Prof: So we want to
understand diffraction patterns.
It's clear that there's some
relationship between the
arrangement of the electrons
that are doing the scattering
and the pattern of the
scattering you get.
Right?
 
At a particular angle there's a
particular intensity,
and it must depend on how those
electrons are arranged.
But how is the
interesting question.
So what we're going to do now
is look just a little bit-- --
no one figures this out,
in a real case;
people have written fancy
computer programs to do this.
But so that it doesn't seem
mysterious, I want you to get
the idea of how it works.
 
Okay, so the first slide we
showed,
or the third slide actually
that we showed last time,
was made of a bunch of these
things we call benzenes,
the collections of six dots,
and they're all oriented the
same way;
that's not very reasonable
except when they get regular in
a crystal, which is where we're
going.
 
But anyhow, if you have a whole
bunch of them,
but randomly distributed,
it turns out the intensity you
get is stronger,
proportional to the number of
them that are doing the
scattering.
Right?
 
But the pattern you get is the
same you'd get from just one.
So that's why we use this funny
slide.
And this was the pattern that
came out,
this thing that looked a little
bit like a snowflake,
and I'd like you to understand
why that hexagon of dots gives a
snowflake.
 
So we take the hexagon of dots,
and what do we look for to see
where scattering might be,
to see-- -- what's the-- --
remember what the trick is?
 
The trick is to look at things
that lie on equally spaced
planes.
 
Try to choose a set of planes
so that as many of the electrons
as there are will fall on or
near that set of evenly spaced
planes.
 
So And this is two dimensions,
so we're looking for lines,
or cross-sections of planes,
right?
So how can we draw lines on
here, evenly spaced lines,
so that all the dots will fall
on them?
Well we can do it this way,
right?
So they're evenly spaced,
three of them,
and all the electrons fall on
or near them.
Notice, they don't have to fall
exactly on,
because if they're displaced a
little bit,
the phase of the wave will
change a little bit,
compared to the others;
but only a little bit.
So it just has to be near,
not exactly on.
Okay, so there we have it.
 
Now, so that means that you'll
get scattering in that
direction, where they will
reinforce one another.
Let's go back to the display we
were looking at before,
here.
 
Okay?
 
We were talking about this just
as class started.
So a wave comes in,
makes these things all move up
and down, the electrons,
here and here,
at the same time.
 
Actually, let's make all three
of them vibrate for current
purposes.
 
Okay.
 
Now we're interested at what
angles light can come out and
all these things will help each
other out, reinforce one
another.
 
So as we go here and look at
that vertical line,
we see that they're not all
in-phase at that angle.
But as we increase the angle,
when we get to say here,
it looks not so bad.
 
The top and the bottom are
actually exactly in-phase,
but the middle one is exactly
out-of-phase,
and it'll cancel out the top
one.
And if there were a whole row
of these,
the second one cancels the
first, the fourth would cancel
the third,
the sixth would cancel the
fifth, and so on;
there wouldn't be anything
coming out.
 
And now we go a little bit away
from that.
And it's worth making a point
here.
Suppose that this were a very,
very, very long line of
scatterers, right?
 
And the first two-- -- let's
get to this place here where
they're all exactly in-phase;
that'll obviously be a very
strong angle.
 
But suppose we looked at
another angle that was close to
that, but just a teeny bit off.
 
So the first two are off by
just 1° in-phase,
and the next one's off by
2° in-phase;
not much off, okay?
 
So are we going to get pretty
strong light coming out from
that?
 
Russell?
 
Student:  If it's a very
long string, you'll eventually
get one that-- --
Prof: If it's a very
long string,
if the first one's 1° out,
by the time we get to the
180th, it'll be 180°
out,
or exactly-- -- so the 180th
will cancel the first one,
the 181st will cancel the
second, the 182nd will cancel
the third,
and so on.
 
Right?
 
So it has to-- -- if you have a
really long string,
it has to be exactly on,
from one to the next,
in order to get them so that
you get something coming out.
And what you get coming out
then will be quite strong
because it'll be the scattering
from the whole bunch of them.
Right?
 
Where did you see that in last
week's lecture,
Friday's lecture?
 
Student:  In the dots.
 
Prof: Right,
you saw it that you got very
fine dots when you had a really
long row of things.
Okay, so back to here.
 
So there'll be things that are
almost exactly in-phase,
and it'll be a certain angle at
which they'll be in-phase,
and that angle is shown by the
end of that arrow,
and you get,
therefore, a dot there,
and also, by symmetry,
the other five of the six dots;
but you could draw the lines
other angles the same way.
Everybody with me on that?
 
Okay, now so we look for
electron density on or near
evenly spaced planes.
 
Now because it only has to be
near, there are other directions
we could draw it,
so we get a set of planes where
they're near but not on;
like here, right?
Now one is a little above the
plane, two are a little below
the plane-- -- right?-- -- or
vice-versa for the second line,
right?
 
