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-- dissolved in water,
such that there are six water

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molecules arranged around the
metal in an octahedral array.

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You will see here that I have
colored one of the water

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00:00:26,000 --> 00:00:30,000
molecule's oxygens in this peach
color because we are going to

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talk about reactions today.
In fact, I am going to

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introduce you to the idea of a
potential energy surface for a

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reaction.
I am taking water substitution

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as an example and as a context
for this, but this very general.

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And the same ideas that we are
going to be talking about today

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would apply to reactions in
organic chemistry or in

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biochemistry.
Let's imagine having this

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system dissolved in liquid
water.

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And let's further imagine that
we have some way,

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00:01:04,000 --> 00:01:09,000
perhaps, for isotopic labeling
of the oxygens in the water to

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differentiate from those that
start out on the metal complex.

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And so I am differentiating
here by using a green color for

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the oxygen in a liquid water
solvent.

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And in this reaction,
it is a very simple reaction in

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which liquid water from
solution, --

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-- we are going to talk about
reaction mechanisms today,

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somehow will come in and bind
to the metal center and displace

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one water molecule from the
metal center.

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And I will represent that here
by saying that we would lose

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that one water molecule that
I've colored this way.

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And that gives us our product,
which is very similar to the

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starting material in this case,
--

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-- except that one water
molecule, the one here colored

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green, has replaced one of the
others.

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00:02:15,000 --> 00:02:19,000
And when we look at this type
of reaction as a function of

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which 3d transition element we
have at the center of the

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octahedron, we find that the
time constant for this reaction

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00:02:29,000 --> 00:02:33,000
taking place varies quite
dramatically.

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00:02:33,000 --> 00:02:38,000
And this is informative with
respect both to the mechanism of

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the reaction,
as we will discuss,

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and to the electronic structure
attributes of the systems that

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control the way a metal binds to
its ligands.

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00:02:50,000 --> 00:02:54,000
And so, for example,
let's make a little table of

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the metal n plus.

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And we will take the log of the
rate constant for the reaction,

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which I will denote with this
k.

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Small k is going to be our rate
constant.

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And the units that we would
find for such a rate constant

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would be reciprocal seconds to
describe how fast this takes

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place.
And, if we start down here at

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chromium three plus,
we would find that the log of

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the rate constant is
approximately minus 6.

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00:03:37,000 --> 00:03:42,000
And then, if we went to
vanadium two plus,

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the rate constant log goes to
minus 2.

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So the reaction is faster than
it is for chromium three.

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And then, if we look at some of

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the other d ions that are
possible here,

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we could have chromium two plus
or copper two plus.

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And, for both of those,

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the log of the rate constant is

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You can see,
because I am taking the log of

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these first order rate
constants, the actual rates of

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these processes differ here by
15 orders of magnitude.

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That is an enormous difference.
So it is not true that whatever

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metal ion you have at the
center, the rate at which water

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comes on and off and it
exchanges with water from

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solution is about the same.
It is not.

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It differs over an enormous
range of actual rates,

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many orders of magnitude,
depending on exactly which

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metal ion you have here.
And so I will classify these as

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either slow or fast.
And we will say that here,

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we are slow.
And here we are fast.

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00:05:01,000 --> 00:05:04,000
And, of course,
there are cases that are

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somewhat in between.
But I will be giving you a

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little bit of nomenclature that
arises from the study of such

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ligand substitution rates in a
little while.

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Here is one type of
substitution reaction.

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Now, I would like you to
consider a different type of

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substitution reaction,
one that does not have a

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product that is essentially
identical to the starting

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material.
And that will be one as shown

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here.
Let's use the metal tungsten at

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the bottom of Group 6,
a 5d transition element.

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This makes a very stable binary
compound with carbon monoxide in

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which six carbon monoxide
ligands bind to the metal center

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at the corners of an octahedron.
And in this tungsten

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hexacarbonyl system,
you are going to find that,

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as you will be seeing in
recitation this week,

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these CO ligands are pi-acids,
they are very high in the

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spectrochemical series.
This leads to a molecule like

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this being colorless.

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It has a large delta O.

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And because ligands are
pi-acidic and the metal center,

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as you will see,
this metal center is a d six

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pi-base,
the metal itself is very

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electronically complimentary to
its set of six carbon monoxide

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ligands.
So a system like this is very

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stable.
And what you might like to do

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is to tag some biomolecule with
a tungsten pentacarbonyl

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fragment, so that you could use
the vibrational spectroscopy

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associated with these CO
oscillators to watch some

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biological process that has that
thing attached to it.

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And one example of a
substitution reaction you might

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wish to use would be the one
shown here.

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This ligand that I am currently
sketching is one in which we

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have a phosphorus with a lone
pair and three phenyl groups

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attached to it.
A phenyl group is a benzene

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ring where one of the positions
is substituted.

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This ligand here,
then, is triphenylphosphine,

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and this is a good ligand for
various types of metals.

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And, in fact,
one of the Nobel laureates that

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I may have mentioned in passing
this semester,

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Sir Jeffrey Wilkinson,
used triphenylphosphine as a

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ligand in the preparation of
what is now universally known as

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Wilkinson's hydrogenation
catalyst.

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So this ligand is quite common,
indeed, and well-known.

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And what we might like to do is
to replace one carbon monoxide

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with this triphenylphosphine
ligand.

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00:08:47,000 --> 00:08:51,000
That would be a simple
substitution reaction.

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We would like one equivalent of
carbon monoxide to be evolved as

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CO gas in this reaction.
But we find that we can put

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00:09:02,000 --> 00:09:07,000
tungsten hexacarbonyl in the
presence of triphenylphosphine,

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and nothing really happens.
They just sit there and look at

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each other.
They do not react.

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So one of the things that
happens when you are learning to

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do synthetic chemistry is that
the outcome of a reaction can be

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either that it proceeded cleanly
onto products.

