This course will cover the topics of a full year, two semester General Chemistry course. We will use a free on-line textbook, Concept Development Studies in Chemistry, available via Rice’s Connexions project. The fundamental concepts in the course will be introduced via the Concept Development Approach developed at Rice University. In this approach, we will develop the concepts you need to know from experimental observations and scientific reasoning rather than simply telling you the concepts and then asking you to simply memorize or apply them. So why use this approach? One reason is that most of us are inductive learners, meaning that we like to make specific observations and then generalize from there. Many of the most significant concepts in Chemistry are counter-intuitive. When we see where those concepts come from, we can more readily accept them, explain them, and apply them. A second reason is that scientific reasoning in general and Chemistry reasoning in particular are inductive processes. This Concept Development approach illustrates those reasoning processes. A third reason is that this is simply more interesting! The structure and reactions of matter are fascinating puzzles to be solved by observation and reasoning. It is more fun intellectually when we can solve those puzzles together, rather than simply have the answers to the riddles revealed at the outset. Recommended Background: The class can be taken by someone with no prior experience in chemistry. However, some prior familiarity with the basics of chemistry is desirable as we will cover some elements only briefly. For example, a prior high school chemistry class would be helpful. Suggested Readings: Readings will be assigned from the on-line textbook “Concept Development Studies in Chemistry”, available via Rice’s Connexions project. In addition, we will suggest readings from any of the standard textbooks in General Chemistry. A particularly good free on-line resource is Dickerson, Gray, and Haight, "Chemical Principles, 3rd Edition". Links to these two texts will be available in the Introduction module.
CDS 8 Molecular Structures I
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普通化学：基本概念与应用
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Bonding and Structures in Covalent Molecules
To understand the types of compounds which can be formed and the properties of those compounds, we have to understand how atoms bond together to form molecules. In this module, we develop a model for the bonding of non-metal atoms to non-metal atoms, called a covalent bond. The model can be used to predict which combinations of atoms are stable and which are unstable. Observations of the structures of the molecules lead to a model to understand molecular geometries and properties related to those geometries. From this, we build a foundation for understanding and predicting how molecular structure is related to molecular reactivity and function.
與講師見面
John Steven HutchinsonDean of Undergraduates and Professor of Chemistry
The Lewis structure model that we be developed in the previous lectures, and
the previous concept development study, is enormously powerful.
One of the most powerful models in all of Chemistry.
In fact, it's fair to say, that the entire field of Organic Chemistry is
based upon drawing, interpreting, and making predictions based upon Lewis
structures, for the properties of organic molecules.
As such it's worth our time to analyze a bit more that both the strength and the
subtlty of this Lewis structure model. And that's the purpose of the next two
lectures. To lead into that, let's consider one
more example similar to the examples we were working in the previous lecture.
This molecule is acetic acid. What I'll tell you is that, the two
carbon atoms in this molecule are bonded to each other.
The two oxygen atoms in this molecule, are both bonded to the same carbon atom.
The four hydrogens, three are attached to one of the carbons, and the fourth one is
attached to one of the two oxygens. Let's work things through the methodology
that we did in the previous lecture. The number of available electrons here is
4 for each carbon, plus 6 for each oxygen, plus 4 for each, sorry 1 each of
the hydrogens. That's a total of 24, electrons.
The number which are needed in order to satisfy octet for each of the main group
elements, is 8, for each carbon, plus 8, for each oxygen, plus 2 for each
hydrogen. If you add up 16 plus 16 plus eight is
40. The number of shared electrons is
therefore equal to 40 - 24 which is 16. And therefore the number of bonds must be
equal to 8. Let's check and see if we've got it.
We have 1, 2, 3, 4, 5, 6, 7 bonds. We're missing a bond.
Where would that go? Well, this carbon is clearly satisfied
its valence of 4. This oxygen has clearly satisfied it's
valence of 2. This oxygen seems to be missing a bond
and this carbon seems to be missing a bond.
So we can satisfy the valences of all of the elements, by adding the eight bond to
form a double bond between the carbon and the oxygen here.
How about the number of electrons? We have 2,4,6,8, 10, 12, 14, 16 electrons
here and as a consequence there should be eight more electrons to go.