But they're pretty close,
and they're at a certain
spacing.
 
Now, that spacing is larger
than the blue spacing.
Does everybody see that?
 
What does that mean about the
angle at which the reflection
will occur, or the scattering,
where they'll be in phase?
Students:  Smaller.
 
Prof: The angle will be
smaller.
So the part of the snowflake
that comes from that is here,
right?
 
A smaller angle because it's a
larger distance.
And you can see immediately why
that is,
if we look at this thing,
that if you look at the top to
the bottom,
where they're far apart,
the lines are far apart,
you get the first reflection
there.
 
But if the lines are closer
together, like these two,
then you have to go to a higher
angle in order to get that first
one, first coming into phase.
 
Right?
 
So that's that reciprocal
relationship.
Okay, so we get a set of six
dots like that,
and that's most of the pattern.
 
But there are other places too
you can see it.
For example,
you can take those lines.
In that set of lines,
every-- -- the center of every
dot is on a line.
 
There happens to be a line in
the middle that doesn't have
anything on it;
but who cares?
Right?
 
The others are on evenly spaced
lines.
Right?
 
So they'll all be in-phase.
 
Where will that one come,
that reflection,
close in or far out?
 
Students:  Far out.
 
Prof: Far out,
and there you see it,
way out at the edge.
 
Can you see it there?
 
Just a little bit with-- -- I
could turn the room lights down,
but I don't think it's worth
it.
You can see that it's there.
 
And so there are six of them
out there as well.
Okay, so you got the idea of
how by looking for evenly spaced
planes you can predict what the
pattern will be for a certain
arrangement of electrons.
 
Now, notice those ones far out
are very weak.
That's because although the
centers of the dots lie right on
the lines,
the lines are getting so
closely spaced that there's part
of the electron density that's
not right on the lines,
that in fact gets out to almost
halfway between the lines.
 
Right?
 
And that part of the electron
density will be canceling.
So most of the electron density
is all in-phase here,
but some of it is out-of-phase,
and as you get the lines closer
and closer together,
the fact that the electron
distribution is finite,
and not tiny points,
means that things will get
weaker and weaker as you go to
higher angles.
 
Okay, so there again we notice
here that the closer spaced
planes give higher angles.
 
This is what we've just been
talking about.
Now, when you take a slide with
a regular lattice of these
benzenes, then-- -- remember,
if you took just one,
you get this snowflake pattern.
 
If you take a whole row,
across the top,
like that-- -- oh pardon me,
I meant to say that one will be
very weak, you can hardly see
it.
What you need is a bunch of
them to cooperate,
to see something that you can
see.
Now first we could take the
whole row along the top,
and we're going to look only at
scattering in the vertical
direction;
so I've narrowed in on the
picture.
 
But if we take that whole row
along the top,
remember, they'll all scatter
in-phase at the specular angle,
because-- -- for scattering up
and down.
So now it gets more intense
because they're all cooperating.
For other directions they
wouldn't all be in-phase from
one hexagon to the next,
but for scattering up and down,
where they're all at exactly
the same level,
they'll all be exactly the same.
 
Okay, now suppose I add the
next row.
Will they be in-phase,
for the same direction of
scattering up and down,
where that top row is all
in-phase?
 
So we would get the scattering
shown here, right?
Stronger than it was before
because it's a bunch of these
things now, right?
 
But still not really strong.
 
But let's add in the next row.
 
Is that going to make them all
stronger again?
No, most of the time they'll be
out-of-phase,
from the next row,
for scattering in this
direction.
 
Right?
 
Most of the time they'll be
out-of-phase,
but for particular
angles they'll be in-phase.
So in fact what that will do,
if we add the next row,
we've made the picture brighter
because we've got twice as many
things scattering now,
but there'll be bars across it,
like that.
 
It's always the same pattern,
getting brighter and brighter
as we add more of them,
but at certain angles they're
in-phase now and at certain
angles they're out-of-phase,
from one pattern to the next,
or from one row to the next.
Now, what's going to happen if
we add the whole sheet of them?
Can you see?
 
Sam?
 
Student:  Well you're
only going to get very thin-- --
Prof: Ah,
there'll only be really
precise angles,
at which they all interfere;
otherwise the 180th will
interfere with the first,
the 181st with the second and
so on.
Right?
 
So if we add the whole phase,
the whole thing,
then if we look at the pattern
it's going to be brighter,
right?
 
But it'll be only dots,
only at very precise angles.
Right?
 
And that's what we saw when we
went to this final picture for
scattering from a lattice.
 
So what this means is that the
lattice repeat concentrates the
snowflake scattering into
tightly focused spots,
on a regular pattern.
 
Right?
 
So this is the diffraction from
a 2-D lattice of benzenes,
which is the same as the
underlying snowflake that came
from just one,
but it's as if it were being
viewed,
as I said last time,
through a pegboard.
 
Right?
 
