Lecture 04. Reactions and Protecting Groups.
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Transcript: English(auto-generated)
00:04
All right, give yourselves a big round of applause. So much, so much of what happens
00:22
in learning organic chemistry is not about me or about Johnny or about Kim, it's about you and actually working problems. And you're one-sixth of the way there with the quizzes and you're a fraction of the way there with the homework.
00:43
You're on board and you're keeping up. Now this quiz may have been a little easy. Chapter 19 wasn't that meaty, but this gives you the flavor. The next one I'll probably ask you some questions about mechanism or about synthesis and curved arrows.
01:01
We'll figure things that develop a good set of skills. But the quizzes are going to build on this problem solving aspect of the course. And the skills that you've cultivated and the work that you put in in preparing for this and doing the homework is exactly what's going
01:21
to set you up to excel in the class. So again, big round of applause for yourself. So today I want to move into the last section of Chapter 20.
01:41
Chapter 20 is kind of interesting because it has a ton of different stuff. And it sort of is all over the map in terms of reactions. And I thought about it and what I've tried to do is really to group things into conceptual themes. And the last conceptual theme that I see
02:03
from this introduction to Chapter 20, and as I said, this is only a part of carbonyl chemistry because we're going to be continuing through four chapters, well, five chapters, six chapters technically, 19 to I think 24 on carbonyl chemistry.
02:22
It's a huge swarth of the course. It's seven weeks. So in breaking out the last part of Chapter 20, what I'd like to do today is to talk about selective reactions of carbonyl groups in the carbonyl family and also about the use of protecting groups in synthesis.
02:42
So I'm going to start with sort of a conundrum I raised in yesterday's class. And that is that we talked about the addition of various types of nucleophiles to esters. We talked about addition of powerful hydride nucleophiles
03:02
like lithium aluminum hydride and powerful carbon-based nucleophiles like organolithium reagents or Grignard reagents. And we said in those reactions you would typically take the ester, remember the ester is plus three oxidation state of carbon, and not move one step down, one notch,
03:24
one pair of steps down, but move two steps down to the primary alcohol oxidation state. In other words, if we take an ester like methyl benzoate, it's the one I wrote in the previous class, and we treat it with lithium aluminum hydride, and then we carry
03:44
out an aqueous workup with water, as your textbook writes, or H3O plus aqueous acid like HCL in water, the product of the reaction is benzyl alcohol.
04:01
And organic chemists are often really bad about writing small molecules and byproducts of reactions. So I'll just remind us, of course, the other product of this reaction is methanol. And if I were thinking about a synthesis, maybe I wouldn't write it. If we go ahead and do the same chemistry,
04:27
the same substrate, methyl benzoate, and now instead of treating it with lithium aluminum hydride, we treat it with say methyl lithium
04:42
or methyl magnesium bromide, and I'll specify that we're talking about two equivalents here, although we really can't get away from problems happening
05:01
from the lack of stopping happening. And then again, we do some type of aqueous workup. Now we've gone down and added two equivalents of the methyl group and gotten this tertiary alcohol.
05:22
And the main point of this example is that we cannot stop these reactions even though they go through the intermediacy of benzaldehyde for the former and for benzophenone on the latter or for acetophenone on the latter, we can't stop the first one at benzaldehyde
05:50
or the second one at acetophenone. And so organic chemists are all about control.
06:02
We're all about being able to choose and to develop tools to make the chemical compounds that we want to make, and we have a conundrum here in the lack of selectivity, the lack of ability to select for the aldehyde or the ketone. I saw a question over there from the gentleman
06:21
in the third row. For the exam, are we going to have to put the byproducts? Great question, very reasonable question. When I'm thinking about a synthesis and I'm thinking about taking my methyl benzoate on through several steps
06:40
to a product that contains that benzene and the carbon and the methyl is just sort of a throwaway, I probably wouldn't write it. But if I'm thinking about writing an equation or a balanced equation in which I show all products of reaction or mechanism, I certainly would. Good question, I know and I know all these little
07:01
variations end up being of tremendous concern to you. Why do I sometimes draw lone pairs? Why do I sometimes draw curved arrows to show mechanisms and how I'm thinking? And other times, we just zip from reactants to products. It's expressing different things. One is expressing a level of detail,
07:22
how the reaction's occurring. The other is what we're forming or in the case where we're emitting the methanol, simply what we're forming as the product. And here, I guess I will, to keep consistent in the first panel here, I will again sort of parenthetically write methanol is the product.