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And one possible outcome is
there was no reaction at all.

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And so this is what we will
call no thermal reaction.

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And that means that just with
the energy supplied by the

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temperature of the room,
the reaction is not proceeding

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in the forward direction.
And what that may mean is that

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there is an energy barrier to
this reaction.

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00:10:00,000 --> 00:10:05,000
And energy barriers are going
to be a topic of this lecture in

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a few moments,
but there may be an energy

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00:10:09,000 --> 00:10:13,000
barrier.
And we don't have sufficient

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00:10:13,000 --> 00:10:18,000
energy to surmount the barrier,
so the substitution reaction is

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00:10:18,000 --> 00:10:23,000
not taking place.
But we find that if we shine

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00:10:23,000 --> 00:10:27,000
light on the system,
UV light, then the reaction

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00:10:27,000 --> 00:10:32,000
takes place.
And we form this desired

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00:10:32,000 --> 00:10:37,000
product that has one of the
carbon monoxide ligands replaced

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00:10:37,000 --> 00:10:42,000
with triphenylphosphine.
That would be the initial

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00:10:42,000 --> 00:10:46,000
product formed in such a
process.

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00:10:51,000 --> 00:10:54,000
Still octahedral,
now five CO ligands instead of

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six, and one of them replaced by
triphenylphosphine.

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00:10:59,000 --> 00:11:03,000
And it occurs under the action
of UV light.

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00:11:03,000 --> 00:11:06,000
If you put a photon into the
system for some reason,

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we can now surmount the energy
barrier to the reaction,

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00:11:10,000 --> 00:11:14,000
blow one of the CO molecules
away as CO gas.

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00:11:14,000 --> 00:11:16,000
We might, in fact,
in practice,

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00:11:16,000 --> 00:11:20,000
possibly be bubbling an inert
gas like N two through

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00:11:20,000 --> 00:11:24,000
the solution to sweep the
evolved carbon monoxide out of

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the system to help the reaction
go to completion as we are

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00:11:29,000 --> 00:11:33,000
carrying out such a prep.
But, in any event,

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the key feature is that it goes
under photolysis in the presence

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of a UV lamp but not simply
through input of normal thermal

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energy.
And, in thinking back to the

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development of the electronic
structure of systems of this

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type, we can actually begin to
understand this.

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Because we have an energy
diagram, a d orbital splitting

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00:12:01,000 --> 00:12:05,000
diagram for simplicity,
where we have our t(2g) set and

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00:12:05,000 --> 00:12:09,000
our e(g)*.
And tungsten is in Group 6,

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the ligands are all neutral,
so this is d six.

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00:12:14,000 --> 00:12:19,000
This is a common electron count
for octahedral systems that are

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quite stable.
And CO is high in the

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spectrochemical series,
so it leads to a large value of

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00:12:27,000 --> 00:12:33,000
delta O.
This splitting here between

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t(2g) and e(g) is large.
And that means we have an s

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00:12:38,000 --> 00:12:43,000
equals zero state,
with the system not being

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00:12:43,000 --> 00:12:48,000
paramagnetic,
all the electrons paired up in

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00:12:48,000 --> 00:12:54,000
t(2g) and this big gap.
And what happens when we add a

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00:12:54,000 --> 00:12:59,000
photon to the system,
we can promote the system into

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00:12:59,000 --> 00:13:05,000
an excited state.
And, in order to satisfy the

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00:13:05,000 --> 00:13:09,000
spin selection rule,
this excited state will have no

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00:13:09,000 --> 00:13:14,000
net spin polarization.
This one is s equals zero,

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00:13:14,000 --> 00:13:19,000
this one is s equals zero to
satisfy the spin selection rule

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that we have talked about.
But notice, now,

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00:13:23,000 --> 00:13:26,000
in the excited state --

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-- we have an electron in e g
star.

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00:13:37,000 --> 00:13:42,000
And, as I was discussing last
time, when you look at the MO

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00:13:42,000 --> 00:13:47,000
theory for a problem like this,
e(g)* is sigma antibonding with

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00:13:47,000 --> 00:13:51,000
respect to the ligands.
I am going to illuminate that

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00:13:51,000 --> 00:13:55,000
for you very clearly in a few
moments with a graphic,

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00:13:55,000 --> 00:14:00,000
but this is sigma star with
respect to the ligands.

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00:14:00,000 --> 00:14:03,000
So what are we doing?
We are taking an electron,

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00:14:03,000 --> 00:14:07,000
here, that is pi bonding
because CO is a pi-acid ligand.

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00:14:07,000 --> 00:14:11,000
In the ground state,
this electron that we are going

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to promote is helping to bind
the carbon monoxide ligands to

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the metal.
And then, when we input that

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00:14:19,000 --> 00:14:24,000
photon and we produce an excited
state, that same electron now

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00:14:24,000 --> 00:14:29,000
has been taken away from our sum
total of bonding.

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00:14:29,000 --> 00:14:33,000
And it has been added to an
antibonding orbital.

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00:14:33,000 --> 00:14:39,000
And the effect of that is to
"labialize" one of the carbon

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00:14:39,000 --> 00:14:44,000
monoxide ligands,
allowing it to dissociate from

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00:14:44,000 --> 00:14:50,000
the metal center producing an
unsaturated intermediate that

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00:14:50,000 --> 00:14:56,000
can be captured by the lone pair
of triphenylphosphine.

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00:14:56,000 --> 00:15:02,000
That word I just used,
labialized, we have from Henry

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00:15:02,000 --> 00:15:04,000
Taube.

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00:15:13,000 --> 00:15:16,000
Henry Taube,
Nobel Prize in 1983.

