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J. MICHAEL MDBRIDE: So, we'll finish up on
NMR spectroscopy, talking
about more dynamic things involving decoupling.
Then a little bit about C-13 NMR and double labeling
And then we'll get onto another reaction,
electrophilic aromatic substitution.
OK, so proton decoupling.
Remember that the proton is four times as strong a magnet
as the Carbon-13, so it precesses much faster.
So it's easy to do a pulse that will tilt the Carbon-13
net magnetization down into the plane, so that it'll
precess and be an antenna that can tell us its local field
without doing the same thing to the proton.
And if we do that, then we see the C-13 NMR spectrum.
And there are two peaks, because of
the adjacent proton.
Right? it can be either up or down, 50:50
or almost exactly 50:50
so you see two peaks.
Now suppose at the same time you're doing that, when you do
this pulse and listen to the C-13 to see what its local
field is, you irradiate at 100 MHz.
That won't do anything to the C-13,
it's the wrong frequency.
But it'll now make the proton precess.
That is, if we look in the rotating frame, we can see a
weak field that's horizontal, about which the
magnetism will precess.
So the protons will go down, and then up, and then down,
and then up.
And what determines how fast it goes up and down?
How strong this field is.
The stronger it is, the faster it will precess.
And if it's fast enough, then you won't see a doublet for
the C-13 anymore, because you'll see the average.
So you'll see a single peak, you'll proton-decouple the
C-13 spectrum and see the C-13 as if the
protons weren't there.
OK, so it depends on how much power you put in.
Now here's a C-13 spectrum of a given compound.
Where we're listening to the C-13 and we see all these
different local fields, right?
But we're irradiating the proton at the same time, but
very, very weakly.
This number up there, 40 decibels, is an inverse ratio
of how strong that horizontal field is for the protons.
So it's 10-4th weaker than what you'll see you
later, 10,000 times weaker.
And if we look at the carbon, we see that some of the carbon
signals are split and some are not split.
The ones that are split are the ones that have
hydrogens on them.
So if we start from the left, we see there's one with no
hydrogens, then one with one hydrogen,
one with one hydrogen.
That red one is the solvent
So we're looking at the carbon of the CDCl3.
But it has a deuterium on it, not a
hydrogen, so it's not split.
Then there's a carbon with two hydrogens, a carbon with three
hydrogens, it's a quartet.
Underneath, a carbon with two hydrogens, it's a triplet.
So very low power in the proton, that's what we see.
Now we're going to reduce the size of that.
And see what happens, is we start increasing the power.
So there's 40 dB, 20 dB,
and now things are beginning to change.
See how the patterns are changing their multiplicity
and their intensity?
Because it turns out that when you irradiate the protons by a
complicated mechanism it strengthens the signal of the
carbons to which they're coupled.
And that's called nuclear Overhauser enhancement, as
OK. So now you see the splitting is going away when
we get down to two decibels or one decibel.
Now we see that each carbon gives just a single peak, we've
decoupled the protons.
So now we have a peak for every carbon.
And notice, incidentally, that the CDCl3 didn't get stronger
as we did this.
Because what strengthened the signal was irradiating the
protons and CDCl3 doesn't have hydrogens on it.
So it doesn't get strengthened.
OK, notice also that the carbon without a hydrogen on
it, on the far left here, didn't get strengthened as
much as the others did.
OK, so that's decoupling, and it can be useful.
And it can be useful in several ways.
There's two things, you get rid of the spin-spin splitting
so you see just a peak for each carbon.
And also you strengthen the signals when
they have nearby protons.
And that could be bad if you're interested in measuring
the size of the peaks to get how many carbons there are,
because you can't count properly anymore because you
get different intensities with different
neighbors of protons.
But it's good if you want to determine structure and see
what hydrogens are how far away, and I'll
show you that shortly.
OK, now bear in mind that carbon-13 is only 1% in
natural abundance of carbon.
Of course, you can synthesize the molecule with expensive
carbon-13-containing precursors and put carbon-13
where are you want it in a molecule, and then the signal
from that carbon will be ever so much stronger.
If you put in pure carbon C-13 in some position, the signal
would be 100 times stronger.
In fact, 200 times stronger than without.
No, pardon me.
It would be 100 times stronger, because it would be
100% instead of 1%.
OK, so here's a compound, neotame, it's
an artificial sweetener.
And it's the proton-decoupled C-13 spectrum.