Fortunately these two oxygens here seem to be deficient.
They both need four more electrons so we could satisfy those last each and also
complete octets for each of the oxygens as follows.
And now if we count up, we have 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24
electrons in our structure, and all of the atoms have satisfied their valences.
Remember that's the methodology that we're going to apply.
Let's apply this now to a different molecule C2H6O.
In C2H6O, let's simply begin by guessing that, as they were in the previous
structure, the two carbon atoms are bonded together, and the oxygen is bonded
to either one of those carbons. The hydrogens, let's see, we could easily
satisfy the valence of the carbon on the left with three of the hydrogens and the
carbon in the middle with two of the hydrogens, and the oxygen on the end with
one hydrogen. That's looking like a pretty good
structure. But, let's check our work here.
The number of available electrons is equal to, let's see, 4 for each of the
carbons, plus 6 for the oxygen, plus 1 each for each of the hydrogens.
That's 8 plus 6 plus 6 is 20 total electrons.
The number needed to satisfy the octet rule for each of the atoms is, let's see,
8 for each of the carbons plus 8 for the oxygen, plus 2 for each of the hydrogens.
That's 16 plus 8 is is 24 plus 12, gives us 36.
So the number of shared electrons should be 16 and the number of bonds should be
equal to 8. Let's see if we got it.
We have one two three four five six seven eight.
We have exactly the right number of bonds.
We have two four six eight ten 12 14 16 electrons.
There needs to be four more electrons because remember the number of available
electrons is 20. The remaining four clearly belong to this
oxygen which otherwise doesn't have an octet.
And we've now completed the structure. Are we done?
Well, remember at the outset of this I made a big assumption.
I said the two carbon atoms are bonded together.
They don't have to be. We could have drawn this by bonding the
oxygen between the two carbon atoms as follows.
Nothing wrong with that, in fact, immediately the oxygen has satisfied its
valence. In addition let's see we have six
hydrogens and we can satisfy the valence of each carbons by simply scattering the
hydrogens, three each on to the two carbons.
We then again have 1, 2, 3, 4, 5, 6, 7, 8, bonds, that's right.
We have 2, 4, 6, 8, 10, 12, 14, 16, electrons, we need 4 more to be in the
diagram, but of course, those belong to the oxygen.
And we in fact have now drawn our structure.
Which is completely consistent, with all the rules we have written for the octet
rule, and for forming covalent bonds, and for all of the valences of the atoms
involved. So, the question is, which one of these
is right? They both meet all of our rules, and it
would seem that a molecule though should only have one structure.
The answer may surprise you. It turns out the answer is both are
correct. Neither one is better than the other.
Rather, rather differently because the structure of these molecules are so
different. In fact they have very, very different
molecular properties. They have very, very different chemical
properties and these two molecules represent two different compounds.
They are not the same compound and they're not sturctures to the same
compounds. They're actually two different chemicals.
When we have two different molecular structures for the same molecular formula
representing two differnt compounds we refer to these as isomers.
Isomers are two different compounds that share the same molecular formula, but
have different molecular structures. Let's illustrate the concept of isomers a
bit more by considerring another molecule, or another molecular formula,
C2H4Cl2. Let's remind ourselves chlorine has seven
valance electrons, just worth remembering since we haven't drawn structures with
chlorine before. And it has a valance of one meaning that
it likes to form one bond. What that means is if I've got two carbon
atoms, I'm going to have to put the two carbon atoms together.
There's no way around that and then let's see.
I've got for C2H4Cl2 I could put the chlorine so let's see, let's put the two
chlorines together, why not. And we have 4 hydrogens, now we see that
we satisfy the valence for each of the carbons and for each of the chlorines.
Let's do our calculation as before, the number of available electrons is 4 for
each carbon And let's see, 7 from each chlorine.
And 1 for each hydrogen, that number comes out to 26.
The number needed to satisfy the octet for each of the main group elements, is 8
for each carbon. Eight for each Chlorine.
Two for each Hydrogen. And that number turns out to be 16 plus
16 plus 8 is equal to 40. As a consequence the number of shared
electrons needs to be 14. And the number of bonds needs to be 7.
Look, we have seven bonds. One two three four five six, seven.