198
00:15:16,000 --> 00:15:20,000
If you want to read more about
Henry Taube and his

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00:15:20,000 --> 00:15:26,000
contributions to chemistry and,
in fact, the study of ligand

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00:15:26,000 --> 00:15:32,000
substitution reactions and the
relationship of reaction rates

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00:15:32,000 --> 00:15:37,000
to the population of the e g
star orbitals,

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00:15:37,000 --> 00:15:43,000
then I would direct you to the
1983 section of the Nobel Prize

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00:15:43,000 --> 00:15:48,000
dot org website.
And you can read his biography,

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00:15:48,000 --> 00:15:51,000
his Nobel Prize lecture and his
banquet address.

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00:15:51,000 --> 00:15:55,000
These are all things that
Professor Schrock from our

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00:15:55,000 --> 00:15:59,000
department is doing this week in
Sweden, so shortly,

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00:15:59,000 --> 00:16:04,000
you will be able to read his
contributions to that website.

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00:16:04,000 --> 00:16:12,000
And I am sure you will all find
that to be pretty exciting,

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00:16:12,000 --> 00:16:15,000
as I do.
But Henry Taube,

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00:16:15,000 --> 00:16:23,000
this Nobel Prize winner,
coined the terms "inert" and

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00:16:23,000 --> 00:16:26,000
"labile."
These terms,

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00:16:26,000 --> 00:16:33,000
inert and labile,
they refer to rates of ligand

213
00:16:33,000 --> 00:16:41,000
substitution reactions.
If a complex is to be thought

214
00:16:41,000 --> 00:16:44,000
of as inert, it isn't really
inert.

215
00:16:44,000 --> 00:16:51,000
What that means is that its
substitution reactions are slow.

216
00:16:51,000 --> 00:16:56,000
And, on the other hand,
if a system is to be referred

217
00:16:56,000 --> 00:17:03,000
to labile, that means that its
ligand substitution reactions

218
00:17:03,000 --> 00:17:07,000
are fast.
And over there on the left,

219
00:17:07,000 --> 00:17:13,000
we looked at a set of four
metal ions whose aqua-complexes

220
00:17:13,000 --> 00:17:18,000
could be classified as either
inert or as labile.

221
00:17:18,000 --> 00:17:23,000
So let me relate that to the
occupation of the eg star

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00:17:23,000 --> 00:17:27,000
energy level.
Here what we have,

223
00:17:27,000 --> 00:17:32,000
t(2g) and e(g)*.
In the case of chromium three

224
00:17:32,000 --> 00:17:37,000
plus or vanadium two
plus,

225
00:17:37,000 --> 00:17:42,000
since vanadium is in Group 5
and chromium is in Group 6,

226
00:17:42,000 --> 00:17:48,000
both of these metal ions have a
d three electron

227
00:17:48,000 --> 00:17:52,000
configuration.
And so, when we populate our

228
00:17:52,000 --> 00:17:57,000
d-orbital splitting diagram for
these two systems,

229
00:17:57,000 --> 00:18:01,000
here is the result.
We have an s equals

230
00:18:01,000 --> 00:18:05,000
three-halves ground state,
three unpaired electrons.

231
00:18:05,000 --> 00:18:10,000
The system is paramagnetic.
You could certainly calculate

232
00:18:10,000 --> 00:18:14,000
the spin-only magnetic moment
for such a system.

233
00:18:14,000 --> 00:18:19,000
And e(g)* is not occupied.
We have three electrons,

234
00:18:19,000 --> 00:18:22,000
here, in t(2g).
t(2g) is basically nonbonding

235
00:18:22,000 --> 00:18:28,000
if the ligand is neither a
pi-base nor a pi-acid.

236
00:18:28,000 --> 00:18:33,000
This is three nonbonding
electrons centered in d(xz),

237
00:18:33,000 --> 00:18:39,000
d(yz), and d(xy),
which is t(2g) on our metal ion

238
00:18:39,000 --> 00:18:45,000
chromium three plus
in the hexa-aquo system or

239
00:18:45,000 --> 00:18:51,000
vanadium two plus in
the hexa-aquo system.

240
00:18:51,000 --> 00:18:57,000
And these are inert and slow.
And then, we have two more

241
00:18:57,000 --> 00:19:02,000
cases to discuss.
And, again, here is our t(2g)

242
00:19:02,000 --> 00:19:07,000
and our e(g)*.
Notice that if we go from

243
00:19:07,000 --> 00:19:11,000
chromium three plus
to chromium two plus,

244
00:19:11,000 --> 00:19:16,000
we go up 15 orders
of magnitude in ligand

245
00:19:16,000 --> 00:19:21,000
substitution rate just by
changing the charge by one unit.

246
00:19:21,000 --> 00:19:25,000
And what is going on here?
Well, now chromium two

247
00:19:25,000 --> 00:19:30,000
is d four,
and the four electrons go in

248
00:19:30,000 --> 00:19:36,000
and populate like this.
So eg* now has an electron in

249
00:19:36,000 --> 00:19:39,000
it.
And that makes it related to

250
00:19:39,000 --> 00:19:43,000
the excited state here of
tungsten hexacarbonyl because

251
00:19:43,000 --> 00:19:47,000
both of those,
if you were to write out their

252
00:19:47,000 --> 00:19:51,000
electronic configuration,
would have eg*1 as the

253
00:19:51,000 --> 00:19:55,000
population of eg*.
In the case of copper two plus,

254
00:19:55,000 --> 00:20:01,000
we have a situation
similar to chromium two plus

255
00:20:01,000 --> 00:20:06,000
in that the complex
is labile.

256
00:20:06,000 --> 00:20:12,000
It is undergoing substitution
associated with a rate constant

257
00:20:12,000 --> 00:20:18,000
here whose log is 9.
This is undergoing exchange of

258
00:20:18,000 --> 00:20:24,000
water molecules from liquid
solution onto the metal very

259
00:20:24,000 --> 00:20:28,000
rapidly.
And this is a d nine

260
00:20:28,000 --> 00:20:32,000
case.
The electrons go in like this.