So you see one peak for each carbon.
There are 20 carbons, there are 20 peaks.
But you can't go by their intensities.
Now, let's see what we can make of this.
It's proton-decoupled, so there's one peak per carbon
and, being carbon, it's pretty well spread out.
Remember, the range of hydrogen chemical shifts is
only about 10 parts per million.
The range for carbon is hundreds of
parts per million here.
200 parts per million are shown.
Now why is there no C-13-C-13 splitting?
There's no proton splitting because it's decoupled.
But why no C-13-C-13 splitting?
STUDENT: The likelihood of two C-13s adjacent in the same
molecule is unlikely.
PROFESSOR: If it's only 1% natural abundance, and you're
a C-13 so that you're giving a signal, the chance that your
neighbor is C-13 is only 1%, so there would be tiny, tiny
peaks from those
you don't see them.
OK, so that's what Chris just said.
OK now, let's see if we can figure out which carbons these
are, that we're looking at.
And of course, this is how it was done empirically.
People put in known compounds and figured out what the
chemical shifts... what peaks
were from one compound to the next.
And then they figured out this comes in this range, this
comes in this range, this comes in this range.
So let us see, knowing what these peaks are.
OK first, there are these ones that are way
down close to 200.
They're numbers 1, 13, and 4.
So what do 1, 13, and 4 have in common that turns out to
make them way far downfield?
Ellen, what do you say?
What do 1, 13, and 4 have in common that makes
them different from the other carbons?
STUDENT: They're almost like esters.
Like, double bonding to an O.
PROFESSOR: Yeah, the double bond to an O. Carbonyl carbons
are way downfield.
OK, now we've got 7 and 8 to 12.
So there's 7 and there's 8, 9, 10, 11, 12.
Pretty clear what those are, right?
Those are the ones on a benzene
ring, aromatic carbons.
OK, then there are ones that are a little bit further down
than the others, and those are those blue ones.
What do they have in common?
Rahul, what do you say they have in common?
PROFESSOR: They're not all tertiary, no. In fact, only
two of them have three things associated with them, the
others have one and two.
Lauren, can you see what the blue ones have in common?
In hydrogen, what makes things come a
little bit further downfield?
STUDENT: Well, the hydrogen would be like deshielding from
an oxygen or something, like nitrogen or oxygen.
PROFESSOR: They all have an electronegative atom next to
Nitrogen or oxygen.
So those are ones. A carbon that's attached to something
And then, finally, you have the others, which are attached
only to carbons.
OK, so you could begin to figure out these, and you can
get tables of what comes where.
And then you can start dealing with unknown compounds and
figure out what it's telling you.
But a nice thing about this proton-decoupled spectrum is
that there's just one peak for each carbon, and they're well
So that it's good to do this kind of thing.
So those are the alkane carbons.
OK, now the power of correlation.
Often when you can measure two things on the same subject
simultaneously, you get a lot more information than
measuring one, or measuring the other, or measuring them
I'll show you that example with C-13 double labeling and
also in two-dimensional.
As it's called, nuclear magnetic resonance.
So first, C-13 double labeling.
And we'll talk about lanosterol biogenesis.
In fact, we already talked about it, if you look back at
OK, so remember we got squalene from isopentenyl
Then it curled around and we had this complicated thing
where it zipped up going one way, and then we had all these
So as we said at that time, this is the source of
cholesterol and the steroid hormones.
And it's a very cute story, how all that rearrangement
OK. But there's a question whether it's true.
And I said at that time, wait for NMR. So now we're at NMR,
so you can get the answer.
OK, so you can use Carbon-13-labeling to figure
So if you feed the plant that makes this stuff Carbon-13
labeled isopentenyl pyrophosphate--
in that particular carbon, that substituted carbon--
then you can see where that would appear in the squalene.
According to the way we linked them together before, it
should be those carbons.
Then when you curl it around and do all this stuff to make
lanosterol, those peaks should be labeled with C-13 and
therefore 100 times stronger than the other
peaks in that spectrum.
And indeed they are.
So that supports it.
Or you could do it a different way.
You could label the methyl group.
And if you label the methyl group, it turns out that those
methyl groups should be labeled in squalene.
And then if you curl it around, it should be those
And indeed, they're 100 times stronger
than the other signals.
But much more interesting than this is to do double labeling.
So if you label both, you get that.
That's just the same information
we got before right?