So, that says five, of course we've only used up 14 of our 26 electrons.
Were are the remaining electrons? Well clearly they belong to the
chlorines, and we can complete the octets on each of the chlorines, by providing
three lone pairs on each chlorine atom. It would seem that we are done.
But I made an assumption along the way. I put both of the chlorines onto the same
carbon. I didn't have to do that.
I could have instead begun this entire exercise by putting the chlorines on
different carbon atoms as shown here. Then the remaining Hydrogens can surround
the two carbons, and this structure, each of the carbons has satisfied it valence,
each of the chlorines has satisfied its valence.
We can get the right number of electrons, by putting the remaining unbonded
electrons in pairs around the chlorine. We should make sure we have the right
number of electrons here. We have 2, 4, 6, 8, 10, 12, 14, 16, 18,
20, 22, 24, 26. Both of these structures satisfy all of
the rules associated with drawing Lewis structures.
They are different compounds. They don't have the same physical and
chemical properties. They can even be synthesized and
purchased separately. These are illustrations of isomers.
Let's do another one. This particular case, we're going to take
C3H6O. Let's see how might we arrange these
molecules. It turns out there's a good number of
ways of doing this, I'm just going to illustrate a few of them.
One way we might do it, is to take the three carbon atoms in a row, and put the
oxygen down on the end here. And let's see we've got six hydrogens
which can go around. And the way I'm going to distribute those
is three on the first carbon, two on the second carbon, leaving only one for the
third carbon. Notice that the carbon and the oxygen
have not yet satisfied their valence. Let's work our way through the math on
this one. The number of available electrons is
let's see 4 for each carbon, 6 for the oxygen, 1 for each hydrogen.
That's 12 plus 6 plus 6 is 24. The number needed is going to be 8 for
each carbon plus 8 for the oxygen plus 2 for each hydrogen.
So we have 24, 32, and 12 more gives us 44.
The number of shared electrons should therefore be equal to 20.
The number of bonds should be equal to 10.
How many do we have? 1,2,3,4,5,6,7,8,9, there's one bond
missing. Now clearly we could attach it or
complete it by adding the double bond on the oxygen here.
How many electrons? 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, that's 20
of the available electrons is 2, 4, 6, 8, 10, 12, 14, 16, 18, 20.
I think I counted them all a minute ago. So we put two more electrons, on the
Oxygens, two more pairs of electrons and we're completed.
That wasn't the only way we could have done this.
We could have also, put the Oxygen on a middle Carbon here as well.
If I had done that. Then I might have put the six hydrogens
around to satisfy the valences of the two carbons on the ends of the molecule.
The extra bond here, then there is still the double bond to the carbon atom in the
middle. Adding the two extra electrons there, we
wind up with a structure in which the double bond, instead of being on the end
of the molecule, is in the middle of the molecule.
There's another opportunity as well. Let's imagine, again, taking the carbons
and putting them in a row, but this time I'm going to go ahead and let that oxygen
on the end have one of the hydrogens. And I'll let the carbon adjacent to that
oxygen have two of the hydrogens. And then where will we put the remaining?
Let's see, we only have three more hydrogens to go.
If I give one here, and the two on the end, it's fairly clear that I'm going to
have to add the double bond on this end.of the molecule.
Now I still have the lone pairs of electrons around the oxygen.
There are actually a few other ways that I could draw this structure.
And I'd recommend that you do so as well. Crucially, what really comes out of this
analysis, is that we wind up with portions of the molecules which are quite
distinct when we draw these diagrams. Here's a double bond in the middle of a
molecule. The first one has a double bond on the
end of the molecule. The last one has no carbon, oxygen double
bond, it has an O-H, on the end. It has a double bond between carbons
elsewhere in the molecule. We wind up drawing lots of different
kinds of structures, representing lots of different groups of atoms in these
molecules. I want us to do one more before we hammer
this point finally home. Lets now consider instead molecule C6H12.
You might immediately recognize that as just a little different than the CnH2n +
2 that we've seen before. That would of come out to be C6H12.
And we got accustomed to that idea that when two of the electrons were missing,
there might actually be a double bond between carbons.
Let's work our way through this. The number of available electrons is,
let's see. For each carbon there are 4, and for each
hydrogen there's 1. So there's 24 plus 12 is 36.
