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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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00:00:22 --> 00:00:26
You will see here that I have
colored one of the water

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

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00:00:30 --> 00:00:34
talk about reactions today.
In fact, I am going to

10
00:00:34 --> 00:00:38
introduce you to the idea of a
potential energy surface for a

11
00:00:38 --> 00:00:41
reaction.
I am taking water substitution

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00:00:41 --> 00:00:46
as an example and as a context
for this, but this very general.

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00:00:46 --> 00:00:50
And the same ideas that we are
going to be talking about today

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00:00:50 --> 00:00:54
would apply to reactions in
organic chemistry or in

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biochemistry.
Let's imagine having this

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00:00:56 --> 00:01:00
system dissolved in liquid
water.

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00:01:00 --> 00:01:04
And let's further imagine that
we have some way,

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

19
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differentiate from those that
start out on the metal complex.

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00:01:14 --> 00:01:19
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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00:01:23 --> 00:01:27
And in this reaction,
it is a very simple reaction in

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which liquid water from
solution, --

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00:01:32 --> 00:01:36
-- we are going to talk about
reaction mechanisms today,

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00:01:36 --> 00:01:40
somehow will come in and bind
to the metal center and displace

26
00:01:40 --> 00:01:43
one water molecule from the
metal center.

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00:01:43 --> 00:01:48
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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00:01:52 --> 00:01:56
And that gives us our product,
which is very similar to the

30
00:01:56 --> 00:02:00
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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00:02:07 --> 00:02:15


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

35
00:02:19 --> 00:02:24
which 3d transition element we
have at the center of the

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00:02:24 --> 00:02:29
octahedron, we find that the
time constant for this reaction

37
00:02:29 --> 00:02:33
taking place varies quite
dramatically.

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

39
00:02:38 --> 00:02:41
the reaction,
as we will discuss,

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00:02:41 --> 00:02:46
and to the electronic structure
attributes of the systems that

41
00:02:46 --> 00:02:50
control the way a metal binds to
its ligands.

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

43
00:02:54 --> 00:02:57
the metal n plus.

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00:02:57 --> 00:03:04


45
00:03:04 --> 00:03:09
And we will take the log of the
rate constant for the reaction,

46
00:03:09 --> 00:03:11
which I will denote with this
k.

47
00:03:11 --> 00:03:15
Small k is going to be our rate
constant.

48
00:03:15 --> 00:03:19
And the units that we would
find for such a rate constant

49
00:03:19 --> 00:03:24
would be reciprocal seconds to
describe how fast this takes

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00:03:24 --> 00:03:28
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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00:03:34 --> 00:03:37
the rate constant is
approximately minus 6.

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

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00:03:42 --> 00:03:45
the rate constant log goes to
minus 2.

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00:03:45 --> 00:03:50
So the reaction is faster than
it is for chromium three.

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00:03:50 --> 00:03:54
And then, if we look at some of

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the other d ions that are
possible here,

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00:03:57 --> 00:04:03
we could have chromium two plus
or copper two plus.

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00:04:03 --> 00:04:08
And, for both of those,

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00:04:08.082 --> 9.
the log of the rate constant is

61
9. --> 00:04:11


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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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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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00:05:19 --> 00:05:23
Now, I would like you to
consider a different type of

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00:05:23 --> 00:05:26
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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00:06:13 --> 00:06:18
as you will be seeing in
recitation this week,

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00:06:18 --> 00:06:24
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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00:06:39 --> 00:06:44


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00:06:44 --> 00:06:50
And because ligands are
pi-acidic and the metal center,

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00:06:50 --> 00:06:54
as you will see,
this metal center is a d six

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00:06:54 --> 00:06:59
pi-base,
the metal itself is very

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00:06:59 --> 00:07:06
electronically complimentary to
its set of six carbon monoxide

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00:07:06 --> 00:07:10
ligands.
So a system like this is very

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stable.
And what you might like to do

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00:07:12 --> 00:07:17
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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00:07:25 --> 00:07:30
biological process that has that
thing attached to it.

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00:07:30 --> 00:07:34
And one example of a
substitution reaction you might

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00:07:34 --> 00:07:38
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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00:07:48 --> 00:07:52
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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00:07:56 --> 00:08:00
This ligand here,
then, is triphenylphosphine,

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00:08:00 --> 00:08:07
and this is a good ligand for
various types of metals.

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00:08:07 --> 00:08:17


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And, in fact,
one of the Nobel laureates that

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00:08:20 --> 00:08:23
I may have mentioned in passing
this semester,

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00:08:23 --> 00:08:27
Sir Jeffrey Wilkinson,
used triphenylphosphine as a

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00:08:27 --> 00:08:31
ligand in the preparation of
what is now universally known as

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00:08:31 --> 00:08:35
Wilkinson's hydrogenation
catalyst.

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00:08:35 --> 00:08:39
So this ligand is quite common,
indeed, and well-known.

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00:08:39 --> 00:08:44
And what we might like to do is
to replace one carbon monoxide

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00:08:44 --> 00:08:47
with this triphenylphosphine
ligand.

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

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00:08:51 --> 00:08:57
We would like one equivalent of
carbon monoxide to be evolved as

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00:08:57 --> 00:09:02
CO gas in this reaction.
But we find that we can put

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

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00:09:07 --> 00:09:12
and nothing really happens.
They just sit there and look at

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00:09:12 --> 00:09:14
each other.
They do not react.