All that scattering gets
concentrated into a few points,
very bright,
and these regular points.
Now some people seemed not to
know what a pegboard was,
so I took a picture of a
pegboard over the weekend.
How many people have never seen
a pegboard before,
just to calibrate me?
 
Okay, most people have seen a
pegboard.
But you can put these hooks in
it and then hang whatever you
want on it, like tools or
pictures or whatever you want to
put on your wall.
 
Okay, so it's like looking
through the holes in that,
at the snowflake pattern.
 
Now, there are two,
quote, spaces that get talked
about: direct space or real
space;
and diffraction space or
reciprocal space.
Okay?
 
So the crystal is the real
thing that you're interested in,
and diffraction photo is in
diffraction space;
you see these dots that Lowy
Laue saw, for example,
for copper sulfate.
 
Right?
 
So in real space you have a
unit cell structure;
that is, the thing,
the pattern that then gets
repeated to make the crystal.
 
But you have the pattern, right?
 
And that gives rise to this
fuzzy pattern,
like a snowflake in reciprocal
space.
And then you have a crystal
lattice, a regular repetition of
these patterns,
right?
And that's as viewing through
the holes in the pegboard.
Okay?
 
And remember that decreased
spacing in real space
corresponds to increased spacing
in reciprocal space;
which is of course why it's
called "reciprocal".
Okay, so that's what we want to
understand about how it works.
Right?
 
There are patterns which give
rise to patterns in reciprocal
space,
that fuzzy thing,
and then if you have a
repeating lattice of that,
it gets concentrated into
particular points;
but it's the same underlying
pattern.
So there's the pattern,
which is the structure of the
molecule, and there's the
lattice, which is the structure
of the crystal.
 
Okay, now let's look at the
light bulb filament,
the last thing we looked at,
and see why the scattering from
that looks the way we do--
-- looks like this X with dots
along it.
 
Okay, so we take-- -- we want
to find if there are electrons
on this helix,
right?
We can find a set of evenly
spaced, parallel planes like the
yellow ones here.
 
So we'll get scattering
perpendicular to those planes.
And sure enough there it is,
that spot.
But we could also take a set of
planes, and it goes not only up
but also down,
both branches;
both directions of that
particular branch of the X,
come from the same planes.
 
Okay?
 
But you could make planes that
are twice as close to one
another.
 
Right?
 
What will those give rise to?
 
Student:  Further out.
 
Prof: Twice as far out;
the sine is twice as big
actually, not the actual
spacing.
But it's that one,
the second one out along the
branch.
 
Okay?
 
Now, will that be-- -- do you
expect that to be brighter or
not as bright as the first one?
 
Student:  Not as bright.
 
Prof: Not-- -- well it
probably will be not as bright,
and the reason is that for-- --
there are two things that have
to do with the brightness.
 
One is how much density there
is on the planes versus how much
there is between the planes.
 
Right?
 
And we see here that-- -- say
take the original yellow one--
-- there's a lot;
all this stuff is nearly on the
plane, and halfway between the
plane there's not very much,
just a thin cross-section of
the wire.
Okay?
 
But as the planes get closer
and closer,
the fact that the wire has a
finite thickness means that some
parts of the wire,
the same part of the wire,
are out-of-phase compared to
others.
So because of the finite
thickness you expect it to
slowly fall off in intensity as
you go out;
but you'll then get the third
one along the branch and so on.
And you can do the same trick
in the other direction.
Right?
 
So that's why it's an X.
 
You can draw the analogous
mirror image lines.
Okay, now what does the X,
the angle of the X tell you?
The X could look like this,
or it could look like this,
right?
 
What do you learn from that,
the angle that the X makes?
Student:  How tightly
wound the helix is.
Prof: How tightly wound
the helix is,
because if the helix were very
tight,
then the planes would be like
this,
and if you grabbed the spring
and stretched it out,
they'd look like this. Right?
 
So the angle of the X tells how
tightly wound the helix is,
tells the pitch of the helix.
 
Okay?
 
What does the spacing of the
spots tell you?
Did you do the homework I
suggested for today,
that you didn't have to hand
in?
Then you'll know what the
spacing of the spots tells you.
If you know the wavelength of
the X-rays, or of the light in
this case, what does the spacing
of the spots tell you?
Student:  The spacing of
the plane that's-- --
Prof: It tells you what
the actual distance is between
successive turns of the helix,
how far apart the planes are.
So it tells the scale of the
helix, right?
The angle you get from the X,
but how big the actual helix is
you get from the spacing of
those dots,
if you know the wavelength and
the distance that you're
magnifying;
in the case of the
demonstration,
the distance from the slide at
the back of the room to the
front of the room,
for seeing this.
 
And the spots weaken
successively because of wire
thickness, as we just explained,
as you go on out.
Now, that looks very much like
Rosalind Franklin's B b-DNA
photograph from 1952.
 
So we see the helix and we see
how angled the successive turns
of the helix are,
what the pitch of the helix is,
and from the spacing of the
dots we get how big the actual
helix is.
 
