Lecture 015b: Test 1 makeup review

Lecture 015b: Test 1 makeup review

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Jean-Claude Bradley: This is a review session for the make-up starting immediately after class. I'm going to go through your questions first. After we go through that, if there's any time left over, I will look at the most commonly misanswered questions. That might give you some feedback as to what concepts that most of you are having trouble with. Again, the same thing, once the make-up period starts, I won't be able to answer your questions. So this is the best time to do it.

I have one question here that is the reaction of 2-methyl butane with bromine. This would have to be in the presence of light and we're looking for a minor product. The answers that are listed here are 1-bromyl pentane; 2-bromyl 2-methyl butane; 2-bromyl 3-methyl butane; 2-bromyl pentane; and, of course, none of the above.

The first thing to do obviously, is to translate the words into an equation, and then to figure out what all the minor products are. So 2-methyl butane, that's the first step. That's going to react with bromine and light. It's not a question about mechanism, so the only thing we want to concern ourselves with in this problem is how many hydrogens do we have, and how many different products, and which one of those products is going to be the minor product.

So I'm going to circle the different hydrogens. How many do we have in this case? Well, let's start by circling them. Let's start from the right-hand side of that molecule. I have a methyl group, so those three hydrogens are going to be one group. Now, is the methyl group on the right the same as the methyl group on the left? I think we, in fact, did this exact problem in class before. The six hydrogens on the left are different than the right. Why? Because it's not symmetrical. I have to start counting from the left-hand side and not the right, because the methyl group is on the second position. That's all I can circle for one group. Now I move down the chain, I have a C-2 group, that's clearly different from any of the other hydrogens Then I have that tertiary hydrogen, CH, that's certainly different from any others, and then finally, I have the six hydrogens at the end. In fact, we have four different products in this reaction.

Although I'm doing one question here, I'm actually doing several at the same time. Because there were several questions, one of the questions asked was how many different products. So the answer is four. One of the questions might be which of the following is a major product? So I'm going to answer all those questions simultaneously. There's a question?

Student: [unintelligible]

Jean-Claude Bradley: The one on the left? Are you asking why those two methyl groups aren't different?

Student: Yes.

Jean-Claude Bradley: Because those two methyl groups are identical and they're on the same tetrahedral carbon. They're on the same carbon. If I add three methyl groups, those three methyl groups would be the same. If I had two chlorines, those chlorines would be the same. That's one way to look at it. Another way to look at it, is if we put a bromine on there, and we name the compound, it'll be the same compound. You can think if I count through this molecule, why should I start counting at that methyl instead of at this methyl? There's no reason. Any way that you can look at it that it makes sense. I like to think of it as if you've got two identical groups that are on the same sp3 hybridized carbon, then they're the same. Sometimes the way you draw it might be a little strange, but that way that's always going to be true.

Four products. Now when we're looking at major products versus minor, the way we define this consistently is we're going to look for tertiary first. Tertiary are more stable. So far, we've only looked at tertiary in this class. Basically we have one tertiary center. We have one secondary center. Again, when I use the term secondary, I'm saying we have a secondary carbon, or I can say we have two secondary hydrogens, but those would generally give rise to secondary alkyl halides. Then we have two different groups, which are primary. Is everybody clear on that?

Student: [unintelligible]

Jean-Claude Bradley: The hydrogen that circled tertiary. In order to figure out what it is, you look at the carbon on to which it's connected. Then you ask the question how many carbons are connected to that carbon? The answer is three. You're always really thinking about the carbons. Think of it that way. Any other questions on that? Any molecule that I've brought you, this is the same process.

This exact question asks about the minor products. Again, we define in this chapter, that the tertiary would be the major, and everything else will be a minor. We would actually have three minor products in this case. You may have come across a question that was almost the same, but with a different selection of options. The same question in this case can have three correct answers. There are three minor products. What we're going to do, is look at what those products would be. I can put the bromine on the left side. I can put it on the secondary carbons. And also I can put it on the right-hand side, if I'm only interested in minor products.

Let's take a look at our options. The first one really has nothing to do with this. First of all, it's a pentane. Hold on a second, let's draw this out. Right. 1-bromyl pentane would be five carbons in a chain. We do have five carbons in the molecule, but it's branched. So, it's not going to be 1-bromyl pentane. 2-bromyl 2-methyl butane. What that one is really talking about is the major product. 2-bromyl 2-methyl butane. The question is about the minor product, so that's not the answer in this case. The next one is 2-bromyl, did I reapeat that? Three. 2-bromyl 3-methyl butane. I see that in the second product I drew I have 2-bromyl 3-methyl butane, so the answer for this particular one would be C.

Whenever you do these problems the most important part is to write out the equation and make sure that the products you draw are drawn correctly and then go through and systematically compare these each. Yes?

