Monday, May 22, 2006

Lecture 015a: Review for test 1

Lecture 015a: Review for test 1

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Jean-Claude Bradley: I'm going to start with a question I have from an online student on the quiz for chapter three, there are two questions here, and they have to do with the most stable conformation of cis, 1-ethyl, 2-methyl cyclohexane, so we're looking at quiz chapter 3.

Whenever you're doing a problem that has to do with conformations of cyclohexane, you have to draw out the two conformations, then figure out which one is more stable, and then answer the question depending on what the options are going to be, so there's no shortcut to doing this question, you really have to go through each one of these steps.

I'm drawing the two conformations, and I'm drawing the axial bonds, the equatorial bonds, for each one of the templates. We're looking at 1-ethyl, 2-methyl, and it's advisable to use the carbons that are facing towards you because you have a little bit more room to use, so I have to put an ethyl group, and the methyl group will be one carbon away, and it's cis, so it has to be on the same side of the ring. If I drew the ethyl group, under the ring, position one, then in position two, I have to put it on the bond that is closer to the bottom of the ring, and that will be this position here, and there we have hydrogens.

To draw the other structure, all we do is flip each one of those groups, so now the ethyl group will go equatorial and the methyl group will go axial. Between these two, let's say that we call the first structure A, and B, which one will be more stable? B. Why? The big group is equatorial. You'll notice that I can't put both groups equatorial. If I could, then clearly I would, but the molecule just can't flip in such a way to put both groups equatorial, what you have to do is look at the least disruption, or the least spheric hindrance, so that will be the one where the largest group is equatorial. The largest group is equatorial, and therefore structure B is more stable. Yes?

Student: What if it is cis 1-2 dimethyl? Would there be any difference between the two?

Jean-Claude Bradley: No, then there wouldn't be any difference between the two in stability, they'd really be the same conformation and energy, so there would only be one conformation.

That's the first step, so you see how long it took me to do this? You can't do these in two seconds; you have to take the time to do the problem. Once you've come to this point, now you look at the options that you have. In the most stable conformation of this compound, the methyl group is axial, so is that true? Yes, that's true. The other options are, the methyl group is equatorial, no. All the other ones basically don't describe the properties of this, so you can't really do these without drawing them out, so in this case the answer would be A, the methyl group is axial.

The next one is the most stable conformation of trans 1-ethyl, 2-methyl cyclohexane. The same relationship between the methyl and the ethyl group, but this time it is trans, if I put the ethyl group under the ring, the methyl has to go on the other side and I just flip these. Now, between A and B, which one would it be? B. This one is easier to tell actually, because it doesn't matter where the groups are. One of them has both groups equatorial; the other one are both axial. B is more stable.

Now we answer the questions. The methyl group is axial, no that's not true. The methyl group is equatorial and the ethyl group is axial, no, that's not true. The methyl and the ethyl group are axial, that's not true, the methyl and the ethyl group are equatorial, that will be true. I'm just identifying where the groups are, so in that case, it would be D.

While we're on the topic, any other questions on cyclohexane conformations?

I have a question here on the Newman Projection of butane, end butane. Which conformation would be the highest in energy?

For a compound like ethane, propane, butane, we did all of those, so when we talk about conformations of butane, we know that it's between C-2 and C-3 unless specified otherwise. If I give you any other molecule I will definitely tell you between which carbons you're going to do the Newman Projection.

To translate the butane into a Newman Projection, we draw circles, and we've got three groups in the front, three groups in the back. You arbitrarily select which three groups are going to be in the front or the back, so let's say that the left side will be in the front. I have a methyl group, I have a hydrogen and a hydrogen, so those three groups are going to be the groups that I put in the front. It doesn't matter where you put them initially. Then I have three groups that're going to be in the back. Again, I'll have a methyl and hydrogen and a hydrogen. The question here asks for both the highest and the lowest energy, so let's draw all the conformations and we can pick out which is which.