But there's something funny.
 
The intensity doesn't just
evenly decease as you go out.
Instead of being strong,
not quite as strong,
less strong,
weaker, very weak,
as you go out,
it goes weak,
strong, strong,
very weak, strong.
Now why is that?
 
Why do you get such a funny
pattern of intensities?
Let's look here at this thing
again.
Remember where,
if we looked at the first,
second, third and so on,
we saw that they come in and
out of phase.
 
But let's suppose-- -- so it
would be,
if you have things that are
twice as close together here,
as in the first case,
what effect does that have on
the actual pattern we see?
 
The pattern we saw,
if we had only every other one,
was here the direct beam,
first reflection,
second reflection,
third reflection,
fourth reflection.
 
How does it differ if we add
one in between?
Okay, here's the first
reflection-- -- can we see that,
if we have all three.
 
That's what would've been the
first reflection if we didn't
have the intermediate one.
 
Do we see a reflection here,
if we have a whole row of them?
No, because every other one
cancels out.
So we won't see that one.
 
How about the next one?
 
Student:  It'll be very
bright.
Prof: Yes,
it'll be very bright,
because there's more doing the
scattering.
How about if we go to the next
one?
No, nothing there,
cancelled, right?
How about the next one?
 
Do you perceive a pattern
emerging?
What is it?
 
What?
 
Student: It's successive;
you see it, then you don't see
it.
 
Prof: Can you say it in
a more mathematical way?
Suppose we numbered them.
 
The one straight ahead we'll
call zero;
then there'll the first,
one, two, three,
four, five.
 
What happens when we put one in
between?
>
Prof: Is one there?
 
This was one here, right?
 
There was one for the top and
bottom.
Is that there?
 
Students:  No.
 
Prof: How about two?
 
Students:  Yes.
 
Prof: How about three?
 
Students:  No.
 
Prof: No.
 
How about four?
 
Students:  Yes.
 
Prof: What's the-- --
can you state it mathematically?
>
Prof: Lexie,
what do you say?
Student:
>
Prof: Can't hear very
well.
Student:  n+2?
 
Prof: Which ones
disappear?
Student: Odd.
 
Student:  The even.
 
Prof: The odd ones,
one, three, five,
seven, nine;
the odd ones disappear but the
even ones are still there,
if we put an extra one in
between.
 
Now let's think about DNA.
 
Here's a helix, right?
 
But what do you say about DNA?
 
Is it just a helix?
 
Students:  Double helix.
 
Prof: It's a double
helix, so there are two helices.
Okay, so suppose we have-- --
there are the successive turns
and we get scattering at a
certain angle--
-- but suppose we make a double
helix,
now what'll happen?
 
Lexie, you're going to help me
again.
Student: Oh.
 
Prof: How different will
it be if we have a second helix
wound halfway between the first
helix?
What will happen to the-- --
what we-- -- from one helix we
had one, two,
three, four,
five, six, seven,
eight, nine angles,
right?
 
What happens now?
 
Student: You're missing
even numbers.
Prof: Can't hear.
 
Student:  You're missing
even numbers.
Prof: Ah, !
 
all the odd ones disappear.
 
Right?
 
So what would you have expected
to see in Rosalind Franklin's
picture then?
 
Right?
 
It would be weak,
strong, weak,
strong, weak,
strong, right?
But generally getting weaker as
you go out because of the finite
thickness of the wire.
 
Is that what we see?
 
No, we saw-- -- the first one
is very weak,
strong, strong,
very weak, strong.
Can you see any-- -- how could
that be?
What is that telling you?
 
Yes?
 
Student:  They're not
perfectly in-phase.
Prof: What do you mean
they're not perfectly in-phase.
Student:  They're not
equally-- --
Prof: Ah.
 
Maybe the second one isn't
halfway between the turns of the
first one.
 
Right?
 
What happens?
 
So the planes are twice as
close there and that would
cancel every other reflection,
all the odd ones.
But suppose it's like that.
 
Did you ever see a situation
like that before where what
repeats is not a line but a pair
of lines?
Angela?
 
Student:  The second
diffraction-- .
Prof: The second
diffraction pattern we looked at
was of that sort,
which showed the same dots but
the intensity got modulated.
 
Remember?
 
Okay, so let's,
instead of doing this where we
have equally spaced ones,
let's do one where they're not
equally spaced,
like this.
Now-- -- oops,
it came loose here;
here we go;
oh no;
there we go, okay.
 
Now consider the top and the
bottom one, which are in
successive turns of the same
helix, and this one is offset in
the middle.
 
But consider just the top and
the bottom one.
Okay, there's the first
reflection, the second
reflection, the third
reflection, the fourth
reflection, everybody see?
 
Between the top and the bottom.
 
Now tell me what happens if I
add that other one offset this
way.
 
How about the first reflection
there, is that going to be
strong?
 
Students:  No.
 
Prof: No,
because this one is canceling
it.
 