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00:09:14 --> 00:09:19
So one of the things that
happens when you are learning to

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00:09:19 --> 00:09:25
do synthetic chemistry is that
the outcome of a reaction can be

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00:09:25 --> 00:09:30
either that it proceeded cleanly
onto products.

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00:09:30 --> 00:09:35
And one possible outcome is
there was no reaction at all.

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00:09:35 --> 00:09:39
And so this is what we will
call no thermal reaction.

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00:09:39 --> 00:09:45
And that means that just with
the energy supplied by the

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00:09:45 --> 00:09:49
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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00:09:55 --> 00:10:00
there is an energy barrier to
this reaction.

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

139
00:10:05 --> 00:10:09
a few moments,
but there may be an energy

140
00:10:09 --> 00:10:13
barrier.
And we don't have sufficient

141
00:10:13 --> 00:10:18
energy to surmount the barrier,
so the substitution reaction is

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

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

144
00:10:27 --> 00:10:32
takes place.
And we form this desired

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

146
00:10:37 --> 00:10:42
with triphenylphosphine.
That would be the initial

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

148
00:10:46 --> 00:10:51


149
00:10:51 --> 00:10:54
Still octahedral,
now five CO ligands instead of

150
00:10:54 --> 00:10:59
six, and one of them replaced by
triphenylphosphine.

151
00:10:59 --> 00:11:03
And it occurs under the action
of UV light.

152
00:11:03 --> 00:11:06
If you put a photon into the
system for some reason,

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00:11:06 --> 00:11:10
we can now surmount the energy
barrier to the reaction,

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

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

156
00:11:16 --> 00:11:20
possibly be bubbling an inert
gas like N two through

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

158
00:11:24 --> 00:11:29
the system to help the reaction
go to completion as we are

159
00:11:29 --> 00:11:33
carrying out such a prep.
But, in any event,

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00:11:33 --> 00:11:38
the key feature is that it goes
under photolysis in the presence

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00:11:38 --> 00:11:43
of a UV lamp but not simply
through input of normal thermal

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00:11:43 --> 00:11:46
energy.
And, in thinking back to the

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00:11:46 --> 00:11:51
development of the electronic
structure of systems of this

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00:11:51 --> 00:11:56
type, we can actually begin to
understand this.

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00:11:56 --> 00:12:01
Because we have an energy
diagram, a d orbital splitting

166
00:12:01 --> 00:12:05
diagram for simplicity,
where we have our t(2g) set and

167
00:12:05 --> 00:12:09
our e(g)*.
And tungsten is in Group 6,

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00:12:09 --> 00:12:14
the ligands are all neutral,
so this is d six.

169
00:12:14 --> 00:12:19
This is a common electron count
for octahedral systems that are

170
00:12:19 --> 00:12:23
quite stable.
And CO is high in the

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00:12:23 --> 00:12:27
spectrochemical series,
so it leads to a large value of

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

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00:12:33 --> 00:12:38
t(2g) and e(g) is large.
And that means we have an s

174
00:12:38 --> 00:12:43
equals zero state,
with the system not being

175
00:12:43 --> 00:12:48
paramagnetic,
all the electrons paired up in

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

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

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

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

180
00:13:09 --> 00:13:14
net spin polarization.
This one is s equals zero,

181
00:13:14 --> 00:13:19
this one is s equals zero to
satisfy the spin selection rule

182
00:13:19 --> 00:13:23
that we have talked about.
But notice, now,

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

184
00:13:26 --> 00:13:34


185
00:13:34 --> 00:13:37
-- we have an electron in e g
star.

186
00:13:37 --> 00:13:42
And, as I was discussing last
time, when you look at the MO

187
00:13:42 --> 00:13:47
theory for a problem like this,
e(g)* is sigma antibonding with

188
00:13:47 --> 00:13:51
respect to the ligands.
I am going to illuminate that

189
00:13:51 --> 00:13:55
for you very clearly in a few
moments with a graphic,

190
00:13:55 --> 00:14:00
but this is sigma star with
respect to the ligands.

191
00:14:00 --> 00:14:03
So what are we doing?
We are taking an electron,

192
00:14:03 --> 00:14:07
here, that is pi bonding
because CO is a pi-acid ligand.

193
00:14:07 --> 00:14:11
In the ground state,
this electron that we are going

194
00:14:11 --> 00:14:16
to promote is helping to bind
the carbon monoxide ligands to

195
00:14:16 --> 00:14:19
the metal.
And then, when we input that

196
00:14:19 --> 00:14:24
photon and we produce an excited
state, that same electron now

197
00:14:24 --> 00:14:29
has been taken away from our sum
total of bonding.

198
00:14:29 --> 00:14:33
And it has been added to an
antibonding orbital.

199
00:14:33 --> 00:14:39
And the effect of that is to
"labialize" one of the carbon

200
00:14:39 --> 00:14:44
monoxide ligands,
allowing it to dissociate from

201
00:14:44 --> 00:14:50
the metal center producing an
unsaturated intermediate that

202
00:14:50 --> 00:14:56
can be captured by the lone pair
of triphenylphosphine.

203
00:14:56 --> 00:15:02
That word I just used,
labialized, we have from Henry

204
00:15:02 --> 00:15:04
Taube.

205
00:15:04 --> 00:15:13


206
00:15:13 --> 00:15:16
Henry Taube,
Nobel Prize in 1983.

