talking about oxidizing and reducing reagents, and about alcohols,
and Grignard reactions to make them. And if we have time at the end,
we’ll get to a little bit about green chemistry. I maybe should apologize a
little bit about the last question on the exam. I got carried away. I wanted to have something that
was relevant, at least historically relevant. So I could have chosen a better
NMR problem, probably. But I wanted to show you why
it’s a neat story. So I showed you, during the
exam, this dissertation from the University of Michigan in
1921. And notice it’s about the color
of free radicals. It was found originally that the
solutions were colored. Have any of you ever had a
colored solution in lab? Have any of you ever not had a
colored solution is lab? Right? They’re always colored. So at first, people paid no
attention to it. But after they got good at doing
it, they observed that the solutions were colored sort
of a reddish yellow. But when you swirled the thing,
the color went away. It was colorless. But then it came back. Then you swirled it again, it
went away. Came back. Went away, came back, went away,
came back, several times. And so this was something that
required an explanation of what the color was and how it
went away. So on page eight of this
dissertation, there’s this thing up on top that has sort of
a complicated scheme– or it seemed so at the time, at
least– to explain the color. So the idea is that A– what compound is A in this? Can you see what it is? STUDENT: Hexaphenylethane. PROFESSOR: A, up at the top left
in the scheme. It’s
triphenylmethyl-triphenylmethyl. It’s hexaphenylethane, right? That’s what Gomberg set out to
make, remember? And then what’s to the right of
it, what’s B? It’s two triphenylmethyl
radicals. That’s what he decided he had
made, on the basis of their reactivity. But he’s drawn something else
here. Down below it, he explains the
color by drawing what we would regard as a resonance structure
of the free radical, which puts the radical out on
the paraposition of one of the benzene rings, and draws other
double bonds there. And the theory then was that
when you had that double bond, and two double bonds conjugated
like that, that caused color. So he explained the color by
saying that it was what we would say a resonance form, but
he said in equilibrium, a form that was colored. And the idea was that the
radical was colored, but very reactive. But it was in equilibrium with
the dimer, and the dimer was not colored, like tetraphenylmethane was not
colored. But the monomer reacts quickly
with oxygen. So when you swirl the flask and
let oxygen get into the solution, it immediately
destroys the colored stuff– that in his opinion, the quinoid
form, these double bonds, right, the resonance
structure we would say. But it destroys the free radical,
then exhausts the oxygen. Right? And then if you let it sit a
while, the color comes back, because equilibrium between the
dimer and the monomer reestablishes itself, so the
color gets back. But if you swirl it again, you
get oxygen and it destroys it. You can keep doing that until
you’ve used up all the radical. So that’s how color got into the
act. Now, the NMR I asked you to
explain, and a number of people got it, is…The dimer of
triphenylmethyl was finally measured by NMR in 1968. And here’s a spectrum of it. And you see, it can’t possibly
be hexaphenylethane, because it’s got one group of 25
hydrogens, another of 2, another of 2, and another of
1. Right? So if we look at
hexaphenylethane, of course, it should have six phenyl
groups. Six times five is thirty. There should just be that one
peak. So this obviously isn’t that. What is it? That hexaphenylethane is the
product of alpha-alpha coupling, the two radical
carbons coming together. But notice what could have
happened. It’s obviously not that. But if you had the
triphenylmethyl radical up there at B, and if we use a line
instead of a dot to denote the extra valence that’s
not being used, and over here, draw the resonance
structure, which has its extra valence out at the right, then
you could imagine–those, of course, are just resonance
structures– but you can imagine two of them
coming together like that to form this dimer. So a different dimer. What advantage would it have