And remember,
there's also one like this
below this one;
remember, it's a really long
row of these things.
 
So it gets almost perfectly
cancelled;
not quite because it's not
quite-- --
there would be exactly the
minimum,
and when the first and the
third are in-phase,
the second one is not exactly
out-of-phase,
but very close;
very weak, that reflection's
going to be.
 
Now let's go to what would be
the second reflection of the
single helix there.
 
How about there?
 
>
Prof: Yes,
the second one now is helping
out a little bit;
not perfectly but it's-- -- in
fact, probably if we go just a
little bit further maybe it'd be
better.
 
But anyhow there'd be a
reflection there.
Okay?
 
Now how about if we go to the
third?
That's as good as the second
one, right?
And how about if we go to the
fourth?
Would the fourth-- -- if they
were evenly spaced,
what would you expect for the
fourth?
Student:  Very strong.
 
Prof: Very strong, right?
 
So let's go to the fourth;
one, two, three, four.
What should it be?
 
Students:  Really weak.
 
Prof: Really weak.
 
And how about the fifth?
 
Students:  Strong.
 
Student:  Really strong.
 
Prof: That's precisely
what she saw,
right?
 
Weak, strong,
strong, very weak,
strong.
 
So that's how much,
this is how much it's offset
from the center,
in order to get that sequence
of intensities.
 
Everybody with me on this?
 
Okay.
 
So that's that repeated pair
pattern that we saw before.
And there you see that the
outer coil of the DNA,
which has phosphorous in it,
with a lot of electrons,
that's very much like this
helix.
Okay, much more electron
density near these planes than
between them,
than halfway between.
But there's also much more
electron density on those,
near those blue planes than in
between,
for the same reason,
and that means that you get
those spots on the side.
 
That tells you the diameter of
the helix.
Okay, and the things top and
bottom tell you the base
stacking distance.
 
Let me back up just a second to
show you what we mean,
right?
 
So there are a bunch of
electrons here,
a bunch of electrons here,
a bunch here,
a bunch here,
and that gives scattering in
this direction.
 
So all that information-- --
the helix diameter,
the fact that it's a helix,
the fact that it's a double
helix,
the fact that it's an offset
double helix,
and the diameter of the
cylinder, are all easily read
out if you know what you're
looking for;
which Crick knew,
because he had been studying
helices of proteins,
so he knew about-- -- he knew
from helix scattering.
Okay, so knowing the molecule's
electron density,
as we knew our artificial sort
of benzene,
or this DNA,
it's easy to calculate the
crystal's diffraction pattern,
especially if you have a
computer to help you out;
but it's not an intellectual
challenge, it's just work.
 
Right?
 
Using pretty heavy- duty math,
people got the Nobel Prize for
developing it;
or a canned program,
(that's what they got the Nobel
Prize for),
you can go the other way and go
from the X-ray pattern to what
it was that was causing it.
 
Right?
 
What we did was go from what's
causing it to the pattern,
but in fact you can go the
other way now when you press a
button on a machine.
 
And you get out the electron
density at every point in space.
Okay, that's the yield from an
X-ray structure.
Why don't you get the nuclear
positions, why do you get the
electron positions?
 
>
Prof: Alex?
 
Can't hear.
 
Student:  It's too big
to diffract.
Prof: Yes,
it's too heavy,
so it doesn't vibrate,
the nuclei, so they don't
function as an antenna,
to give off the new waves the
way the electrons do.
 
Okay, so you get a result,
which you can plot on a map.
Now in the olden days when I
started doing this kind of
thing,
the kind of map you got out was
a big sheet of computer output
paper,
and it had numbers on it,
and a given piece of paper was
a given slice through the
"unit cell',
through the pattern,
and numbers were printed out
that said what was the electron
density at that particular point
on that level.
 
So you had an x axis,
y axis, and at this point it
was seventy-five,
forty-two, fifty-five,
eighty-seven,
or whatever the number is,
and so on.
 
And what you do is take a felt
tip pen and connect dots of
equal size, and you'd get a
contour of that particular
electron density.
 
And here's where that was done
for a particular compound.
This is taken from a book
written at that time by Stout
and Jensen.
 
And you drew these by hands
because you didn't have a
computer that could draw-- --
that could figure out where to
draw it and so on.
 
So here were atoms in a
particular crystal structure.
Then you had a whole bunch of
sheets of paper for different
slices, through the electron
density;
stack them up, right?
 
Now sometimes the particular
slice you're working in went
right through the middle of an
atom, so you got very high
electron density.
 
Sometime it just barely touched
the atom, right?
So you see here that that one,
that atom, this particular
slice, was very near the
nucleus, so we got high electron
density.
 
But that one,
the nucleus was not in this
plane, so you just got a little
glancing blow at it.
Okay?
 
But you could draw these things
on acetate or plastic sheets and
stack them up,
right?
So this was the model made for
the first determination of the
structure of penicillin;
in fact, the potassium salt of
penicillin.
 