207
00:15:16 --> 00:15:20
If you want to read more about
Henry Taube and his

208
00:15:20 --> 00:15:26
contributions to chemistry and,
in fact, the study of ligand

209
00:15:26 --> 00:15:32
substitution reactions and the
relationship of reaction rates

210
00:15:32 --> 00:15:37
to the population of the e g
star orbitals,

211
00:15:37 --> 00:15:43
then I would direct you to the
1983 section of the Nobel Prize

212
00:15:43 --> 00:15:48
dot org website.
And you can read his biography,

213
00:15:48 --> 00:15:51
his Nobel Prize lecture and his
banquet address.

214
00:15:51 --> 00:15:55
These are all things that
Professor Schrock from our

215
00:15:55 --> 00:15:59
department is doing this week in
Sweden, so shortly,

216
00:15:59 --> 00:16:04
you will be able to read his
contributions to that website.

217
00:16:04 --> 00:16:12
And I am sure you will all find
that to be pretty exciting,

218
00:16:12 --> 00:16:15
as I do.
But Henry Taube,

219
00:16:15 --> 00:16:23
this Nobel Prize winner,
coined the terms "inert" and

220
00:16:23 --> 00:16:26
"labile."
These terms,

221
00:16:26 --> 00:16:33
inert and labile,
they refer to rates of ligand

222
00:16:33 --> 00:16:41
substitution reactions.
If a complex is to be thought

223
00:16:41 --> 00:16:44
of as inert, it isn't really
inert.

224
00:16:44 --> 00:16:51
What that means is that its
substitution reactions are slow.

225
00:16:51 --> 00:16:56
And, on the other hand,
if a system is to be referred

226
00:16:56 --> 00:17:03
to labile, that means that its
ligand substitution reactions

227
00:17:03 --> 00:17:07
are fast.
And over there on the left,

228
00:17:07 --> 00:17:13
we looked at a set of four
metal ions whose aqua-complexes

229
00:17:13 --> 00:17:18
could be classified as either
inert or as labile.

230
00:17:18 --> 00:17:23
So let me relate that to the
occupation of the eg star

231
00:17:23 --> 00:17:27
energy level.
Here what we have,

232
00:17:27 --> 00:17:32
t(2g) and e(g)*.
In the case of chromium three

233
00:17:32 --> 00:17:37
plus or vanadium two
plus,

234
00:17:37 --> 00:17:42
since vanadium is in Group 5
and chromium is in Group 6,

235
00:17:42 --> 00:17:48
both of these metal ions have a
d three electron

236
00:17:48 --> 00:17:52
configuration.
And so, when we populate our

237
00:17:52 --> 00:17:57
d-orbital splitting diagram for
these two systems,

238
00:17:57 --> 00:18:01
here is the result.
We have an s equals

239
00:18:01 --> 00:18:05
three-halves ground state,
three unpaired electrons.

240
00:18:05 --> 00:18:10
The system is paramagnetic.
You could certainly calculate

241
00:18:10 --> 00:18:14
the spin-only magnetic moment
for such a system.

242
00:18:14 --> 00:18:19
And e(g)* is not occupied.
We have three electrons,

243
00:18:19 --> 00:18:22
here, in t(2g).
t(2g) is basically nonbonding

244
00:18:22 --> 00:18:28
if the ligand is neither a
pi-base nor a pi-acid.

245
00:18:28 --> 00:18:33
This is three nonbonding
electrons centered in d(xz),

246
00:18:33 --> 00:18:39
d(yz), and d(xy),
which is t(2g) on our metal ion

247
00:18:39 --> 00:18:45
chromium three plus
in the hexa-aquo system or

248
00:18:45 --> 00:18:51
vanadium two plus in
the hexa-aquo system.

249
00:18:51 --> 00:18:57
And these are inert and slow.
And then, we have two more

250
00:18:57 --> 00:19:02
cases to discuss.
And, again, here is our t(2g)

251
00:19:02 --> 00:19:07
and our e(g)*.
Notice that if we go from

252
00:19:07 --> 00:19:11
chromium three plus
to chromium two plus,

253
00:19:11 --> 00:19:16
we go up 15 orders
of magnitude in ligand

254
00:19:16 --> 00:19:21
substitution rate just by
changing the charge by one unit.

255
00:19:21 --> 00:19:25
And what is going on here?
Well, now chromium two

256
00:19:25 --> 00:19:30
is d four,
and the four electrons go in

257
00:19:30 --> 00:19:36
and populate like this.
So eg* now has an electron in

258
00:19:36 --> 00:19:39
it.
And that makes it related to

259
00:19:39 --> 00:19:43
the excited state here of
tungsten hexacarbonyl because

260
00:19:43 --> 00:19:47
both of those,
if you were to write out their

261
00:19:47 --> 00:19:51
electronic configuration,
would have eg*1 as the

262
00:19:51 --> 00:19:55
population of eg*.
In the case of copper two plus,

263
00:19:55 --> 00:20:01
we have a situation
similar to chromium two plus

264
00:20:01 --> 00:20:06
in that the complex
is labile.

265
00:20:06 --> 00:20:12
It is undergoing substitution
associated with a rate constant

266
00:20:12 --> 00:20:18
here whose log is 9.
This is undergoing exchange of

267
00:20:18 --> 00:20:24
water molecules from liquid
solution onto the metal very

268
00:20:24 --> 00:20:28
rapidly.
And this is a d nine

269
00:20:28 --> 00:20:32
case.
The electrons go in like this.

