level of detail. And the problem is that I've -- the problem is that I've not broken this down into elementary reaction steps. That's the number one problem. But a true sign that I wasn't breaking this

down into elementary reaction steps is that I have all these crazy arrows. So, there's a simple rule in this class. Don't have more than three curved arrows because if you do, when you're first starting off in the reaction arrow pushing and drawing arrow pushing mechanisms, if you have more

than three arrows like this kind of weird scheme right here, you're probably not breaking things down into elementary reactions. So, what happens -- there will be cases where you just want to abbreviate. Maybe you did this because you didn't feel

like you had enough space on the page to draw the whole mechanism. Well, that doesn't make it okay to draw a wrong mechanism. So let me give you one suggestion that can be used by experts for drawing arrow pushing mechanisms. And you're not an expert yet.

Maybe by the end of this class you'll be an expert. So what do we do if we want to abbreviate things that we think are obvious? This is not the right way. This is fundamentally incorrect and implies it's all concerted. So the correct way would be first

to draw arrows for the first step. So draw arrows for step one. And then draw a series of stack reaction arrows. Not curved arrows, but reaction arrows. I'll give you an example of how that works. So let's just suppose I don't have enough space to draw

out a correct arrow pushing mechanism here. And I'll come back to this simple hydration reaction that we considered earlier. So suppose I have a hydration reaction in which I'm adding water across this C-O pi bond. And I want to give people some idea

of what the mechanism is for this reaction. It's not a one-step reaction. This is an acid-catalyzed reaction. What I can do, if I want to give people an idea of what's going is I can draw the arrow pushing for the first step.

And that's protonation of this carbonyl to give an oxonium ion, a protonated carbonyl. And then to show that there's two more steps in the mechanism I can abbreviate it like this. This gives all the information that we need to give. I've broken it down into a series

of three elementary reaction steps. I've shown the arrow pushing for the first step without adding incorrectly other arrows that would suggest concerted reactions. So ultimately, later on, when you start to become very good at arrow pushing and you don't feel like you need

to show everything to the viewer or the reader, you can use this kind of a shortcut. But for now, you need to draw every single step and every single intermediate and every single curved arrow. There will be times when you will desperately want to draw more than three curved arrows.

And in fact, there are many mechanisms in which the correct mechanism is drawn with more than three curved arrows. But for this course, there is always a way to avoid that. Let me give you an example of a reaction that would be correctly depicted with more

than three curved arrows. And so this would be after Chem 201, after you've taken this course. How might you depict this reaction of electrophilic aromatic substitution with pyrrole, or with some sort of a pyrrole species?

Pyrrole is aromatic, but it's very nucleophilic because these lone pairs here on nitrogen are donating into this pi system. And so let me go ahead and draw out the electrophilic aromatic substitution of pyrrole with an acylium ion.

So this is nucleophilic simply because the lone pairs on nitrogen are donating down. And I can draw this pair of electrons forming a double bond here. And then I can use this pi bond to attack my acylium ion. This is the correct mechanism.

And that's the correct way to depict that. [ Writing on Board ] But I've used four curved arrows to do this. How could I have satisfied my three arrow rule and still drawn out this mechanism? What I could have done is I could have started off

with a resonance structure of pyrrole. I could use arrow pushing to depict the resonance structure. And if I start with this resonance structure, I can use curved arrows to get to this resonance structure. Now I don't need to violate this three arrow rule in order to show this bond-forming event.

Now I've only got two arrows in my mechanism, where I'm now using this lone pair on my resonance structure of pyrrole to attack the acylium ion. And taking the time to draw the resonance structure between your initial intermediate and between this really does inform you

about why pyrrole is nucleophilic. So, it's a good idea to draw resonance structures as a way of gaining information about reactivity. And it will help you maintain this or stay within the bounds of our three arrow rule that is strictly for this class. It's a safety device to keep you from going out of control.

[ Erasing Board ] Okay, rule number five. [ Writing on Board ] No termolecular elementary reactions. Let me give you an example. We said that the goal of mechanistic arrow pushing is

to break reactions down into a series of elementary reaction steps. Here's a common transformation in organic chemistry, and that's to take phenolic hydroxyl groups and O-methylate them. There's many natural products

and biologically active molecules that have O-methyl groups. And the way you typically do this in the lab is you take methyl iodide and your phenolic compound. And potassium carbonate is a base

in DMF, it's a polar solvent. And that's how you O-methylate phenolic groups. So the potassium carbonate is clearly the base. And so you might have a tendency to want to depict that in the mechanism. If I ask you the mechanism first off,

we're starting off wrong here by drawing a lone pair without drawing the Lewis structure. But that's okay. We'll just not worry about that right now. The purpose of the base is to deprotonate the phenol. And I think everybody can see that.