Where's the potassium atom?
 
How can you distinguish
potassium from nitrogen or
carbon or hydrogen or oxygen?
 
Student:  Electron
density.
Prof: It'll have a lot
more electron density.
So do you see it?
 
There's the potassium,
lots of density.
Okay, now here's a molecule
that happened to be flat.
So it was possible to slice
through practically all the
nuclei.
 
It's not a very common type of
molecule, but this one is,
rubofusarin.
 
Okay, now these contours are
drawn at intervals of one
electron per cubic angstrom.
 
Right?
 
So the first one is one
electron per cubic angstrom,
two electrons per cubic
angstrom, three electrons per
cubic angstrom and so on.
 
Now there are both carbons and
oxygens in this.
How can you tell which ones are
oxygens?
>
Prof: Yeah, Yoounjoou?
 
Student:  They are more
electronegative,
so they should have more
contour lines.
Prof: They should have
more contour lines.
So if we look here,
we see that that one goes up to
five electrons per cubic
angstrom;
one, two, three,
four, five rings,
right?
 
That one goes up to seven
electrons per cubic angstrom,
and so do those.
 
Right?
 
So those are the oxygens.
 
Okay?
 
And there's a picture of the
structure and you see indeed
those atoms are oxygens.
 
Now, what is our fundamental
goal of looking at this stuff,
what do we want to see?
 
Students:  Bonds.
 
Prof: Bonds.
 
Do you see bonds?
 
What does it look like?
 
It looks like just a bunch of
balls, just atoms put there.
Was Lewis right?
 
Are there electron pairs
halfway between the bonded
atoms?
 
No, it's just atoms.
 
Okay.
 
Why are there no hydrogens?
 
>
Prof: Because they have
hardly any electrons compared to
the others, right?
 
So they're a very weak thing.
 
Okay, now those are bonds.
 
You can see them there.
 
Where did those lines come from?
 
>
Prof: Angela,
what did you say?
Student:  Someone drew
them.
Prof: Yes, you drew them.
 
How did you know where to draw
the bonds?
Student:  You connect
the dots.
Prof: You connect the
centers of the rings,
right?
 
Those are the bonds.
 
Now how do you know what's a
single bond, what's a double
bond, what's a triple bond?
 
Student: More density?
 
Prof: You notice in the
structure here,
on the bottom right,
there are double bonds and
single bonds.
 
How do I know which is which?
 
Is there any way of knowing?
 
Elizabeth?
 
Student:  Maybe the
distance between.
Prof: Ah, the distance!
 
Double bonds should be shorter
than single bonds.
And, in fact,
notice that here we have short,
long and intermediate.
 
Why intermediate?
 
>
Prof: Why do I draw a
dotted line there?
Student: 
>
Prof: Because we have
this, look here;
resonance, right?
 
So there are one- and- a- half
bonds.
Right?
 
Okay, but fundamentally what we
see are spherical atoms,
right?
 
No bonds, no unshared pairs on
oxygen, just atoms.
So was Lewis way off base?
 
Actually you can see the bonds
with X-ray, but you don't look
at total electron density the
way we're looking at it.
What you look at is difference
density;
you see how different is
the picture I actually determine
from what I would've expected if
they were truly just
spherical atoms,
undistorted.
How do you find-- -- how would
you make such a map?
So you have these pieces of
paper that say how much the
electron density would be at all
different points;
experimentally,
how much it is experimentally.
Now I want to see how different
is that from if I just had
spheres for the atoms.
 
What would I do?
 
I have to find out what the
spheres would look like,
what density they would predict
at all these different points,
and then do what?
 
Student:  Subtract.
 
Prof: Subtract, right?
 
So that's the difference
density.
Sometimes it's called
deformation because it's
how much the electron
distribution gets deformed when
bonds get made.
 
Okay, so you take the observed
electron density,
which is from X-ray,
which is experimental,
and you subtract atomic
electron density,
which you calculate somehow.
 
You know what spherical atoms
would look like;
we'll get to that next,
how you would know what that
is.
 
But anyhow, how do you know
where to--
-- when you're going to do the
subtraction you have to make
sure you position the atoms
properly,
to be subtracted, right?
 
How do you know where to put
them?
How do you know where to put
your spherical atoms so that you
subtract it?
 
Yes?
 
Student:  On the center
of the rings.
Prof: You put it on the
center of those rings,
right?
 
Okay, so let's look at a case
like this.
This was done by Leiserowitz
and Ziva Berkovitch-Yellin
thirty-three years ago.
 
So this is an interesting
molecule because it has lots of
different kinds of bonds in it:
single bonds,
double bonds-- -- in fact three
double bonds in a row--
-- and also benzene rings that
have these resonance bonds.
Okay, so we're going to look
first at one of the benzene
rings.
 
Okay?
 
So this map that I'm showing
you here is a difference density
map.
 
It's the total,
minus what it would be for the
atoms.
 
But notice what was subtracted
is just the carbon density;
didn't subtract hydrogens.
 