261
00:20:32,000 --> 00:20:36,000
And this is an s equals
one-half system with a single

262
00:20:36,000 --> 00:20:41,000
unpaired electron.
Both of these are paramagnetic.

263
00:20:41,000 --> 00:20:45,000
Both of these have a population
of e g star.

264
00:20:45,000 --> 00:20:50,000
And the two ions that have
e(g)* that contain electrons

265
00:20:50,000 --> 00:20:55,000
have fast ligand exchange rates.
When e(g)* is not occupied,

266
00:20:55,000 --> 00:21:00,000
the ligand substitution rates
are very slow.

267
00:21:00,000 --> 00:21:05,000
And we can understand that
because of the antibonding

268
00:21:05,000 --> 00:21:11,000
character of e(g)*.
There is an additional nuance,

269
00:21:11,000 --> 00:21:17,000
which is when e(g)* is occupied
by an odd number of electrons,

270
00:21:17,000 --> 00:21:23,000
we can get what is called a
Jahn-Teller distortion.

271
00:21:23,000 --> 00:21:29,000
And that distortion is a
structural response to the odd

272
00:21:29,000 --> 00:21:35,000
electronic population of e(g)*.
It is a distortion of the

273
00:21:35,000 --> 00:21:39,000
molecule that makes these
systems, the chromium two

274
00:21:39,000 --> 00:21:43,000
and copper two
the fastest of all

275
00:21:43,000 --> 00:21:46,000
the 3d metal ions for exchanging
water molecules.

276
00:21:46,000 --> 00:21:50,000
Let's look at a structure of
one of these systems briefly.

277
00:21:50,000 --> 00:21:54,000
Here is the result of a quantum
chemical calculation optimizing

278
00:21:54,000 --> 00:21:57,000
the structure of chromium two
hexaaquo two plus.

279
00:21:57,000 --> 00:22:01,000
This system as a spin

280
00:22:01,000 --> 00:22:04,000
polarization of four,
meaning we do have the four

281
00:22:04,000 --> 00:22:08,000
unpaired electrons,
three of which are in t(2g) and

282
00:22:08,000 --> 00:22:12,000
one of which is in e(g)*.
This is how the molecule

283
00:22:12,000 --> 00:22:16,000
optimizes without using any
symmetry constraints based on

284
00:22:16,000 --> 00:22:20,000
the density functional theory
model that is typically in use

285
00:22:20,000 --> 00:22:23,000
these days for optimizing
geometries of molecules.

286
00:22:23,000 --> 00:22:27,000
And what you can see is that
this program I am using to

287
00:22:27,000 --> 00:22:31,000
display the structure is drawing
four of the lines from the

288
00:22:31,000 --> 00:22:37,000
chromium ion to the oxygens.
And the other two it is not

289
00:22:37,000 --> 00:22:39,000
drawing.
And the reason for that,

290
00:22:39,000 --> 00:22:44,000
if I go ahead and use the tools
in this program to find out just

291
00:22:44,000 --> 00:22:49,000
what are those chromium oxygen
distances in this hexaaquo ion,

292
00:22:49,000 --> 00:22:54,000
what we find out is that four
of them, the four where it is

293
00:22:54,000 --> 00:22:58,000
choosing to draw the lines
between the metal and the

294
00:22:58,000 --> 00:23:04,000
oxygen, are about 2.1 angstroms.
Whereas, to the two oxygens

295
00:23:04,000 --> 00:23:08,000
where it is not drawing the
metal oxygen line,

296
00:23:08,000 --> 00:23:11,000
the distance is about 2.45
angstroms.

297
00:23:11,000 --> 00:23:14,000
This molecule,
this hexaaquo ion,

298
00:23:14,000 --> 00:23:20,000
has decided to deviate from a
pure octahedral geometry because

299
00:23:20,000 --> 00:23:24,000
what it has done is it has
stretched two of the oxygens

300
00:23:24,000 --> 00:23:30,000
along an axis and made those two
water molecules move farther

301
00:23:30,000 --> 00:23:36,000
away from the metal center.
And that has to do with the

302
00:23:36,000 --> 00:23:40,000
Jahn-Teller distortion that I
mentioned a moment ago,

303
00:23:40,000 --> 00:23:45,000
when e(g)* is occupied by an
odd number of electrons.

304
00:23:45,000 --> 00:23:50,000
Now, if you see that two waters
have moved far away from the

305
00:23:50,000 --> 00:23:55,000
metal and four are in closer as
a result of how e(g)* is

306
00:23:55,000 --> 00:24:00,000
populated, then which d orbital
do you think has the e(g)*

307
00:24:00,000 --> 00:24:03,000
electron?
d z squared*,

308
00:24:03,000 --> 00:24:06,000
that is right.
And that means the two

309
00:24:06,000 --> 00:24:11,000
molecules that can interact with
the two big lobes of d z squared

310
00:24:11,000 --> 00:24:14,000
along the axis,
those are repelled away from

311
00:24:14,000 --> 00:24:19,000
the metal more than the other
four that are in the x,y-plane

312
00:24:19,000 --> 00:24:23,000
because we don't have any
electron in x squared minus y

313
00:24:23,000 --> 00:24:28,000
squared,
that other component of e(g)*.

314
00:24:28,000 --> 00:24:33,000
This molecule achieves its most
stable structure by distorting

315
00:24:33,000 --> 00:24:37,000
in response to the unequal
occupation of e(g)*.

316
00:24:37,000 --> 00:24:42,000
And I will just show you what
that orbital looks like by

317
00:24:42,000 --> 00:24:45,000
quitting out of there.

318
00:24:50,000 --> 00:24:53,000
Now this graphic,
sadly, isn't as nice as the one

319
00:24:53,000 --> 00:24:57,000
I could have shown you using my
laptop and VMD,

320
00:24:57,000 --> 00:25:00,000
but it will have to serve the
purpose.

321
00:25:00,000 --> 00:25:03,000
And this is just a snapshot of
the orbital.