Except done in one experiment rather than two.
And, in a sense this is not as valuable as these
two independent label experiments, because you don't
know but what two of those carbons might, in principle,
have changed places.
But the double labeling is much more important
if you do it dilute.
Now what do I mean dilute?
I've been talking as if we put 100% C-13 in here.
But suppose we don't put nearly 100%, suppose we only
put 10% in.
But suppose that those 10% that are
labeled are double labeled.
So we prepare a sample that's 100% labeled in two positions,
but then we mix it with 90% percent of unlabeled material
and feed it to the organism and get the stuff out.
OK, so what we're going to do is a dilute double label.
Now why do it dilute?
Because then, when label goes in, two labels go in.
But it's unlikely that any given squalene will have two
different positions double labeled, or two sets of
positions double labeled.
Because only 10% of the stuff is double labeled.
So when a squalene is double labeled, it'll be labeled in
just one pair of positions, a given molecule.
So those two come labeled together.
The signal is 10 times stronger because
remember, it was 10%.
It's 10 times stronger than it was for the
ones that aren't labeled.
But what's special about it?
Can anybody see what's going to be special about the signal
we get from these molecules that look like that?
Ayesha, have you got an idea?
STUDENT: They're going to be the same more or less
because they're just an extra--
PROFESSOR: They're going to be the same what?
STUDENT: They're going to be the same more or less
because they're just extra carbons and
there's no special function.
PROFESSOR: I didn't quite understand you, but I was
listening for special words and I didn't hear them.
Sebastian, have you got a special word for us?
STUDENT: You get splitting?
PROFESSOR: Ah, they'll split each other, because now the
C-13s are next to one another.
Because they're not coming in at random 10%, they're coming
in as a pair.
So now when we look at the signal from that, it's going
to be a double doublet.
The blue and the red one are going to split each other.
Now other molecules in the same sample will turn out to
have been labeled elsewhere.
For example, so that proves that those entered as a unit.
Others will have come in there, and those will also be
a double doublet, show that they came in as a unit.
And if we go over to the other end of the molecule, those
will be a double doublet.
And those singles will be a double doublet.
What's going to be different?
Suppose we had that molecule where we had a methide shift.
So the carbon got away from its original neighbor.
Now it won't be next to a C-13 and it'll be just a pair of
singlets instead of a double doublet in a
And the same thing will be true there.
So this double labeling absolutely proves that these
rearrangements took place.
So those two, incidentally, are
labeled in the same molecule.
But they're not adjacent to one another anymore, so they
don't show splitting.
And those are both labeled, the ones that are adjacent to
one another, but not in the same molecule.
So they don't do splitting either.
So this dilute double label experiment enhances the same
12 peaks as the single label experiments did, the two
single label experiments.
But only eight of them show spin-spin splitting because
their bonds stay intact.
So this strongly confirms the story I told you before about
the rearrangement scheme.
OK, so this is the power of correlation.
And let's look at a different kind of power of correlation
in what's called two-dimensional NMR spectroscopy.
Now here's a two-dimensional NMR spectrum.
And the way it's collected is rather technical and we don't
have time to go into it.
But it shows that there are two different frequencies
And if you go along the diagonal of this plot, those
two frequencies are the same.
So what you see there, when you have just a certain
frequency going in, is the normal spectrum you see along
But this is the spectrum of a protein.
So it's got a whole bunch of amino acids linked
together like that.
And we're in the range of the spectrum between 6 and 9,
where the N-H protons show up In this
particular kind of molecule.
So there are a whole bunch of them in this long chain and
they all overlap with one another, and it's very hard to
tell what's going on.
But if you don't have these two frequencies you're using
the same, you're looking at one while you're irradiating
And that affects the intensity of the one you're looking at.
That's this nuclear Overhauser enhancement that we talked
about, where you irradiate one, other things that it's
coupled to will change their intensity.
So what we're going to do is plot on the off-diagonal how
one frequency is influenced by having other frequencies in
there, in its intensity.
So that's called NOE, or nuclear Overhauser
And it comes, incidentally, by through-space magnetic
Not the stuff like we talked about coupling where it goes
This works through space.
So it works if the two protons you're looking at-- the one
you're looking at and the one you're tickling to see if it
influences this one--
if they're within about 6A of one another.
So as I said, this is mostly these N-H protons.
