J. MICHAEL MCBRIDE: So We've been talking about
oxidation and reduction.
And today we're going to talk about a very special kind of
oxidation by iodic acid.
Periodate, that is to say.
Periodate cleavage of diols.
Carbonyl with an alcohol, then. Let's try that.
Carbonyl with two alcohol molecules.
STUDENT: The chromium, which is double-bonded with the
oxygens, and then--
PROFESSOR: Don't think about the chromium.
Just for the terms of the analog, suppose that was a
carbonyl group.
Instead of I=O, suppose it was C=O
What reaction do you get?
STUDENT: Is it a esterification?
PROFESSOR: Not esterification.
You could imagine esterification of an
acid with an ester[correction: alcohol].
But I said not...Suppose it's not a--
suppose it were a carbonyl group, C=O. No OH
on the group up at top.
What reaction do you get between a ketone--
yeah, Matt?
STUDENT: Form a ketal.
PROFESSOR: A ketale, right?
So you get a reaction that's just like that.
A ketale forms. And now we have a thing that's a little
bit like ozone, because this sort of ketale can unzip.
That's what reminds you a little bit of ozone,
that kind of thing.
So it can come apart that way.
And notice what's happened is that the diol has become two
carbonyls, and of course that's an oxidation.
The iodine started in oxidation state +7, right,
seven bonds to oxygen, and it ended in +5.
So the iodine is reduced, and at the same time, the carbon
is oxidized.
It's +1 here and it's +2 over here.
And two of them get oxidized. Two carbons get oxidized.
Now, this might remind you of this reaction, where you get
two carbonyls.
And what reaction is that?
What reagent will do that?
STUDENT: Is that ozone?
PROFESSOR: That's ozone, right?
What reagent does it?
Ozone.
So it's nice to think about, when you have a certain
functional group you want to get, what reaction
will lead to it.
So there's one that will lead to that.
We could also think, what reaction would come back from
two ketones to give a diol with a
carbon-carbon bond in between?
That's a more exotic reaction.
Does anybody remember that one?
That's the pinacol reduction that we talked about, where
metals give electrons to the ketones, the radicals couple,
and then you have two O's that get
hydrogens on them.
So that's the pinocol reduction that we mentioned.
And then you might wonder if there's a reaction that does
that, right?
And we'll see an indirect version of that
later in this lecture.
That involves the Wittig reaction, which is familiar to
you from lab.
So periodic acid, cleavage of carbohydrates, is
a diagnostic tool.
This is a useful reaction.
It's actually used to prepare compounds.
But it also was historically important in finding out the
ring size of sugars that we talked about
earlier as a problem.
Remember, nowadays you do that kind of thing by NMR, but in
the early days, you had to do chemical transformations to
known compounds in order to figure out what something was.
So if you have a diol, as we just showed, you could react
it with periodic acid and get two aldehydes.
In particular for this one, formaldehyde.
So remember what sugars are, are a bunch of
carbons in a row.
They're carbohydrates.
So it's carbon-hydrate, OHH, and then the carbons link
together, and one end is CH2OH, typically, and the
other end is an aldehyde.
Or the carbonyl could be other
positions down the chain.
But how do you figure out what they are?
Suppose you had three in a row and did periodic acid.
OK, so you react with one of them, and get the formaldehyde
out from the end.
But from the bottom part there, you get another
aldehyde and OH.
And that reacts again with periodic acid.
Now, why?
Because remember, the aldehydes form hydrates, so
they're equivalent in water to alcohols.
So now we have a diol again, and periodic acid oxidizes and
cleaves such diols.
So from this reaction, then, you get two
formaldehyde molecules.
And the one in the middle, which had only one hydrogen on
the carbon, gives formic acid, So formaldehyde arises from
the primary alcohols at the end of the chain, and formic
acid from the ones in the middle of the chain, that have
just an H on them.
And incidentally, I've taken these slides from lectures
given in the course last year by Professor Ziegler, because
he had nicely animated them already.
So here's that reaction we just talked about.
Now, suppose we had an aldehyde at the end of the
chain and we did this.
Now the one...not only the one in the middle of the chain has
only one H on it, but the one at the top has only one H.
So from that one, you'd get a formaldehyde and two formic
acids, right?
Or if the carbonyl were in the middle of the chain,
you're going to get formaldehyde and the acid in
the middle.
But that acid in the middle itself can hydrate and be
oxidized and give CO2.