322
00:25:03,000 --> 00:25:08,000
It may not be possible to
increase the size of this.

323
00:25:15,000 --> 00:25:17,000
Here we go.
I think it should be clear,

324
00:25:17,000 --> 00:25:21,000
from your inspection of this
diagram, that we actually

325
00:25:21,000 --> 00:25:26,000
superposed a couple of different
representations of the system on

326
00:25:26,000 --> 00:25:28,000
top of one another.
First of all,

327
00:25:28,000 --> 00:25:32,000
this program has drawn balls
and sticks to connect the atoms.

328
00:25:32,000 --> 00:25:35,000
You have a red ball here,
a red ball.

329
00:25:35,000 --> 00:25:38,000
Each place there is a red ball,
that is an oxygen.

330
00:25:38,000 --> 00:25:42,000
And at the center,
we have our chromium plus two

331
00:25:42,000 --> 00:25:44,000
ion.
So these are our water

332
00:25:44,000 --> 00:25:49,000
molecules.
And what you should be able to

333
00:25:49,000 --> 00:25:53,000
see here is that this orbital,
on this water molecule up here

334
00:25:53,000 --> 00:25:56,000
along positive z,
has an appearance that would

335
00:25:56,000 --> 00:26:01,000
suggest that that is the highest
occupied molecular orbital of

336
00:26:01,000 --> 00:26:04,000
the water molecule,
interacting with the big lobe

337
00:26:04,000 --> 00:26:08,000
of d z squared in
this antibonding fashion where

338
00:26:08,000 --> 00:26:12,000
we go from blue phase to red
phase as we go along the sigma

339
00:26:12,000 --> 00:26:17,000
axis between the oxygen and the
metal center.

340
00:26:17,000 --> 00:26:19,000
And then, in a better
representation,

341
00:26:19,000 --> 00:26:22,000
you would be able to see that
there are small contributions

342
00:26:22,000 --> 00:26:26,000
from the four water molecules
that are interacting here with

343
00:26:26,000 --> 00:26:29,000
the torus of the d z squared
orbital.

344
00:26:29,000 --> 00:26:33,000
And then what you have down on
negative z is equal to what you

345
00:26:33,000 --> 00:26:37,000
have on positive z.
Namely, you have an oxygen lone

346
00:26:37,000 --> 00:26:41,000
pair here that is being repelled
by the electron occupying d z

347
00:26:41,000 --> 00:26:43,000
squared,
which is one of our e(g)

348
00:26:43,000 --> 00:26:46,000
components.
The idea is that systems like

349
00:26:46,000 --> 00:26:50,000
this exchange water so very
rapidly, 15 orders of magnitude

350
00:26:50,000 --> 00:26:53,000
more rapidly than a simple d
three system,

351
00:26:53,000 --> 00:26:57,000
because two of the water
molecules are repelled away from

352
00:26:57,000 --> 00:27:01,000
the metal center quite a bit.
And it makes this structure

353
00:27:01,000 --> 00:27:05,000
close to the transition state
for just losing a water molecule

354
00:27:05,000 --> 00:27:09,000
from the coordination sphere and
dissociating it from the metal

355
00:27:09,000 --> 00:27:11,000
center.
And that is what we are going

356
00:27:11,000 --> 00:27:13,000
to talk about next.

357
00:27:36,000 --> 00:27:41,000
Now, we are going to talk about
a reaction like that in terms of

358
00:27:41,000 --> 00:27:44,000
a potential energy surface
diagram.

359
00:28:06,000 --> 00:28:10,000
We have seen how d orbital
occupation affects color and

360
00:28:10,000 --> 00:28:13,000
magnetism and bonding between
metals and ligands.

361
00:28:13,000 --> 00:28:17,000
And now we are seeing how
actually we can interpret

362
00:28:17,000 --> 00:28:21,000
chemical reaction rates on the
basis of these electronic

363
00:28:21,000 --> 00:28:24,000
structure considerations.

364
00:28:31,000 --> 00:28:35,000
We like to draw energy level
diagrams that have orbitals on

365
00:28:35,000 --> 00:28:39,000
them or states on them,
but sometimes what we like to

366
00:28:39,000 --> 00:28:43,000
do is to try to describe the
total energy of the system,

367
00:28:43,000 --> 00:28:47,000
fold all these different
coordinates into one and look at

368
00:28:47,000 --> 00:28:51,000
the total energy of the system
as a function of where we are

369
00:28:51,000 --> 00:28:55,000
along a chemical reaction
coordinate.

370
00:28:55,000 --> 00:29:00,000
That means as we progress
through a sequence of elementary

371
00:29:00,000 --> 00:29:05,000
steps corresponding to a
chemical reaction.

372
00:29:05,000 --> 00:29:11,000
A complete description of any
reaction implies a knowledge of

373
00:29:11,000 --> 00:29:17,000
the potential energy surface for
the reaction.

374
00:29:22,000 --> 00:29:25,000
We have our reaction
coordinate, and we desire some

375
00:29:25,000 --> 00:29:29,000
kind of a plot that could
represent this reaction

376
00:29:29,000 --> 00:29:32,000
coordinate.
And what that plot looks like

377
00:29:32,000 --> 00:29:36,000
depends on the particular
reaction mechanism of the

378
00:29:36,000 --> 00:29:40,000
reaction that you are
scrutinizing.

379
00:29:40,000 --> 00:29:43,000
If you have a reaction and you
want to know how it works,

380
00:29:43,000 --> 00:29:45,000
the first thing you do is make
a hypothesis.

381
00:29:45,000 --> 00:29:49,000
And from your hypothesis,
you can build predictions about

382
00:29:49,000 --> 00:29:52,000
the reaction coordinate,
and then you can test those

383
00:29:52,000 --> 00:29:55,000
experimentally.
And either your hypothesis will

384
00:29:55,000 --> 00:29:59,000
turn out to be consistent with
the predictions or not.