And for one thing, one advantage of having these off-
diagonal peaks is they're much less congested than what's
along the diagonal than the normal spectrum.
So for example, we're looking here at the NH proton that's
at 7.25 parts per million.
But you can see that it interacts with
several other protons.
It's within 6A of signals that are at
8.9, 8.3, 8.25, 7.7.
Now you don't measure the exact distance, all you know
is that it's close.
But this allows you to make a three-
dimensional map of the protein.
And the way you do it, is first you have to know which
NH goes with which R group.
So you see that that NH--
the one you're looking at, the red one--
is close to that blue one.
So it'll influence it.
But the blue one is then close to Hs that are in the R group,
so you can tell which R group that that blue H is nearby,
and which one the red H is near.
So the red is near the blue is near the R. So you know which
NH is near which R. And now the way you make a map, then,
in this way--
Professor Loria, who does this kind of
thing, uses this analogy--
it's like making a map of a town if all you have is a
telephone and a telephone book.
Sometimes there's a front page that has a map of the town,
then that's easy.
But if you don't have the map of the town, then what you do
is you just call someone at random.
Say, Hi, how are you today?
Blah blah blah.
Could you tell me who three of your near neighbors are?
And they tell you three of their near neighbors.
Then you call those people, and you say Hi, can you tell
me who three of your near neighbors are?
And they tell you.
And then you call the next one, Who are three of your
So once you know who is near what, then you can see that
there will be a map that will allow you to figure
out what's near what.
Especially if you have some help, because you know in the
case of the protons, they have to fit with normal bond
distances, normal bond angles, and so on.
So if you do molecular mechanics together with this
knowing what's near what, you can get a very good three-
dimensional structure of the protein.
And the nice thing about this, from the point of view of
people who are actually working with proteins-- what's
the best way of getting the three-dimensional structure of
something like a protein or anything?
X-ray, but you need a crystal.
For this, you don't need a crystal.
So you can find the structure that way.
So with molecular mechanics constraint, you get the 3D
structure without a crystal.
Now here's a different kind of 2D correlation with NMR. It's
correlation not in space, but in time.
So this is the spectrum of a cation, that cation that's
shown at the right.
Notice it's benzene with methyl groups all around it,
with one extra methyl group and a positive charge.
Now in fact, you can get this in solution as long as you
don't have anything that reacts with cations in there,
that reacts with low LUMO, so you need very, very non-basic
solvents to do this.
So again, the diagonal running a different way this time--
the increase in parts per million is down to the right,
rather than what we usually see, increasing to the left.
But anyhow, it's color coded, according to these peaks that
show what they are in the molecule.
So the scale is slanted and backwards.
So you see that if you look at the resonance structures--
that one, that one, that one--
you can see that the blue and the green are near where
there's positive charge in the resonance structure.
So they're far downfield, very deshielded
compared to the others.
Now this can rearrange.
A methide can shift, so one of the red methyls on the top
shifted to the right.
So now you've changed which methyl groups
are in which locations.
So now the red one at the top, this one and this one are now
C, and will be shifted way downfield.
This one will be shifted downfield--
what was blue before in the D
position, and so on.
Then this blue one, which was downfield before is now an A,
it's up here.
So we've changed the locations, the local
frequencies, the local fields.
So that's a methide shift.
But is it a 1,2 methide shift?
Because it's possible that the methyl, instead of going just
to its neighbor, would move to the center of the ring, up
above the center of the ring and then it
could come down anywhere.
Wouldn't have to come down adjacent to where it started.
So you could have a 1,anywhere shift.
Now you can distinguish that and measure the rates with
this NMR spectrum, this two- dimensional NMR spectrum.
So again, we're seeing when one
frequency influences another.
Now if a given set of protons moves from one position to
another, then having been irradiating one influences
what happens where it ended up.
Because those same protons are now here.
So here you see off-diagonal peaks where
one has become another.
So we can see here, for example, that the proton that
was in position C becomes A. And that's what happened in
the blue, here, when we had that 1,2 rearrangement.
We can also see that D becomes B, and that you can see
And you can see that B becomes C. But you notice there's not
A becomes B, A becomes D, or C becomes D. So you don't have
peaks for some of these patterns that
you could have done.
Now if you go through a whole sequence of the these, then
you can start any place and end any place else.
But it takes a longer time.
So depending on the time scale of this thing, you can tell
which things happen quickly.