07:42
But good question and a very, very legitimate concern, particularly when you're starting out learning and trying to figure out what people are expecting of you. So, chemists have developed a number of different reagents
08:05
to achieve selectivity in these types of reactions. Another question. Ah, another good question. So your textbook sometimes shows things as covalent bonds
08:24
and sometimes as ionic bonds. And we're going to see that today to some extent. In general, lithium and magnesium when bonded to carbon are primarily covalent in their bonding. Sodium's a little more electropositive.
08:42
It's more ionic in its bonding. Sometimes we'll write an ionic resonance structure. I mentioned in the last lecture, we'll sometimes in our mind think of methyl lithium as methyl minus with a lone pair, lithium plus. That's a non-bond resonance structure. But the main resonance structure is the covalent one.
09:03
All right, so on to selectivity and control. So your textbook presents one reagent to do this reaction selectively. That reagent is dibal, diisobutyl aluminum hydride. So first let me write this as a synthetic reaction.
09:23
And so if you take the reagent dibal, this reagent has three different names that people call it depending on the source you go to.
09:41
The reagent is diisobutyl aluminum hydride. That's two isobutyl groups and an aluminum on hydride and a hydride on a hydrogen on aluminum. And you will see it written sometimes as dibal. With an H on the end to indicate there's a hydride.
10:04
Sometimes as dibal without the H on the end. And sometimes as dibah just because people want to catch the notion of the H. Anyway, if you treat your ester with dibal
10:21
and then do an aqueous workup. And again, you could add water or I would probably add aqueous acid. The product of reaction is the aldehyde. You can stop the reaction at the aldehyde. And again, the byproduct
10:41
of the reaction would be bimethanol as a byproduct. So the gist here is just as we've dealt with other sorts of nucleophiles like methyl lithium adding to a carbonyl, here we have a hydrogen nucleophile, hydride nucleophile on aluminum.
11:03
And the overall result or partway through you get a tetrahedral intermediate. So your methyl benzoate reacts with diisobutyl aluminum hydride to form a tetrahedral intermediate with your aluminum
11:26
and your two isobutyls on there. But the main thing is because the aluminum oxygen bond is very strong, stronger
11:41
and more covalent, again, it's this notion of degree of covalency, degree of ionicness. Because aluminum is a little bit less electropositive than say a lithium, you have more covalent bond character. You form a very strong bond to oxygen. And you have a tetrahedral intermediate that's stable enough
12:06
to not break down. I'll write it as more stable tetrahedral intermediate. In other words, your tetrahedral intermediate is
12:22
sufficiently stable that until you carry out the aqueous work up, until you add water or you add aqueous acid, it sticks around. And remember the mechanism for this double reaction with lithium aluminum hydride. Hydride adds.
12:41
You get a tetrahedral intermediate. The tetrahedral intermediate breaks down, kicking out a methoxide leaving group. That gives you the aldehyde. Another mole of hydride adds giving you a new tetrahedral species, your final alkoxide that protonates on work up. And so at this point, you can stop the reaction.
13:03
Now in practice, sometimes a chemist would choose to use dibal as a reducing agent to stop at the aldehyde. And sometimes a chemist would say, all right, look, I'll just use lithium aluminum hydride, reduce it all the way down to the primary alcohol and then oxidize the benzyl alcohol back up to the aldehyde.
13:25
Sometimes in practice it's a little more efficient one way or another. As in many cases, there are multiple ways, multiple right ways of doing things. And in the laboratory, a chemist is often concerned about what's easiest or what gives the best overall chemical
13:45
yield, the best amount of product in the end. All right, so that takes care of achieving selectivity in our reaction to add a hydride nucleophile. Let's now take a look at achieving selectivity
14:02
in adding a alkyl nucleophile.
14:20
Now, as I said before, carboxylic acids are part of a big family, a family in which your carbon is in the plus three oxidation state. And all of these species are easily interconverted. You'll be learning more about these interconversion reactions in the subsequent chapters.
