Friday, May 19, 2006

Lecture 007: Alkane Introduction

Lecture 007: Alkane Introduction
So we're going to resume the polarity issue and talk about intermolecular forces. So we spent a lot of time building up since last week to this. Lewis structures determining the shape and determining if the molecules were polar or not.

The reason for that is now you can use that information and predict some relative properties such as melting point and boiling point. In order to do that you need to understand what forces keep molecules of the same compound together.

Okay, so there are three major forces, and we're going to start with the most powerful force and go down to the weakest.

The first force is called hydrogen bonding. It occurs whenever you have an X-H... :X pattern like this, where X is nitrogen, oxygen or fluorine. So you need some fairly electro-negative elements to form a hydrogen bond.

So an example of this, the example you are probably most familiar with, is water. So water basically has an X-H component. The oxygen is the X in this case, and we find that we can line up another water molecule next to the first one. Through the lone pair of the oxygen, on that second molecule we can form a hydrogen bond.

So that hydrogen bond is always drawn from the lone pair to the hydrogen. It is basically that lone pair is being shared a little bit between the oxygen and the hydrogen. So it is not a full bond. It's a very partial bond, but it has a strong enough that that's the first thing we need to consider when we talk about intermolecular forces.

There are three main forces and you go in that order. So if you find that your compound has hydrogen bonding, then you are done. The major force holding water molecules together is hydrogen bonding- the strongest force. You don't need to go to the other ones because they will always have it, but it will be less strong and so it won't contribute as much to the intermolecular forces.

Okay, so that's the strongest. The next force here, number 2, would be dipole-dipole. And you will get this when first, you have no hydrogen bonding so you go and ask, "Is there any dipole-dipole interaction?" Whenever you have a polar compound then yeah, you will have dipole-dipole interaction.

Let's pick something we've already used before, chloromethane. And you recall this has a dipole. There is a polar bond from the carbon to the chlorine making the chlorine slightly negative and the carbon slightly positive.

So again, I'm using the delta symbol. That's not a formal charge. Okay, a formal charge would just be a plus and a minus. This is a delta or a slight minus and a slight plus.

So the way the dipole-dipole works is the two molecules line up anti-parallel, like this. So in blue here I can show that it still comes down to a plus minus attraction, but it's partial plus to partial minus.

Okay, so that's how dipole-dipoles line up and that is why chloromethane will be attracted to itself. The stronger the interaction, the harder it will be to pull those molecules apart. So it's easier to pull chloromethane molecules apart than it is to water molecules, for example.

Okay, the third one, which is the default force, is the Van der Waals interaction. So I'm not going to go into detail here about the differences here with London dispersion and all that. It's basically hydrophobic interactions.

When you don't have hydrogen bonding, when you don't have dipole-dipole, the only thing that would be left would be the Van der Waals or dispersion forces. So let's pick an example here like methane. So methane has no significant dipole moment. So you would say that would be the force responsible for the intermolecular attraction.

So this is the weakest force, which means that it will be easier to pull apart two methane molecules than two chloromethane molecules or two water molecules, of course.

Okay, so that's what we're getting at in terms of talking about the forces. Let me give you a specific example of how you would use this and the type of questions you might come across.

You may be asked to predict which one would have a higher boiling point, ethanol or dimethylether. So drawing them in a condensed form is not terribly useful because we can't tell exactly what's happening. So I am going to redraw those molecules in their full Lewis structure with the geometry and everything correct, well at least correct to the oxygen.

Okay so I'm trying to draw the tetrahedron for the oxygen on ethanol. So that will have two long pairs. And let's do the same thing for the ether. So the lone pairs are important because we're going to need them if we are going to worry about the dipole moment.

I didn't draw, for ethanol, every tetrahedral structure, you know for the CH3 and the CH2. The reason for that is that there is no significant dipole moment between carbon and hydrogen. So it doesn't really matter what they look like, there won't be any vectors to draw there. The important part of this molecule is the oxygen, because there are significant dipoles and you have to add them up, at least to confirm that they don't cancel out.