And what happens quickly is a 1,2 shift, not
a 1 anywhere shift.
So again, that's information on mechanism.
It comes from two-dimensional NMR spectroscopy.
And it's proton versus proton correlation in time.
That's, incidentally, just another visualization of the
same map I showed before, that shows the peaks on the
diagonals and the off- diagonal peaks.
OK, now we've finished with NMR. We'll refer to it again,
but we've finished discussing it.
And we'll go on to the next important reaction, which is
electrophilic aromatic substitution.
We talked about electrophilic things before.
Remember, electro philic addition to an alkene.
Now you might think benzene is an alkene, so you can get
electro philic addition to the double bonds of benzene, but
you don't get addition.
You get substitution.
So here's an example of substituting
deuterium for hydrogen.
You put benzene and concentrated sulfuric acid
that has deuterium in it, and you get deuterium first in one
position then, of course, in other
positions in the benzene.
So you can go all the way to C6D6, that's how you prepare
deuterated benzene if you want to use it for a solvent for
something, for example.
You can use other electrophiles as well, not
just D+, you can use NO2+, Br+, sulphur+-, R+
All these things can substitute for hydrogen, and
we're going to talk about those.
Now, the mechanism is that the deuterium, D+, the
deuteron, adds to the ring.
But that, notice, changes the aromatic ring into just a
regular pentadienyl system.
So you lose a lot of resonance stabilization.
It happens, but it's not very stable and it'll easily come
off again if there's anything that can take a proton.
But if you have a very, very non-nucleophilic, non-basic
solvent, nothing to take it off, then it can stay on and
in fact it's observable.
You can get spectra of it.
And so in electrophilic addition to alkenes, the next thing that
happens is a nucleophile would add, and you'd
lose the double bond.
But here it's easier to lose the deuteron or, more
interestingly, the proton.
Losing the deuteron would take you back to
the starting material.
But losing the proton, the H, then takes you to the
And that's easier than adding a nucleophile, because you get
back to the aromatic stabilization, 30
And remember, we just saw a spectrum
that looked like that.
When you added not H, but CH3+, in this case, to the
So it's possible to see these things in NMR. So there's the
benzene, the D+ comes on, we get that.
Notice that it converts the aromatic ring into a chain,
destroying the aromaticity.
Let's look at the orbitals that are involved, just using
this simple Huckel program.
So benzene, we've looked at the orbitals of that on the
left, and pentadienyl we looked at before, too
and we see it on the right.
Now, the most interesting orbital in pentadienyl is the
one that's indicated in yellow over there.
The highest of the low three orbitals, and it's at 0.
And you know it's at 0, the same energy as a P orbital.
How do you know the energy of that molecular orbital,
consisting of three P orbitals is the same energy
as a normal P orbital?
Remember, if you make a double bond and put the two next to
one another, they overlap, one goes down, one goes up, it's
not to same energy.
But this one has the same energy as
an isolated P orbital.
How do you know that?
Notice how many nodes are there in the lowest one, the
one that's about minus two?
How many nodes for the pi orbitals?
One, two, three, four, five P orbitals connected in a chain.
How many nodes in the lowest combination?
PROFESSOR: Zero, except for the node of the p
orbitals, of course.
Then the next one, the one that's at about -1?
How about the ours?
PROFESSOR: OK, and so there are two nodes.
Red, white, red on the top.
Now, how do we know the energy of that?
How much overlap is there?
PROFESSOR: None, they're not next to one another.
So it's the same as an isolated p orbital at the
OK, so that one is very interesting, that SOMO.
It's non-bonding in the case where we have a single
electron in there, which is what's shown.
So that's the locus of that odd
electron in the free radical.
If we have 5 pi electrons, the fifth, the one that's in a
pair, is in those three positions.
Now suppose we're interested in the cation with five P
orbitals in a row.
So what we're going to do is just take that electron out.
And where we take it out is now going to
be a positive charge.
So that's the same as if we put another electron in, in
the anion, that would be the locus of the negative charge.
And in the cation, the locus of the positive charge.
And that's the same thing we'd get by drawing resonance
structures that show those same positions having charges
or free radicals in them.
So those are the positions of positive charge.
And we get the same thing drawing resonance pictures.
Now consider the influence on rate of substituting not
benzene, but a benzene that already has
some X group on it.
Now, a very common thing to have studied with this was
nitration, because it's such an easy reaction to do.