So let's look at the features
that are big here.
This is how the electron
density changed in forming the
molecule from atoms.
 
Okay?
 
So first look up there.
 
There are some very dense
contours, right in the middle up
there, and that's the hydrogen
which was not subtracted,
right?
 
So it's the biggest game left
in town, once you've subtracted
the spherical carbons.
 
Okay, so that's-- -- and the
total amount of intensity in
that region is one electron,
which is what you expect.
Now the total amount of
deformation density in this
region, where there's a single
carbon-carbon bond,
is only 1/10th of an electron.
 
What would Lewis have said?
 
Two electrons,
it's 1/10th of an electron.
Okay, now let's slice that-- --
and within the carbon-carbon
bonds of the ring,
the aromatic ones,
it's bigger.
 
Why bigger?
 
Because it's partially double
bond in the resonance
structures, right?
 
Okay, so it's 0.2 electrons for
those ones in the ring.
Now let's slice it and turn it
so we can see its cross-section.
And it's round;
no surprise there.
But let's slice the ones in the
ring and turn them.
It's oval, not round,
the bonds are not round.
Why?
 
Students: Ï€ bonds.
 
Prof: Ah,
because they're partially made
up of these p orbitals
that point in and out of the
ring,
that distort it from being
round.
 
Okay, now let's look at the
double bonds that are in the
middle of this molecule,
okay?
First that one;
we've already seen that bond
before that was 0.1,
had a round cross-section.
But let's look at one of the
double bonds.
Not too surprisingly it has
more electron deformation
density in it;
0.2, right?
And the one in the middle
actually has 0.3.
Now they're both double bonds,
so probably they look similar,
and in a sense they do.
 
If we slice the first one,
and turn it,
it's oval for the same reason
as before;
there's a p orbital pointing in
and out of the board,
and if we look at it below,
part of the bond is caused by
overlapping,
as we'll see,
of those p orbitals.
 
So it's got an oval,
up and down shape.
How about the next one?
 
If we slice it and turn it,
it's also oval in
cross-section,
but perpendicular.
Right?
 
What do you expect the next one
would look like?
>
Prof: It's symmetrical;
it'll look like the first one
again.
 
But here's something
interesting.
That cross-section of the
middle one showed the
deformation due to the pink but
not the deformation showing-- --
due to those neighboring blue
ones.
There must be something
different about the interaction
at the end and the interaction
in the middle.
Does everybody see that problem?
 
You might think the middle one
would be round or clover-leaf
shaped, but it's not,
it's distorted in the in-plane
direction.
 
But to understand this,
you have to be patient and wait
till next week,
or later this week,
when we start getting into
quantum mechanics.
But notice that it's as if the
carbons formed bent bonds.
So the bonds from those central
carbons at the end go up and
down, but the ones in the middle
are in- plane.
Now how many electrons are
there in a bond?
So here we're going to plot how
many electrons vertically,
and how long the bond is,
which is related to whether
it's single, double,
triple.
In fact, a single bond is there;
a double bond and a triple bond
have those distances.
 
Now if you-- -- and in fact you
could also do a one-and-a- and a
half bond, those bonds in the
benzene ring.
Now what would you expect this
to be?
How many electrons-- -- if you
were Lewis, how many electrons
would you have in a single bond?
 
Students:  Two.
 
Prof: Two.
 
Double bond?
 
Students:  Four
Prof: Four.
Students:  Six.
 
Prof: Six.
 
Students:  Three.
 
Prof: Three,
okay one and a -and-a-half.
Okay, so there's what you would
expect for the plot,
if Lewis was right:;
two, four, six.
Okay?
 
Now we can take this molecule
that we just looked at and see
how much electron density there
is in each of those bonds--
-- and I've color-coded it-- --
and it looks pretty darn good.
Right?
 
Hooray for Lewis.
 
And in fact the people who did
the work,
Leiserowitz and
Berkovitch-Yellin,
looked at a bunch of other
structures too and added those
to the same plot,
and they're not bad,
except the scale is wrong.
 
Instead of being two,
four, six,, which I've faded
out here, it's 0.1,0.2,0.3.
 
So bonds are a very minor game
in the distribution of
electrons.
 
Lewis, in a sense,
was sort of right,
there is electron
density shifted between the
atoms to hold them together,
right?
But much less than pairs of
electrons.
In fact, this amount of bonding
density is about 1/20th of what
Lewis might have said,
but never did say,
because he wasn't that
naïve.
Okay?
 
So instead of getting two,
you get a 20th of two;
0.1.
 
Okay?
 
So bonding density is about
1/20th of a Lewis.
Incidentally,
the unit "Lewis"
is used only within this class.
 
Okay?
 
So here's another example done
by Dunitz in Switzerland and
these people who were the--
-- Schuyler Seiler and
Schweizer, who were Swiss,
who were the progeny of
watchmakers and people that do
very precise work;
because to do this you have to
have really precise
experimental work,
if you're going to subtract
something from something else
and have the difference be
meaningful,
when the numbers are almost
precisely the same,
right?
Remember we looked at the total
electron density.
It looked just like spheres.
 