over…what disadvantage would it have compared to hexaphenylethane? How about this ring here? We’ve lost the aromatic
stability of the ring. What have we gained? So that’s a way in which this
dimer would be worse than hexaphenylethane. Is there any way in
which it would be better? STUDENT: Less spheric hindrance. PROFESSOR: It’s less spherically
hindered, right? And spheric hindrance counts for
a lot in these things, although that wasn’t much
appreciated, really, until after 1950 or so. I mean, people had talked about
it since the time of Viktor Meyer, but it really
didn’t convince anyone that you should think about another
dimer rather than hexaphenylethane. So anyhow, if you consider this
structure, notice that you have the explanation for two
hydrogens here, the blue one, two hydrogens here, the
green ones, a hydrogen here, the red one, and you have
coupling between these two of about 10 Hz, so it’s a double
doublet, right? And it’s distorted, though,
because this 60 MHz spectrum, the chemical shift
difference, isn’t large compared to the J-coupling. So the inner peaks are large,
and the outer peaks are small. But if you look even more
carefully, you can see that there’s a 3 Hz coupling between
the red hydrogen and the green. There’s a little bit of doublet
splitting here and a little bit of doublet splitting
here. It’s hard to see. And in fact, in some related
samples, you can see a little coupling between this hydrogen
and that one, as well. And these, therefore, are a
little bit split, but not enough that you can see it. Why don’t you see the splitting
in that one? If A splits B, B should split A,
right? Because this hydrogen is being
split into a triplet by these, and into a small triplet by
these, only 3 Hz, and then into even smaller triplets
by these. So it’s just a whole bunch of
peaks on top of one another that just look like a lump out
here. So it’s quite clear what this
structure is, that it’s this quinoid dimer, once you have the
NMR to see it. So it’s an alpha to para
coupling. The central carbon, alpha,
right, with the para position to one of the benzene rings,
which avoids the styric hindrance. Now, how about the color? Remember, the reason Gomberg
thought that it was this, was that the radical was colored,
the stuff that went away rapidly when you swirled it. But the stuff that was in
equilibrium with that was not colored. That’s why the color went away,
but came back again when the equilibrium was
reestablished. So the stuff it was in
equilibrium with was not colored. This, he thought, should be
colored. Right? So he thought it couldn’t be
that dimer. And he was almost right. This is the UV spectrum of that
stuff. So it has quite a strong
absorbance here, at 318 nm. But it peters out just at 395
nm. And as it turns out, the
ultraviolet, the invisible part of the spectrum, ends at
395 nm. If it’d been 5 nm more, he would
have seen the color of that with his eye. So he was close but no cigar on
that one, right? And because of that, for
whatever it was, 46 years, people had the wrong structure
of this important compound. Now in fact, that’s at minus 60
degrees where the equilibrium lies toward the
dimer, away from the monomer. But if you warm to 21 degrees,
now you see the absorption from the monomer. There’s enough of it to see. That’s at 513 nm. And the reason it’s colored is
not because it has this quinoid structure, which was his
theory, but because it has a non-bonding SOMO. So normally you have bonding and
anti-bonding orbitals. pi and pi* are conjugated ones,
right? But this one has a non-bonding
orbital right in the middle, so a much smaller gap, so the
energy is much lower of the quantum that gets absorbed by
the free radical. Now, this was published, that
thesis that appeared here, was published the next year in the
Journal of the American Chemical Society, almost word
for word, about the color of the free radicals. And this is a passage of that. It says, “The relationship
between color and constitution of triphenylmethyl are, briefly
stated, the following. First, that you have a colored
free radical in equilibrium with the dimer”– hexaphenylethane, right?– “and second, the color is
attributed to the quinoid tautomer”– we would now say resonance
structure– “C.” So the color was attributed