14:41
Carboxylic acids, esters, acid chlorides, acid anhydrides, amides, and even nitriles end up being all in this plus three oxidation state family of carbon. So in order to achieve selective addition of an alkyl group, we're going to use a different member
15:05
of the carboxylic acid family, an acid chloride. And you'll learn about how to form acid chlorides later on. Now, if you just treated almost any member of the carboxylic acid family with an organometallic,
15:23
oops, an organometallic, cool iPhone, with an organometallic nucleophile. So if we treated it with the same methyl lithium,
15:41
or methyl magnesium bromide, we would have, and again, I'll indicate that we're talking about two equivalents, but the main point is that we can't really stop at one equivalent. If we, in other words, if you added one equivalent, it wouldn't stop at the ketone in the other case.
16:02
We'd just get a mixture of the alcohol and unreacted ester in some ketone. It wouldn't be a controlled reaction. So if we imagine treating with methyl lithium and then in a second step, again, our aqueous workup, and I'll just write this as H3O plus because in my lab,
16:21
we'd certainly throw aqueous acid in here. The problem is we'd be in the exact same situation that we were with methyl benzoate. We would have two equivalents of methyl nucleophile adding here. We really don't have control.
16:40
And this is essentially the same theme. There are little variations. Your textbook, this is kind of the same theme with lithium aluminum hydride. If we treated lithium aluminum hydride with an acid chloride or carboxylic acid or an ester, it would always reduce it down. There are little variations in reaction of Grignard's
17:01
and organolithiums with carboxylic acid. There are little variations which your textbook gives you an example on in the reaction of lithium aluminum hydride with amides to give you a means. But for the overall general principle, it's members of carboxylic acid family
17:20
with these powerful nucleophiles, lithium aluminum hydride, Grignard reagents, and organolithium compounds all get addition of two equivalents and all kick out the leaving group. Question? No. And you could easily write excess.
17:40
In my own laboratory, people would probably use two and a half or three equivalents in order to do this to make sure that there was enough of the reagent to achieve the reaction. But yeah, it would be a minimum of two equivalents. But as I said, if you used less, you still wouldn't be able to stop and controllably have a ketone product.
18:02
All right. The reagents that people have developed to make this reaction selective, because what this is all about is selectivity. It's achieving the reaction that you want to occur. The reagent that people have developed to make this selective is organocuprate reagents.
18:24
I'll write the reagent first, and then I'll show you how you make it. So the reagent here is lithium dimethyl cuprate, CH3, 2, C-U-L-I. It's a carbon nucleophile again. Now we have carbon bound to copper.
18:42
Copper is even less electropositive than lithium or aluminum. And so it's even more of a covalent bond. The exact mechanism of the reaction goes beyond the scope of this class, but in the very last lecture, if we go on schedule and talk about 26, chapter 26,
19:03
I'll be giving you a little bit of a flavor of the very special reactivity of transition metals like copper and palladium. But that will be later on if we have time. Anyway, the overall result of this reaction is that you can controllably add one equivalent
19:24
of your methyl group to your member of the carboxylic acid family. In this case, specifically, your acid chloride for reasons of the mechanism. And organocuprate reagents can be made by the reaction
19:42
of two equivalents of methyl lithium with copper iodide to give you CH3, 2, C-U-L-I, which as I said, is called lithium dimethyl cuprate.
20:08
And you'll also, if you want to write a balanced equation, which as I said, organic chemists tend to be very bad at, you'll get lithium iodide as the other byproduct of the reaction.
20:22
There are many subtleties in this chemistry, many of which go beyond the scope of the course. Certain groups like ethyl groups due to what's called beta hydride elimination don't work as well in this reaction. Your textbook gives you examples of a vinyl cuprate.
20:40
Those work well. Certain cuprates like acetylenes on copper don't transfer. At this point, I'm just giving you and your textbook is just giving you a teeny taste of this big wide world of selective reactions here. Another way one can actually achieve this reaction
21:01
selectively is to add a little bit of copper to a Grignard reagent. But again, we're not going to focus heavily on any of that right now. I just want to give you a little flavor as your textbook does of this notion of control in a reaction.