The number which is needed to satisfy the octet for each of the atoms here is,
let's see, I need 8 for each carbon. And I need 2 for each oxygen.
So that's 46 and 24 is 72. The consequence of that is the number of
shared electrons needs to be 36, and the number of bonds therefore needs to be
equal to 18 bonds altogether. See if we can draw a structure that has
that many bonds. Let's draw the carbons all lined up in a
row. Let's start adding hydrogens to satisfy
the valance of each of the carbons. But let's be careful not to be over
zealous and add too many hydrogens along the way.
Let's see, how many have I added altogether?
That's nine of them. now I only got three more to go to
satisfy the last two on the end. I've gotten a little bit used to this, so
I have a feeling that I need to distribute them in this way.
And let's see, if I count the number of bonds, I have 1, 2, 3, 4, 5, 6, 7, 8, 9,
10, 11, 12, 13, 14, 15, 16, 17 bonds. There's one bond missing.
Which is in fact that bond there forming a double bond along the way.
That's just like we did with ethylene earlier on.
And we actually looked at a number of such molecules.
That's one way we can deal with the problem.
It turns out there's another way that might be subtle.
There are lots of others of these that we can draw by the way.
No reason why the double bond has to be at the end.
It might be in various positions in the molecule.
I would encourage you to actually draw, all the different ways, in which you
might put this molecule together. There's actually not even a reason why
the carbon atoms have to be lined up in a row.
They might be branched in some type of a structure or one way they might be drawn,
is in fact, to draw A ring, of carbon atoms.
Let them circle back around onto themselves.
If this is the case, there are enough hydrogen atoms here, in fact to put two
hydrogens on each of the carbons, in this ring.
And now let's count everything up, and see how many bonds do we have?
1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, if I've got them
all the way around. Which I think that I did.
So we've go the 18 bonds here. Every carbon has satisfied its valence
every hydrogen has satisfied its valence. So this is a perfectly acceptable
molecule, here. this one over here we're going to refer
to as hexene, the hex refers to six carbon atoms, the e refers to the double
bond. This one we're going to call
cyclo-hexane, indicating that we actually have a cycle or a ring.
The important part of this message is that we can get different kinds of
functionalities out of molecules, by drawing different Lewis structures
representing different compounds. In order to Organic Chemistry we refer to
these as Functional groups. Functional groups are different patterns
of, of atoms, that show up in different molecules, and often have very similar
properties from one molecule to a next, and alcohol has the groups C-O-H.
That one is in alcohol. We've seen examples of alcohols as we've
drawn them along so far. And ether is the is the functure of
C-O-C, we've seen one of those before. An aldehyde, particularly fragrant
molecules, have the group that we just drew just a little while ago, with the
carbon on the end of a molecule, attached to a hydrogen, and a doubly bonded
oxygen. A keytone is a molecule that contains a
structure that we saw a little while ago, with a carbon between two carbon atoms
doubly bonded to an oxygen. In general a carbonyl group is the group
CO anywhere in a molecule; a carboxylic acid molecule is one like we saw at the
beginning when we had acetic acid. It's a molecule in which we see the group
with a carbon atom, doubly bonded to an oxygen, also singly bonded to an oxygen
and attached to a hydrogen atom. And amine, which we have not drawn today,
is a molecule in which we see a nitrogen attached to two hydrogens, attached to
the end of a molecule somewhere along the line.
And the cyclic ring, of course, is something that we've just seen a little
while ago. The important point here is that we start
to see patterns in the molecules. We start to see arrangements of atoms.
That start to look familiar. And it turns out that chemistry of these
patterns of atoms look very similar, so every molecule that looks like this or
contains this group of atoms, regardless of which[UNKNOWN] over here turns out to
be acidic. And that actually is something very
useful for us to do. Everything here that has the C-O-H group
in it, turns out to be an alcohol and most of these alcohols are highly
soluable in water for example, and have very similar kinds of properties.
You learned a lot more about functional groups in organic chemistry.
This was just to give us a taste of why we might want to draw these Lewis
structures, so that we can start to look for these types of functional groups and
molecules. We're going to pick this up again in the
next lecture.