When you're determining the various Newman projections, you want to keep either the front of the back constant, it doesn't matter which, let's say in this case we keep the back constant, so all my structures are going to start off looking like this. Then I'm going to rotate clockwise, you can do it either way, but I like to do it clockwise. We're going to rotate that until we reach another conformation. The next conformation we're going to reach is when they methyl group is on top of the hydrogen. Then we continue to rotate, keeping the back constant. Finally, we can draw one more, where the two methyl groups are on top of each other.

In this problem, I think your selections are anti, gauche, totally eclipsed, partially eclipsed, so why don't we label them, and that'll be general for any question. The one where the two methyl groups are opposite is anti; the one where the methyl group is on top of a hydrogen is an eclipse conformation, specifically for n-butane it is partially eclipsed, that's the name of it. Then we have a staggered conformation where the two methyl groups are next to each other, that's gauche. Finally, we have the totally eclipsed conformation.

The one that's going to have the highest energy will be the one that has the highest spheric interaction between the two methyl groups. The highest energy is always going to be an eclipsed conformation, but here we have two eclipsed conformations, so we're going to look at the one with two large groups on top of each other, so that will be our highest energy. Another way of saying highest energy is most unstable, you can think of it that way. The one that has the lowest energy, or that is the most stable, will be the one where the two groups are as far away from each other as possible, so that would be the anti. I think I've covered every aspect of this question. Anything else on that?

Student: [unintelligible]

Jean-Claude Bradley: Yes. Only n-butane can be named specifically with these terms, anti, partially eclipsed, gauche, totally eclipsed. What I can say for any conformation is that the two methyl groups are anti, or that the two methyl groups are gauche, or that the methyl group is gauche to a propyl group. There should've been a couple of questions that used that kind of reference, especially when I ask you to find the most stable Newman Projection on a bigger molecule. I'll say the ethyl group is gauche to the methyl group or something like that. I don't have a question in writing on that, but I can certainly do one if we need to.

Now I have a question on the number of Newman Projections. How many different Newman projections can be drawn for n-pentane using C1-C2 axis? The C1-C2 axis will be this in yellow. Now we do it the same way we always do it; we start with a circle. We're going to have three groups attached, let's say to the back carbon, and then there'll always be three groups attached to the other carbon. That would be a hydrogen, a hydrogen, and an n-propyl group.

If I want to figure out how many different groups I have, what I have to ask is, if I keep rotating either the front or the back, how long before I regenerate the same projection? You'll notice that one of them have three groups that are the same, so hopefully, it will be obvious that I can only draw one more. All I can really do is draw the eclipsed conformation of that. Let's say that I keep the front constant. If I kept rotating the back, I would just go back to the first structure, so the answer to the question is two. That's pretty much it for this number of projections, you just count them.

Next question is actually draw a Newman Projection. If you have three hydrogens in the back, one methyl in the front; I have to have three groups in the front and three groups in the back. I don't know how the question was originally posed, but let's say that it means one methyl and two hydrogens in the front, so three Hs in the back, and one methyl in front. There are a number of things I can ask you about, this might be a question where I'm asking you what molecule is it, so you work backwards. Three Hs in the back would be like this, and in the front, we have a methyl and two hydrogens. Again, working backwards, the yellow rectangle will correspond to two of the carbons that consist of the circle. On one of those carbons you're going to have H-H-H, and on the other carbon you're going to have H-H-methyl, so the molecule that has that Newman Projection would be propane.

Next I have a reaction of propane with bromine gives how many monobrominated products? We're looking for monobrominated, which just means that you only have one bromine, which is the standard the way we've always done this problem; we've only put one bromine on the alkene. So, monobrominated or monohaloginated would mean the same thing, just put one halogen on there.. This is where we have to identify how many different hydrogens we have. We've done this problem before. We've got six hydrogens at the ends that are the same, then we have the two hydrogens in the middle which are the same, so there're only two different kinds of hydrogens in this molecule. These would be the two products. The question doesn't ask about major or minor, but we can do that as well. The first one would be 2-bromopropane, and the second one would be 1-bromopropane. The one that would be the major one would be from the secondary carbon. Anything that's not major is obviously going to be a minor product. I think that's about all that we can do with that question. Anything else?