270
00:20:32 --> 00:20:36
And this is an s equals
one-half system with a single

271
00:20:36 --> 00:20:41
unpaired electron.
Both of these are paramagnetic.

272
00:20:41 --> 00:20:45
Both of these have a population
of e g star.

273
00:20:45 --> 00:20:50
And the two ions that have
e(g)* that contain electrons

274
00:20:50 --> 00:20:55
have fast ligand exchange rates.
When e(g)* is not occupied,

275
00:20:55 --> 00:21:00
the ligand substitution rates
are very slow.

276
00:21:00 --> 00:21:05
And we can understand that
because of the antibonding

277
00:21:05 --> 00:21:11
character of e(g)*.
There is an additional nuance,

278
00:21:11 --> 00:21:17
which is when e(g)* is occupied
by an odd number of electrons,

279
00:21:17 --> 00:21:23
we can get what is called a
Jahn-Teller distortion.

280
00:21:23 --> 00:21:29
And that distortion is a
structural response to the odd

281
00:21:29 --> 00:21:35
electronic population of e(g)*.
It is a distortion of the

282
00:21:35 --> 00:21:39
molecule that makes these
systems, the chromium two

283
00:21:39 --> 00:21:43
and copper two
the fastest of all

284
00:21:43 --> 00:21:46
the 3d metal ions for exchanging
water molecules.

285
00:21:46 --> 00:21:50
Let's look at a structure of
one of these systems briefly.

286
00:21:50 --> 00:21:54
Here is the result of a quantum
chemical calculation optimizing

287
00:21:54 --> 00:21:57
the structure of chromium two
hexaaquo two plus.

288
00:21:57 --> 00:22:01
This system as a spin

289
00:22:01 --> 00:22:04
polarization of four,
meaning we do have the four

290
00:22:04 --> 00:22:08
unpaired electrons,
three of which are in t(2g) and

291
00:22:08 --> 00:22:12
one of which is in e(g)*.
This is how the molecule

292
00:22:12 --> 00:22:16
optimizes without using any
symmetry constraints based on

293
00:22:16 --> 00:22:20
the density functional theory
model that is typically in use

294
00:22:20 --> 00:22:23
these days for optimizing
geometries of molecules.

295
00:22:23 --> 00:22:27
And what you can see is that
this program I am using to

296
00:22:27 --> 00:22:31
display the structure is drawing
four of the lines from the

297
00:22:31 --> 00:22:37
chromium ion to the oxygens.
And the other two it is not

298
00:22:37 --> 00:22:39
drawing.
And the reason for that,

299
00:22:39 --> 00:22:44
if I go ahead and use the tools
in this program to find out just

300
00:22:44 --> 00:22:49
what are those chromium oxygen
distances in this hexaaquo ion,

301
00:22:49 --> 00:22:54
what we find out is that four
of them, the four where it is

302
00:22:54 --> 00:22:58
choosing to draw the lines
between the metal and the

303
00:22:58 --> 00:23:04
oxygen, are about 2.1 angstroms.
Whereas, to the two oxygens

304
00:23:04 --> 00:23:08
where it is not drawing the
metal oxygen line,

305
00:23:08 --> 00:23:11
the distance is about 2.45
angstroms.

306
00:23:11 --> 00:23:14
This molecule,
this hexaaquo ion,

307
00:23:14 --> 00:23:20
has decided to deviate from a
pure octahedral geometry because

308
00:23:20 --> 00:23:24
what it has done is it has
stretched two of the oxygens

309
00:23:24 --> 00:23:30
along an axis and made those two
water molecules move farther

310
00:23:30 --> 00:23:36
away from the metal center.
And that has to do with the

311
00:23:36 --> 00:23:40
Jahn-Teller distortion that I
mentioned a moment ago,

312
00:23:40 --> 00:23:45
when e(g)* is occupied by an
odd number of electrons.

313
00:23:45 --> 00:23:50
Now, if you see that two waters
have moved far away from the

314
00:23:50 --> 00:23:55
metal and four are in closer as
a result of how e(g)* is

315
00:23:55 --> 00:24:00
populated, then which d orbital
do you think has the e(g)*

316
00:24:00 --> 00:24:03
electron?
d z squared*,

317
00:24:03 --> 00:24:06
that is right.
And that means the two

318
00:24:06 --> 00:24:11
molecules that can interact with
the two big lobes of d z squared

319
00:24:11 --> 00:24:14
along the axis,
those are repelled away from

320
00:24:14 --> 00:24:19
the metal more than the other
four that are in the x,y-plane

321
00:24:19 --> 00:24:23
because we don't have any
electron in x squared minus y

322
00:24:23 --> 00:24:28
squared,
that other component of e(g)*.

323
00:24:28 --> 00:24:33
This molecule achieves its most
stable structure by distorting

324
00:24:33 --> 00:24:37
in response to the unequal
occupation of e(g)*.

325
00:24:37 --> 00:24:42
And I will just show you what
that orbital looks like by

326
00:24:42 --> 00:24:45
quitting out of there.

327
00:24:45 --> 00:24:50


328
00:24:50 --> 00:24:53
Now this graphic,
sadly, isn't as nice as the one

329
00:24:53 --> 00:24:57
I could have shown you using my
laptop and VMD,

330
00:24:57 --> 00:25:00
but it will have to serve the
purpose.

331
00:25:00 --> 00:25:03
And this is just a snapshot of
the orbital.

