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I am going to start now by telling you a little bit more
about Lewis theory. Last time we went through
aspects of the cube theory of Lewis and showed how that could
account for single bonds and double bonds,
but not triple bonds. And that when that was
superseded by the electron pair theory, the idea being that
these electron pairs would, in fact, be oriented at the
vertices of a tetrahedron. And I showed you that
tetrahedron, as described in a cube last time,
that you could then account for triple bonds.
And, if you looked at the notes, you also saw that certain
aspects of the dynamic behavior of certain bond types,
rotation around single bonds, hindered restricted rotation
around double bonds, and so forth.
Those were accounted for by this electron pair theory in
which the four pairs of electrons were oriented at these
vertices of a tetrahedron. So, that was another quite
important triumph of that part of Lewis's theory.
And the problem was, although that electron pair
theory initially put forward by Lewis was so successful at
accounting for the properties of many different kinds of
molecules and was a good description of their electronic
structure, there was one class of molecules,
a very important class of molecules, namely those known as
aromatic.
And the most important member of the class of aromatic
compounds is, in fact, the benzene molecule.
Benzene has the formula C six H six.
And we can easily calculate the number of valence electrons in
benzene if we just say 4 for carbon times 6,
plus 1 times 6 for the number of valance electrons from each
of the six hydrogens. And that is 30 electrons.
So, in trying to understand how these 30 electrons in a valence
shell of benzene hold this molecule together,
it is known to be a planar molecule, how do those electrons
not only hold it together, --
-- but how do they account for the structure of benzene,
and how do they account for its amazing stability?
And this picture that I have drawn here is a representation
of benzene, I will explain to you in a moment,
but it comes from a person by the name of Ernest C.
Crocker.
Ernest C. Crocker was an MIT Bachelor of
Science degree holder, who earned that degree in 1917.
He was an MIT undergraduate like you.
And he published a paper. Let me write down the reference
for that paper.
This is the Journal of the American Chemical Society,
1922, Volume 44, Page 1618.
That paper was the first paper in which the Lewis electron pair
theory was applied to an understanding of the electronic
structure of aromatic molecules, in particular,
benzene. And the amazing thing about
that paper is -- Well, there are many amazing
things, but one of the amazing things about that paper is that
there is only one author on that paper, Ernest C.
Crocker. And Ernest, who was a very
bright individual, nonetheless did not go on in
graduate school to earn a Ph.D. in chemistry,
had many various chemistry-related interests.
He explained the chemistry of many different kinds of
fragrances and odors. I think he was referred to as
"the man with the million dollar nose."
So, this person had quite a variety of interests.
He worked in what was then the Applied Chemistry Laboratory at
MIT after his graduation with his Bachelor of Science degree.
And one of the things he was thinking about was how to use
modern electronic structure descriptions,
such as Lewis theory, to explain molecules like
benzene. And so, if you go and read this
paper, you are going to find a very lucid discussion of how the
Lewis electron pair theory could represent benzene according to
this formula that I have drawn here.
And in this formula, there is considered to be an
electron pair between each carbon and each of the hydrogen
nuclei. There is an electron pair along
each carbon-carbon axis, as shown here,
and here, and so on, all the way around the ring.
And then, finally, you have six more electrons to
come up to the number of 30, which is the number of
electrons in benzene. And so, the question really is
what to do with this remaining six electrons.
And I have shown them around the outside of the ring,
here, which is where Ernest arranged them in his work.
And I just want to point out that he put forward the idea
that these six electrons were circulating around the plane of
the ring and involved a net one-half bond between each pair
of adjacent carbon atoms in the benzene ring.
So, in effect, when we draw benzene this way,
with a circle in the middle, we know that that circle
represents the circulating six electrons in what we are going
to call the pi system. But Ernest C.
Crocker was the man who put the circle into the middle of
benzene. And he was an MIT undergrad.
And this was a sole-authored paper.
Aromaticity has a vast history in chemistry,
and it is still a very active and unfolding history because of
the problems, to our understanding,
posed by electrons that seem to be circulating around a whole
molecule rather than localized between pairs of nuclei.