So from this one, we get two aldehydes, two formaldehydes,
and a CO2 molecule.
So now we could go down and see what groups where that R
includes an adjacent H, right, and what products
you'd get from them.
Formaldehyde, formic acid, and carbon dioxide.
So that just summarizes what we just looked at.
Now suppose you had glucose.
Now, the top would give formic acid, the next one would be
formic acid, formic acid, formic acid, formic acid.
The bottom would be formaldehyde, because there
were two Hs on it.
If you had mannitol, a reduced form related to glucose, then
the top would be formaldehyde, two Hs, then acid, acid, acid,
acid, and at the bottom would be formaldehyde.
So you see, the ratio of these various well-known products
you get out tell you something about what the structure of
the sugar was.
Or if you had fructose, which has a ketone in it, you get
formaldehyde from the top, CO2 now from the one that has no
hydrogens at all.
Acid, acid, acid, formaldehyde.
And you can also apply it to the question of the ketals
that are formed when you form a ring.
So remember, we spoke before about this ketal being a
protecting group.
Both for the aldehyde that was formerly here--
and that's its oxygen--
and for the alcohol that was here.
OK?
Now, we talked about the problem with
determining the ring size.
Think what would happen with this with HIO4.
So it would cleave those two bonds where there
are adjacent OHs.
Right?
Vicinal.
Those are the only places where there are adjacent OHs.
So it would give formic acid-- there was one H on
that central carbon--
and aldehydes on the end.
And now if you treated that with acid and heat, then you'd
open up the ketal, right?
Ketals could be opened by acid.
Incidentally, people don't typically notice the necessity
for heat in this, but it must require heat.
Because if you can get that product, which hasn't opened
the ketal in the first reaction, which is done at
room temperature for a day, periodic
acid is a strong acid.
So here's water--
and it's done in water.
So it's water with an acid catalyst.
You'd think it would take the ketal apart.
But in fact taking the ketal apart requires some heat.
So that's done in the second step.
And then you get the aldehyde taken apart.
So the pieces you get are--
from the piece up here, you get OH, OH, and aldehyde.
So you get D-glyceraldehyde.
From the piece over here you get aldehyde
and aldehyde again.
And finally, of course, there's methanol that came
along, because it was a ketal involving that alcohol.
So by telling what the products are, you can tell
what the ring size was.
Now, here's a problem for you to think about.
Suppose you could imagine these ketals being formed with
other-- instead of this OH being the one that's involved
to make a six-membered ring, it could have been this OH and
made a five-membered ring.
Or in principle, this one that made a four, or that one that
made a seven-membered ring.
I'd like you to make a table that shows what the products
would have been for periodic oxidation of those others to
see whether, in fact, you could prove what the ring size
was by what products come out from this.
This reaction was developed in 1928 and
applied for that purpose.
That analytical purpose.
OK, so that's oxidation.
Now we're going to talk about a little bit of reduction, and
in particular, about synthesizing alcohols and the
idea of retrosynthesis a little bit.
Which is, you look at the product-- usually it's a
product you want to have, and you're trying to figure out
what starting materials and reagents would have given it.
So really, you're working backwards, rather than
forwards, in a synthesis.
So analyzing such problems is called retrosynthesis, and
we'll engage in a little bit of this here, using a problem
that occurs in the Jones textbook.
OK.
So here's an alcohol we want to get.
We want to put R on an alcohol that already has two methyls.
Now, we can start with acetone.
There are the two methyls.
And now we have to put the R on.
Now, what kind of R do we need in order to attack a carboile?
What will attack a carbonyl to make a
carbon-carbon bond for R?
What form of R will we use?
For example, could you use an alkyl halide, RCl?
Would that be something that would react in order to give
this product?
What is it that will react with a carbonyl?
What makes a carbonyl reactive? Ruoyi?
STUDENT: You need a nucleophile.
PROFESSOR: You need a nucleophile.
So you need a carbon that has a high HOMO on it.
Can you think of any such things?
Where do you got a high HOMO on a carbon?
STUDENT: From a Grignard.
PROFESSOR: From a Grignard reaction, or an alkyl lithium.
Something where R is attached to a metal ion,
so it'll be R minus.
OK.
So you have R with a metal, like a Grignard reagent or an
alkyl lithium, right?
You do this reaction.
Then, of course, those electrons could be used to
form a bond to the metal, and you get that intermediate.
And in fact, as we've already discussed, it's not so obvious
what the order of those additions are, the R added to
carbon and the metal added to oxygen.