332
00:25:03 --> 00:25:08
It may not be possible to
increase the size of this.

333
00:25:08 --> 00:25:15


334
00:25:15 --> 00:25:17
Here we go.
I think it should be clear,

335
00:25:17 --> 00:25:21
from your inspection of this
diagram, that we actually

336
00:25:21 --> 00:25:26
superposed a couple of different
representations of the system on

337
00:25:26 --> 00:25:28
top of one another.
First of all,

338
00:25:28 --> 00:25:32
this program has drawn balls
and sticks to connect the atoms.

339
00:25:32 --> 00:25:35
You have a red ball here,
a red ball.

340
00:25:35 --> 00:25:38
Each place there is a red ball,
that is an oxygen.

341
00:25:38 --> 00:25:42
And at the center,
we have our chromium plus two

342
00:25:42 --> 00:25:44
ion.
So these are our water

343
00:25:44 --> 00:25:49
molecules.
And what you should be able to

344
00:25:49 --> 00:25:53
see here is that this orbital,
on this water molecule up here

345
00:25:53 --> 00:25:56
along positive z,
has an appearance that would

346
00:25:56 --> 00:26:01
suggest that that is the highest
occupied molecular orbital of

347
00:26:01 --> 00:26:04
the water molecule,
interacting with the big lobe

348
00:26:04 --> 00:26:08
of d z squared in
this antibonding fashion where

349
00:26:08 --> 00:26:12
we go from blue phase to red
phase as we go along the sigma

350
00:26:12 --> 00:26:17
axis between the oxygen and the
metal center.

351
00:26:17 --> 00:26:19
And then, in a better
representation,

352
00:26:19 --> 00:26:22
you would be able to see that
there are small contributions

353
00:26:22 --> 00:26:26
from the four water molecules
that are interacting here with

354
00:26:26 --> 00:26:29
the torus of the d z squared
orbital.

355
00:26:29 --> 00:26:33
And then what you have down on
negative z is equal to what you

356
00:26:33 --> 00:26:37
have on positive z.
Namely, you have an oxygen lone

357
00:26:37 --> 00:26:41
pair here that is being repelled
by the electron occupying d z

358
00:26:41 --> 00:26:43
squared,
which is one of our e(g)

359
00:26:43 --> 00:26:46
components.
The idea is that systems like

360
00:26:46 --> 00:26:50
this exchange water so very
rapidly, 15 orders of magnitude

361
00:26:50 --> 00:26:53
more rapidly than a simple d
three system,

362
00:26:53 --> 00:26:57
because two of the water
molecules are repelled away from

363
00:26:57 --> 00:27:01
the metal center quite a bit.
And it makes this structure

364
00:27:01 --> 00:27:05
close to the transition state
for just losing a water molecule

365
00:27:05 --> 00:27:09
from the coordination sphere and
dissociating it from the metal

366
00:27:09 --> 00:27:11
center.
And that is what we are going

367
00:27:11 --> 00:27:13
to talk about next.

368
00:27:13 --> 00:27:36


369
00:27:36 --> 00:27:41
Now, we are going to talk about
a reaction like that in terms of

370
00:27:41 --> 00:27:44
a potential energy surface
diagram.

371
00:27:44 --> 00:28:06


372
00:28:06 --> 00:28:10
We have seen how d orbital
occupation affects color and

373
00:28:10 --> 00:28:13
magnetism and bonding between
metals and ligands.

374
00:28:13 --> 00:28:17
And now we are seeing how
actually we can interpret

375
00:28:17 --> 00:28:21
chemical reaction rates on the
basis of these electronic

376
00:28:21 --> 00:28:24
structure considerations.

377
00:28:24 --> 00:28:31


378
00:28:31 --> 00:28:35
We like to draw energy level
diagrams that have orbitals on

379
00:28:35 --> 00:28:39
them or states on them,
but sometimes what we like to

380
00:28:39 --> 00:28:43
do is to try to describe the
total energy of the system,

381
00:28:43 --> 00:28:47
fold all these different
coordinates into one and look at

382
00:28:47 --> 00:28:51
the total energy of the system
as a function of where we are

383
00:28:51 --> 00:28:55
along a chemical reaction
coordinate.

384
00:28:55 --> 00:29:00
That means as we progress
through a sequence of elementary

385
00:29:00 --> 00:29:05
steps corresponding to a
chemical reaction.

386
00:29:05 --> 00:29:11
A complete description of any
reaction implies a knowledge of

387
00:29:11 --> 00:29:17
the potential energy surface for
the reaction.

388
00:29:17 --> 00:29:22


389
00:29:22 --> 00:29:25
We have our reaction
coordinate, and we desire some

390
00:29:25 --> 00:29:29
kind of a plot that could
represent this reaction

391
00:29:29 --> 00:29:32
coordinate.
And what that plot looks like

392
00:29:32 --> 00:29:36
depends on the particular
reaction mechanism of the

393
00:29:36 --> 00:29:40
reaction that you are
scrutinizing.

394
00:29:40 --> 00:29:43
If you have a reaction and you
want to know how it works,

395
00:29:43 --> 00:29:45
the first thing you do is make
a hypothesis.

396
00:29:45 --> 00:29:49
And from your hypothesis,
you can build predictions about

397
00:29:49 --> 00:29:52
the reaction coordinate,
and then you can test those

398
00:29:52 --> 00:29:55
experimentally.
And either your hypothesis will

399
00:29:55 --> 00:29:59
turn out to be consistent with
the predictions or not.