21:26
Question, for the exam, it is a very good idea to know arrow pushing for organolithium reagents
21:40
and Grignard reagents and lithium aluminum hydride. And this overall principle of reaction of the carboxylic acid family, the addition elimination reaction of the first nucleophile followed by the addition of the second nucleophile. But for more specialized reagents like the organocuprate reagent, as I said,
22:03
the mechanism goes a little beyond the scope of the course. All right, so this brings us to a notion here of control and specifically the notion of chemoselectivity that is achieving control for the reaction
22:23
of one group but not another. And in a way, we've already seen that notion in these two reactions and that we're controlling the reaction to get just to the aldehyde or ketone oxidation state. But really, the best sort of the archetypal examples
22:40
of chemoselectivity are when you have two functional groups and you want to carry out a reaction on one of those functional groups to the exclusion of the other functional group. So let us take, and again, like many of my examples in the class, they'll be a little bit artificial in the sense that you might not immediately have a reason
23:02
to do this. Let's take a derivative of a compound called terephthalic acid where we have half of it as the aldehyde and half of it as the ester. And if we imagine treating this bicarbonyl compound, this dicarbonyl compound with lithium aluminum hydride
23:23
and then with aqueous acid, lithium aluminum hydride just blasts down all of the members of the carboxylic acid family. It reduces all of them. It's a super powerful source of hydride nucleophile.
23:41
And so our ester group gets reduced, and I'll just write it as CH2OH to tell us that we have the alcohol. And our aldehyde gets reduced to the alcohol. In other words, this reaction is not chemoselective.
24:01
It doesn't hit just one group. Now, conversely, if we take a milder reducing agent like sodium borohydride, then we do get selectivity. So I'll just write ditto mark for the reactant here, and then I'll write sodium borohydride.
24:22
Typically, people use alcohol solvents like methanol for sodium borohydride reduction. If we carry out the reaction of this ester aldehyde in with sodium borohydride,
24:40
then we will selectively reduce the aldehyde group without reducing the ester group. We say that sodium borohydride is chemoselective.
25:05
It reduces aldehydes, and I might add ketones, but not esters.
25:32
And so that principle of control is something that's very useful because it allows you to make a product that you desire.
25:42
And as I said, so much of organic chemistry is about making molecules that often are useful, for example, as drugs.
26:06
I want to give you a couple of other examples of selectivity, and I borrowed these directly from your textbook. There are many ways to do these types of reactions, and I thought this example, which is straight
26:20
from your textbook, would actually illustrate these points very nicely. So if we take a compound like cyclohexanone, which has both a carbon-oxygen double bond, a carbonyl group, and a carbon-carbon double bond, we have two different groups that can be reduced,
26:40
the carbon-oxygen double bond and the carbon-carbon double bond. Reagents like sodium borohydride or lithium aluminum hydride, and I'm not going to write all the conditions. I'll just write sodium borohydride or lithium aluminum hydride reduction,
27:00
reduce the carbonyl group. These reagents are nucleophilic toward carbonyls. They add to the carbonyl group. If we take this same cyclohexanone and instead perform catalytic hydrogenation on it,
27:21
your textbook writes one equivalent of hydrogen, and I will add or low pressure. In other words, the way you could do this reaction is either to go ahead and control the amount of hydrogen that you're adding or carry out the hydrogenation quickly
27:43
at with just a balloon of hydrogen at atmospheric pressure monitoring by TLC. And you do this in the presence of a palladium on carbon catalyst or a platinum catalyst. Your textbook uses palladium on carbon as an example. You can selectively reduce the carbon-carbon double bond
28:06
and so go from cyclohexanone to cyclohexanone. Hydrogenations of carbonyls are more difficult. It requires more effort to reduce the carbon-oxygen bond.
28:23
In other words, hydrogenation is chemoselective for the carbon-carbon double bond. But if you beat on it, your textbook writes it as H2XS.
28:41
That might be a big balloon of hydrogen in monitoring the reaction for a long time. Or sometimes people will carry out these reactions at higher pressure. I'll say high pressure. By high pressure it can mean anything
29:00
from bicycle tire tube pressure of like 75 PSI in a special glass bottle to an apparatus that's steel with a tank that can hold thousands of PSI aptly called a bomb. If one hydrogenates under high pressure, again with a catalyst it can be palladium on carbon
29:22
or platinum oxide or any of that platinum family of metal. Then you can reduce both the carbon-carbon double bond and the carbon-oxygen bond. And again, one is achieving control. One is achieving chemoselectivity in this type of reaction.
29:47
There are entire books on the topic of catalytic hydrogenation. In fact, a book by that title. So there's a lot of different ways to do this reaction. And again, your textbook is just giving you a little taste
30:02
which is fine because it gives you a lot of the needed thought processes. Now, in these reactions on this blackboard we've looked at taking the oxidation state down. In first case we took the oxidation state of the carbon-oxygen double bond down.