It could be either first or in fact probably
both at the same time.
So anyhow, we can make that new carbon-carbon bond, and
then treat the salt, the O minus M plus, with H plus, and
get the alcohol.
So there's a simple synthesis.
Now, suppose we wanted to put an H on instead of an R. What
form of H do we need?
Can we do this with an acid, for example, H plus?
Is that going to be a good idea? Megan?
What type of the reagent, what kind of H,
would attack a carbonyl?
What makes the carbonyl reactive?
STUDENT: The low LUMO
PROFESSOR: Low LUMO.
So you need an H with a high HOMO.
How about proton?
STUDENT: Oh, no.
PROFESSOR: No, protons have no electrons at all.
So what form of H do you need?
STUDENT: H minus.
PROFESSOR: H minus.
Where are you going to get H minus?
Where did you get R minus?
STUDENT: From R-Metal.
PROFESSOR: R-metal.
So what do we do now?
STUDENT: H-Metal.
PROFESSOR: H-metal.
So this is going to be a hydride reduction, OK?
So we have H with a metal, doing exactly the same thing
we showed above, except it's H minus instead of R
minus doing the trick.
And we get it from something like H attached to aluminum
with a minus charge on the aluminum, which means it's
really going to be a high HOMO.
Or H attached to boron with a minus charge on it.
So those reagents are lithium aluminum hydride, sodium
borohydride, which can reduce an alcohol to
put the H on there.
And we also talked about the biological equivalent of these
hydride donors, which is NADH, where the unshared pair on the
nitrogen can get into the, act forming an aromatic system and
losing H minus, which can go onto the carbonyl group.
So we have R minus or H minus as ways to
attack carbonyl groups.
Now let's look a little bit at the synthetic versatility of
Grignard reagents.
So here's a problem.
To suggest high yield syntheses where the carbons
come only from alcohols with no more than three carbons,
but you can use any other reagent.
Now, these are very artificial.
These are not a problem you'd be faced with in the
laboratory, because many of these you could buy.
Not all of them, but many of them you could buy cheaper
than you could make them.
Right?
But in a way of thinking about how you do syntheses in an
elementary way, this is a good exercise.
And as I mentioned here, you can find that problem in the
Jones textbook.
So let's think first about the first molecule in this list.
Now, what problem do we face in making this from alcohols with
no more than three carbons?
There are four carbons in it.
So we've got to make a carbon-carbon bond somehow.
So let's think about what carbon-carbon
bonds we might make.
We could make the first carbon-carbon bond, nearest
the OH, or the next one, or the next one.
There are three that we could make.
And let's see how we would go about that.
First let's try to make the one adjacent to the H. So we
need to have two carbons that will react with one another,
one a high HOMO the other the low LUMO, that will
establish that bond.
So we need a nucleophile, a high HOMO, and an
electrophile, the low LUMO.
So what will the nucleophile be?
Anybody got an idea?
It could be either the piece on the left or the
piece on the right.
If the electrophile were the piece on the left, the three
carbon, what kind of molecule would you do use?
If it were three carbons, and you wanted it to be attacked
by a high HOMO?
STUDENT: R-Cl.
PROFESSOR: A halide would do it, right?
The sigma* of an alkyl halide would do that.
Now suppose you wanted it to be the high HOMO.
Then what would you do?
Megan, you just told us about that.
Suppose you wanted R to be a high HOMO.
STUDENT: A metal.
PROFESSOR: You'd use a metal, an alkyl metal.
OK.
Now, how about the electrophile on the right?
Here's a great example of use- the-low-LUMO-of-carbonyl.
So now what we want on the left is the high HOMO.
We want the alkyl metal.
But we can only start from alcohols.
In fact, we can't even start with this, because that's a
carbon that goes into the product.
We have to start from an alcohol.
So how are we going to make that from an alcohol?
How is it related to an alcohol?
We can have a H on the carbon, H on here.
That would be an alcohol methanol.
What kind of reagent do we need to get from
the alcohol to this?
Is it an oxidation?
Is it a reduction?
Is it neither?
STUDENT: Oxidation.
PROFESSOR: We have to remove two hydrogens.
It's an oxidation.
There's a problem, though.
Because the oxidizing agents that oxidize alcohols also
react with aldehydes.
If you have any water there, water will add, make the
aldehyde into a diol,
and then that will react.
That will get oxidized by the reagent.
How do you avoid that?
We talked about this last time. You don't have any water
there, right?