to that. But notice what’s interesting
about taking this paper from the dissertation to the printed
form in the Journal of the American Chemical Society,
the published form? What got left out? The correct structure of the
dimer. So it was intentionally left out
because of the color thing. The correct dimer was dropped
for 46 years because they thought it should be colored. And it just almost was. So I think that’s an interesting
application of spectroscopy. OK. Now, third thoughts on the
Friedel-Crafts reaction. Remember, we had thoughts, we
had second thoughts, now we have third thoughts. And the second thoughts, just to
remind you, was that when you do a Friedel-Crafts
reaction, you could imagine Benzene being a nucleophile and
displacing chloride from n-propyl to give n-propyl
benzene, and that would be an SN2 reaction. Or you could imagine if it broke
the bond a little more that the hydride could shift,
and you’d get isopropyl, either with an aluminum chloride
tightly associated with it, or since it’s a
secondary cation, maybe much looser. So that would be like an SN1
reaction that allowed the rearrangement. But then we wondered, what would
happen if we rotated it and had a methide shift instead of a
hydride shift? Then, if we had the
rearrangement, we’d get the same thing back again. It would still give n-propyl. So the question is, is it SN2 or
is it SN1? And then there was this
additional complication that if the methyl got stuck halfway
in a cyclopropyl–protonated cyclopropane structure, that
still would give the n-propyl benzene. And in fact, Dino found in 1968
that you did that, and that it was randomly
distributed, because the hydrogen can zip around the
periphery of the protonated cyclopropane. OK. So then I posed to you a way to
become famous, a problem. How to decide whether the
Friedel-Crafts reaction really is an SN2, where benzene is a
nucleophile and displaces chloride from a thing like this,
either before or after the methide shift, or whether
it’s an SN1 reaction. Did anyone have an idea how to
solve that? Yeah, Matt? STUDENT: Use deuterium. PROFESSOR: We could label to see
where the deuterium ends up in the n-propyl. And what I’ve found since I
posed you this way to become famous was that somebody
actually did that experiment. Right? It was done two years later, in
1970, and it didn’t make them famous. The paper has been cited only
twice in the succeeding 40 years or whatever. And it was done by Lee and
Woodcock in Saskatchewan in 1970. So they prepared the one with
deuterium here, and they did the reaction with aluminum
chloride right at 5 degrees centigrade for 45
minutes. That’s near the freezing point
of benzene, so it’s about as cold as you can get, if you’re
going to use benzene is a solvent. They also did it with another
inert solvent. The idea is that the nucleophile
would be more dilute the benzene that’s doing
the thing. So it would slow down the
displacement reaction if it were an SN2. OK. So what would you expect if it
had been an SN2 reaction without any rearrangement? If it had just been benzene
being a nucleophile. The only thing aluminum chloride
does is help the chloride be a good leaving
group. Then where would the deuterium
end up? Here, here, here; one, two, three. Ashvin, what do you say? You just did the exam. You should be primed on this. STUDENT: Next to the ring. PROFESSOR: It should be here,
right? Because if benzene just comes
and attacks this carbon, and chloride leaves, then the
deuterium should all be there. What if it undergoes a methide
shift? Then where would it be, Ashvin?
The deuterium, remember, started here. And then the methide shifted, if
that’s the mechanism. Yeah? STUDENT: In the middle. PROFESSOR: Then it would be
here, right? And what if it went through a
protonated cyclopropane? STUDENT: Scrambled. PROFESSOR: Then it would be
scrambled. We already know what protonated cyclopropane would be. So now by studying where the
deuterium is, we should be able to figure out what the
mechanism is to give n-propyl benzene with n-propyl chloride. And what they found was that 90%
of the product was dideuterated there, and none of
it was dideuterated elsewhere. So it really is the case, then,
that there’s less than 2% of partial or full methide