385
00:29:59,000 --> 00:30:02,000
And, if not,
you reject the hypothesis and

386
00:30:02,000 --> 00:30:04,000
start again.
It has been said,

387
00:30:04,000 --> 00:30:08,000
concerning reaction mechanisms,
that the closest you can get to

388
00:30:08,000 --> 00:30:11,000
the truth is your own best
guess.

389
00:30:11,000 --> 00:30:15,000
But the way that people test
reaction mechanisms is through

390
00:30:15,000 --> 00:30:19,000
studying reaction rates under
various conditions and comparing

391
00:30:19,000 --> 00:30:23,000
the results to those predicted
by your hypothesis.

392
00:30:23,000 --> 00:30:28,000
And so here is a hypothesis for
a dissociative mechanism for the

393
00:30:28,000 --> 00:30:32,000
type of ligand substitution
reaction that we are talking

394
00:30:32,000 --> 00:30:37,000
about.
A dissociative mechanism means

395
00:30:37,000 --> 00:30:44,000
that the mechanism proceeds by
dissociation of one of the

396
00:30:44,000 --> 00:30:52,000
ligands from the metal center as
the first important step of the

397
00:30:52,000 --> 00:30:56,000
reaction.
We have down here some ML five

398
00:30:56,000 --> 00:31:02,000
X compound.
And let's say we have it in a

399
00:31:02,000 --> 00:31:08,000
solvent Y, where the solvent can
also act as a ligand.

400
00:31:08,000 --> 00:31:14,000
And we are interested in what
rate profile will we have if,

401
00:31:14,000 --> 00:31:19,000
at the end of the reaction,
down here in this well,

402
00:31:19,000 --> 00:31:25,000
this is meant to represent a
potential energy well where now

403
00:31:25,000 --> 00:31:29,000
we have X replaced by Y,
--

404
00:31:35,000 --> 00:31:38,000
-- as illustrated there.
That is the overall result of

405
00:31:38,000 --> 00:31:41,000
the reaction.
But how does it take place?

406
00:31:41,000 --> 00:31:44,000
Are there any intermediates in
the reaction?

407
00:31:44,000 --> 00:31:48,000
And so, in a dissociative
mechanism, what happens is that

408
00:31:48,000 --> 00:31:52,000
there is an intermediate.
And it occurs here in this

409
00:31:52,000 --> 00:31:55,000
intermediate well.
One thing that you are going to

410
00:31:55,000 --> 00:32:00,000
be interested in distinguishing
between, on potential energy

411
00:32:00,000 --> 00:32:04,000
diagrams, will be potential
energy minima.

412
00:32:04,000 --> 00:32:08,000
Because these potential energy
minima that I have just colored

413
00:32:08,000 --> 00:32:11,000
in, those correspond to starting
materials, products,

414
00:32:11,000 --> 00:32:14,000
or intermediates.
And then, alternatively,

415
00:32:14,000 --> 00:32:18,000
you can have potential energy
maxima on the potential energy

416
00:32:18,000 --> 00:32:20,000
surface.
This is a two-dimensional

417
00:32:20,000 --> 00:32:22,000
diagram.
And what it really represents

418
00:32:22,000 --> 00:32:26,000
is a two-dimensional slice
through a three-dimensional

419
00:32:26,000 --> 00:32:30,000
surface that has hills and
valleys on it.

420
00:32:30,000 --> 00:32:34,000
And these hills on the
potential energy surface

421
00:32:34,000 --> 00:32:38,000
represent energy barriers to a
chemical reaction.

422
00:32:38,000 --> 00:32:44,000
What we are going to have at
these points of high energy will

423
00:32:44,000 --> 00:32:49,000
be things that we call
transition states.

424
00:32:59,000 --> 00:33:03,000
And you start out down here at
the starting materials,

425
00:33:03,000 --> 00:33:07,000
the reaction will proceed to
the right, we are assuming in

426
00:33:07,000 --> 00:33:11,000
this case, and you go up hill in
energy first.

427
00:33:11,000 --> 00:33:15,000
You have to surmount this
energy barrier associated with

428
00:33:15,000 --> 00:33:18,000
this first transition state,
or TS1.

429
00:33:18,000 --> 00:33:22,000
And when you system has gone
over that first barrier,

430
00:33:22,000 --> 00:33:26,000
if it is a dissociative
mechanism, that produces an

431
00:33:26,000 --> 00:33:32,000
intermediate ML five.
Because X has dissociated from

432
00:33:32,000 --> 00:33:36,000
the metal center and Y,
of course, being the solvent,

433
00:33:36,000 --> 00:33:40,000
is still there.
And then, once we get down into

434
00:33:40,000 --> 00:33:45,000
this well, this potential energy
minimum, this valley associated

435
00:33:45,000 --> 00:33:50,000
with the intermediate ML five,
it is a five coordinate

436
00:33:50,000 --> 00:33:53,000
intermediate,
so it should have a structure

437
00:33:53,000 --> 00:34:00,000
of the trigonal bipyramid or the
structure of a square pyramid.

438
00:34:00,000 --> 00:34:02,000
And, over there,
I was talking about the

439
00:34:02,000 --> 00:34:06,000
distortion of that molecule
based on its occupation of e(g)*

440
00:34:06,000 --> 00:34:10,000
having repelled two of the water
molecules farther away from the

441
00:34:10,000 --> 00:34:14,000
metal than the other four.
And what that does is it begins

442
00:34:14,000 --> 00:34:18,000
to lower the energy barrier here
to get over to TS1.

443
00:34:18,000 --> 00:34:22,000
It makes it easier for one
ligand to fall off the metal

444
00:34:22,000 --> 00:34:25,000
center because it is being
repelled by that d z squared

445
00:34:25,000 --> 00:34:30,000
electron.
And what we will see is that we

446
00:34:30,000 --> 00:34:33,000
can associate rate constants
with each of these barriers.