So you're going to go from plus to minus. All right so from the carbon, towards the oxygen. Carbon and hydrogen have very similar electro-negativity. So if you predict that there will be a dipole moment between carbon and oxygen, you would also predict that this would be the case between the hydrogen and the oxygen. So I'll draw another arrow going in that direction.

And for oxygen and nitrogen you typically take into account the lone pairs, drawing a vector from the oxygen or the nitrogen through the long pairs like this.

Okay, so we have a tetrahedral structure that has four arrows on it. The key thing that I want to point out to you is that they definitely don't cancel out. Right? Everything is going up.

So I'm not trying to get you to draw the exact place just to be able to draw it well enough to understand whether or not it cancels out. And clearly in this case it doesn't. So let me draw the overall vector up and to the right in this case.

And for the dimethylether we're going to do a very similar analysis. Draw two vectors from the carbons to the oxygen and then through the long pairs. Okay, so once again we have a dipole moment for both of these molecules, ethanol and dimethyl ether.

Now, now that we have this information. Let's step through these forces. For the first force, what we want to ask is "Is there hydrogen bonding?" So, is there hydrogen bonding in ethanol? What do you think? It's possible.

So again, let me draw this out. Now we're looking for an X-H X pattern. Okay, so in this case the X-H will be O-H. We also have CH3, and CH2. All right and we have the other oxygen. Again worry about the lone pair, because you need the long pair to do the hydrogen bonding.

Draw the hydrogen bond like this.

Okay, so in being able to draw this X-H X pattern, we confirm that we have hydrogen bonding. So the answer to the first one. We stop there. It will be hydrogen bonding. It will be the major force.

Okay, so do we have hydrogen bonding for the ether? We have an X. Right? And we have hydrogens, but that's not enough. We need to have a situation where we have an X directly connected with a covalent bond to hydrogen. And we don't have that here. The oxygen is only connected to carbons so there is no hydrogen bonding for the structure on the right.

Okay, so we then go to the next question- "Do we have dipole-dipole attraction?" Yeah, if you have a dipole moment, you'll always have a dipole-dipole. So the answer to the one on the right, to the question "What is the dominant force?" would be dipole-dipole.

Okay, but I worked out the fact that the ethanol has a dipole moment as well. It also has dipole-dipole attraction, but it's not important because it is overshadowed by the hydrogen bonding. Okay, so that's why I have specified the dominant, not the only.

Okay, so with this information now we know the intermolecular attraction for ethanol will be higher than for dimethyl ether. Okay, so it's going to take more energy to separate two ethanol molecules, to take them from a liquid state to a gas state, than the ether molecule. So we predict that we have a higher boiling point for ethanol on the left.

Okay, and the boiling point for ethanol is 78 Celsius, whereas the boiling point for diamethyl ether is -25 Celsius. Okay, so there is almost a hundred degree difference between the two. So hydrogen bonding is a really important force, as you can see. It has a huge impact.

Now we can't compare any two molecules when we do this comparison because you know what happens if I have a big molecule on one side and a small one on the other and they have a different groups on them. Well you can't really do the comparison in that case.

The only reason we can confidentially say that ethanol has a higher boiling point is that they are the same size, or roughly the same size. So if that criterion is met then you can do all the comparisons for their respective boiling points.

Does anyone have any questions on this?

Okay so melting point is the same principle. When you are melting a compound, you are basically separating molecules, except that in this case you are separating them from a solid state to a liquid state, but it is the same principle. So the higher the force the higher the melting point, as well.

And we'll do some examples of that. I gave you some problems to do and we can see that later.

Okay, the other physical property I want to cover is solubility. Ok, where like dissolves like. Okay, but let's think a little bit about what it means for one compound to dissolve into another.