So you have to be really good
with this if you want to have
this be meaningful;
and those guys were really good.
This was done,
what, twenty-five years ago.
So they took this molecule,
which is highly symmetric,
which is great.
 
Remember it's got its resonance
structure, so all the bonds
around the middle are the same,
top and bottom.
Okay, so we need only look at a
quarter of the molecule,
or for convenience a little bit
more than a quarter.
So I'll blow it up here.
 
And there is the electron
difference density.
So this is how the electrons
shifted during formation of the
bonds.
 
Okay, so there's a
carbon-carbon aromatic bond,
a bond and a half,
and if you turn it and look at
it,
it's distorted the way you
expect, just as in the previous
experiment we talked about.
Here's another one, same deal.
 
Now here is a single bond,
carbon-carbon single bond.
What's it going to look like in
cross-section?
Prof: Round. Good.
 
Now here is a carbon-nitrogen
triple bond.
This is something new.
 
What's it going to look like in
cross-section?
Maybe like a clover-leaf, huh?
 
Like that, it looks like that.
 
It's round.
 
It's not clover-leaf,
and it's not a diamond,
it's round in cross-section.
 
So we'll address that later.
 
What's that?
 
That's not a bond.
 
Students:  Unshared pair.
 
Prof: That's the
unshared pair on Nitrogen.
Now what's special here?
 
What are we looking at?
 
We're looking at the
carbon-fluorine bond.
Does it look like the
carbon-carbon bonds,
or carbon-nitrogen?
 
How's it different?
 
There's no there, there. Right?
 
There aren't any electrons
holding carbon to fluorine.
What the heck is going on?
 
Where is the C-F bond?
 
This brings up the second great
question in the course.
What's the first question?
 
Students:
 How do you know?
Prof: How do you
know, right?
And the second question is
illustrated by this film clip.
>
Come on baby.
 
I worked hard on this.
 
Okay, I'll tell you what it
says.
>
 
>
Prof: Okay,
I can do it myself.
(This is a disaster.)
 
"Catch some thieves
Hobson, that's what we're here
for." 
"Very well sir."
Click, click,
click, click,
click, click,
click, click,
click, click.
> 
"What do you think of
him?" 
"Compared with what
sir?" 
"Exactly." 
Okay.
 
>
"What do you think of
him?"
"Compared with what
sir?"
"Exactly."
 
That's the second question,
compared with what? In
anything quantitative,
even many qualitative things,
that's the question,
and that's why we're in
problems here looking for the
C-F bond.
We already saw this question
when we talked about resonance.
Remember, I asserted that when
you have resonance the molecule
is unusually stable.
 
What should you have asked?
 
Student:  Compared to
what?
Prof: Compared to what?
 
Right?
 
In that case it turns out it's
compared to observed or
calculated energy.
 
Compare the observed or
calculated energy,
(what really happens),
to what you would expect if it
were not resonant.
 
And we'll talk later about how,
on what basis,
you might have an expectation.
 
Now we're talking about
difference electron density.
So what do we compare?
 
>
Prof: We compare
experimental electron density,
(or it could be calculated,
if we can use quantum mechanics
to calculate it)--
, but compare experimental
electron density to what?
 
What have we compared it to?
 
Student: 
>
Prof: What it would be
if it were just atoms.
 
Right?
 
But what kind of atoms, right?
 
We compare the observed or
calculated total electron
density to the sum of electron
densities for a set of
undistorted atoms.
 
Now what's an undistorted atom?
 
Okay.
 
Now, to avoid Pauli problems--
-- and this isn't the time to
discuss what those are,
we need to subtract not an
unbiased spherical fluorine
atom,
which if you count up how many
electrons there are on fluorine,
each quadrant of space around
fluorine--
-- of which there will be
four-- --
each quadrant will have one and
three three-quarters electrons
in it.
 
Right?
 
You multiply that by four and
you get seven electrons,
which is the valence electrons
of fluorine, right?
But you can't have one and
three three-quarters electrons
from fluorine,
and also an electron from
carbon in the same place.
 
That's too many electrons in
the same place,
right?
 
So what you have to start with
is not an undistorted spherical
fluorine atom,
but a fluorine atom that's
prepared to make a bond,
which has only one electron in
the region where it's going to
make a bond.
So if you subtract a spherical
one, you're subtracting too much
from the region of the bond,
right?
So that's why you don't see any
electron density in there.
So, and if you want to
understand this better than I've
explained it,
go on the Web and look at
Professor Dunitz explaining it.
 
We have a little movie of him
talking about this particular
problem.
 
Okay?
 
But at any rate the conclusion
of all this is that bonding
density is about 1/20th of a
Lewis.
All right?
 
And it's time for me to quit.
 
We'll go on with this next time.
 