30:22
We reduced the oxidation state of the carbon. In the second reaction we've reduced the oxidation state of these two carbons. And in the third reaction we've reduced them both. Just as there are many ways to reduce the oxidation state of carbon.
30:41
There are many ways to increase the oxidation state of carbon to oxidize carbon. And your textbook gives you a little taste. They give you three related sets of oxidation conditions to oxidize a primary alcohol up to a carboxylic acid. All the ones that they use
31:01
that they show you involve chromium. It's a classic. Potassium chromate or sodium, potassium dichromate or sodium dichromate or chromium trioxide in sulfuric acid and water are all essentially equivalent. All of these oxidize primary alcohols
31:23
up to carboxylic acids. There are newer, more modern reagents that don't use toxic carcinogenic chromium. Your textbook leaves those out and that's fine. As I said, you can write an entire book. Ditto. There are ways to stop at the aldehyde oxidation state.
31:43
And your textbook gives you one such set of reagents. The use of silver oxide in aqueous ammonia. And that will stop you at the aldehyde oxidation state.
32:07
And again, what this is all about is about control in chemoselectivity. All right, so I guess by conclusion, I will just write, there are many other selective oxidizing agents.
33:01
All right, at this point, I would like to turn my attention to another problem. So we've looked at a problem of chemoselectivity.
33:22
And at this point, I'd like to turn my attention to the problem of regioselectivity.
33:40
And we've kind of seen this issue already with the hydrogenation question. And I'd like to give us the same type of example and bring back our cuprate reagents. So here's our cyclohexanone again. And if we treat our cyclohexanone
34:02
with a powerful nucleophile, we already saw lithium aluminum hydride. And I'll now give you an example of methyl lithium or Grignard reagent. I'll choose methyl magnesium bromide just as an example.
34:23
And then we carry out an aqueous workup. We add, just like the lithium aluminum hydride did, we add to the carbonyl group. The powerful nucleophile attacks the carbonyl group.
34:42
And what's interesting with this alpha beta unsaturated carbonyl compound is if we treat it with an organocuprate reagent. So I'll just write a ditto mark. And then I'll bring back our lithium dimethyl cuprate.
35:01
And again, we'll carry out some type of aqueous workup. Typically, you would use acid. Now we get a very different sort of addition. The addition is not to the carbon oxygen bond, but rather to the carbon bond in what we call a 1,4 addition.
35:31
And I'm going to contrast that to the previous example, which we call a 1,2 addition.
35:41
And I'd like us to think, as I said, we're not going to really fuss about the mechanism of the cuprate reaction in detail. And why, or to put it more specifically, why copper does what it does. The reasons for the details of the difference are rather subtle.
36:03
But we're going to focus on the hows and the what's, the hows and the what's that are happening. So in the 1,2 addition, here's our carbonyl compound. Here's our alpha, beta unsaturated carbonyl compound. I call it an alpha, beta unsaturated carbonyl compound
36:23
because the carbon that's next to the carbonyl group we call the alpha carbon. And the carbon that's 1 over we call the beta carbon. So in the first example, our nucleophile, which we could think of as methyl minus or we could think of it
36:41
as a methyl bound to a lithium, but I'll just write it as NU minus, our nucleophile attacks the carbonyl group. And I'm going to draw in lone pairs of electrons to help us think about this. And I'll draw a curved arrow from the nucleophile,
37:01
from the thing with the electrons, to the electrophile, to the thing that wants the electrons. And I will push electrons up onto the carbonyl oxygen because of course if I stopped at this point, I would have 10 electrons around the carbon and we would be in trouble.
37:21
So at this point in the mechanism, we have formed an alkoxide anion, the nucleophile has added. And that's how your first reaction goes. And one of the things that was guiding our thinking in thinking
37:42
about the reactivity of the carbonyl group is we recognize that although the carbon oxygen bond is a covalent bond, it is a polar covalent bond. We embodied that last time or the other time when we talked about this, we embodied this in two ways.
38:02
One was writing a delta minus on the oxygen and a delta positive on the carbon. The other way of embodying or thinking about this is to write two different resonance structures. And so I can write one resonance structure like so, the main resonance structure or major resonance structure.
38:26
And then I can write a second resonance structure that reflects the polarization of the bond where the carbon doesn't have a complete octet.