shift en route to n-propyl benzene. It’s just the simplest thing you
think, that it forms by a benzene nucleophile doing an SN2
reaction. Now, why isn’t it 100%? Because other cations are able
to abstract hydride from this position to make this cation
here, and then the deuteriums get washed out of the molecule. But anyhow, that’s what they
found. So you’ll have to find out some
other way to become famous as an organic chemist. Now, at the end last time, we
were doing oxidation and reduction as bookkeeping. Not because there’s anything
real about it, but because it can help us to classify
reagents. We’ve been classifying reagents
as high HOMOs, low LUMOs. Weak bonds, for example, if you
want to get a free-radical reaction. But there’s a completely
different way of classifying them, in a sense, a very
artificial way, as to whether they’re oxidizing or reducing
agents. So we went through a whole bunch
of reagents and classified them. As oxidizing reagents, the
elemental halogen, disulfide, chromium trioxide. Reducing agents, methane is a
reducing agent. But it’s not very reactive,
because it doesn’t have good HOMOs or LUMOs. Free radicals are what we need
to get a foothold on that. Or lithium aluminum hydride, H
minus. NaH, potassium metal, which can
give up electrons. And RSH can give up H
minus. And we saw that when you
oxidized back and forth between disulfide and thiol. But HCl and KCl, the elements
are in their normal oxidation states, so they’re neither
oxidizing or reducing. And just at the end, we got to
water and showed that in water, hydrogen is +1 and oxygen
is -2. So it’s neither. And in fact you can see that in
the reaction of water with ethylene, catalyzed by acid. The carbon is -2, -2 for
ethylene. After you’ve added, it’s -1, -3. So one carbon got oxidized, but
the other carbon got reduced. And there’s no net change, so
you don’t need either an oxidizing or a reducing agent,
because there’s no net change in the oxidation step. So water can do that. But water does work in
oxidizing-reducing scenarios. And in particular, a very
important one, photosynthesis, changes- oxidizes water to O2, but at the
same time reduces the hydrogen in water to the
equivalent of H minus. And that’s NAD takes up the H to
become NADH. And remember, we’ve talked about
that being a source of hydride, H minus. So it gets both oxidized and
reduced. So not all reducing agents would
work satisfactorily for any given reaction. But it’s typically futile to try
a reagent from the wrong class. You don’t usually use bromine if
you need to do reduction. Right? Or hydrogen chloride if you need
to do either oxidation or reduction. So it can be a big help in
narrowing the field, in choosing reagents. Now, I wanted to go to the board
a little bit to show oxidation states. This was done in the class last
year by Professor Siegel from Zurich. But I thought it was handy to
do, and I will do it again on the board here. So let me just put this screen
up. OK. So we’ve talked about
arranging– there’s one other set. There we go. Now, he chose to use CH4, CH3OH,
CH2O formic acid and CO2 as examples of oxidation levels
of carbon. Let’s just do it a little
differently, not that it’s better or worse, but let’s just
have an R on there. So we could have RCH3. What’s the oxidation level of
the carbon? Remember how we do the
bookkeeping? If you have another carbon
attached to it, they share the pair of electrons evenly– no contribution to oxidation
state. But if it’s hydrogen, hydrogen
gives its electron to carbon in making the bond, so this
carbon will be +3. STUDENT: -3 PROFESSOR: Oh, -3 pardon me.
Thanks for being on your toes. Then we could have RCH2OH. So the oxygen cancels one of the
carbons, so this will be -1. Notice we’re going in steps of
two here, right? It’s the same when up there, but
I’ve decided to put an R on, instead of one of the Hs,
just for variety. So now RCH=O. Now it’s +1,
because you have two oxygens, one hydrogen. Then we could go RCOOH. A carboxylic acid. Now it’s +3. And at the very end, you can go
to CO2. Then you don’t have the R in
there. And that’s +4, as it says up
there. OK. Now let’s look at different
classes of compounds at their levels of oxidation. So we could have alkanes. So we could have– well, we’ve already got RCH3. That’s -3, so we’ll check that