So, Ernest Crocker, Bachelor of Science,
1917, MIT, had a huge hand in that.
I thought you might find that interesting.
And, having looked at benzene rings like that,
I will now draw them perhaps another way that is also useful,
which comes from KekulÈ.
Because I want to continue our discussion of Lewis acid-base
theory. And I am going to draw two
molecules here that are going to be pretty similar.
When I write a molecule, as I have done here on the
left, I haven't explicitly indicated each of the hydrogens
that are present on the periphery of this substituted
benzene ring. But you should understand from
a formula like this that this is the molecule boron C 18 H 15.
And over here I am drawing
explicitly, at each of these peripheral positions on the
substituted benzene rings, fluorine atoms in place of the
hydrogens. So, this is a different
molecule with formula B C 18 F
These are both Lewis acids.
And based on our discussion, last time, if you were to add
ammonia to one of these molecules, where would the lone
pair of electrons on the ammonia bind?
The boron. So, yes, you have here a
trigonal planer boron center. And, if you were to add an
ammonia molecule, the lone pair of electrons
would come in and stick to the boron --
-- because the boron is electron deficient.
Just like the aluminum we discussed last time,
it has only six electrons around it, and it wants eight.
But what if we were to add one ammonia molecule to a flask
containing both of those Lewis acids that would then be
competing for the ammonia molecule?
What I am asking you to do is something I will ask you to do
throughout this semester, and that is to analyze a
molecule's properties based on its structure and its
composition.
Exactly. She said that it would
preferentially stick to the one with the fluorines because these
very electronegative fluorines are drawing electron density
away from the boron. This is one of our most
electronegative elements.
So you have a whole bunch of fluorines in that molecule.
The whole thing, we call it perfluorinated.
It is a perfluorinated triaryl boron reagent.
These, in fact, are great Lewis acids,
really powerful Lewis acids. And they are modern Lewis acids
whose implementation in chemical processes has come about really
in the last 10, 15 years.
And, in fact we were talking a little bit about Professor
Schrock last time. In some of his research,
he has used that perfluorinated Lewis acid as an activator in
catalysis to get catalytic polymerization reactions to
work. And that is a very popular
approach these days in Lewis acid chemistry.
The design of new kinds of Lewis acids with interesting
molecular architectures is something that is very much a
current topic of interest in research in chemistry,
because you can make lots of chemical processes happen when
you have something that can tug on electron pairs.
And this one tugs a lot harder than that one because this one
has very electronegative fluorines to pull electron
density away from that boron and to adjust the distribution of
the electron density in the molecule.
And we will be talking more about electron density and
distribution in a few minutes in connection with what I am going
to show you now. And that is --
That has to do with this molecule, which is the SO two
molecule. Anyone know where SO two
comes from in nature? Or in the environment,
I should say?
Volcanoes, absolutely. And also coal-burning power
plants. Coal is a very dirty fuel,
and it contains a lot of sulfur.
And, when you burn that coal without controlling the way you
burn it, you emit SO two into the atmosphere.
So, that can be a big problem. And we will try to understand
why. One of the things that SO two
can do when it gets into the atmosphere is it can react
with dioxygen. And that can give you --
-- SO three.
And if SO two and SO three are present in the
atmosphere and if there is also water present in the atmosphere,
acid rain, that is exactly the type of process that we will be
talking about here. SO two and SO three
are examples of what we call anhydrides.
Anhydride is a word that means without water.
And so you should not be surprised that SO two
can react with water. And when it reacts with water,
it takes up water. And the product of that
reaction will look like this.
I draw it like that. Okay.
So, H two O plus SO two going to H two SO three.
The name of this molecule is
sulfurous acid. And, alternatively,
when SO three reacts with water --
I have one extra electron pair that should not have been there.
So, both SO two and SO three,
as they react with water, are going from a situation in
which they are electron deficient to a situation in
which the sulfur attains an octet.
Okay? And you should verify that the
number of electrons that I have drawn up here actually is
consistent with the elements that I am using with the stated
charge that I am using. But when SO three
reacts with H two O to give H two SO four,
we have now got sulfuric acid.