38:42
And it's not as important a resonance structure because a resonance structure in which the carbon doesn't have a complete octet is less of a contributor. But I can also write a third resonance structure. And the third resonance structure is to continue to move the electrons of the double bond,
39:04
much as you do in an allylic cation, which you learned about in the previous quarter. And now you have your positive charge over at the beta position. And in order to help avoid confusion, I want to draw in my key hydrogens here so that you can see
39:23
that just like the resonance structure at which you have the positive charge on the carbonyl carbon, this last resonance structure, this other minor resonance structure has a positive charge on the beta carbon, which has one other hydrogen. In other words, it doesn't have a complete octet either.
39:43
So collectively, these three resonance structures make up a more complete picture that help us think about the reactivity. And in the case of the 1, 4 addition, again, we're going to skip the Y's of it at this point
40:01
in your learning of organic chemistry. But in order to understand the reactivity on the 1, 4 addition reaction, you can say, okay, now I can see why in some cases a nucleophile is going to react at the carbonyl carbon.
40:23
And at other cases, because there's also a partial positive charge at the beta position, at some cases, the nucleophile is going to add to the beta carbon. And now we just push our electrons like so to represent
40:46
and think about the reactivity here.
41:12
Now, that kind of took us halfway through the conjugate addition through the 1, 4 addition of the cuprate. Because we did a second step where we did an aqueous workup.
41:24
And the product of that reaction, which I've just erased, it was on the bottom of your right-hand blackboard, was a ketone. And at this point, we're halfway through. We don't have a ketone, and we're about to add a source of aqueous acid.
41:40
Now, the species that we've just generated here is called an enolate anion. And this is your first exposure to the enolate anion. You're going to be learning more about it in a moment. More specifically, you're going to be learning more about enolate anions in Chapter 23.
42:04
So here is your enolate anion. And just like your ketone, you can think
42:21
about it having multiple resonance structures. Again, this is going to come back more later on. So we can think about one resonance structure. Indeed, it's going to be a major resonance structure, the major resonance structure, which I've written on the left. And a second resonance structure,
42:45
which I've written on the right. And you're going to, as I said, be seeing more about these in Chapter 3. But for now, what you can think about is that in the workup step of the organocuprate reagent,
43:02
you protonate the enolate anion. You can think of it occurring in two different ways. I probably like to think of it more in the way that I'm writing over here. You can think about the proton going onto the alpha carbon to give rise to your ketone product.
43:27
And in terms of Occam's razor, in terms of the greatest simplicity, that's probably the way I like to think about it. Your textbook writes the mechanism slightly differently, and that's okay, too.
43:41
Your textbook likes to think about the proton going onto the oxygen, like so. And this product is called an enol.
44:01
And the enol is in equilibrium or goes to form the ketone by a reaction, which again you will be learning more about in Chapter 23, called tautomerization.
44:27
I'm going to write that again. We call the ketone and the enol forms tautomers, and we call their conversion tautomerization.
45:13
This issue of selectivity and achieving control is one
45:24
that cannot always be done by picking the right reagent. In other words, sometimes you can say, oh, yeah, I want a Grignard reagent here. I want a cuprate reagent here. I want lithium aluminum hydride here.
45:40
I want dibal there. But there are cases where one gets into a situation where no amount of choice of the right reagent alone is going to do something, allow you to do something that you want to do. And let me show you an example of this and how chemists work around it.
46:01
Imagine that I wanted to generate a Grignard reagent from 3-bromo-1-propanol. And I eagerly envision saying, okay, I'm going to go ahead and treat my 3-bromo-1-propanol with magnesium and diethyl ether in hopes or in expectations
46:27
of getting this Grignard reagent. We wouldn't get it. The problem with this Grignard reagent is it's inherently unstable. It can't be formed. And I gave you an example in our previous lecture
46:44
that sows the seeds for destruction here. If we take a Grignard reagent and treat it with an alcohol like ethanol, this is the example from the previous class, I pointed out that Grignard reagents are very strongly basic
47:02
and alcohols are sufficiently acidic that they protonate them. If we take butyl magnesium bromide and treat it with ethanol, we get butane plus ethoxy magnesium bromide, basically the magnesium salt, magnesium alkoxide salt of ethanol.