Student: [unintelligible]

Jean-Claude Bradley: I'm not sure I understand. Are you asking me why is it that the 2-bromyl 2-methyl butane is not a minor product?

Student: Yes.

Jean-Claude Bradley: 2-bromyl 2-methyl butane, which I can draw down here, this is a major product. It's a major product because it arises from the most stable radical, from the tertiary radical. I'm not sure how else to explain it. Basically, it's the definition that we use in this chapter, we basically going to name our major products just from being from the most stable radicals. In this case it's only one of the products. I actually covered probably a dozen questions with this, that were related.

We have a Newman Projection question here. Three methyls in the back and three ethyls in the front. The first thing is we'll just affix the groups. We draw a tmplate. Why am I drawing it staggered? It's a conformation question, it doesn't matter, it's just easier to draw a staggered conformation and than an eclipse. Ultimately, we just want to position the groups. Then I randomly put the groups, the three methyls in the back, so my back three carbons are here. Three ethyls in the front. That's what that Newman Projection would look like. We want to retranslate that. I think the question is about naming it. I'm going to retranslate that, and I'm going to do the reverse, I'm going to draw the axis of the Newman Projection in a yellow rectangle. Then we're going to fix the groups. So, it's really the same thing now. You've got the three methyls on the left and the three ethyls on the right.

Now I don't have any options for the answers, but we can go ahead and name this thing. We're going to pick the longest chain, so you notice on the left that I've got three methyls, it doesn't matter where I start from, I might as well start from the middle one, on the right also, we've got three ethyl groups so they're all going to give you the same length chain. If I number this guy, I'm going to number it from the side that's going to give me the lower numbers. I find that my longest chain has five carbons, so it's going to be a pentane. What is it in addition to a pentane? It's got 2-2-dimethyl and 3-3-diethyl, so if we want to name it alphabetically, we would say that we have 3-3-diethyl 2-2-dimethyl pentane. If the order is different, that shouldn't prevent you from answering the question because you're going to see from the list of options given, I don't know what the options were in this particular case, but you would be able to draw out from each of the options given and if it correlated with the molecule that you drew, it would be correct. The key thing is to take your time and draw out your options. Use up all the time to do review. A lot of you are finishing really quick, and are not doing that well, on the make-up take a little bit more time to review, to make sure everything is correct, that you have the right number of carbons and everything. Do I have any other questions, these are all the paper questions that I have. Yes?

Student: [unintelligible]

Jean-Claude Bradley: That's the question we just did, that was the first question. The minor product of the reaction of 2-methyl butane with bromine. Like I said, you may have had a slightly different selection of answers, but from those answers you would see one of those three products and if you don't, it would be none of the above. Okay, anything else?

Student: [unintelligible]

Jean-Claude Bradley: Because in chapter five we covered enantiomers and diastereomers, I'm telling you not to worry about counting enantiomers and diastereomers, because that's not the way we learned it in chapter four. There are actually more products than the way we do it in chapter four, if you count mirror images and diastereomers. The way you're doing the problem is exactly the same way you did it on the quiz in chapter four. All you're doing is looking at the different kinds of hydrogens.

If the question is where we are reacting pentane with bromine in the presence of light, how many products can you have? Let's take a look at the number of hydrogens. What do you think, how many different kinds of hydrogens in this molecule?

Student: Three.

Jean-Claude Bradley: Three. Let's start with the ends, we have a methyl group, so those three hydrogens are the same, but unlike the first problem we did, this molecule is symmetrical. There isn't any reason that that would be the one position. If I go on the left side, that just as easily can be the one position. Another way to look at it, if I put a bromine on the right, it would be called 1-bromyl pentane, if I put it on the left, it's still called 1-bromyl pentane. Same color here, six hydrogens are the same. I continue to make that argument, if I go inward, the CH2, that's the two position if I start to count from the right and the same thing on the left. Then we end up in the middle, and those would be different. The total number of products would be three, and let me do a little more work on this in case we have a similar question. Those are the only three products we can make, 1-bromo, 2-bromo, and 3-bromo pentane. How many minor products do we have? Let's label primary, secondary, and tertiary, and let's see. The ones at the end are what?

Student: Primary.

Jean-Claude Bradley: Primary, and these are secondary, and the middle ones?

Student: Tertiary.

Jean-Claude Bradley: Not tertiary. Look at the hydrogen, how many carbons is that connected to? Two, so that is also secondary. Based on our definition of major and minor we have how many minor? Just one. The 1-bromo would be minor, and the other two would be major. In other words, if you do this reaction you can expect an even distribution of the two middle ones, or a significant amount of the two middle ones. Yes?