Okay. And, as I did over there,
I have two acids that I want here now to compare in terms of
their relative strengths. Here I have a Lewis acid SO
three and a Lewis acid SO two engaging in a
hydration reaction, which produces sulfurous acid
and sulfuric acid. And the type of acids that
these are on the bottom is Bronsted acids,
--
-- distinguished from Lewis acids in that the way that they
behave as acids is through ionization that produces a
proton. They are also Lewis acids
because the Lewis definition of acidity is far more general,
in saying that acids are simply entities that can accept a pair
of electrons. Protons can accept a pair of
electrons, so they are Lewis acids.
But if you are talking about Bronsted acids,
you are talking exclusively about protons that are produced
by ionization of some kind of a Bronsted acid.
Which one of these is the stronger acid?
Sulfuric acid. And why?
Yes, down here. You got the other one right.
Exactly. When this ionizes,
you get SO four minus.
I will draw is a slightly different way that is quicker.
You get HSO four minus.
There is your ionization. And the idea now is that this O
minus that you have, the negative charge can
actually be shared among a greater number of
electronegative oxygens here, namely four,
as compared to here, where we have only three
electronegative oxygens. It is a consideration of the
very electronegative elements in your molecule that will help you
understand the properties that these molecules will have in
terms of acid-base chemistry. Now, how many electrons do we
have in the valance shell of SO three?
And, that being the case,
what molecule from last time does that remind you of?
Maybe seeing a picture of it will help refresh your memory.
AlCl three. Exactly.
I am going to show it to you anyway.
And this is going to be faster than the last time,
if I set this up right. Just to remind you.
And if we could have the lights down just a little bit for a
moment.
I want to refresh your memory of the electron density
distribution here in AlCl three.
This is an electron density isosurface of AlCl three.
And what you are noticing is
that the electron density drops to a low value in between the
central aluminum and the radially disposed chlorides,
the three chlorides that surround that central aluminum
ion. And the coloring in this is
such that the blue regions represent regions in space where
there is a high probability of finding paired electrons.
So, basically you have three Cl minus's that are packed tightly
around an Al three plus.
This is a very ionic compound. See the empty region in space
between aluminum and chloride, and the polarization of that
otherwise spherical cloud of electrons around the chloride in
the direction of that positively charged aluminum?
There is your electron density distribution for that.
And now I want you to keep that in mind, and we will compare the
isoelectronic SO three molecule to it.
SO three. Here is another case where we
have 3 times 8, 24 valance electrons in a
system. But the character of this
molecule in terms of electron density distribution is very
different. While on the blackboard,
I am not able to really tell you very much about the
difference between SO three and AlCl three,
here I think you can see that indeed they are quite
different. This is an electron density
isosurface at the same contour level as what we were looking at
for AlCl three. Now, to explain this and to
understand just what is going on here, you need to remember that
the electronegativity difference between the central sulfur and
the peripheral oxygens is not very great compared to the
electronegativity difference between aluminum,
which is a very electropositive and metallic element,
and chlorine, which is a very electronegative
halogen. And so, what that results in,
as shown here, is a much more equal sharing of
the electrons between that central sulfur and these
peripheral oxygens. So, even though these things
are both Lewis acids and they both have 24 valance electrons,
the electron density distribution in
three-dimensional space for these molecules and the covalent
verses ionic character of these molecules, is really quite
different.
Our section that is going to be devoted to bonding has not
really kicked into gear yet, but the nice thing is that
Lewis theory applies both to acid-base chemistry and to
bonding, so we are able to talk a little bit about that.
In a few moments, I will tell you a little more
about an issue that is very important in chemistry as
regards bonding.
And it has to do with what happens when acids,
like sulfuric acid, ionize in water.
When Bronsted acids ionize in water, we get this ion produced.
H three O plus, which is the hydronium ion.
That is to say that if you ionize in water some Bronsted
acid, the protons that are produced through that ionization
are not floating around freely, naturally, because they are
positively charged and they are attracted to negatively charged
electrons. So, they look around in
solution and find the next source of an electron that they
can. And you know that if we draw
out a molecule like water, according to the Lewis dot
structure, it has two extra pairs of electrons,
in addition to those two pairs of electrons it is using in
making bonds to the two hydrogens that are on the oxygen
of the water molecule. So, H plus is not just
isolated around by itself in solution.