It's true that generally like will dissolve like, but what happens in the situation where two compounds don't dissolve. If we look at it from the situation of intermolecular forces, what you are comparing is "If I replace a molecule, interacting with itself, is that going to require energy? Is it going to give off energy? Or, is the energy going to be pretty much the same?'

Okay, so let's do a couple of examples and see if that becomes clear.

We want to predict if water and methanol will be soluble in each other. There's methanol. There's water. So, let's go a little bit more quickly over this. Methanol is really very similar to ethanol. Its dominant intermolecular force will be hydrogen bonding. With itself, it will hydrogen bond.

For water it's the same thing. Its dominant intermolecular forces will be hydrogen bonding as well. So let me draw a little water molecule bonding to it self with a hydrogen bond through the lone pair. The same thing will happen with the methanol.

So if you mix these two together, when one dissolves in the other, there is going to be an interaction. So the question we now want to as is: "What is the interaction between the methanol and the water? Can they hydrogen bond to each other?"

Well yeah, because one will have an X-H and one will have an X.

So let me draw the mixture. Here's the methanol. And here's the water.

So the dominant intermolecular force between methanol and water is also hydrogen bonding.

So what it means is that I am replacing hydrogen bonding with hydrogen bonding. So I'm not really losing that much in energy. Therefore, we would expect those two compounds to be soluble in each other. And they are. In fact, they're miscible, meaning they're soluble in all proportions.

Now let's take another example of two compounds that are different from each other. So let's look at, for example, CH4 and water. So water, we just determined, was hydrogen bonding. And for methane, does it have hydrogen bonding? No. Does it have dipole-dipole? No, because it doesn't have significant dipole, therefore it has to be Van der Waals interactions.

So what will be the force between methane and water? I have a methane molecule. I have a water molecule. Is there any force between the two?

Well, there's always a force. Okay. You can never have no force. So it's not hydrogen bonding. It's not dipole-dipole. Because for dipole-dipole you need two dipoles. So the water has dipole, methane doesn't. So the only thing left is Van der Waals.

So yes, there actually is a positive interaction between the two and it's attractive. So for the mixture, the dominant force will be Van der Waals.

So what you might think is; "Well if it has an attractive interaction, what's the difference? Why doesn't methane dissolve?"

And the reason for this is that in order for one methane molecule to go into the water, it has to displace a water molecule, to do that it has to break a hydrogen bond. If it didn't have to do that, there would be no reason for methane not to go into the water because it's an attractive interaction. So that's basically the reasons that compounds are not soluble in each other. Okay, because you're replacing a strong force with a weak one. So we would predict that these two would not be soluble, and they're not.

Okay, now solubility is a graded kind of thing. You can have miscible compounds that are soluble in all proportions like water and methanol. I mean methane has a measurable solubility in water, it's just so small that we say that it's insoluble. But keep in mind, with all of these things, there is a certain amount that will dissolve. It's just that we have to draw the line as "is it a reasonable amount?" And in this case its not.

What I want to take a look at now are functional groups. So to make things simpler for chemists, instead of considering each molecule different from another, chemists have figured out that certain kinds of functionalities on compounds give similar kinds of reactions. So you don't have to memorize reactions for every compound in organic chemistry. You can put them into classes. And for compounds that have never been made before you can predict how they are going to behave, pretty much. Okay, so that is what the purpose of functional groups are.

We are going to take a look at maybe a dozen functional groups that I'm going to expect you to know and be able to recognize. And there are going to be some examples that I'm going to ask you what functional groups are in a molecule.

So if you have any questions this is a great ask because you might have a few questions.

The first one is not really a functional group. It's kind of the default. If you have no functional group, you'll have an alkane. So an example of an alkane will be CH4, methane.

And later on we're going to take a look at some of these functional groups in greater detail. Okay, so right now I just want to list them, but we're definitely going to look at alkanes shortly.

The next one is alkenes. So alkenes have carbon-carbon double bonds. So I'm going to list the example ethene or ethylene. It has a carbon-carbon double bond with four hydrogens. So you can have an alkene if you have different kinds of groups on the carbon-carbon double bond. So if you have alkyl groups, aromatic groups, whatever, but it doesn't change the fact that this will still be an alkene.