400
00:29:59 --> 00:30:02
And, if not,
you reject the hypothesis and

401
00:30:02 --> 00:30:04
start again.
It has been said,

402
00:30:04 --> 00:30:08
concerning reaction mechanisms,
that the closest you can get to

403
00:30:08 --> 00:30:11
the truth is your own best
guess.

404
00:30:11 --> 00:30:15
But the way that people test
reaction mechanisms is through

405
00:30:15 --> 00:30:19
studying reaction rates under
various conditions and comparing

406
00:30:19 --> 00:30:23
the results to those predicted
by your hypothesis.

407
00:30:23 --> 00:30:28
And so here is a hypothesis for
a dissociative mechanism for the

408
00:30:28 --> 00:30:32
type of ligand substitution
reaction that we are talking

409
00:30:32 --> 00:30:37
about.
A dissociative mechanism means

410
00:30:37 --> 00:30:44
that the mechanism proceeds by
dissociation of one of the

411
00:30:44 --> 00:30:52
ligands from the metal center as
the first important step of the

412
00:30:52 --> 00:30:56
reaction.
We have down here some ML five

413
00:30:56 --> 00:31:02
X compound.
And let's say we have it in a

414
00:31:02 --> 00:31:08
solvent Y, where the solvent can
also act as a ligand.

415
00:31:08 --> 00:31:14
And we are interested in what
rate profile will we have if,

416
00:31:14 --> 00:31:19
at the end of the reaction,
down here in this well,

417
00:31:19 --> 00:31:25
this is meant to represent a
potential energy well where now

418
00:31:25 --> 00:31:29
we have X replaced by Y,
--

419
00:31:29 --> 00:31:35


420
00:31:35 --> 00:31:38
-- as illustrated there.
That is the overall result of

421
00:31:38 --> 00:31:41
the reaction.
But how does it take place?

422
00:31:41 --> 00:31:44
Are there any intermediates in
the reaction?

423
00:31:44 --> 00:31:48
And so, in a dissociative
mechanism, what happens is that

424
00:31:48 --> 00:31:52
there is an intermediate.
And it occurs here in this

425
00:31:52 --> 00:31:55
intermediate well.
One thing that you are going to

426
00:31:55 --> 00:32:00
be interested in distinguishing
between, on potential energy

427
00:32:00 --> 00:32:04
diagrams, will be potential
energy minima.

428
00:32:04 --> 00:32:08
Because these potential energy
minima that I have just colored

429
00:32:08 --> 00:32:11
in, those correspond to starting
materials, products,

430
00:32:11 --> 00:32:14
or intermediates.
And then, alternatively,

431
00:32:14 --> 00:32:18
you can have potential energy
maxima on the potential energy

432
00:32:18 --> 00:32:20
surface.
This is a two-dimensional

433
00:32:20 --> 00:32:22
diagram.
And what it really represents

434
00:32:22 --> 00:32:26
is a two-dimensional slice
through a three-dimensional

435
00:32:26 --> 00:32:30
surface that has hills and
valleys on it.

436
00:32:30 --> 00:32:34
And these hills on the
potential energy surface

437
00:32:34 --> 00:32:38
represent energy barriers to a
chemical reaction.

438
00:32:38 --> 00:32:44
What we are going to have at
these points of high energy will

439
00:32:44 --> 00:32:49
be things that we call
transition states.

440
00:32:49 --> 00:32:59


441
00:32:59 --> 00:33:03
And you start out down here at
the starting materials,

442
00:33:03 --> 00:33:07
the reaction will proceed to
the right, we are assuming in

443
00:33:07 --> 00:33:11
this case, and you go up hill in
energy first.

444
00:33:11 --> 00:33:15
You have to surmount this
energy barrier associated with

445
00:33:15 --> 00:33:18
this first transition state,
or TS1.

446
00:33:18 --> 00:33:22
And when you system has gone
over that first barrier,

447
00:33:22 --> 00:33:26
if it is a dissociative
mechanism, that produces an

448
00:33:26 --> 00:33:32
intermediate ML five.
Because X has dissociated from

449
00:33:32 --> 00:33:36
the metal center and Y,
of course, being the solvent,

450
00:33:36 --> 00:33:40
is still there.
And then, once we get down into

451
00:33:40 --> 00:33:45
this well, this potential energy
minimum, this valley associated

452
00:33:45 --> 00:33:50
with the intermediate ML five,
it is a five coordinate

453
00:33:50 --> 00:33:53
intermediate,
so it should have a structure

454
00:33:53 --> 00:34:00
of the trigonal bipyramid or the
structure of a square pyramid.

455
00:34:00 --> 00:34:02
And, over there,
I was talking about the

456
00:34:02 --> 00:34:06
distortion of that molecule
based on its occupation of e(g)*

457
00:34:06 --> 00:34:10
having repelled two of the water
molecules farther away from the

458
00:34:10 --> 00:34:14
metal than the other four.
And what that does is it begins

459
00:34:14 --> 00:34:18
to lower the energy barrier here
to get over to TS1.

460
00:34:18 --> 00:34:22
It makes it easier for one
ligand to fall off the metal

461
00:34:22 --> 00:34:25
center because it is being
repelled by that d z squared

462
00:34:25 --> 00:34:30
electron.
And what we will see is that we

463
00:34:30 --> 00:34:33
can associate rate constants
with each of these barriers.