one. Now, can we have an alkane which
has the oxidation state of carbon -1? It would be a carbon that has
all bonds to carbon, right, except one bond to hydrogen. So the way to do that would be
to have RCH=CHR. So an alkene.[correction: alkane
would be R3CH] And then we could go to alkyne. R, C, C, R. So this would be alkyne. And each of these is -1. Overall it’s -2. And each of these is zero. Now, what this means is that if
you wanted to get an alkene, it has the same
oxidation level as the alcohol. I got that right? Actually, I should have worked
this through better ahead of time. That one works better in his
scheme. But notice that you can… or
another alkene would be RCH=CH2. So this one, you can get from
here just by a reaction we know. What’s the reaction that would
go from the alcohol to the alkene? How do you get there from here? What do you call that reaction? To start with an alkane… Pardon me. This will be R’, because this C
is part of the R here. So this one involves an
elimination reaction. You just treat it with acid. You don’t need an oxidizing or
reducing agent. So the idea is that you can go
down this way with neither, but if you have to go back and
forth this way, you do oxidation or reduction. So you could have alkyl halides. So here you would have RX. And the carbon in the RX would
be -1. The carbon that’s attached to
the X, that is, the X…X-ness of it is -1.[correction: the
X-ness is +1 for carbon] I should make it RCH2, sorry. And how do we get, say, the
bromide from the alcohol? What reagent do we use? PBr3, or the chloride? SOCl2. These are not oxidizing or
reducing agents, right? The elements are in their normal
thing. So we can go down without using
an oxidizing or reducing agent. Now, how could we get from here,
from an alkane, over to here? How would you put bromine in,
say? What reagent would you use? If
you had an alkane and wanted to make an alkyl bromide? What reagent? STUDENT: PBr3 PROFESSOR: No. That’s very good. If you have an alkane, so C-H
bonds, and you want to make C-Br– Anybody remember a reaction to
do that? What’s the problem in
reacting an alkane? STUDENT: They’re not reactive PROFESSOR: They’re not reactive. Why not? No high HOMOs, low LUMOs. No functional group. How do you get at something
that’s not a functional group? How do you get at the hydrogens? You can’t use HOMOs and LUMOs. How would you change methane
into methyl bromide? STUDENT: Br2 plus initiator. PROFESSOR: Br2, Cl2. It’s a free-radical chain
reaction. But notice the things you’re
using. Br2, Cl2 are oxidizing agents. So to go across in this
direction, from one column to the next, you need oxidation. Or you could do another
oxidation, and get RCHX2. And that would require from
here, doing a second free-radical substitution. But if you came down from the
aldehyde to get there, it’s the same oxidation level. So you don’t need something like
bromine to get from here down to here. We could do alcohols, too. And of course, we could go also
to the tribromide, RCX3, say. And that we could get from an
acid, without using an oxidizing or reducing agent. But if we have to get it from
anything over here, from an alcohol or from an aldehyde, we
have to do oxidation to get that, somewhere. And the ultimate one is like
CO2, would be carbon tetrachloride, for example. That’s the completely oxidized
form. Or if we have alkyl
halides, alcohols, here we had an
alcohol. But suppose we made a diol. R, C, (OH)2, and with H in here. RCH(OH)2. If we’ve started from an alcohol
to get that, we have to do an oxidation, choose an
oxidizing agent. But if we could make it from an
aldehyde without using an oxidizing or reducing agent, can
you think of any way of going from an aldehyde to a
carbon that has two alcohols on it? Do you remember what you call a
carbon that has two OHs on it? STUDENT: A gem-diol. PROFESSOR: A gem-diol. Where do you get it? STUDENT: Umm… PROFESSOR: So we need to do an
oxidation, like bromine or something like that? No. We immediately see you don’t,
because they’re in the same column, right? It’s the same oxidation level. So you remember how you make a
diol from a– let’s do that here. R, C, OH, and I want to get R,
C, OH,OH, H. What do I do first?