Sometimes if you put another group on, it will change the functional group. And I'll let you know when that is, but in the case of alkenes, there is not much you can do to it to turn it into something else.

The same thing for alkynes, which is whenever you have a carbon-carbon triple bond. So just as an example, here I'm going to put a propyne. And there at each end, there could be a hydrogen, there could be an alkyl group, could be an aromatic, it doesn't really matter, it's not going to change the fact that you would call that an alkyne.

Alcohols, methanol is the one that we've talked about before, CH3-OH. So you'll have an alcohol whenever you have an R-OH pattern. Where R is an alkyl group, an aromatic, whatever, it's going to be an alcohol. But R can't be certain things. If R is a carbonyl group for example, then you no longer have an alcohol, you have a carboxylic acid. So you can't put anything you want under R, and that's why I'm saying, if you come across something and you're not sure, definitely ask me. But any alkyl group will certainly stay as an alcohol.

An ether would be similar to an alcohol, but instead of having H, you would have another alkyl or aryl group. For example diamethyl ether. So in this case we have R-O-R'. This is Where R and R' are alkyl or aryl groups.

So an aryl group would be like benzene. Benzene, benzene ring. So one thing that R can't be is a hydrogen because if R or R prime are hydrogens then I have an alcohol. That's the kind of limitations that we have to put on these functionalities.

Aldehydes. So aldehydes will have this format. C double bond O, -H. And you can, you can also have a hydrogen. So formaldahyde H C double bond O, H is also an aldehyde. And aldehydes will have the general formula R- C double bond O H where R can be a hydrogen, it can be an alkyl group, or an aromatic group. I don't think we can change the functionality. If you have a carbon at R, you're pretty much going to have an aldehyde.

Ketones are related. In this case you don't have a hydrogen on one side. You have C double bond O and you have two alkyl or aryl groups on either side. So general formula would be R, C double bond O R'. And here R and R' can't be hydrogens. The have to be an alkyl or aromatic group.

A lot of this might seem arbitrary. You know, what difference does a hydrogen make? And it really comes from the reactivity. It turns out that aldehydes are similar to keytones, as we will see, but they have enough difference that it's worth it to call them a separate functional group. Okay, for example aldehydes oxidize really easily into carboxylic acids. Keytones don't do that.

The naming comes from how they behave.

Carboxylic acids. So an example is acetic acid. Basically that is what is in your vinegar. It's just a solution of acetic acid. It has the formula R, the general formula R -C- double bond O, OH. So here in our the example. If instead of C double bond O we had a CH 2, this would be an alcohol. By putting a carbonyl group, it makes a different functional group.

Okay, if it didn't change the functionality, the behavior of the compound, we call these alcohols. But this change actually makes these compounds, as their name says, very acidic compared to alcohols. They have all kinds of other reactivity.

Acid Chloride. Acid Chlorides are very similar to carboxylic acids. In fact, you make them from carboxylic acids. They have the general formula R,C- double bond O Cl. And these are very reactive compounds. You'll see that we use them as reagents. They'll react very easily with some compounds.

Esters are similar to Carboxylic acids. We still have the C double bond O, O pattern, but now we have to have another alkyl or aryl group on the end of the oxygen. So the general formula is, R- C- double bond O, O, R', where R' here can't be hydrogen, it can be an alkyl or aryl. Okay, so in the case where R' is a hydrogen, then I don't have an ester, I have a carboxylic acid.

Next we have amides. So amides will have C- double bond O and then a nitrogen. And there's all kind of things we can put on the nitrogen on the other side of the carbonyl group. So we can generalize that into R, C double bond O, N and then R' and R''. And here you can have hydrogens, or alkyl, or aryl anywhere. So that won't change the fact that it's an amide.