Student: [unintelligible]

Jean-Claude Bradley: Okay, I can do hexane. Any other questions for propane?

Student: [unintelligible] major or minor group?

Jean-Claude Bradley: When we look through the radicals, in order to make 2-bromopropane, you have to go through a secondary radical, and to go through 1-bromopropane, you have to go through a primary radical, so secondary radicals are more stable than primary radicals. Without having to draw the mechanism every time, the way that you know that is you look at the molecule, and you have to be able to answer, the red hydrogens are primary, the green hydrogens are secondary. If I had tertiary, the tertiary would be the major products. If I had two different tertiary hydrogens, then I would have two major products. Anything else, primary or secondary, would be major products. That's how we've consistently used the terms major and minor in this chapter. When we do other chapters with different reactions, I'll be very specific about what is a major product and a minor one, but in this chapter, this is definitely how we are defining it.

I can do hexane, let's see what happens. What do you think, how many products do you think you have?

Student: [unintelligible] two primary.

Jean-Claude Bradley: Let's start with the primary ones. This is a symmetrical molecule, so I can't tell one end from the other, those six hydrogen in red would be the same, and you're right, they would be primary. Now all the other hydrogens are secondary, but they're not all the same. We need to determine the different hydrogens, so how many different hydrogens are left?

Student: Two.

Jean-Claude Bradley: Right. Let me use a different color here, green. Those two are different, but they're really the same as the two hydrogens on the other side. Those green ones are all the same. Then let's put some blue ones, and the blue ones will be the same, so the answer is we would have secondary hydrogens, primary hydrogens, we would have a total of three products. How many major products?

Student: Two.

Jean-Claude Bradley: Two. Exactly, because we have two secondaries and no tertiaries. Two major, one minor.

Student: Why, exactly, are those two secondaries different?

Jean-Claude Bradley: Let's draw the products, because if I draw the products and the products are the same, then the hydrogens would be the same, so let's see. If we exchange the red hydrogen, we would end up with 1-bromohexane. That would be the blue hydrogens being replaced, so do you see that 2-bromohexane is different from 3-bromohexane? That's where you get your three products. Let's say that you couldn't really tell, and you drew it, when you name the molecule, you should end up with the same name, if you start numbering from the side that would give you the smallest number.

Next, what is the intermolecular force between dimethylamine and water? First thing we look at for intermolecular force is we look at the strongest force, and we have hydrogen bonding between those two. We need to have HxH, and in amine we have an NH, and in the water we have an OH, so there's actually a couple of ways to do hydrogen bonding between these two. Just one of these ways would be through the nitrogen, so the answer is H-bonding.

Next I have bromomethane and methanol. The first question is, can we have hydrogen bonding between those two? Methanol has an OH, it can hydrogen bond to itself, but it can't hydrogen bond to bromomethane. Bromomethane doesn't have a nitrogen or an oxygen that it can bond to. The first question is answered no, cannot hydrogen bond. Do we have dipole-dipole, is that possible? To have dipole-dipole, we need to have dipole moments in both molecules. Well, let's figure it out, do we have a polar bond in bromomethane?

Student: [unintelligible]

Jean-Claude Bradley: Carbon bromine, right. In fact, that's how we learned about dipole moments.

Bromine is more electronegative, so that's the only bond we have in there, therefore it adds up to a positive dipole moment for the molecule, so it's polar. Do we have a dipole moment in methanol?

Student: [unintelligible]

Jean-Claude Bradley: Carbon oxygen?

Student: And the oxygen.

Jean-Claude Bradley: Yes, and the oxygen-carbon. It's towards the oxygen. That oxygen is an sp3 hybridized center, let me redraw this in a way we can more clearly see. So, oxygen is sp3 hybridized, we have dipole moment from the hydrogen to the oxygen, from the carbon to the oxygen, and we can consider through the lone pairs. We have four arrows and they're all pointing towards the left, so they definitely don't cancel out. We would say that methanol would definitely be polar, so we have two polar molecules, so dipole-dipole would be the dominant intermolecular force.