464
00:34:33 --> 00:34:37
This is the second instance
today that I have used the term

465
00:34:37 --> 00:34:40
rate constant.
And we are going to want to

466
00:34:40 --> 00:34:43
distinguish the term "rate
constant" from the term

467
00:34:43 --> 00:34:46
"reaction rate."
And one of our goals today is

468
00:34:46 --> 00:34:50
to understand the difference
between a reaction rate and a

469
00:34:50 --> 00:34:53
rate constant.
Going up this first hill,

470
00:34:53 --> 00:34:56
over TS1 and down into the
intermediate state,

471
00:34:56 --> 00:35:00
where you have a five
coordinate species and X has

472
00:35:00 --> 00:35:03
become freely dissociated from
the metal center,

473
00:35:03 --> 00:35:06
that is k1.
And then, when you are down in

474
00:35:06 --> 00:35:09
the valley, there are two ways
out of the valley.

475
00:35:09 --> 00:35:12
You can either go back in the
direction of TS1.

476
00:35:12 --> 00:35:16
We call that back reaction k
minus one.

477
00:35:16 --> 00:35:19
We associate the rate constant
k minus one with the back

478
00:35:19 --> 00:35:24
reaction that would take us back
over the same transition state

479
00:35:24 --> 00:35:28
and back to the starting
materials.

480
00:35:28 --> 00:35:32
That would be a nonproductive
step because it is going in the

481
00:35:32 --> 00:35:35
wrong direction.
And then, over here,

482
00:35:35 --> 00:35:39
ML five has this
branching option.

483
00:35:39 --> 00:35:42
When you are at the stage of
the intermediate,

484
00:35:42 --> 00:35:47
you can go either back to
starting materials or you can go

485
00:35:47 --> 00:35:51
onto products.
And this rate constant we will

486
00:35:51 --> 00:35:55
call k two for the
reaction in which the molecule Y

487
00:35:55 --> 00:36:02
that is doing the substituting
would add to ML five.

488
00:36:02 --> 00:36:04
That is associated with some
barrier.

489
00:36:04 --> 00:36:08
And it produces the six
coordinate ML five Y and X

490
00:36:08 --> 00:36:12
dissociated.
And then, finally,

491
00:36:12 --> 00:36:17
in this last step we would call
the rate constant associated

492
00:36:17 --> 00:36:21
with going back into the
intermediate from the products,

493
00:36:21 --> 00:36:25
we would call that k minus two
for that rate

494
00:36:25 --> 00:36:28
constant.
This is how you begin to write

495
00:36:28 --> 00:36:35
out the math associated with a
hypothetical reaction mechanism.

496
00:36:35 --> 00:36:38
And so, for those of you who
are gearing up to take 18.03

497
00:36:38 --> 00:36:41
next semester,
Differential Equations,

498
00:36:41 --> 00:36:45
you are going to find that this
is one area in chemistry where

499
00:36:45 --> 00:36:48
differential equations become
extremely valuable.

500
00:36:48 --> 00:36:51
Because you are going to
generate differential equations

501
00:36:51 --> 00:36:55
that describe the kinetics of a
chemical reaction system.

502
00:36:55 --> 00:36:59
And those equations will differ
depending on the hypothetical

503
00:36:59 --> 00:37:03
mechanism.
And you are going to want to

504
00:37:03 --> 00:37:09
figure out which hypothetical
mechanism actually fits your

505
00:37:09 --> 00:37:12
data.
Here is one limiting case,

506
00:37:12 --> 00:37:15
but let me first just --

507
00:37:15 --> 00:37:25


508
00:37:25 --> 00:37:29
I am going to leave that,
given the amount of time here.

509
00:37:29 --> 00:37:34
We are going to go ahead and
talk about a limiting case.

510
00:37:34 --> 00:38:00


511
00:38:00 --> 00:38:04
When I use square brackets
here, I am going to be talking

512
00:38:04 --> 00:38:07
about concentration,
just like we did when we were

513
00:38:07 --> 00:38:10
talking about acids and bases
and so forth.

514
00:38:10 --> 00:38:14
Square brackets denote
concentrations.

515
00:38:14 --> 00:38:19
If k2 times Y,
and I will stick with the green

516
00:38:19 --> 00:38:26
color for Y, is much greater
than k minus one times the

517
00:38:26 --> 00:38:34
concentration of X,
hat corresponds to

518
00:38:34 --> 00:38:39
a limiting case.
We are also going to make

519
00:38:39 --> 00:38:42
another assumption,
here.

520
00:38:42 --> 00:38:50
We are going to say assuming
complete reaction.

521
00:38:50 --> 00:38:58


522
00:38:58 --> 00:39:01
What does that mean?
That means that k minus two

523
00:39:01 --> 00:39:06
times the concentration of ML
five Y times the concentration

524
00:39:06 --> 00:39:11
of X at the end
corresponds to essentially

525
00:39:11 --> 00:39:16
a negligible back reaction such
that when we get to products we

526
00:39:16 --> 00:39:21
stop, this is a very deep well,
and we don't go back anymore

527
00:39:21 --> 00:39:24
once we get over to the
products.

528
00:39:24 --> 00:39:27
Under those conditions,
we have made two

529
00:39:27 --> 00:39:32
simplifications.
Then, the reaction rate.