Don't confuse amides with amines. Amines are similar. They also contain nitrogen, but there's no carbonyl group. So trimethylamine is an example. So in general R, R', R" can be anything. It can be hydrogen, an alkyl or an aryl group.

Okay, questions?

Well I'm sure you come across some things when you start to do the problems. You're asked to identify different functional groups in molecules so we'll see what you run into when you get to that.

Well I do want to talk a little bit more about the hydrocarbons, go into a little bit more detail about that.

Okay, so hydrocarbons basically mean is that all you have is carbon and hydrogen as the name suggests. But from that, there are different kinds of arrangements.

We can have an aliphatic. Aliphatic just means that we have an alkane. We can have aromatic, which means that you have compounds similar to benzene. And we'll get to that in another chapter. Or we can have alkenes, which we saw previously, which have carbon-carbon double bonding. So what we want to do is to look at the aliphatic hydrocarbons or the alkanes in this section.

Okay, so what you need to learn how to do here is to count to ten, in organic chemistry because those first ten names of compounds you are going to use as prefixes for any molecule that you want to name later.

Okay, so we're going to learn, for example that one carbon, example CH four is called methane. So the key ingredient here is the meth that means one carbon. So whenever you come across meth, you know it's one carbon. You know that this is methane and that it's an alkane because it has the -ane ending. But if I put a different ending, for example methanol, I know that it's an alcohol that only has one carbon. So that's what I mean by counting to ten in organic chemistry.

Two carbons, we have ethane. Three carbons propane. Four carbons, butane. Now when we hit butane, there is actually more than one way of connecting those carbon atoms together as we saw when we did one of the problem sets. So in order not to create millions of compounds here, I'm only going to draw the compounds that have a straight chain like this.

So if you start to do branching, then you'd have to use another terminology, but if I want to specify the butane that has all the carbons in one line, I put a small n in front of it, which means normal.

So if you see butyl group and it doesn't have the n, it is implied that it is the normal, that it will be a straight chain.

Five carbons. So in this case, we have a pentane. Six carbons hex, hexane. Seven carbons we have a heptanes, When we reach eight, we have octane. Nine, nonane, and finally ten decane.

Okay, so a lot of these you won't really have to memorize, because we're going to use so much. You know heptane, octane, nonane, decane you'll probably use a little bit less, but you're certainly going to come across the first ones a lot more. And that just tells you how many carbons you have.

Okay, so I had mentioned that there are different butanes that you can make. So if I connect the carbons in a straight chain I would have n butane. And if I make a branch structure, like this you go to isobutane. Okay, now with four carbons there are two ways of connecting them together.

With Three carbons, unless you make a ring out of them, there is only one way you can connect them together. But there's more than one place that you could link to that three-carbon fragment. So, although there's only one propane, there are two different ways that we can connect a propyl group to another subsistent.

So these are going to be fragments, not molecules. Okay, so if I write CH2, CH2, CH3, and put a line here that CH2 has to have one more thing connected to it but the group, outside of that connection, is a fragment. And that's called the n-propyl fragment.

Okay, now if I connect to the propyl group through the middle carbon, like this, now I have an isopropyl fragment. So again, these are not molecules. You have to add something to them to make them real molecules.

So an example would be, lets say I put an alcohol group. So if I put the alcohol on the straight chain group that would be n-propanol. Now if I add something else, let's say a chlorine, to the isopropyl, that would be isopropylchloride.

There are other ways of naming these compounds and we'll get to that. We'll get to that when we get to alkylhalides in detail. But basically isopropylchloride is a real molecule. Isopropyl is just the fragment. There are a couple of these fragments that I'm going to ask you to memorize. I'll ask you for three carbons and for four and then we'll stop.

There are names for the five carbons, but they're not used as much and you know, I don't think it will help you that much. But definitely butyl, you'll need to know.