So we talked about that last time.
Pyridinium chlorochromate, done in methylene chloride
solvent, so there's no water there, you can do that trick.
So we have the electrophile.
Now we need the nucleophile.
We need a C3 group with a metal on it.
But we can only start with an alcohol.
So if we had normal propyl Grignard reagent, that would
do the trick.
Where do you get a Grignard reagent from?
RMgBr.
What's the starting material?
So you guys need to go through this to get a little drill
under your belts.
You get Grignard reagents from alkyl halides.
So we could get RMgBr from magnesium and alkyl bromide.
But where do we get the alkyl bromide?
Night The rules of the game has us starting with alcohol.
Yea, Amy?
STUDENT: You can start with PBr3.
PROFESSOR: PBr3 on the alcohol.
So we got it.
OK.
So we could make that first one.
Now how about the second one?
Now we need an electrophile and a nucleophile.
The electrophile, now on the left, could be the
ethylmagnesium bromide, just two carbons.
And we would make it the same way here, starting with
ethanol instead of propanol.
So that one's OK.
Now, what will the group on the right be?
We want to be able to attack this carbon and end up with OH
on the adjacent carbon.
Now, you haven't done a lot of problems like this, but you
need to think them through
and whatever textbook you're using, you'll find
problems of this sort.
So try some of them.
But we talked about such a reaction, which is an epoxide.
Remember?
The strained rings could be attacked by a nucleophile to
open here to form this ring, where you attack this carbon
and the O comes away as the leaving group, protonated.
Now, where do we get that from?
Where does an epoxide come from?
We have to start with an alcohol.
Any reactions that give epoxides?
They come by adding oxygen to an alkene.
Anybody remember the reagents that'll add an oxygen?
STUDENT: Osmium?
PROFESSOR: What?
STUDENT: Osmium something, tetrachloride?
PROFESSOR: Say again?
STUDENT: Osmium?
PROFESSOR: No.
The osmium petroxide takes an alkene and puts two Os on it.
That'll put two Os.
We want to just put one O in the middle.
And you'll kick yourself when--
oh yeah, Chris?
STUDENT: HOCl, I think you do it with something else.
PROFESSOR: You can add Cl and OH and then remove HCl.
That was one way of doing it.
There's another way where you do it in one step.
STUDENT: mCPBA.
PROFESSOR: mCPBA,
meta-chloroperbenzoic acid.
Remember that cyclic butterfly mechanism that transferred the
oxygen from the peroxy acid?
OK, so we got that.
But now we need the alkene, but we only
have available alcohols.
But that's a trivial one.
Just dehydrate[correction: eliminate from] an alcohol, or
dehydrate an alkyl bromide.
So you treat with strong base, and remove HBr,
and you got the thing.
OK.
So we've got that.
We can do that one with heat and NaOH.
Now, how about the bottom one?
That one is quite a problem.
The problem is that you could have methyl Grignard, but what
are you going to attack?
Because if you had a four-membered ring here with
oxygen, it wouldn't open.
It's not strained like the three-membered ring would be.
And if you try to make the nucleophile be on the left,
that top right carbon being an anion, and this one being,
say, formaldehyde, then you have the problem
that you'd get an OH here.
And that's a problem, because you have to get rid of just
one of the two OHs.
Or you would have to have a Grignard reagent, or something
like here, and that would react with its own H.
Now, you could imagine complicated schemes where you
use protecting groups, and stuff like that.
So it's conceivable, but it's hard work.
because this is not a particularly good carbon to be
to be reactive.
So it's not in an activated position.
This one was in a position where it could have been
attached to that oxygen.
This one was in a position that was a carbonyl.
But this one has no special virtues that make it reactive,
so that one is the tough one to make.
OK.
Now let's think about the second one.
Let's try to make that bond, for example, right?
Now we need a nucleophile and an electrophile.
Does anybody have an idea what the low LUMO is that could be
attacked in order to make that red bond?
Yigit, you have an idea?
We want to make this carbon-carbon bond, and one of
the two carbons has to be a high HOMO, and the other an
unusually low LUMO.
And notice, incidentally, that this one that needs to be one
of them has an oxygen next door.
STUDENT: Yeah, that's carbon double bond with O; pi*.
PROFESSOR: Carbon double bond with O, right?
So then what would you use on the right?
STUDENT: You need a nucleophile, so,
RCl.
PROFESSOR: RCl is like R plus, right?
You want R minus.