Student: [unintelligible]

Jean-Claude Bradley: When you're thinking about major versus minor, the major will be the products that come from the most stable radicals. We have no way no way of distinguishing between the stability of those two secondary radicals, so we're going to assume they're the same stability. If there were a tertiary radical, then that would be the major product, but we don't, so we have to look at the most stable that we have. And that's secondary. Yes?

Student: [unintelligible]

Jean-Claude Bradley: Okay, let me make sure we don't have any more questions on this. You have a question on this?

Student: [unintelligible]

Jean-Claude Bradley: Well, the first one is minor because it comes from a primary center. The primary radicals are less stable than the secondary radicals.

Methyl cyclopentane reacts with bromine in light. The first thing is we're going to make sure we draw all the hydrogens and not miss any and then we will circle the different ones. The easy ones to spot are the three Hs on the methyl. Those are clearly different from anything else. The hydrogen right underneath the methyl, there's nothing else that looks like that, so that would be the second one. Now you've got to be a little bit careful, as we go around the ring, there may be some that are the same. If I go down and look at the two hydrogens that are on the carbon next to that, are they the same? They're different. They're different because one of them is cis to the methyl group and one of them is trans to the methyl group. They're not the same relative to each other, but if I go on the other side of the ring, then those would be the same. If I circle the top hydrogen in blue, then those two would be the same. That's why we're not consideringenantiomers and diastereomers, because that would complicate this further. The way we did these problems, those two are the same. The ones that are underneath are going to be the same as each other. Then we go down to two carbons away. Same deal, same side, opposite side.

One of the questions you may have had from this is how many different products do we have from methyl cyclopentane? We have six products in this reaction. To determine the major ones versus the minor ones, we have to determine primary, secondary and tertiary. The Hs on the CH3 would be what? Primary, and this H, tertiary. It's connected to three carbons. Then as we move down the chain, secondary, in fact, all of these are going to be secondary. Again, using our definition of major and minor, we have a tertiary, so the tertiary will be the major product, if I put the bromine there. Anything else is going to be minor. There are five possibilities for that question. We could have cis 1-bromo 2-methyl, trans1-bromo 2-methyl, whatever. Those would be all the minor products in this case. Yes?

Student: [unintelligible]

Jean-Claude Bradley: In total we have six? One major product and five minor products. Let me at least draw the major product. The major would be 1-bromo 1-methyl cyclopentane. All right, what else?

Student: [unintelligible]

Jean-Claude Bradley: The question is this, 2-2- 3-3- tetra-methyl butane reacts with bromine and light in this case. How many products do we have from that? Let's write out the formula for that. Here you have the butane. We have at the two position, we have two methyl groups, at the three position we also have another two methyl groups. This looks awful familiar, I think this may be straight out of one of the problems. The point is, we have to determine how many different products we have. So, how many different hydrogens do we have in this case? Let's start from the right side. Clearly these three hydrogens are the same. What about the other methyl groups? The one that's here and one that's here. They're the same. Remember that rule? They're on the same tetrahedral carbon, so those would be the same. If I keep going on the other side of the molecule, you would notice this molecule is completely symmetrical. There's no reason I would number it from the right side versus the left side. The nine hydrogens on the left are also the same. There would be only one product from this reaction. It would be 1-bromo 2-2-3-3-tetratmethyl butane. No matter where you put the bromine, it's going to be in the one position.

Student: [unintelligible]

Jean-Claude Bradley: Because all of the methyl groups are the same. You can see that the three on the right are connected to the same sp3 center, so those three methyl groups are the same. You see the right side of the molecule is the same as the left side, so there's perfect symmetry there. Let me use another argument. Let's say we actually do put a bromine in there, and let's say we try to name our molecule so it's actually one product. Let's say you became convinced that hydrogen was different from the one on the right. What I would do now, now that I think that I have two different compounds, I would try to name them. If I try to name them, the bromine has got to be on a position on the main chain. When I count my longest chain, I'm going to include the bromine in it. If I start from that position I find that I have a butane. I would have a 1-bromo 2-2-3-3-tetramethyl butane. Now you go into the compound on the right, I have a bromine and I have to include that bromine as part of the main chain and I have to find the longest chain. Now my one position is the same, so I would end up with the same name. If you're really confused, you can name the thing to see if it comes out to the same name, alternatively if you use the strategy I just described, where are the groups that are on the same tetra-carbon are the same, and if there is a perfect symmetry, then each half has to be the same.

Student: [unintelligible]

Jean-Claude Bradley: When I say symmetry, I did refer to a plane. If you have perfect symmetry on all sides, there is only going to be one product, like cyclo-hexane, cyclo-pentane, methane. Any time you have perfect symmetry, there's only one bromo-cyclo-hexane. There's only one bromo-benzene. When you start to introduce a symmetry into the product, if I put a methyl group on a cyclopentane, like we did up here in the last problem, if I have cyclopentane, there's only one product, because it looks all the same. As soon as I introduce a methyl group, I now break the symmetry. Although I have symmetry vertically here, I have a mirror plane, that's why the left-hand hydrogens are the same as the right-hand hydrogens. However, horizontally, I've lost my symmetry, so now I've got six products instead of one. It's not fair to say if you have symmetry you have only one product. You have to have symmetry along all planes to have only one product.