447
00:34:33,000 --> 00:34:37,000
This is the second instance
today that I have used the term

448
00:34:37,000 --> 00:34:40,000
rate constant.
And we are going to want to

449
00:34:40,000 --> 00:34:43,000
distinguish the term "rate
constant" from the term

450
00:34:43,000 --> 00:34:46,000
"reaction rate."
And one of our goals today is

451
00:34:46,000 --> 00:34:50,000
to understand the difference
between a reaction rate and a

452
00:34:50,000 --> 00:34:53,000
rate constant.
Going up this first hill,

453
00:34:53,000 --> 00:34:56,000
over TS1 and down into the
intermediate state,

454
00:34:56,000 --> 00:35:00,000
where you have a five
coordinate species and X has

455
00:35:00,000 --> 00:35:03,000
become freely dissociated from
the metal center,

456
00:35:03,000 --> 00:35:06,000
that is k1.
And then, when you are down in

457
00:35:06,000 --> 00:35:09,000
the valley, there are two ways
out of the valley.

458
00:35:09,000 --> 00:35:12,000
You can either go back in the
direction of TS1.

459
00:35:12,000 --> 00:35:16,000
We call that back reaction k
minus one.

460
00:35:16,000 --> 00:35:19,000
We associate the rate constant
k minus one with the back

461
00:35:19,000 --> 00:35:24,000
reaction that would take us back
over the same transition state

462
00:35:24,000 --> 00:35:28,000
and back to the starting
materials.

463
00:35:28,000 --> 00:35:32,000
That would be a nonproductive
step because it is going in the

464
00:35:32,000 --> 00:35:35,000
wrong direction.
And then, over here,

465
00:35:35,000 --> 00:35:39,000
ML five has this
branching option.

466
00:35:39,000 --> 00:35:42,000
When you are at the stage of
the intermediate,

467
00:35:42,000 --> 00:35:47,000
you can go either back to
starting materials or you can go

468
00:35:47,000 --> 00:35:51,000
onto products.
And this rate constant we will

469
00:35:51,000 --> 00:35:55,000
call k two for the
reaction in which the molecule Y

470
00:35:55,000 --> 00:36:02,000
that is doing the substituting
would add to ML five.

471
00:36:02,000 --> 00:36:04,000
That is associated with some
barrier.

472
00:36:04,000 --> 00:36:08,000
And it produces the six
coordinate ML five Y and X

473
00:36:08,000 --> 00:36:12,000
dissociated.
And then, finally,

474
00:36:12,000 --> 00:36:17,000
in this last step we would call
the rate constant associated

475
00:36:17,000 --> 00:36:21,000
with going back into the
intermediate from the products,

476
00:36:21,000 --> 00:36:25,000
we would call that k minus two
for that rate

477
00:36:25,000 --> 00:36:28,000
constant.
This is how you begin to write

478
00:36:28,000 --> 00:36:35,000
out the math associated with a
hypothetical reaction mechanism.

479
00:36:35,000 --> 00:36:38,000
And so, for those of you who
are gearing up to take 18.03

480
00:36:38,000 --> 00:36:41,000
next semester,
Differential Equations,

481
00:36:41,000 --> 00:36:45,000
you are going to find that this
is one area in chemistry where

482
00:36:45,000 --> 00:36:48,000
differential equations become
extremely valuable.

483
00:36:48,000 --> 00:36:51,000
Because you are going to
generate differential equations

484
00:36:51,000 --> 00:36:55,000
that describe the kinetics of a
chemical reaction system.

485
00:36:55,000 --> 00:36:59,000
And those equations will differ
depending on the hypothetical

486
00:36:59,000 --> 00:37:03,000
mechanism.
And you are going to want to

487
00:37:03,000 --> 00:37:09,000
figure out which hypothetical
mechanism actually fits your

488
00:37:09,000 --> 00:37:12,000
data.
Here is one limiting case,

489
00:37:12,000 --> 00:37:15,000
but let me first just --

490
00:37:25,000 --> 00:37:29,000
I am going to leave that,
given the amount of time here.

491
00:37:29,000 --> 00:37:34,000
We are going to go ahead and
talk about a limiting case.

492
00:38:00,000 --> 00:38:04,000
When I use square brackets
here, I am going to be talking

493
00:38:04,000 --> 00:38:07,000
about concentration,
just like we did when we were

494
00:38:07,000 --> 00:38:10,000
talking about acids and bases
and so forth.

495
00:38:10,000 --> 00:38:14,000
Square brackets denote
concentrations.

496
00:38:14,000 --> 00:38:19,000
If k2 times Y,
and I will stick with the green

497
00:38:19,000 --> 00:38:26,000
color for Y, is much greater
than k minus one times the

498
00:38:26,000 --> 00:38:34,000
concentration of X,
hat corresponds to

499
00:38:34,000 --> 00:38:39,000
a limiting case.
We are also going to make

500
00:38:39,000 --> 00:38:42,000
another assumption,
here.

501
00:38:42,000 --> 00:38:50,000
We are going to say assuming
complete reaction.

502
00:38:58,000 --> 00:39:01,000
What does that mean?
That means that k minus two

503
00:39:01,000 --> 00:39:06,000
times the concentration of ML
five Y times the concentration

504
00:39:06,000 --> 00:39:11,000
of X at the end
corresponds to essentially

505
00:39:11,000 --> 00:39:16,000
a negligible back reaction such
that when we get to products we

506
00:39:16,000 --> 00:39:21,000
stop, this is a very deep well,
and we don't go back anymore

507
00:39:21,000 --> 00:39:24,000
once we get over to the
products.

508
00:39:24,000 --> 00:39:27,000
Under those conditions,
we have made two

509
00:39:27,000 --> 00:39:32,000
simplifications.
Then, the reaction rate.