You want a high HOMO, not the low LUMO.
Not sigma*.
And what would you use?
STUDENT: R-Metal.
PROFESSOR: It's a metal, right.
OK.
So we need a nucleophile and an electrophile.
We use the three carbons on the right
as a Grignard reagent.
And now we need that aldehyde, right?
That four-carbon aldehyde.
We can only start with three carbon things, so we can't
oxidize one of the things they give us,
and make that aldehyde.
How are we going to make that four carbon aldehyde?
Any ideas?
You don't have to look far.
Linda do you have an idea?
You need four carbons that terminate in oxygen, double
bond O. And you can only start with three carbons.
Can you think of any way of making four carbons when you
start with three or less?
That was the first problem, right?
So you've already made this.
Can you think of any way of making the aldehyde?
How do you go from the alcohol to the aldehyde?
What do you need to do?
You need to remove two Hs.
Right?
It's an oxidation reaction.
So think only of oxidizing agents.
Can you think of any oxidizing agent that the changes an
alcohol to an aldehyde?
STUDENT: [very quietly]PCC.
PROFESSOR: Sebastian?
STUDENT: Br2.
PROFESSOR: Br2 can do it.
But you're going to have problems, depending on how you
do it, of over oxidation.
The aldehydes can go on to acids.
STUDENT: PCC.
PROFESSOR: PCC, the same one we used on the last slide.
So you treat that one as PCC in the
methylene chloride solvent.
You got that aldehyde.
So we've solved the second one.
PCC and methylene chloride.
Now, the next one that they suggest you try to
make is this one.
Suppose you wanted to make that bond.
Can you think of any way of doing it?
Ellen?
What's going to be the nucleophile?
STUDENT: the R-Metal.
PROFESSOR: Yeah.
What's the R going to be?
Is it on the right or the left, the one we're going to--
that's going to be the R-Metal that we're going to deal with?
Is it going to be this carbon or this carbon?
The bottom-right or the top-left carbon?
What's the top-left carbon going to be in order
to make that bond?
Amy?
STUDENT: It'll be another C=O.
PROFESSOR: C=O. When we see the O next to the
carbon, we think aldehyde.
That's going to be the LUMO that gets attacked.
And the HOMO? Ellen?
STUDENT: The one on the right.
PROFESSOR: The one on the right is going to be a Grignard.
But this time, instead of n-propyl, it's isopropyl.
And this is the aldehyde we need.
Any way of getting that aldehyde Ellen?
STUDENT: Yeah, it's the same one.
PROFESSOR: It's the same one above.
So we got that one.
OK.
Now how about this one?
Any ideas for that?
So let's make that bond.
So what gets attacked on the left?
Amy you want to help us again?
STUDENT: Well, it would be...isn't it the to...
or the second one?
PROFESSOR: How many carbons on the left?
One, two, three, four, five, six, seven.
We can only start with three carbon alcohols.
STUDENT: Could you make the second product and then there's
some kind of shift?
PROFESSOR: Ah.
But the second product has the seven with an
alcohol in the middle.
So what do you do?
You need a ketone in the middle.
You've made the alcohol.
You need the ketone.
So you have to oxidize it.
So you could use the same reagent again, but you don't
need to worry, because it's a ketone this time, so you're
not so worried about overoxidation, so you could
use one of those chromate things that we talked about,
or bromine, or anything.
And on the right, it's just the n-propyl again.
We've already made that.
OK.
So we got that one made.
What if we wanted to make the next bond, the
vertical bond, instead?
Oh.
I didn't show it here.
Well, what if we needed to make the vertical bond?
It's the same thing.
Right?
The thing is symmetrical.
It has three propyl groups on there.
OK.
So here's another one.
How are we going to make this one?
Well, we could make that bond.
Anybody see how to do that?
Mary? What are we going to use on the right?
STUDENT: We're going to use the same reagents.
PROFESSOR: The same one as here, isopropyl.
OK.
And what's on the left?
STUDENT: We're going to oxidize the oxygen...
PROFESSOR: Right.
It's going to be the same one as above...
STUDENT: Yeah, and then we're going to add isopropyl.
PROFESSOR: that seven-carbon.
So we've got that one.
And now finally, how about if we wanted to
make that one instead?
Then what would the Grignard reagent be?
It would be n-propyl instead of isopropyl.
And what would the ketone be?
It would be this.
And where would we get that?
By oxidizing this one.
And finally this one.
Now, we could imagine making this bond, or making this
bond, or making this bond.