Student: What's the difference between [unintelligible]

Jean-Claude Bradley: A halide is fluorine, chlorine, bromine, iodine. Mono means one, so basically what it means is don't put more than one halogen. When we say mono halogenated, and it happens to be bromine, it's the same as mono brominated. It just means standard conditions. Standard questions that you did on all the quizzes that we did on all the problems. What else?

Student: [unintelligible]

Jean-Claude Bradley: The question is how many Newman Projections? 2-methyl propane. If I didn't specify which two carbons, C1, C2, but even if I hadn't, I could've asked the question without specifying, because you have no choice, there's only one way to do this one. I would have to pick two carbons, C1 C2. We're going to draw two projections for that. There's many ways to draw that, I just chose to draw it like this, the two methyls in the front. What I'm going to do is I'm going to rotate the back. Because I've got three identical groups it's a little bit easier to see the problem that way. I'm going to rotate the back clockwise. If I continue to rotate the back clockwise, I will just get back the first one. Basically, it's three of the same groups. The only thing I can really do is just draw them staggered or draw them eclipsed. But overall, I only have two conformations. That's the answer. If we had a different product, we could end up with more projections, but in this particular case, we only have two.

Student: [unintelligible]

Jean-Claude Bradley: We have a cyclohexane conformation question. What's the cyclohexane?

Student: [unintelligible]

Jean-Claude Bradley: Trans. We do these problems always the same way. We draw out the two conformations, the templates at least. You have your choices as to where you want to put them. I suggest you always put them in the front because it's easier to draw. We have 1-ethyl and 4-methyl, so if I count along here I look at four. The four position would be all the way across and I have to be careful to make it trans. The way that I put the ethyl is underneath the ring, so I want to make sure the methyl is on top of the ring, which is easy to see from the axial position. All I have to do is to draw the other conformation is to just flip each one. This one is a lot harder visually to see that they're trans because they're both equatorial, but they are definitely trans. In this case, which one would be more stable? Left or right? The right one. We want to put as many of the groups as possible on equatorial. Sometimes, you have to make a tradeoff, in this case we don't. Both groups are either equatorial or axial, so it doesn't matter what the groups are. The right one is going to be more stable.

We want to note here that the ethyl is equatorial, the methyl is equatorial. If you had this question, there's a number of different possible answers, but if you have an answer where the ethyl is equatorial and the methyl is equatorial, that would be correct in this case.

Student: [unintelligible]

Jean-Claude Bradley: If you could have them all equatorial that would be good, but sometimes you can't. Sometimes you can have one equatorial and one axial. In that case, you're going to pick the structure where the largest group is equatorial.

Student: [unintelligible]

Jean-Claude Bradley: Do you see all the bonds I drew straight up and down? That's axial. The ones that are kind of horizontal are equatorial. I'm not a great artist, so if you want to see perfect angles, look at your book. It's much easier to see the axial ones because they're definitely exactly up and down..

Student: Why would one be axial and one equatorial?

Jean-Claude Bradley: Because they're trans. Actually, that's what I'm testing you on, is that you're able to see that when they're trans they're on opposite sides of the ring. Trans is just opposite sides of the ring. So methyl is underneath and ethyl is on top. They're both trans, it's the same molecule. Are there any questions about the make-up or anything else related to it?

Student: [unintelligible]

Jean-Claude Bradley: All the same rules apply for the make-up. Make sure you read the FAQ carefully, find out which rooms you're supposed to do it in and all that stuff, you don't have as much time, I think. Does anybody know how long it is?

Student: [unintelligible]

Jean-Claude Bradley: It's always going to be more questions, I'll put more questions in, in the same amount of time.

Student: [unintelligible]

Jean-Claude Bradley: I will take the second score, no matter what it is. This is an opportunity if you really messed up the first one, it's an opportunity for you to do better, but you don't want to take it lightly. It's just an opportunity. You don't have to tell me you're going to do it, you just do it. Everybody can do it if they want to. But, by the way, the average, do you see the average when you log in? The average is about 81 percent, so if you want to relate yourself to that, that's what it is. You have a question?

Student: [unintelligible]

Jean-Claude Bradley: The same thing, 90 minutes.

Student: [unintelligible.

Jean-Claude Bradley: Well, I think it's four days, you've got to log in and see what period of time. I think the next make-up, I can't give you as much time. But this one, you've got plenty of time. Anything else? I wish you luck, if you're doing it again.

Transcription by CastingWords