Then this species becomes more nucleophilic. It turns into a phenoxide ion that then attacks the carbon atom that then releases the iodide. And I didn't violate the three arrow rule here. This still has only three curved arrows, one,

two and three curved arrows. But this is wrong. The problem is, by drawing all these three arrows in a single reaction step, it implies that it is concerted. What does that mean? What that implies is a series of --

not a series, but a set of simultaneous collisions. It implies that this base had to collide with this proton at exactly the same point in time that this oxygen atom was colliding with this carbon atom and releasing iodide. That's what it implies.

It is extremely unlikely that any two species in organic chemistry will collide with enough energy and the proper trajectory to make or break bonds. Usually it takes hundreds of thousands of collisions for things to form a bond, to have just the right energy, just the right angle, just the right trajectory.

The chance that any three molecules will simultaneously collide, there is no chance. There are no known reactions in organic chemistry that occur in the solution phase where three things simultaneously collide and in one elementary reaction step generate products.

The problem here is I didn't break this down into a series of elementary reaction steps. The true mechanism for this reaction involves initial deprotonation. And that's fast and reversible. The first step is deprotonation.

And that will happen millions of times before this phenoxide anion finally collides with a methyl iodide with sufficient energy and just the right trajectory to displace the iodide. So the correct mechanism involves two steps. So don't draw elementary reaction steps that have more

than three -- two components colliding. At most, only two things can collide with just the right geometry and trajectory and energy to form bonds or break bonds. Okay, our sixth rule. You know, it may seem like a lot to keep track of.

Gee, all these rules. Seven rules and their sub-rules. Most of you have already been obeying these rules. From the very get go you were probably already obeying these rules and you didn't know it. So I'm simply codifying probably what you've already been doing

in organic chemistry. I need to switch pens here for just a moment. [ Writing on Board ] Okay, this is a very simple rule, and that is that H is always attached to something. And the problem is, you're going to be misled because I'm going

to use terms that mislead you. I'm going to talk about proton transfer. I'm going to talk about hydride reductions. I'm going to talk about hydrogen transfer reactions. Those are all describing functional groups. So when I say proton, I'm not talking

about some free-floating H plus. I'm talking about a functional group. When I say hydride, I'm not talking about these species here, H plus. I'm not talking about H minus and I'm not talking about H dot. These species have no role whatsoever

in solution phase organic chemistry. The place you find species like this is maybe if you fly into the center of the sun, you'll find these kinds of free species floating around. That's the kinds of energies it takes to create free H plus or hydride.

Here's -- so in other words, how should we draw these mechanisms that involve what look like species like this? So here's what you'll often see written in the literature, something like this. When you want to talk about protonation, for example, of this ketal, some people will draw out this H as H plus.

There is no such thing as H plus in any organic solution reaction. But the correct way to draw this out, protonation of this ketal, is to show that the H was attached to something. And if you don't know what that H was attached to, just represent it as some generic acid.

I'll talk a little bit about that in a second. So this would be the correct mechanistic depiction. So let me urge you not to do anything like this, because there is no such thing as H plus. There is no bottle of H plus that I can get in the lab. So it's not that much easier to write this than to write a bond

to an A and break the bond. And that's the correct way for it to be depicted. So similarly, you might find some reactions in which -- that might be depicted like this, incorrectly. There is no such nucleophilic hydride that floats around in solution that looks like this.

There's no such thing. I might talk about a hydride reduction. But what I'm really talking about is a hydride functional group attached to something else. So here's an example of a tetrahydridoaluminate species.

We would refer to this reagent as a hydride donor. The functional group it's donating is a hydride with a pair of electrons. But nowhere in solution did I ever have H minus floating around free in solution. And likewise, I might have hydrogen atom

transfer reactions. So here's a ketyl species with a carbon atom that has seven electrons. There's just a single electron on that carbon. And it's reactive. And this will pick up hydrogen atoms from places

in your reaction, but it never picks up a hydrogen atom by simply joining to an H dot. There's no such thing as this that floats around in solution. For radicals, we're not going to cover radicals this quarter, but for radicals, you use these stylized fishhook arrows that have their own meaning.