So let's start with the one you already know n butyl. So, I can link to four carbons in a line by going at the end, or by going at one of the internal carbons, the second carbon. So if I go to the end it's n butyl, if I go to the second carbon it's sec butyl. By rearranging the carbon atom in a branch structure there are again two ways that I can link to that. If I link to it by one of the end carbons like this, that will by isobutyl. And finally I can link the branch structure through the middle carbon and that's called tert butyl, or t. Same thing, you can just put t butyl.

Okay, so with these tools you can construct a lot of the names of the simpler molecules. Questions? Okay.

So, another thing you have to be able to name on molecules is whether your carbons and hydrogens are primary, secondary, tertiary, or quaternary. So, let me pick an example. I'm going to draw propane and if we look at a carbon, and we and we want to ask whether it's primary, secondary, tertiary, or quaternary, the question is how many carbons are directly connected to that carbon. So, in this case, it's just one, so that would be a primary carbon. If I look at the middle carbon it's connected to two carbons, and that would be a secondary, and the bottom one, just like the top one, would be primary.

So we're going to be referring to, you know, a primary center and a secondary center. You always figure it out by counting the number of carbons on that particular group. Now, when I refer to the primary center I'm also going to be referring to the hydrogens on here. So, these three hydrogens on the top, we're also going to say that they're primary. So, the designations for primary will go for the carbon or it will go for the hydrogens as well. Okay, and that's something that you'll need in order to predict the outcomes of reactions. So, if I draw isobutane, this carbon would be what? Primary, secondary, tertiary...primary? Right?

Because I look at it and ask how many carbons are directly connected to it and the answer is one. And the carbon in the middle...it's got one, two, three carbons attached to it, so it's tertiary. I'm going to write these out too. Okay, so the other example is quaternary. So if I have this compound, I have four carbons connected to the central carbon, so that would be quaternary. Okay, so that's basically all there is to primary, secondary, tertiary, and quaternary.

And again, just by using the terms you're going to get to know them pretty well.

Okay, so we're talking about alkanes, let's look at some reactions of alkanes. So the first one is combustion, probably the most important reaction. Example, CH4 + O2 would be C02 and water. So, you get combustion whenever you burn something, and you burn something by combining oxygen with it to generate a lot of energy. So, that's the basic source of our energy and if that didn't happen, we couldn't drive cars, couldn't heat our houses and all of that stuff, so it's a very important reaction.

Now I'm not going to go into detail on this reaction, I will go into detail on some reactions, but not this one. So in this case, all you're really expected to know is that you generate carbon dioxide and water, but you don't need to know how. Okay? Also this reaction doesn't necessarily go all of the way to carbon dioxide. If you have limited combustion you can go to carbon monoxide.

That's also part of combustion. So this basically gives you some sort of an idea, you know, of how combustion works. The key point is that combination of oxygen with the alkane to generate heat.

The second one is cracking, where we take a long carbon chain and break it down into smaller pieces. So, a simple example of cracking would be starting with propane and then reacting it with some catalyst, like palladium or platinum, and generally this is done at high temperatures. What you do is you break the molecule, so I can go from a three carbon fragment into a two carbon fragment and a one carbon fragment.

I'm going to make an alkene in this case. Okay, so I've broken down propane into ethylene and methane, the key thing being that I have reduced the number of carbons. This is not something that's necessarily terribly useful for propane, but it's more useful for long chains.

So for example if you have fifteen chains you want to break them down to pieces of eight to make octane, which is used in gasoline, for example. So that's how a lot of the octane is made, it's not just found in the ground like that. Oil is found. Oil has longer chains, those longer chains are cracked into smaller pieces, and that's how we make most of the octane.

Student:Why is it not ethane.

Professor: Well the reason it's not ethane is I don't have enough hydrogens. Again, cracking is a very complicated subject, yes, you can take hydrogen and do the cracking, or if you do it without the hydrogen you're going to end up with an alkene. But both of them are considered cracking. Any time you break a long chain into smaller pieces.

