Lecture 034: review for test2 make-up

Lecture 034: review for test2 make-up

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Jean-Claude Bradley: I'm going to first go through the email questions. Many of these will be redundant, so we can get through these.


The first question is how many different molecules can be drawn for 1, 2 dibromopropane? Lets draw 1, 2 dibromopropane; when I draw it like this I'm not showing any chiral information, I'm just trying to see where the atoms are connected. The key thing is that we are going to have to count the chiral centers. How many chiral centers do we have in this molecule?

Student: One.

Jean-Claude Bradley: Just one? The carbon on the left has two identical groups, so that is not chiral; the right is a methyl group, thats not chiral. Only the one in the middle has four different groups on it. Now that we've determined how many chiral centers we have, with one it's easy; you're going to have an R and you're going to have an S. If we were to draw Fischer projection for that, we would have H and Br and we would have a set of enantiomers like this, so, the answer is two. Basically that's how you do these "how many molecules are there."


Next we have (R) 2-bromobutane reacts with hydroxide by an sn2 reaction. There's 2-bromobutane, we have to draw an (R)-2-bromobutane and use a Fischer projection for that. That's clockwise, and clockwise is normally R, but the lowest part of it is in the front. I actually drew the S isomer, so the one that I want to work with then would be the mirror image. This is the R isomer.


The question is we're doing sn2 with hydroxide. sn2 is one of the easier mechanisms, there is only one step. The nucleophile will attack the center and will kick out the R-. One thing you need to remember to do, when you have an sn2 reaction, is the Walden inversion. We will invert the chirality at the carbon where the attack is taking place. The OH will then go in the Fischer projection on the right side. If we determine the product, we have three clockwise, normally R, the lowest part of the group is in the, front so that's S. We have (S)2-butanol, which in the particular example I'm looking at was not listed, so it would be none of the above. We did have (R)-2-butanol listed, but (R)-2-butanol, we don't get any of that because of the inversion.


We have an E2 mechanism, CH2-CH, CH-Br, CH3. We are doing an E2 with hydroxide. Again, when you see that number 2, you should be happy because it means there's going to be a very short mechanism, it's going to be one step. We're doing E2, so we are going to lose HBr and generate a set of alkenes. The key thing to do here is to figure out how many different ways can we lose HBr. A convenient way to do it to start at the bromine and count to three until you hit a hydrogen. I can count one, two, three, and that's the candidate for the E2 elimination. I can also go on the left side; one, two, three, and there is no other way I can count to three to generate another another alkene, so there are two pathways that we will follow. I'll do them on the same molecule here in different colors.


By the red mechanism, we would generate an alkene just at the end of the molecule. That's one possiblity and there's no cis-trans products here, so that's all we get. If you go by the blue arrow, then we would have the double bond in a different position. On the left side I have a methyl and an ethyl group, and on the right side I have a hydrogen and a methyl. In this case, we do have cis-trans isomers, so we have to draw the cis isomer.


We look at the branching on the left structure and we have a branching of three; on the right structure we have a branching of one. The major products will be on the left hand side. This particular question asks for the major product, so we're going to have CH3-CH2-C, so it looks like it's B in this case. I'm not sure what quiz question this is, but this would be B.


We have the same molecule reacting by an sn1. Sn1 would always start off the same way; we would lose the R- and generate a carbocation in that position. We generate a carbocation, now we want to do a flip and see if we can do a 1-2 shift. We start a carbocation and count to two. I can move this hydride over and transform the secondary carbocation to a tertiary. Anything else I move, the methyl or the ethyl group, would just give me secondary carbocation. If I went to the other side, and moved one of these hydrides, I would I would go to primary, which definitely won't happen; there is only possible 1-2 shift, this one here. In sn1, the hydroxide acts as the nucleophile and finishes this. We end up with this tertiary alcohol. It looks like it's C in the answer set on the quiz.


Which of the following are impossible structures? We have cis cyclopentine. There's nothing stopping you from having a cis double bond in a ring of any size. That's fine; that certainly is a possible structure. In B, we have trans-cyclomelamine. It's harder to put a trans double bond in a cyclic structure; you need at least eight carbons to do that. We have cyclomelamine in this cause, so nine is more than eight, so that is also a possible structure.


Cis-cyclopropane. We know that three membered rings are going to be strained, but that doesn't mean that they are impossible, so cyclopropane would not necessarily be the most favorable molecule, but it would certainly be a possible structure because I have cis double bond in the ring.