47:23
And so this hypothetical Grignard reagent that we would like to generate is unstable. It can't be formed. It can't be isolated. It can't be generated. It immediately reacts with itself
47:45
with the alcohol quenching the Grignard reagent. And so to come around, to work around problems like this, organic chemists will use something called a
48:00
protecting group. Protecting group is an idea that you're going to hide, you're going to mask one part reactive part of the molecule. You're going to protect it from a reagent or from conditions that would otherwise destroy it. And at this point in your textbook,
48:22
they introduce one class of protecting groups. You'll learn about more later. Protecting groups that they introduce are silyl ethers. And so the general gist of protecting an alcohol
48:43
as a silyl ether, particularly a tert-butyl dimethyl silyl ether is to take your alcohol, treat it with a chlorosilane, tert-butyl dimethyl dichlorosilane is particularly popular and a base.
49:02
The base that's widely used is called imidazole. It's a good base for this particular reaction. I guess I won't draw in lone pairs. And the product of this reaction is our silyl ether.
49:23
By the way, we call this reagent TBDMS chloride. And then we're going to call our ether the TBDMS ether. I'll write it out here as R-O-T.
49:43
Actually, I will write it first showing you the silicon SI dimethyl tert-butyl. And we also get, if you want to write a balanced equation, we'll also get protonated imidazole, the imidazole hydrochloride.
50:09
But again, organic chemists typically don't pay attention to these byproducts of reaction. Now, the big idea behind protecting groups is
50:26
that they are easy to put on, they hide the alcohol, and then they're easy to take off. So before I show you a whole sequence of reactions, let me show you how to take the protecting group off.
50:41
So we have our TBDMS ether. And I'll just write this as R-O-T-B-D-M-S. If you notice me slurring my words here, many organic chemists abbreviated further, just like we said dibal H, they abbreviate further. Many abbreviated further as a TBS ether.
51:03
Anyway, if you treat your TBDMS ether with tert-butyl, with tetrabutyl ammonium fluoride, again, now we get sort of into the lands, the land of acronyms. Tetrabutyl ammonium fluoride is a big fat tetrabutyl ammonium
51:26
cation with fluoride as an anion. It's a source of fluoride that dissolves in organic solvents like tetrahydrofuran. And fluoride loves silicon.
51:41
Oxygen likes silicon a lot, fluoride loves silicon. It forms a really, really, really strong bond to silicon. When we treat our TBS DMS ether with TBAF, we get back our alcohol. And again, organic chemists are terrible
52:01
about writing byproducts of reactions. But I will for this one time write tert-butyl dimethyl silyl fluoride as a product. So this brings us to an application of protecting groups. Because the reason that people want to put protecting groups
52:23
onto alcohols are to solve synthetic problems and do something they couldn't otherwise. So let me show you an example of doing something we couldn't otherwise. We want to generate the equivalent of this Grignard reagent
52:40
that I just said you couldn't make. So imagine that we take 3-bromo-1-propanol, just as we did before, but instead of treating it with magnesium and ether, we treat it with TB DMS chloride and imidazole.
53:05
Now we would get the TB DMS ether. As I said, Johnny would write it as the TBS ether to save room.
53:21
Now if we treat our TB DMS, see I can't even write all of those letters, TB DMS ether with magnesium and diethyl ether, now we can generate a Grignard reagent
53:46
that's the equivalent of the Grignard reagent that we couldn't do.
54:02
And if we then, let's say, added that Grignard reagent, so I'll just continue my equation over here. If we imagine adding that Grignard reagent to let's say a generic ketone or aldehyde, I'll just write it as R carbonyl R prime to indicate
54:22
that we're not dealing with any specific ketone or aldehyde. Now we could generate this otherwise impossible to achieve product and the only thing that would be different
54:58
about this product than having generated
55:01
that Grignard reagent that we couldn't generate is we have the TB DMS group on that, but we know what to do with that. We simply take this and we treat it with TBAF and the silo group comes off and we've generated a product
55:34
that we couldn't otherwise get.
55:56
One of the tremendous intellectual triumphs
56:01
of organic chemistry is the ability to think backwards about reactions. It's retrosynthetic analysis and it's one of the reasons that my academic grandfather and the academic father of the textbook, E.J. Corey, received the Nobel Prize in chemistry.