Okay, next we have halogenation. Example, CH4 reacts with bromine in the presence of light. Okay, so I'm using hυ (H nu)to say that I have light, that's the common abbreviation in organic chemistry. I'm going to write this once. Okay, so what we're doing is we're going to swap a bromine and a hydrogen. So this is called halogenation because I'm using a halogen to do an exchange reaction like this.

Now this reaction we're going to look at in very great detail. So here you're just expected to know that this happens but we're going to look at it in enough detail that you can predict what's going to happen with ethane, propane, benzene, does it still work?

But in order to be able to do that it takes time to go into that in detail. Okay, so there's not a lot of reactions here. Primary alkanes are pretty inert. So there's really not that much you can do to them. These three reactions are probably the most important. Okay, the one thing we want to worry about in alkanes is how do we predict the shapes they're in, not reactions, but how they twist.

So let's consider methane first. So methane has a carbon that's sp3 hybridized, and has four sigma bonds, right, and what we know about sigma bonds is that you can rotate around a sigma bond. Now hydrogen is just a sphere. When I rotate it, it still looks like a sphere. Okay, so basically, although I have sigma bonds, where rotation is possible, it doesn't change the shape of the molecule at all. So methane you really just have one confirmation, just the way that I drew it here.

But if I look at ethane the situation is different. So I'm drawing ethane showing two tetrahedral structures connected to each other. Again, if I look at the carbon hydrogen bonds, if I rotate around the sigma bonds nothing will happen, but if I rotate between the two carbons, then actually it will look very different.

The way that I drew the molecule here, remember that a bond that is neither a wedge going up or down is in the plane. All right, so the way that I drew this, this hydrogen is in the plane with this carbon, with this carbon, and this hydrogen. But if I rotate around the sigma bond, I can move the hydrogens away from each other.

So, it's really difficult to draw that using this kind of representation. Okay, so what we do is we're going to learn a new representation, it's called Newman projection that will let you see what the confirmation of an ethane molecule is like.

So, it works like this. So, we want to do the Newman projection for ethane. First thing we do is we write it out, and then draw a circle. That circle represents the two carbon atoms and ethane that I just put a box around, so this circle represents to two carbons.

And what we're doing is we're looking at the molecule from the side, like this. So if you look at the molecule from the side, you're going to see three hydrogens close towards you and three hydrogens far away. The hydrogens that are close to you we draw lines all the way to the middle like this. Okay, so the lines go all the way to the center of the circle, and it kind of makes sense because you should be able to sort of see these bonds to the carbon that's closest to you. To represent the carbon that's in the back we draw lines that go to the periphery of the circle like this. So, this hydrogen in blue would be the same as one of the hydrogens over here, and hydrogens in purple would be one of the hydrogens in the front. That's how we're labeling these.

So, the reason that these are useful for confirmation is that it's very easy to turn that sigma bond now. All I have to do is keep either the front or the back constant and rotate the other part. So for example if I keep the front constant, and it really doesn't matter which you do, and now I'm going to rotate that bond clockwise, so this hydrogen is now going to move over until it's right behind the hydrogen in the front, and of course they all move like that.

Now I want to say that they're exactly on top of each other, you can't of course draw them on top of each other so whenever you draw a hydrogen that's real close, like this, it implies that they are exactly on top of each other. That's the convention, okay?

When they're in between, it's very easy to draw that, because they don't overlap. So these are the two extremes in confirmations. The hydrogens are next to each other or they are on top.

So each confirmation will have its own name. The one where the hydrogens are next to each other will be called staggered, and when they're on top of each other it's called eclipsed. So, those words should kind of make sense. Eclipsed, it's like a planetary eclipse, go on top of each other.

Now there are actually an infinite number of confirmations here because I can have any angle between the front and the back, but for convenience sake we only name the extremes of that, so we name the staggered we name the eclipse, but you have to understand that there are really an infinite number. So, if you want to get an idea what that looks like, we can draw an energy diagram, to show how the energy changes as we go from staggered to eclipsed.