It perches on an oxygen lone pair.
So, H three O plus is what you get when Bronsted acids
ionize in water. And furthermore,
when you put these things in solution you find that you
organize the water molecules that are close to the hydronium
ion. Let's draw here a neighboring
water molecule.
And another one.
And I think you can imagine that throughout a solution,
we might have many of the kinds of interactions that I have
drawn here as dotted peach-colored lines.
Those lines represent what we call hydrogen bonds.
And hydrogen bonds are enormously important in
chemistry. Later, when we talk about the
structure of proteins in DNA, in particular,
you may be aware that the DNA double helix is held together by
a network of hydrogen bonds between complimentary base
pairs. So hydrogen bonds are not only
restricted to the hydronium ion in aqueous solution.
There are many other types of molecules that can form what we
call hydrogen bonds. Another really interesting
thing is that in water, the hydronium ion can move
around really rapidly, much more rapidly than
molecules normally diffuse through aqueous solution.
And the reason for that is if you look at the arrangement of
electrons and nuclei here, all I have to do is,
without even moving the nuclei much at all, reorganize the
hydrogen bonding network as such.
And now you can see that through just a slight set of
motions, our hydronium ion has moved from the left-hand side of
this hydrogen bonded network, where you can see that it is
indicated with the positive charge and the three solid lines
drawn to the oxygen, over to the right-hand side.
But it did that not by coming off and moving over,
but rather through just redistribution of the electron
density, so that the positively charged part ends up down on the
other side. And so this way of propagating
hydronium ions in aqueous solution is one of the really
special aspects of Bronsted acid chemistry that takes place in
water. And I think I will also show
you what a hydrogen bond looks like from the standpoint of
electron density.
First, I am just going to show you the position of the nuclei
in a very simple hydrogen bonded system.
Here what you can see, the oxygens are drawn in red as
spheres and the hydrogens are drawn in white as spheres.
You can see that the geometry around the oxygen atoms is
slightly pyramidal. And that is due,
of course, to the presence of that extra lone pair here,
up above one oxygen and here, up above the other oxygen.
And what we have now is a hydrogen serving in a bridging
fashion. And the number of electrons in
this system is exactly 2 times 8, because we have two water
molecules and we have an H plus.
So this is a positively charged ion in which a hydronium ion,
and you can pick either side, actually, is interacting with
one of the lone pairs of the other water molecule.
And you could imagine lots of different types of water
clusters like this that are singly positively charged.
And people have done a lot of work to study such clusters in
solution. What you should remember,
though, is that the size of the spheres that I have drawn there
to represent those oxygens and hydrogens is somewhat arbitrary.
But what is not arbitrary is the way that the electron
density represents a molecule like this.
So, we will show that to you next.
Here it is. And if we could have the lights
down just a little bit, please, since this one is a
little harder to see. What you should see here is
that we have the same structure now surrounding that
representation of the water molecule hydrogen bonded to the
hydronium ion that I drew a moment ago.
We now have this sort of mesh, which is exactly what we have
been looking at with these other molecules, namely,
an electron density isosurface. And what you can see is that
the electron density is falling to a pretty small value in the
middle, here, where we have the proton that
is the connecting glue binding together these two water
molecules in this 16 valance electron system.
And after one more representation of that,
we will be onto our next topic.
And this one is a solid display of the electron density
isosurface associated with this hydrogen bonded cluster.
And it is, once again, color mapped with this function
that tells us about the probability of finding electrons
paired up together in space. There is H three O plus
hydrogen bonded to H two O.
The blue color represents those regions in space where you are
most likely to find pairs of electrons.
And you can see that the two OH bonds over here are nicely
colored blue. The lone pair of electrons up
here is nicely colored blue. And then we have an interesting
situation where there is some blue in between that bridging H
plus and the two lone pairs that are pointed at it
that produces, in fact, our hydrogen bond.
So, there is a picture of hydrogen bonding in terms of
electron density. And it is a type of bonding
that compliments the straight, covalent, and ionic bonding
that I was talking about in terms of SO three and
AlCl three. So, we have added this third
type of hydrogen bonding to our list of bonding interests.