Derek? STUDENT: Protonate the oxygen. PROFESSOR: Protonate the oxygen. Plus charge here. STUDENT: And then an OH minus
comes and attacks this. PROFESSOR: OH minus. So we’ve added H plus, OH minus. It’s more likely, if you’re
using acid catalysis, that there’s very little OH minus
around, right? Because the acid would have
reacted with it. So what would you normally use
to come in to make the C-O bond down here? STUDENT: Water. PROFESSOR: Water. And then lose a proton. So the proton is just a
catalyst. But notice what we’re using is
just water. And H plus catalysis. Or OH minus catalysis, too. You could have hydroxide add
first, and then this pulls a proton off water. OK. So you go up and down using just
water and acid. Things that aren’t oxidizing or
reducing agents. So you can see how that might
work out, that if you that create on your study sheet a
list of oxidation levels of things, then you could see, when
you’re faced with a given problem, whether–what kind of
reagent you need to choose. And if you can simplify to a
third of the reagents you know, that’s a big help. OK, now. So that’s just the
bookkeeping. There’s also the question of the
mechanism by which you do this. So let’s just consider BR2 as an
oxidizing agent. So if you have an alkane, RCH3,
say, and you react it with BR2 to get an oxidized
form, RCH2Br, what’s the mechanism by which you do it? We just did this. Chris? STUDENT: Free Radical. PROFESSOR: It’s a free-radical
one, right? So we get Br*, taking the
hydrogen, and we’ve done this before. Get the radical, and then free-radical chain reactions. So bromine is an oxidizing
agent, but acting as SOMO. It’s different with alcohols. Suppose we have Br2, and we want
to react it with an alcohol. How does the reaction begin? I mean, it would be possible to
do it by a free-radical reaction, right? If you put in an initiator, and
get change going, it would be possible to do it. But there’s a better reaction to
react Br2 with ethanol. And how does that begin? Amy, what makes Br2 reactive? High HOMO, low LUMO? What’s high HOMO? STUDENT: Unshared pairs. PROFESSOR: Unshared pairs. What’s low LUMO? There are not many
orbitals there. STUDENT: Sigma*. PROFESSOR: Sigma*. OK. So we’ve got sigma* here. And what’s high HOMO, then, of
the alcohol? Amy? STUDENT: The lone pair on the O. PROFESSOR: OK. So lone pair on
the O. Let’s see how these work. STUDENT: It attacks the sigma*
and takes one of the Br’s. PROFESSOR: Right. I want to do this with a red… Actually, that’s no better red
than the other one. OK. So this comes in here, and these
leave. And we’ve seen things like this forever. So we have HOBr. And let me write this with this
unshared pair, just to be cumbersome about it. And bromide came away, and
there’s a plus charge here. How do we get rid of the plus
charge? Amy, you’re doing this here. STUDENT: Will the H leave? PROFESSOR: Yeah. The proton will go away. So now we’ve got O, Br– I was going to make that– CH2CH3. And here we’ve got this, that
pair. And now can you see anything
that this might do? What reactions have we seen when
we have atoms next to one another, bromine on one,
hydrogen on the other? STUDENT: Undergo E2. PROFESSOR: Pardon me? STUDENT: Undergo E2. PROFESSOR: An E2 reaction. So we have some base. It doesn’t have to be a strong
base. It could just be the alcohol. Or it could be the bromide that
we lost here. So this can take the H, this
goes in, and these go on the bromine. So now we have double bond O.
This H is still there. CH3. And bromide, which has that pair
of electrons. So we’ve oxidized the alcohol to
the aldehyde. We’ve gone from here to here. And notice what the trick was. The first oxidations and
reductions we talked about were electron transfer, an
actual electron transfer, like from a metal to a sigma* bond of
an alkyl halide, right? Here we’re ferring electrons,
but in pairs. The electrons start belonging to
the oxygen. Then they’re shared between the
oxygen and the bromine, and then they leave with the
bromine. So the bromine came in and
shared a pair of electrons, then went away, carrying them
away. So it removed this pair of
electrons. That’s how the oxidation
occurred. So these reactions are just…
This is an SN2 reaction and this is an E2 reaction. So they’re just normal
mechanisms, but because of the way we look at it, it’s an
oxidation-reduction reaction. And now CrO3, Cr6. It reminds you a little bit of a
ketone. Right? this Cr=O. So you can imagine adding water
to give Cr, O, O, OH, OH. What’s the oxidation state? If the chromium is oxidation