Okay, so even though I use terms like proton and hydride and hydrogen atom, that doesn't mean you should ever draw a free species like this. At the very beginning of the reaction, this H starts off attached to a conjugate acid. It starts off attached to something.

In the transition state, it's attached to two things. And in the end, it's attached to this oxygen atom. It's never attached to nothing, and at most it's attached to two things. So avoid drawing species like this even though I use language like that.

So likewise, don't simply have protons pop off like this and release free protons. Because that's completely incorrect. The reason why this bond will break is because something adds to that antibonding orbital and breaks the bond. Okay, so avoid that --

avoid drawing these types of free species, because they don't exist in typical solutions. [ Erasing Board ] So the second part of that rule, of rule number six -- and you'll see violations of this throughout the literature in textbooks.

Don't write this as a reagent, either. So you might see reactions like this drawn where somebody wants to hydrolyze an acetal. If you wanted to hydrolyze this acetal, you might see it depicted as this sometimes in the literature.

There is no reagent H plus that you can get in the lab. It is not that much harder, if you wanted to show how you hydrolyze this acetal, it is not that much harder to draw this, catalytic HA. You don't have to catalytic if you don't want to, but just simply to draw that H is attached to something.

And better yet, show the actual reagent that was used, because that affects the rate and speed of the reaction. So, let me just put an arrow through this. Don't use this kind of a depiction either in your mechanism or as in a reagent above a reaction arrow. So if you really had to put H plus like this,

put quotation marks around it to show that you're sophisticated enough to know that there is no such thing as H plus as a reagent that you can add. Okay, I'll come back to this acid concept of how to represent acids in just a moment here. We're down to our last rule, and that's rule seven.

[ Writing on Board ] Avoid proton transfers through four-membered transition states. Very often in organic chemistry you're going to run into species that have tetrahedral atoms like this. So this might be a species that you run into when you're hydrolyzing an amide

or cyclizing to make a lactam. So if I were just trying to cyclize this in order to make a lactam, at some point I might want to have some sort of a proton transfer step occur. Or if I were to do the reverse and add a methoxide anion to open a lactam.

So at some point I might want to move a proton from here over to this oxygen in order to complete my mechanism. And the important point is do not draw this. If I count up the atoms in this transition state, there's four atoms. That means that my transition state has

to be the shape of a square. That requires this oxygen to attack that proton at a 90 degree angle. And that's simply implausible. That's energetically disfavored. The thing that you ought to do in these cases --

I'll show you that in just a little bit -- is have something else come along from exactly the back side and pluck off the hydrogen atom and then redeliver it to that oxygen in two steps. Not a one-step mechanism through a four-membered transition state, but a two-step mechanism involving some sort

of a conjugate base that arises from an acid. So I'm simply telling you that that's slower than a two-step alternative. It can be much worse than that. Here's an enol. It's a tautomer of a ketone.

If I wanted to show how that proton ends up at the alpha position, then I would need to somehow move that proton over to here on carbon. And you cannot draw a plausible mechanism that looks like this. That violates a concept called -- the concept of orbital symmetry.

And there is no pericyclic concerted reaction that looks like that. So that's implausible. Okay, so what's our choice? What are the alternatives for that? Here's the alternative.

Instead of doing these types of one-step proton transfers through four-membered transition states, something that's slow versus something that's simply impossible. And I'll label this as impossible as a concerted reaction. What are the choices here?

And the choice is to first and foremost determine are your reactions taking place under acidic or basic conditions? I can hear my phone going off. Okay. [ Writing on Board ]

So the first thing you need to do, and we'll use this tautomerization as an example. If you wanted to avoid some four-membered transition is first decide whether this enolization is occurring under acidic or basic conditions. If it's occurring under acidic conditions,

instead of moving this proton over in a single step, do it in two steps. First add the proton and then pluck the proton off. So if you know the conditions are acidic, use a combination of -- so if it's acid, use a combination of HA for your acids and A minus for your bases.