And we have the means to do all these things now.
But one wonders, is there a preferred order?
Is one way of doing this better than the others?
And in this case, I suspect that they're
all about the same.
It might come down to what's cheap and available and so on,
which alcohols are cheaper.
And in fact, of course, you would never do this, because
simple things like this, you wouldn't start from the
alcohol, you'd buy the bromide or something like that.
But when you talk about versatility of Grignard
reagents, there are other things they can do, too.
For example, if you try to make the diisopropyl methyl
version of the carbinol here to make this one, you find
that you can first make the salt, then treat it with acid,
you get that alcohol, and it's 95% yield.
Great.
Now, suppose you want to use t-butyl instead of methyl.
Right?
So you want to get that one.
So you try that reaction, and you get 0% yield of that.
The product is this.
So what was it that added to the carbonyl?
It wasn't the t-butyl anion.
What anion was it?
What did the addition?
It was hydride.
But where did the hydride come from?
So it was a reduction by H minus, or the
equivalent of H minus.
What you had was the t-butyl Grignard.
Now, what problem did t-butyl Grignard have in trying to
form this alcohol? Yeah Cassie?
STUDENT: It's sterically hindered.
PROFESSOR: It's sterically hindered.
The carbon can't get close to the carbon.
But a hydrogen can get close to the carbon.
That beta hydrogen can get close to the carbon.
So now we can do this.
And you see that the carbon part that was t-butyl loses
magnesium and loses H to become the alkene, and the
other product, then, is that 65% yield alcohol.
And this reminds you, this hydrogen transfer from
t-butyl, although it's happening here in the context
of an anion, it reminds you of two t-butyl radicals having
trouble forming a carbon-carbon bond and
transferring a hydrogen atom instead for the same reason,
to avoid steric hindrance.
OK.
Now, if you use t-butyl CH2 magnesium bromide, then you
don't have to worry about this hydride transfer.
Because it has CH2 in this position, and three methyls
attached to that carbon.
There is no hydrogen in the beta position, so it looks
like we're home free, here.
But you only get a 4% yield of the Grignard, right?
Because it's still sterically hindered.
What you get instead is protonation of
the neopentyl group.
So remember that the ketone itself has alpha hydrogens.
So you can form enolates, you can give up
a proton from those.
So this thing which is a nuclophile is also a strong
base, and can take that off, and you get that one plus the
enolate, which lost the proton.
But then when you work it up with acid, you protonate it
again, so the ketone gets recovered, 90% of it, and your
carefully made Grignard reagent just becomes the
hydrocarbon.
No use there.
OK.
So those are other things that can happen
with Grignard reagents.
And that same thing, incidentally, can happen here,
because the alpha hydrogen was present here, and this one was
a strong base, and in fact 35% of that one was that plus the
starting ketone after you put the proton back on.
Now, if you try to make this bond, you could make it with--
notice the difference this time is that it's isopropyl
instead of t-butyl group here.
So you could try doing this.
So that has an H beta and steric hindrance, so there's
the risk that it would give reduction.
But if you started with that ketone and added methyl
Grignard, then the methyl doesn't have any hydrogen beta
that it could give in reduction, so now you get a
good yield of the product that you wanted.
So sometimes it's helpful to choose one of
the bonds to make.
So in trying to make this alcohol, the thing you get
when you protonate here, you could make this bond or this
bond or this bond.
But the bonds to methyl are much more practical to make
than the bond to the CH2 because
of the risk of reduction.
OK.
Now, the Wittig reaction we can zoom through, because you
all know that from lab.
I don't remember what base you used to
make the Wittig reagent.
Do you remember?
STUDENT: Sodium hydroxide.
PROFESSOR: Sodium what?
STUDENT: Hydroxide.
PROFESSOR: Hydroxide.
So you didn't have such a simple Wittig
reagent as this one.
Notice that here you make the reagent by reacting with
bromide, just like nitrogen would do-- an SN2 reaction.
Now you have that cation, and you treat it, in this case,
with a very strong base.
Butyl lithium.
So a carbon anion.
What's the pH[correction: pK3] of a
carbon-hydrogen bond, typically?
Just like butane losing a proton?
STUDENT: 50.
PROFESSOR: Order of 50, right?
Very weak acid.
So the butyl lithium is a very strong base, and pulls that
proton off the CH2.
Now in this case, the pKa is about 30 of the CH3 group to
lose a proton.
You're never going to get that off with hydroxide.