It's a very vigorous reagent, NO2+.
You make it by just having a mixture of nitric
and sulfuric acid.
You see how you get NO2+ from that mixture?
It's quite simple.
You get a proton from sulfuric acid, and it protonates the
OH, and then you lose water.
So you've got NO2+.
So it could be the thing that comes in and then a proton
comes off from the same position, and you've got the
But now, when you have X in there, you can get three
Ortho, meta, and para, they're called.
Notice, incidentally, that NO2+ is an interesting
molecule because it's exactly the same as CO2, except it
has one more proton in the nucleus of the central atom.
Nitrogen instead of carbon.
So CO2 has a pair of electrons in that double bond, a pair of
electrons on the oxygen that are in the y orbitals,
pointing up and down.
It's also got x orbitals coming in and out that have
the same kind of thing.
But if we consider just the y orbitals, that'll be the LUMO.
There are two pairs of electrons, so the one with no
nodes has two electrons, the one with one
node has two electrons.
This is the one with two nodes, no electronics.
So that's the vacant orbital, the LUMO of NO2+
just like CO2.
So NO2+- attacks, and we can measure the rate.
Now what we're going to do to measure the rate--we could do
careful work with a stopwatch and so on, but there's a
clever way to do it.
Which is to compare two molecules by doing them both.
Put both in there, and do the reaction for just a little
while and then kill it.
If you add base, you won't have the strong acid and the
reaction will stop.
And now see how much of each one becomes product.
So we get relative rates that way.
So we're doing relative rates compared to hydrogen being 1,
that is, benzene when X is hydrogen.
If X is methyl, it's 25 times faster.
That is, if you have 25 times as much benzene as toluene--
the one with methyl on it-- you'll get equal amounts of
the products at first, until you've
used up all the toluene.
So it's 25 times faster.
So methyl helps the reaction.
OH really helps it, if you use phenol it's 1,000 times faster
at getting the sum of all the products.
Now why are those so fast?
Because electron donation eases the formation of the
carbon cation intermediates.
Or, looked at a different way, the carbon cations are stable
because there are other electrons that become
stabilized when you get a cation there.
For example, the unshared pair on oxygen.
OK, now that's because we have positive
charge at those locations.
When X is there, that helps out.
Now if you put NO2 as X, so it's already there, then it's
6 times 10-(-8), it's 10- (-7), 10
million times slower.
And the same if you have trimethyl ammonium with its
You don't want to make those cations if there's another
cation there, or if there's an NO2 group there.
So those are electron withdrawing, and they retard
the formation of cation intermediates.
The most interesting, as many times, is the halogen.
The halogen is slower than benzene, but not much slower.
Only 30 times slower.
And that's because both things are happening with halogen.
It's withdrawing electrons in the simga bond, but it has an
unshared pair that donates back, so
that it mostly cancels.
Why is NO2 electron withdrawing when OH is
O is more electron negative than nitrogen, so you might
think that OH should be more electron withdrawing.
An unshared pair on nitrogen is higher energy than the
unshared pair on oxygen
should be more electron donating.
So this seems, at first glance, to be sort of nuts,
that OH is 1,000 times faster and NO2 is 10
million times slower.
But it's pretty easy to understand when you look at
the orbitals that are involved.
So in that case of oxygen, there's this unshared pair on
the oxygen, which can overlap with the p orbitals on the
It turns out, indeed, that the energy of the oxygen unshared
pair is about the same as pi electrons of a
So that the unshared pair on oxygen has about that energy,
about -1 on the scale we're talking about here.
OK, so that's unusually high, as pi are, unusually high.
So when you have a vacant orbital next door, those
electrons get stabilized.
So it's a high HOMO, a good overlap with phenyl, and
therefore OH can donate its unshared pair.
OK, that's good.
How about NO2?
Notice that NO2 is allylic.
It's got three p orbitals on each
oxygen and on the nitrogen.
So it's going to have a low occupied molecular orbital
that gives good overlap with phenyl.
So that one is able to donate electrons because it has good
overlap, but it's not willing to share its electrons.
Its electrons are very low in energy.
They're not going to be much stabilized by having a vacant
orbital next door.
So you're not going to get anything out of that,
from the low one.
The next orbital of the NO2 allylic system has a node in
That one has the energy shown in yellow there, it's going to
be very good.
It's got a good, high HOMO.