state here, what’s its oxidation state here? STUDENT: +6 PROFESSOR: +6. What’s its oxidation state here? STUDENT: Still +6 PROFESSOR: How do you know? You could add up all the bonds. But if you used water, it’s
going to be the same oxidation state, because water’s not an
oxidizing agent. So it’s still +6. Or you could imagine having two
of these– so these are a little bit like alcohols. You could imagine to make an
ether from that. So you could have CrOCr, right,
with OOHOH– wait a second, O here– O, O, OH. So this is chromium trioxide,
chromic acid, dichromate, and so on. So there are lots of different
forms. Or you can get, if you reacted it with HCl, you could
protonate, lose water, add the chlorine instead. See, you could get Cr, O, O, O,
Cl, OH. But all of these are just
different disguises of the same thing, Cr6. Now let’s consider the reaction
of any of these with an alcohol. Like H, O, CH2, CH3. So we can have the unshared
pair– and I’ll do it in red again– attack the chromium. So we’ve got H, O, Cr=O, double
bond O, O minus, or OH. Let’s do the OH. And now notice that we’re in the
same business we were with bromine. We have now– let me write this out so that it
looks like the one over there. H, H. So now we could bring up some
base, take this H, this goes in here. And the chromium leaves with
that pair of electrons. So again, just as before, we’ll
have lost the proton. It looks just like what happened
over there. The chromium came in, shared the
pair of electrons, and then left with them in the
elimination reaction. Right? So again, it’s an oxidation. And if we could look at the
chromium afterwards. It’s Cr, O, O, O, OH. And what’s– am I right here? Minus. So it’s got one, two, three,
four, five that it’s giving to oxygen, but it gets an electron
back with the minus. So this is +4. So chromium goes from +6 to +4
in the process of doing this oxidation. Now, there’s a problem in doing this to make an aldehyde. And the problem is, when you’re
working in water, if the product you’re going to get
here from this elimination– you’ve lost this proton, you
lose this H, break this bond, so you have C=O. Now, if that’s what you want to
make, there’s a problem. Because aldehydes, you know,
have an equilibrium constant about one with water. So this is going to add water to
make a diol. And why is that a problem? Because alcohols react with
chromium trioxide to get oxidized. So you could lose this. So this oxygen can give up its
pair of electrons, get oxidized. You lose this H and this H. And
then you would have OH, C=O. So you overoxidize all the
way to the carboxylic acid. We go on another step up there. Now, how can you avoid that
problem? Let me get this back down. You could do it by making this,
the reagent shown here. Pyridinium chlorochromate. So notice, you added HCl to
CrO3, which made chlorochromic acid. And then the proton is given to
pyridine to make a pyridinium cation. Now, that is organic enough that
it’s readily soluble in organic solvents, which the
other forms of chromium that we talked about are not. Now, can you see what advantage
that might provide, to do it in an organic solvent
rather than in water? So the reagent is called
pyridinium chlorochromate. And here’s an example where you
get a 92% yield of the aldehyde from a straight chain
alcohol using pyridinium chlorochromate. The crucial thing is the solvent
is methylene chloride, CH2, Cl2. Right? No water, no gem-diol, and
therefore, no overoxidation. Because the mechanism requires
the alcohol to react with. I took that example from
Lauden’s textbook. However, so textbooks often– I have no doubt that that
happened, and that they got 92% yield. But as a general reaction, there
can be a problem. If you look at what Wikipedia
says about this, there seems to be the voice of experience
writing. It says, “In practice, the
chromium byproduct deposits with pyridine as a sticky black
tar, which can complicate matters.” You’ve gotten a rose-colored
picture in your lab. I mean, you’ve often had
frustrations, I know. But those reactions are
carefully chosen so you don’t get sticky black tars. But you find very often, when
you try to do something in organic chemistry, you get
something that’s very difficult to filter or deal
with. And that can happen in this
reaction. But anyhow, pyridinium
chlorochromate is what you use on an exam when it’s asked, How
are you going to oxidize an alcohol to an aldehyde
without getting a carboxylic acid? Now, you can have– Oh, sorry. We’re going to have to carry on
with this next time. Thanks.