56:21
And so I want to show you an example of this very important intellectual process of thinking backwards about making organic molecules. And so I'm going to give you again a little bit of a contrived problem that gives you the flavor of synthesis. Let us ask how would we synthesize the molecule 1,
56:46
5-octanediol from compounds containing 4 carbon atoms
57:05
or fewer? So I'll say compounds containing 4 carbon atoms or fewer.
57:26
And again, just like in my previous example, I've done this to give you sort of a small catalog of organic compounds that you can think of without necessarily saying there aren't bigger compounds available.
57:41
But since we've already seen molecules like imidazole and TBAF and so forth are reagents that may have in TBDMS chloride are reagents that may have many carbon atoms, I will open up our catalog and say and any reagents that you require.
58:18
And while to my knowledge 1,
58:20
5-octanediol isn't an insect pheromone and it isn't a drug, it's certainly representative of the types of molecules that synthetic organic chemists make. So, of course, one of the things is knowing what 1, 5-octanediol is and so I will draw
58:42
out the structure of it. And of course, you have to think, all right, there's an octane chain, so there's an 8-carbon chain with an alcohol at the 1 position. So 1, 2, 3, 4, 5, 6, 7, 8 and an alcohol at the 5 position.
59:13
Now, we've started to learn to recognize that different groups can come
59:21
from different sorts of connections. And often there are lots, lots and lots of correct ways to put together molecules. We're looking to build a molecule here. And so we know, for example, since we're looking to make it
59:41
from smaller pieces, that we probably want to put it together by carbon-carbon bond forming reactions. Alcohols could come from ketones. Alcohols can come from addition of Grignard reagents to ketones or aldehydes. So we could imagine.
01:00:00
breaking this molecule apart and in this view that I mentioned from 30,000 feet last time, you can look at this molecule and say okay, I could envision forming this carbon-carbon bond by some form of addition of an organometallic reagent
01:00:20
to a carbonyl compound, in this case formaldehyde. I could envision forming this carbon-carbon bond from a molecule where we add a carbon nucleophile like a Grignard reagent or an organolithium compound
01:00:41
to an aldehyde. These are both perfectly good ways to put the molecule together. Both of them, however, involve a fragment in one case that's seven carbon atoms, in one case that's five carbon atoms.
01:01:01
But if I envision forming this carbon-carbon bond, now things look really, really attractive. We've split the molecule into two equal pieces. My hypothetical example gave you only up to four carbons to work with. And so that's really, really convenient here. And so we could think in terms of
01:01:25
retrosynthetic analysis and say yeah, a good retrosynthetic disconnection would be to break this molecule into this four carbon Grignard reagent,
01:01:46
this four carbon aldehyde. And remember, this is the view from 30,000 feet. And we already know that we can't make this four carbon Grignard reagent because we know that Grignard reagents containing alcohols
01:02:05
and Grignard groups are inherently unstable. But now we have all of the tools that we need in our toolbox to solve this problem because we know that we're clever enough to work with compounds that are equivalent to this Grignard reagent
01:02:24
by use of protecting groups. And so we can approach this problem with confidence and now work our solution through in a forward direction.
01:02:46
We start with four bromo, one butanol. We protect it with TBDMS chloride and imidazole
01:03:04
to get the corresponding TBDMS ether. We take our TBDMS ether,
01:03:20
we treat it with magnesium in diethyl ether. And then we add to that Grignard reagent, if you've generated one in the laboratory, you'll typically take metal turnings, magnesium turnings, put them in a dry flask, maybe flame dry the flask,
01:03:44
you'll add your ether, you'll add your Grignard reagent, you'll crush the magnesium turnings until the reaction starts and you generate your Grignard reagent. Once you've generated your Grignard reagent, you will immediately add to it
01:04:02
butanol. You'll probably distill your butanol or your TA will distill your butanol so it's nicely pure. You'll carry out an aqueous workup with aqueous, let me choose an acid here, I should commit. Let's say aqueous HCl.
01:04:20
You literally, literally could do this at the chemistry stockroom. You would go to them and ask them for all of these reagents. A little short on space here, I've got to crunch in my TBDMS group
01:04:43
and now in the final step of our synthesis you simply treat this TBDMS ether with TBAF
01:05:05
and lo and behold you've generated your 1,5-octanediol. All right, well this is the way we think about carbonyl chemistry. We've learned a lot of reactions of them.
01:05:21
As we continue into the next chapter we're going to learn new reactions of carbonyl groups. Thank you.
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