510
00:39:37,000 --> 00:39:40,000
And, when we talk about
reaction rate,

511
00:39:40,000 --> 00:39:44,000
we usually define it in terms
of the disappearance of the

512
00:39:44,000 --> 00:39:49,000
starting materials or
alternatively the appearance of

513
00:39:49,000 --> 00:39:52,000
the products,
the rate of appearance of the

514
00:39:52,000 --> 00:39:57,000
products in a system like this.
And, under the conditions of

515
00:39:57,000 --> 00:40:01,000
those assumptions,
the reaction rate is going to

516
00:40:01,000 --> 00:40:06,000
depend on k1 times the
concentration of the starting

517
00:40:06,000 --> 00:40:08,000
material.

518
00:40:14,000 --> 00:40:19,000
Times the concentration of our
entering group Y.

519
00:40:19,000 --> 00:40:25,000
And that says that every time
we get over the first barrier,

520
00:40:25,000 --> 00:40:32,000
we rapidly get over the second
barrier without going back.

521
00:40:32,000 --> 00:40:36,000
That is the assumption here
that k two times the

522
00:40:36,000 --> 00:40:41,000
concentration of Y is much
greater than k minus one times

523
00:40:41,000 --> 00:40:46,000
the concentration of X.
This is

524
00:40:46,000 --> 00:40:50,000
often true when the
concentration of Y is very

525
00:40:50,000 --> 00:40:52,000
great.

526
00:40:58,000 --> 00:41:01,000
i.e., Y is the solvent.

527
00:41:06,000 --> 00:41:10,000
In the case of water,
you have essentially 12 molar

528
00:41:10,000 --> 00:41:11,000
Y.
Y is your solvent.

529
00:41:11,000 --> 00:41:15,000
And, because of that,
under these circumstances,

530
00:41:15,000 --> 00:41:19,000
the concentration of Y,
which is very large,

531
00:41:19,000 --> 00:41:23,000
does not change effectively
during the reaction.

532
00:41:23,000 --> 00:41:28,000
It is a constant that can be
folded into together with the

533
00:41:28,000 --> 00:41:33,000
constant k1.
So we can describe this as k

534
00:41:33,000 --> 00:41:38,000
observed.
K observed is equal to k1 times

535
00:41:38,000 --> 00:41:43,000
Y times ML five X. And

536
00:41:43,000 --> 00:41:50,000
these are what we call
pseudo-first order conditions.

537
00:42:16,000 --> 00:42:18,000
That is a particularly simple
rate law.

538
00:42:18,000 --> 00:42:22,000
And a lot of times,
experimentalists will engender

539
00:42:22,000 --> 00:42:27,000
pseudo-first order conditions by
purposely using an entering

540
00:42:27,000 --> 00:42:31,000
group concentration that is very
large.

541
00:42:31,000 --> 00:42:36,000
And, when you do that,
you will find that,

542
00:42:36,000 --> 00:42:40,000
for example,
the change in concentration

543
00:42:40,000 --> 00:42:47,000
with time of ML five X
is, upon integration,

544
00:42:47,000 --> 00:42:53,000
when you integrate differential
equations such as this,

545
00:42:53,000 --> 00:43:00,000
you can obtain expressions for
the concentrations of these

546
00:43:00,000 --> 00:43:07,000
species of interest as a
function of time.

547
00:43:07,000 --> 00:43:17,000
So we can obtain ML five X of t
divided by ML five X at t equals

548
00:43:17,000 --> 00:43:26,000
zero is equal to e to the minus
k observed, that contains that

549
00:43:26,000 --> 00:43:36,000
large invariant concentration of
y folded into it times t.

550
00:43:44,000 --> 00:43:48,000
Hence, this single exponential
decay of the starting materials

551
00:43:48,000 --> 00:43:53,000
predicted by this mechanism
under these conditions for the

552
00:43:53,000 --> 00:43:58,000
change in concentration with
time of your starting material.

553
00:44:04,000 --> 00:44:12,000
Finally, what you do seek is to
follow, as a function of time,

554
00:44:12,000 --> 00:44:18,000
the concentration of the
species involved such that,

555
00:44:18,000 --> 00:44:23,000
for example,
you may be following,

556
00:44:23,000 --> 00:44:29,000
here, the decay of your
starting species ML five X.

557
00:44:29,000 --> 00:44:33,000
As a function of time,

558
00:44:33,000 --> 00:44:36,000
you seek an expression for
that.

559
00:44:36,000 --> 00:44:41,000
There are different methods to
carry out the integration

560
00:44:41,000 --> 00:44:46,000
required for more complicated
rate laws than did appear there

561
00:44:46,000 --> 00:44:49,000
for the pseudo first-order
mechanism.

562
00:44:49,000 --> 00:44:53,000
But, simultaneously,
as ML five X decays,

563
00:44:53,000 --> 00:45:00,000
the concentration of ML five Y
would be increasing.

564
00:45:00,000 --> 00:45:06,000
And under the simple case shown
here, where the intermediate

565
00:45:06,000 --> 00:45:12,000
does not build up to any
substantial concentration,

566
00:45:12,000 --> 00:45:16,000
this is no buildup of
intermediate,

567
00:45:16,000 --> 00:45:22,000
meaning whenever the
intermediate is formed it either

568
00:45:22,000 --> 00:45:30,000
goes back to starting materials
or goes on to products.

569
00:45:35,000 --> 00:46:03,000
In other words,
the intermediate never attains

570
00:46:03,000 --> 00:46:38,000
significant concentrations
relative to starting materials

571
00:46:38,000 --> 00:47:06,000
and final products.
This is the basis for the

572
00:47:06,000 --> 00:47:43,000
steady state approximation that
is explained in your notes.

573
00:47:43,000 --> 00:48:22,000
And I will go through that also
at the beginning of the hour on

574
00:48:22,000 --> 00:48:25,000
Friday.
Have a nice day.