Those are all the written questions I have. Anything else?

Student: [unintelligible]

Jean-Claude Bradley: Resonance structure. It will have consequences in respect to figuring out whether the molecule is polar or not. It will have consequences with respect to figuring out the average charges on atoms. If you don't consider resonance, then SO3 looks like it's really polar in this resonance form. It also looks like these two oxygens are charged -1 and the other oxygen is charged zero, but, in fact, that's not the case. SO3 is not polar and the charge on each oxygen is minus two-thirds. That's a consequence of resonance. If you don't take into account resonance, then you won't get the right answers for those kinds of questions.

By considering all three resonance forms of SO3, we see that we get three arrows, and the three arrows cancel out. It's more likely to be questions that'll get you into trouble if you don't know understand resonance, rather than a direct question about resonance.

Student: [unintelligible]

Jean-Claude Bradley: The charge of the oxygen will be, in this case, each resonance form contributes equally, so one of the resonance forms has -1, one has -1, and one of them is zero, so the average charge on the the oxygen will be minus two-thirds. The average on sulfur would be +2, because it doesn't change. What else?

Student: [unintelligible]

Jean-Claude Bradley: Will it have dipole-dipole bonding?

Student: Yes.

Jean-Claude Bradley: It's not polar, so no. But you could get hydrogen bonding, with water, for example. Anything else?

Student: [unintelligible]

Jean-Claude Bradley: Acids and bases, we didn't do too much with that. Basically all I did was I defined a Lewis acid and a Lewis base.

Student: [unintelligible]

Jean-Claude Bradley: The test is going to be similar in format to the quizzes, that's the whole point of the quizzes. It doesn't mean there might not be a couple of questions that are a little different, but if we haven't spent a great deal of time on a particular concept, then it's unlikely that I will have many questions on it. The acid and base thing, again, is because we're going to use that concept throughout the Lewis acid and Lewis base, we just didn't spend that much time on it. Are there any questions about the format of the test?

Student: [unintelligible]

Jean-Claude Bradley: How many questions are there? There are 25 questions, which you can know just by looking at the test that's been posted for several weeks. It tells you how many questions, it tells you how long it is, it tells you when the start time and the end time it; that's under 'quizzes and tests'. That tells you what the tests are, so you know pretty much the format and everything.

Student: I understand the times [unintelligible]

Jean-Claude Bradley: If you have any suggestions on how to make the FAQ clearer, I would appreciate it because you just walk in and do it; you have four days to do it. There's no need to reserve or tell me when you're going to do it.

Student: [unintelligible]

Jean-Claude Bradley: No. I get that question so much, I'm really curious why that's being missed. Question nine on the FAQ gives you four rooms where you can do it in. It's pretty clear, four rooms. What else?

Student: [unintelligible] sigma bonds and pi bonds [unintelligible]

Jean-Claude Bradley: Why we did the hybridization?

Student: [unintelligible]

Jean-Claude Bradley: You basically need to understand, like for ethylene, it's anatomy. It has double bonds, but when we labeled stuff, we said this is a sigma bond, this is a pi bond, you need to know there's one pi bond in ethylene. You would need to know how many sigma bonds there are, you would need to know that when you draw those two lines indicating double bond, one of those is a sigma bond and one of them is a pi bond. You do need to know that, I'm not going to ask you to draw a picture obviously, but you need to understand that and to understand the consequence that it prevents the rotation between the two. Yes?

Student: [unintelligible]

Jean-Claude Bradley: Entropy? Yes, there's a lot of other stuff in the chapter that I didn't talk about. The syllabus basically gives you a checklist of the topics that I cover, so I don't think I talked about entropy once, so I really wouldn't worry about it.

Student: [unintelligible]

Jean-Claude Bradley: We definitely covered that, and that's part of the syllabus. All the functional groups, we drew them out. Yes, so you definitely need to know the differences between the functional groups. Any other questions? Gong once, twice; all right, good luck on the test!

Transcription by CastingWords


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