And now, we will talk more about what we can do when we
consider Bronsted acids ionizing.
Here is a generic representation of the formula of
a Bronsted acid, HA.
A might be, for example, the HSO four minus
ion that we showed over there. When we put a Bronsted acid in
aqueous solution, as I said before,
we can get ionization into H plus and A minus.
But we know that it is not just
H plus. It is actually H three O plus
**H3O^+**. And H three O plus is
further hydrogen bonded in networks in the water system.
But, for simplicity, I will just write it as H plus
right here. Recently, in your crash review
of thermodynamics, you were talking about
equilibria and equilibrium constants.
And we are going to make use of some of that right here because
we are going to talk about the acidity constant,
Ka. And that is going to be defined
as equal to the hydrogen or hydronium ion concentration
times the concentration of the conjugate base,
A minus -- When a Bronsted acid ionizes it
produces what we call the conjugate base of the acid.
Here is conjugate base --
-- divided by the concentration of the acid.
And this is at equilibrium.
And let me just emphasize something so that you don't
forget. This is an important piece of
nomenclature. These square brackets here
refer to concentration, usually in molarity.
Okay? So that is what we are talking
about. And concentration is something
that can be measured. You may be familiar,
for example, with the pH meter invented by
Arnold O. Beckman and its utility in
measuring the concentration of hydrogen ions in solution.
Well, we can make use of information like that to talk
about the properties of our Bronsted acids.
How can we do that? Well, let's say we are going to
take a particular acid such as this one, which is acetic acid.
You know the smell of acetic acid if you have ever been in an
establishment where they were making barbecued chicken wings.
That is the smell of acetic acid.
A beautiful smell. Anyway, what you do here is you
are trying to figure out what is going on.
You have some concentration of the acid HA, which is,
we are going to talk about, acetic acid.
And in solution there may also be a hydrogen ion or a hydronium
ion. And then there may also be A
minus, which in the case of acetic acid would be acetate,
--
-- where we have two electronegative oxygens,
among which the negative charge can be shared and the acetate
ion, which is the conjugate base of acidic acid.
So we make a table. We need to have some initial
concentration. That is to say let's consider,
for example, tenth molar acetic acid.
We are just choosing tenth molar as a concentration for
acetic acid solution. What that means is you have
pure acetic acid. And then you dissolve it in
water and bring it up to a total volume such that the
concentration was 0.1 molar, assuming that none of it had
been ionized yet. And so that means we have an
initial concentration of acetic acid of 0.1 molar.
And initially, before the ionization,
we have zero H plus or hydronium and zero A minus.
And then, the concentration
changes. And it changes because the HA
ionizes to some particular extent, depending on the KA
value for the acetic acid. And what is going to happen is
that some of the HA ionizes. And the amount of the HA that
is undergoing ionization is x, so we are going to lose x.
And then, for every HA that ionizes, we get that same amount
of H plus produced and that same amount of A minus.
And so then,
after the system reaches equilibrium, we will finally
have 0.1 minus x as our concentration of HA
and x and x will be our concentrations,
respectively, of H plus and A minus.
And so let me put this board
all the way up.
Therefore, we can write the following, that Ka is equal to x
squared over 0.1 over x by
substituting into the expression for the acidity constant.
Ka is our acidity constant.
And we can go to a table and look up the acidity constant for
acetic acid because it is a known quantity.
And it turns out that that is 1.8x10^-5.
And now that we have this equation for the acidity
constant and we know what the acidity constant is,
we can solve this for x. Of course, this is a cubic
equation. We are going to get two roots.
You will see that you get a positive root and a negative
root. The negative root is
meaningless because concentration cannot be
negative, so you pick the positive root.
And when you have done that, you can then go ahead and
answer questions, like, what is the pH of the
solution? And what is the percent
ionization?
And we can talk about that. So next time,
at the beginning of class, we will do that calculation.
We will find what the pH of a tenth molar solution of acetic
acid would be. We will also go on and talk
about pH and the pKa scale, and also a general equation for
discussing titrations and buffers.