530
00:39:32 --> 00:39:37


531
00:39:37 --> 00:39:40
And, when we talk about
reaction rate,

532
00:39:40 --> 00:39:44
we usually define it in terms
of the disappearance of the

533
00:39:44 --> 00:39:49
starting materials or
alternatively the appearance of

534
00:39:49 --> 00:39:52
the products,
the rate of appearance of the

535
00:39:52 --> 00:39:57
products in a system like this.
And, under the conditions of

536
00:39:57 --> 00:40:01
those assumptions,
the reaction rate is going to

537
00:40:01 --> 00:40:06
depend on k1 times the
concentration of the starting

538
00:40:06 --> 00:40:08
material.

539
00:40:08 --> 00:40:14


540
00:40:14 --> 00:40:19
Times the concentration of our
entering group Y.

541
00:40:19 --> 00:40:25
And that says that every time
we get over the first barrier,

542
00:40:25 --> 00:40:32
we rapidly get over the second
barrier without going back.

543
00:40:32 --> 00:40:36
That is the assumption here
that k two times the

544
00:40:36 --> 00:40:41
concentration of Y is much
greater than k minus one times

545
00:40:41 --> 00:40:46
the concentration of X.
This is

546
00:40:46 --> 00:40:50
often true when the
concentration of Y is very

547
00:40:50 --> 00:40:52
great.

548
00:40:52 --> 00:40:58


549
00:40:58 --> 00:41:01
i.e., Y is the solvent.

550
00:41:01 --> 00:41:06


551
00:41:06 --> 00:41:10
In the case of water,
you have essentially 12 molar

552
00:41:10 --> 00:41:11
Y.
Y is your solvent.

553
00:41:11 --> 00:41:15
And, because of that,
under these circumstances,

554
00:41:15 --> 00:41:19
the concentration of Y,
which is very large,

555
00:41:19 --> 00:41:23
does not change effectively
during the reaction.

556
00:41:23 --> 00:41:28
It is a constant that can be
folded into together with the

557
00:41:28 --> 00:41:33
constant k1.
So we can describe this as k

558
00:41:33 --> 00:41:38
observed.
K observed is equal to k1 times

559
00:41:38 --> 00:41:43
Y times ML five X. And

560
00:41:43 --> 00:41:50
these are what we call
pseudo-first order conditions.

561
00:41:50 --> 00:42:16


562
00:42:16 --> 00:42:18
That is a particularly simple
rate law.

563
00:42:18 --> 00:42:22
And a lot of times,
experimentalists will engender

564
00:42:22 --> 00:42:27
pseudo-first order conditions by
purposely using an entering

565
00:42:27 --> 00:42:31
group concentration that is very
large.

566
00:42:31 --> 00:42:36
And, when you do that,
you will find that,

567
00:42:36 --> 00:42:40
for example,
the change in concentration

568
00:42:40 --> 00:42:47
with time of ML five X
is, upon integration,

569
00:42:47 --> 00:42:53
when you integrate differential
equations such as this,

570
00:42:53 --> 00:43:00
you can obtain expressions for
the concentrations of these

571
00:43:00 --> 00:43:07
species of interest as a
function of time.

572
00:43:07 --> 00:43:17
So we can obtain ML five X of t
divided by ML five X at t equals

573
00:43:17 --> 00:43:26
zero is equal to e to the minus
k observed, that contains that

574
00:43:26 --> 00:43:36
large invariant concentration of
y folded into it times t.

575
00:43:36 --> 00:43:44


576
00:43:44 --> 00:43:48
Hence, this single exponential
decay of the starting materials

577
00:43:48 --> 00:43:53
predicted by this mechanism
under these conditions for the

578
00:43:53 --> 00:43:58
change in concentration with
time of your starting material.

579
00:43:58 --> 00:44:04


580
00:44:04 --> 00:44:12
Finally, what you do seek is to
follow, as a function of time,

581
00:44:12 --> 00:44:18
the concentration of the
species involved such that,

582
00:44:18 --> 00:44:23
for example,
you may be following,

583
00:44:23 --> 00:44:29
here, the decay of your
starting species ML five X.

584
00:44:29 --> 00:44:33
As a function of time,

585
00:44:33 --> 00:44:36
you seek an expression for
that.

586
00:44:36 --> 00:44:41
There are different methods to
carry out the integration

587
00:44:41 --> 00:44:46
required for more complicated
rate laws than did appear there

588
00:44:46 --> 00:44:49
for the pseudo first-order
mechanism.

589
00:44:49 --> 00:44:53
But, simultaneously,
as ML five X decays,

590
00:44:53 --> 00:45:00
the concentration of ML five Y
would be increasing.

591
00:45:00 --> 00:45:06
And under the simple case shown
here, where the intermediate

592
00:45:06 --> 00:45:12
does not build up to any
substantial concentration,

593
00:45:12 --> 00:45:16
this is no buildup of
intermediate,

594
00:45:16 --> 00:45:22
meaning whenever the
intermediate is formed it either

595
00:45:22 --> 00:45:30
goes back to starting materials
or goes on to products.

596
00:45:30 --> 00:45:35


597
00:45:35 --> 00:46:03
In other words,
the intermediate never attains

598
00:46:03 --> 00:46:38
significant concentrations
relative to starting materials

599
00:46:38 --> 00:47:06
and final products.
This is the basis for the

600
00:47:06 --> 00:47:43
steady state approximation that
is explained in your notes.

601
00:47:43 --> 00:48:22
And I will go through that also
at the beginning of the hour on

602
00:48:22 --> 48:25
Friday.
Have a nice day.