As high as the p orbital on oxygen, so it should be good
at donating electrons, at having its electrons
stabilized by the vacant orbital in the benzene ring.
But it's not, why not?
Can you see?
Why doesn't this orbital be stabilized by mixing with
orbitals that are on the benzene ring?
Because there's no overlap.
It doesn't have any size where it's adjacent
to the benzene ring.
So it has no overlap, so it's willing to give its electrons,
but it's not able because there's no overlap.
And finally, you have the third of the allylic
orbitals, that one.
Which is a low LUMO.
And it has good overlap with phenyl.
So it withdraws electrons, it's willing to take electrons
because it's unusually low, and it has overlap so it's
able to accept electrons.
So that's why nitro is electron withdrawing when OH
is electron donating.
it's because of that node in the middle that makes it
unable to give electrons to the benzene ring.
So NO2 is a pi accepter, whereas OH is a pi donor.
OK now, we just looked at the rates of these reactions, of
forming all three products together.
But we might be interested in the individual products,
ortho, meta, and para.
So let's look at that.
In the case of hydrogen, we'll call each of these 1.
And now we'll do mixtures of hydrogen with others--
but now we'll analyze the ones that already have an X in
them-- we'll analyze whether it's orth, meta, or para, and
how much of each one.
Now if it's methyl, you see that two of them are very fast
but one of them is not fast, compared to what it was when
only hydrogen is on the ring.
How do you understand the two that are fast?
Why are those fast when X is methyl?
Linda, do you have an idea?
When X up here is methyl, this is good and this is good.
39 and 46.
But this one is not any better.
Methyl wants to be where the cation is.
So it's especially good here, and especially good here.
But not here.
There are only hydrogens here, here, and here.
And it has the same rate as hydrogen.
OK, that's fine.
If you look at t-butyl, now again it's fast over here.
But it's not fast here.
So why don't you get this product when you have the R
group here, when it's t-butyl?
You get it when you have methyl, but
not when it's t-butyl.
Nathan, you got an idea?
STUDENT: Steric hindrance.
PROFESSOR: It's steric hindrance, it's too big.
It's hard to put this extra group here when you have
something big there already.
OK, so that's good.
Now there are two things about a substituent, X. Does it make
it go faster, or does it make it go slower, is it activating
So it's activating, these are faster than benzene.
And how is it directing, what does it direct?
Its ortho and para, for the reason we saw.
Except that in the t-butyl case, it's not very much ortho
because of steric hindrance, as Nathan told us.
OK, so those are electron donating, and like to be
ortho/para, other things being equal.
Now here's one that's the same size as that neopentyl...
or t-butyl group,
the carbon with three methyls on it.
But notice it's very much slower.
And we saw that before.
But it's selectively slower, it's slowed much more here--
0.6 times 10-(-8) than it is here...
3 times 10-(-8).
So it's slow to get product, but when you get product this
is the one you get.
You don't get ortho and para.
Especially you don't get ortho because
it's sterically hindered.
But you see, that's the same thing.
If you want to form these things when X has a plus
charge, don't put it here and don't put it here.
If you're going to do it, put it there.
Make it as far away as possible.
So that's meta.
And the same thing is true of nitro.
It's slowed down everywhere.
But it's not nearly as slowed down meta as it is here-- it's
slowed 10 times more here for ortho and whatever--
300 times, right?
Yeah, 300 times slower when it's para.
And if you have an ester group, the same
kind of thing is true.
Here it's 10 times slower here than here, and twice slower
here than there.
So those are deactivating, for the reasons we talked about
before, and they're meta directing.
But again, halogen, is especially interesting.
Notice it's slowed down, it's deactivating, but it's
So it shows both these characters, deactivating and
It's deactivating because of the electron
withdrawal, which is sigma.
But if you're going to make it, at least make it where the
halogen's unshared pair can be stabilized by the vacant
orbital next door.
Now, how can you make the reaction work better, the
You can make the electrophile better.
So C+ would be a great nucleophile,
but you don't buy Cl+,
you have Cl2.
So what's the low LUMO here, in Cl2?
I've asked this maybe 15 times before in lecture.
PROFESSOR: Sigma*, OK.
But it's not nearly as low as Cl+.
One way to make it better is to treat it with a Lewis-acid
catalyst, like AlCl3, which has a vacant orbital on
aluminum, which can attack an unshared pair on the
chlorine like that.
And now you have a minus formal charge on the aluminum