Let me go back to this diagram here. Which do you think would have a lower energy, be more stable between these two? The staggered or the eclipsed? What's the reason that the staggered is more stable? More space?

The way we say that in organic chemistry is it has less steric hindrance. Steric just means filling up space. It just means the two groups don't want to be near each other, because they take up space. That's exactly right. It turns out that, according to the picture, they are farthest away from each other when they are in the staggered confirmation. Okay, so we'll call that a steric interaction, or actually steric hindrance would be a little bit better in this case.

Okay, so that will be the reason that that will have lower energy.

Let's go back to our energy diagram then. Let's start at that confirmation. So, as we rotate we'll reach a point where we reach the eclipsed structure. And if we continue to rotate, we'll find that we wind up in the same position as we did originally.

So, we can draw the energy diagram like this. So we have energy versus angle. And of course I can keep spinning this forever, so I'm basically going to have something that looks like a sine wave, it's just going to be up and down like that. And, the importance of this is that you can calculate it, you can run a computer program that will tell you exactly what the energy will be at any given point.

The reason that's important is that it's these energies that are going to determine that shape of a molecule, any molecule. Now, for ethane, that's it. There's only one sigma bond that you can rotate around.

But, for a complicated molecule, like a drug for example, that has lots of sigma bonds, they can rotate in all kinds of different angles. So you need to basically draw these energy surfaces and figure out where the lowest energy point will be, you know, at the temperature of your body, basically. So you can predict that the molecule will have this shape, which is important because it's the shape of the molecule that determines if it will fit in a receptor or not.

It's important, because it enables you to predict the confirmation, the most stable confirmation for a molecule. All right, let's do one more here. If we go from ethane to propane how does that change the Newman projection? The thing about the Newman projection is that you can only do two carbons at one time. So you have to decide which carbons you're going to do.

In the case of propane, it doesn't matter. If I take the left two carbons or the right two carbons, it'll give me exactly the same thing. So, I shouldn't have to tell you which carbons to use. If there's more than one possibility, I'll tell you, you know, do C2 to C3, so that will tell you between the second carbon and the third carbon. Okay, so I'll need to specify that, but in this case we don't have to because there's only one possibility.

So that's what we start with is a box between the two carbons. And, what you'll find with all these Newman projections is that you'll have three things in front, three things in the back. You notice here that it's three hydrogens over here, and on the right side it is two hydrogens and a methyl. So, to translate this, again, draw a circle, three lines, pointing to the middle. And it doesn't matter which you put in the front and which you put in the back, okay, but if you start to put one in the front or in the back, you need to continue that in all of your projections. So, you can't change your mind once you start. That's really important.

So let's say I decide to put the methyl group in front. I'd have methyl, and then H and H. In the back I would have three things, H, H, and H. So I started with a staggered confirmation, just because it's easier to draw, but once we start we can then draw all of the other confirmations. If I keep, in this case, the back constant, and I rotate the front clockwise, this is what will happen. Here's the back, rotating clockwise. The methyl group will now be on top of this hydrogen back.

Okay, so again, I have a staggered and an eclipsed confirmation. It looks pretty similar to ethane, basically. If I continue to rotate this, I'm going to put the methyl group between two hydrogens and that is identical to what I started out with. So there are only two confirmations here, two extreme confirmations. The staggered and the eclipsed.

So, the energy diagram will look very similar to this, but most likely the difference in energy will be higher, because methyl group is larger than the hydrogen, so the effect will be more pronounced in terms of the steric hindrance. Okay, so the next one we have to do is butane, and there's a lot more confirmations in there, that's going to take a little more time, so I think for today we'll stop here.

On Fridays we have a little more time reserved. It's a great time to come see me after class. If you wanted to talk in smaller groups I'd be happy with that.

2 Comments:

At 11:37 AM, Blogger NIRUPA said...

it wil be more clear if you explain with structures of the compounds..

 
At 11:37 AM, Blogger NIRUPA said...

This comment has been removed by the author.

 

Post a Comment

<< Home