But you can do it in this case with a stronger base.
But what is it that...you couldn't do this if, instead
of phosphorus, you had nitrogen.
Now, what's good about forming this anion?
Why is that anion a reasonable one to try to make, where you
don't pull hydrogens off carbons willy-nilly, not off
normal hydrocarbons?
What does the neighboring phosphorus have to do with it?
How does it make it easy to form the anion?
Well, the phosphorus is a cation.
It's easy to form an anion near the cation.
You could do that with nitrogen, "It could have been
N plus, the next
neighbor up in the periodic table, right?
But phosphorus is even better.
Because when you get that anion to be stabilized,
phosophorus, down in the next row of the periodic table, has
vacant d-orbitals that are able to
stabilize that electron.
So it's especially acidic, but not so acidit that you're
going to pull it off with OH minus.
So you could draw a resonance structure where you have a
double bond, where not only the phosphorus has given
electrons to the carbon in making the bond, but the
carbon has given electrons back to the phosphorus
through this d-orbital or interaction.
Anyhow, you have this the so-called ylide, and then it
can act as a nucleophile with a carbonyl, and now the
oxygen can add to the phosphorus, which can form
five bonds, because it has extra valence orbitals, these
d-orbitals.
So you get the four-membered ring.
But that very easily unzips because the PO double bond is
unusually stable.
The NO bond of a trivalent nitrogen that then adds
oxygen, is only seventy kilocalories per mole, but the
phosphorus version is 100 kilocalories per mole, because
of those d-orbitals making it especially stable.
OK.
So you can get that triphenylphosphene oxide and
the alkene, and as you know, what the Wittig reaction
results in is replacing double bond O by double bond CH2, or
in other cases, double bond C something else, depending on
what you started with up here.
And I say directly--
because there are other ways of going roundabout to do
this, rather than to do it directly.
For example, you could add a methyl Grignard to this ketone
to make the alcohol.
Treat that with acid, lose the water to make the primary
cation, and it could then lose a proton.
And that would replace--
here's what our CH2 is coming from, right?
And then we lose the H from the CH3 to get that.
So that replaces the double bond O with double bond CH2.
The same thing the Wittig reaction would do, except
that's not the major product.
It's a minor product.
What's the other product?
The other product is losing a different alpha hydrogen.
One of those.
That's the major product.
Why?
It's a more stable alkene, more highly substituted.
So you can see the advantage of the Wittig reaction, which
directly gives the replacement of double bond O by double
bond CH2, in this case, or double bond CHR,
or double bond CR2.
Now, I was going to say a little bit about green
chemistry in connection with some of these reactions we're
talking about.
So pharmaceuticals are a big chemical industry nowadays.
A hundred years ago, it would still have been dye industry.
And then petroleum, pharmaceuticals, plastics.
But now, pharmaceuticals generate only 0.2% of the
chemical industry's product, measured by mass, but 25% of
its value in money.
Which means that it's worth doing a lot of work on those
to do a lot of reactions in order to get these
pharmaceuticals.
There's a lot of value added.
But in the process of doing all these manipulations, the
pharmaceutical industry generates more than 50% of the
chemical waste.
So the idea of green chemistry is to find new ways of doing
things that won't generate so much waste, or will be safer,
for example.
So there's a journal called Green Chemistry, and in 2000,
they published a paper called "Key Green Chemistry Research
Areas: A Perspective from Pharmaceutical Manufacturers."
So representatives of the biggest pharmaceutical
companies, AstraZeneca, GlaxoSmithKline, Lilly,
Pfizer, Merck, Schering-Plough, which is now
part of Merck, got together to talk about what they saw as
the places where you could make advances to make
chemistry more friendly to the environment and to the people
practicing it.
And obviously from the perspective of the
pharmaceutical industry.
So this is gazing in a crystal ball, and there's nothing as
hard to predict as the future.
But at least these are their ideas, and they're people that
have a great practical interest in it, not only in
the sense of there are things they want to happen, but they're
willing to give money to people to try to
help make it happen,
to do research projects, to try to develop new
ways of doing it.
So they were trying to decide what they would most like to
have. So it involves frequency of use.
If there's a reaction you hardly ever use, you don't
want to spend a lot of trouble on that one.
But if there's something that you use all the time that you
could make better, then that's important.
So frequency of use.
How much volume is involved of the stuff.
If it's only a milligram, who cares?
But if it's tons, that's a big deal.
And of course, safety is another one.