Don't use the symbol B, stick with these two symbols. You have an acid and you have a base. So how would I use those? So for example here, if I was under acidic conditions, I would protonate that double bond first. And then in a second step, after I protonate that double bond --

let me use the electrons here on this oxygen -- after I protonate that double bond, now I can come along with my A minus, my conjugate base of my acid, and pull the proton back off. Two steps. And mechanistically, we'll talk more

about proton transfers later in this class, but both of these types of reactions, proton transfer reactions, are very fast as opposed to a slow four-membered transition state type process. So if I know the conditions are acidic, then use this combination of a symbol HA and A minus.

Or if my conditions are basic, then I should use a combination of BH as an acid and then B as a base. And you can add pluses or minus on there as you would like, depending on whether you believe that your base is neutral or not neutral. So here's how I would do this type of proton --

do a proton transfer using this combination if I knew that my conditions were basic. Let's suppose I wanted to transfer a proton in this tetrahedral intermediate. So if somehow wanted to move a proton from this oxygen to this amine so that the amine's a better leaving group,

and I know my conditions are basic, I would start off with my symbol B. And I don't need to say whether that's an alkoxide or an enolate or just a water molecule. I'm just using this as a symbol. And then I'll end up with this intermediate.

So it's a two-step proton transfer that's entirely plausible. And now I use this new protonated conjugate base that I generated. And now I can transfer in the second step that proton. So this would be the correct way to depict these types

of proton transfers after I first decide whether my conditions are acidic or basic. And now the lone pairs in the alkoxide are ready to push out that amine leaving group. Okay, so that's the end of our seven rules for mechanistic arrow pushing.

So let me just remind you of the rules that we covered today. So rule number one, draw correct Lewis structures. If you don't start with correct Lewis structures, your chances of drawing the correct mechanism are substantially diminished. Rule number two, make arrows start with bonds or lone pairs. Bonds and lone pairs are the way

that we represent filled canonical orbitals. And it's the interaction of filled orbitals with unfilled orbitals that is the goal of arrow pushing. Rule number three, make arrows end on atoms or bonds. And when we correct Lewis structures, we don't draw the unfilled orbitals.

But each bond you draw is also associated with an antibonding orbital. Rule number four, obey the three arrow rule. This is not a fundamental rule or law in organic chemistry. It's just a safety device to make sure that you break mechanisms down into elementary reaction steps.

Rule number five, don't draw termolecular elementary reactions. Three things never collide simultaneously to react in an elementary reaction step. Rule number six, H is always attached to something. So don't draw free protons, H plus floating

around in your solution or as reagents above a reaction arrow. And then finally, rule number seven, avoid proton transfer through four-membered transition states. Use HA as a symbol for acid and show the breakage of the HA bond, or use B as a symbol for a base and then BH as the acid under those conditions.

Okay, so when we come back for our next lecture, we're going to be in our regular classroom setting. And we'll talk more about basic concepts like energy and molecular orbitals and bonding. And I'll see you then. ------------------------------4a6f2f03aafb--

Video Description

UCI Chem 201 Organic Reaction Mechanisms I (Fall 2012)
Lec 01. Organic Reaction Mechanism -- Arrow Pushing -- Part 2
View the complete course: http://ocw.uci.edu/courses/chem_201_organic_reactions_mechanisms_i.html
Instructor: David Van Vranken, Ph.D.

License: Creative Commons BY-NC-SA
Terms of Use: http://ocw.uci.edu/info.
More courses at http://ocw.uci.edu

Description: Advanced treatment of basic mechanistic principles of modern organic chemistry. Topics include molecular orbital theory, orbital symmetry control of organic reactions, aromaticity, carbonium ion chemistry, free radical chemistry, the chemistry of carbenes and carbanions, photochemistry, electrophilic substitutions, aromatic chemistry.

Organic Reaction Mechanisms I (Chem 201) is part of OpenChem: http://ocw.uci.edu/collections/open_chemistry.html
This video is part of a 20-lecture graduate-level course titled "Organic Reaction Mechanisms I" taught at UC Irvine by Professor David Van Vranken.

Recorded December 12, 2012

Required attribution: Van Vranken, David Organic Reaction Mechanisms 201 (UCI OpenCourseWare: University of California, Irvine), http://ocw.uci.edu/courses/chem_201_organic_reactions_mechanisms_i.html [Access date]. License: Creative Commons Attribution-ShareAlike 3.0 United States License (http://creativecommons.org/licenses/by-sa/3.0/us/deed.en_US).