So one of the top priorities is solvents, because that's
what they use the most of.
And something that wouldn't occur to the outsider,
probably, is a very important thing is to clean the-- they
have these big reaction vessels and they have to clean
them so that stuff that's left over from the previous
reaction doesn't get into some other drug.
So they have to be scrupulously cleaned.
And two thirds of the solvent they use is used in cleaning
the reactors to make sure that they're clean.
But there might be other ways of doing it.
For example, use water and an oxidizing agent like hydrogen
peroxide maybe to do it.
So anyhow, they're interested in that as one of the big
steps they could take.
Or replacements for what are called aprotic solvents.
So these are organic solvents.
They'll dissolve organic compounds.
But they also typically dissolve salt.
So you can have salts react with organic compounds in
these solvents.
So ones like NMP there on the left, N-Methylpyrrolidone or
dimethylacetamide, or dimethylformamide are very
heavily used solvents, because you can mix organic and salt
reagents in them.
But the problem is that they're very difficult--
because they're miscible with water,
they're very hard to get out of the water at
the end of the day.
They're worried about polluting water streams. So
they would like to get replacements for that.
Now, you could wish for anything.
You could wish for $2 billion.
I'd like to have $2 billion now;
I'm willing to split it with all of you guys.
That doesn't mean it's going to happen.
And so you can see a real problem with this.
The reason these solvents are good is because they're
organic and they're miscible with water.
But the misciblity with water is what makes them bad,
because it's hard to get out of the water again.
So I don't know what progress has been made on this.
But anyhow, it's something that they would really like.
But then they thought of thirteen processes, that is,
not just solvents but actual reactions that need improving.
And the ones that they already use heavily.
And fourteen, they suggested new processes that there isn't
a good way to do now, but would really be handy if you
could do them.
And then they voted.
So there were six companies represented doing the voting,
and each of them could cast five votes to choose among the
thirteen processes that need improving, and another five
votes among the fourteen processes that they would like
to have available.
So this is the perspective from the people who really do
it about what would be useful to do.
So current processes that need reviewing.
All six of them voted that a way of forming amides,
avoiding poor atom economy reagents.
Now, what that means is you're just trying to put this
nitrogen onto a carboxylic acid to make an amide.
But if you use other reagents that have an enormous number
of atoms in it, then necessarily, you're going to
have an enormous amount of waste generated.
But if you could do it with atom-economy, so you have a
minimum number of atoms going in, in order to--
ideally would be just the element you want to have added
to the molecule, right?
For example, a catalyst might allow you to do that, where a
stoichiometric reagent, where you have to add a whole mole
of it, wouldn't do it.
So then OH activation for nucleophilic substitution.
This is something we've talked about a lot.
You make the OH, you protonate it, you make it into an acid
chloride, you make it into a tosylate
with the leaving group.
But this is something that five of these industries still
say is one of the most important things to develop.
A better way of activating OH for
nucleus[correction: nucleophilic] substitution.
Reduction of amides, the carbonyl group in amides,
without using hydrides.
So they say, lithium aluminum hydride, having a molecular
weight of 38 and four hydrides per molecule, has the highest
hydride density.
So from the terms of atom-economy, it's very good.
But it co-generates an inorganic byproduct which is
difficult to separate from the product, and that makes the
filtration slow, you lose product when you're trying to
filter it, get absorbed in this extra junk
that's being made.
So they need something better than the
available hydride reagents.
Or to be able to do oxidation or epoxidation without using a
chlorinated solvent.
Four of them said that was a top priority.
And they noticed, in connection with the oxidation,
that using the kinds of metals, like chromium or
osmium, that we've talked about, or manganese, these
have virtually been eliminated from pharmaceutical processes,
because small amounts of these metals could be injurious to
the health.
So the FDA requires that there be none in there.
So better is to use oxidizing agents that don't involve such
metals when you're making pharmaceuticals.
OK.
Then safer and more environmentally friendly
Mitsunobu reactions.
Now, we're going to talk about the Mitsunobu reaction next.
But three said that was important, so it's
worth knowing what the
Mitsunobu reaction is, if they think it's so important.
Friedel-Crafts reactions on unactivated systems. We talked
about substituents that were activating or deactivating.
But there are not good ways, they say, of doing their
Friedel-Crafts reactions on unactivated systems. At least
two of them said that.
And two of them said nitrations was a problem,
not because the reaction's hard to do,
but because of the danger of explosion with nitric acid.