Trans-cyclodecene. We have ten carbons, so it's very easy to put a double bond in the ring. In this particular question, the answer is none of the above; I dont have any impossible structures.


Propene reacts with reacts with HBr in the presence of peroxide. We need to determine what kind of reaction this is, HBr peroxide is anti Markovnikov addition of HBr. We went through the entire mechanism; we don't have to go through the mechanism to get the final product in this case, though you are responsible for the mechanism. Here, all we need to know is that HBr will add where the hydrogen will go - where there are fewer hydrogens. That means that the Br will go on the end and the H will go in the middle, so one bromopropane. That would be A.


Same question but now with H-Cl and peroxide, what happens in this case?

Student: Just a Markovnikov addition.

Jean-Claude Bradley: A normal Markovnikov addition. Just because there's peroxide there, it doesn't mean you will get anti; you'll only get anti Markovnikov with HBr, the energetics don't work out for H-Cl. Whether you have a perioxide or not will not change anything you'll just get a normal Markovnikov addition, so the answer is 2-chloropropane, which is D.


1-butene reacts with methylene, iodide, and zinc. This is the Simmons-Smith reaction. What kinds of products comes from that? This is how you add aCH2 group to a double bond to make a cycloprpane; wherever the double bond is, you just add a CH2. I would end up with this cyclopropane, so this is ethyl cyclopropane, and that's actually answer C in the quiz.


I think that's all I have for email questions; let me go on to the paper. I have a question here from the chapter seven quiz.


Which one of the followings can be used in a Wittig synthesis of 2-methylpropene? When your doing a Wittig synthesis, you're working backwards from the alkene. If it's unsymmetrical, like this one, there will be two different routes that you will have to work out. One of the things I could do is put the Wittig reagent on the left, and have an aldehyde on the right. You normally don't have a Wittig reagent laying around the lab, so you normally have to make it; you're going to make it from an alkyl halide. To do that, you would replace the C double bond P with an H and a Br. It's going to be two steps to that; first we have to treat it with triphenylphosphine, and then with n-butyl lithium. We went through that whole mechanism here; we don't need to do that, we just need to figure out the starting materials.


That would be one route; the other way would be to put the Wittig reagent on the right. That means that I would have acetone and ketone on the left, and again, we make the Wittig reagent the same way. We have four potential candidates as starting material for this Wittig synthesis. We have bromomethane, acetone, formaldehyde or 2-bromopropane. I see here in this particular example that B is formaldehyde, HCsO, so formaldehyde would be a canadite. But, any of these four products would be a correct answer to this question, so you really do have to work out both ways everytime you do these problems.


We're doing an E2 on this molecule. What could I do to abstarct an E2 in this molecule? I have g ot to count to three from the halide to a hydrogen. One, two; this one is too close. One, two, three; anywhere I count to three, I hit a methyl group. You will never lose a methyl group as the result of an acid base reaction. In E2 elimination, you can only lose a proton, so basically there's no way you can do an E2 on this molecule; it's just not possible, so there would be no products. E1 might be possible becaue I make a carbocation, then I get an rearrangement; then I might be able to an E1 but I can't do an E2 on this.

Student: [unintelligible]

Jean-Claude Bradley: Yes. This is my first step, do I do a 1-2shift on this? On the secondary, if I count to two from the carbocation, there's only one possible 1-2 shift; I have six identical methyl groups that can move. The question is, if they move do I end up with something more stable? I end up with a tertiary carbocation. So yes, you would defintely get a 1-2 shift. Now, you see what the 1-2 shift does for you, because without the 1-2 shift, you wouldn't do an E1 as well. There would be no way to do an elimination in this molecule. The shift enables a reaction to happen, so lets do that 1-2 shift.


Now I have tertiary carbocation, and now we will lose a proton. How many different ways can I lose a proton in this carbocation? Two. I can lose one of these hydrogens. These two methyl groups are the same; that would give the same product so we don't count them twice. On the right, I could lose this proton. Let's do the color thing here. In red, I'm going to lose a proton like this, so what do I have left? I have a methyl group and I have CH, CH3 and then a t-buty, so that would be one product. I don't have cis-trans isomers here, so that's just one. Alternatively, I can lose a proton in this direction. Two methyls on the left, a methyl and a t-butyl on the right. Again, this is trans isomer, so there're only two products from this reaction, and the right product has a branching of four, so that would be the major product.


How many different molecules can be drawn for one 3-dibromocyclohexane? I figure out that I have two chiral centers, I draw four possibilities. It doesn't mean that we have four molecules here. Any of these two the same and how can you tell? To be able to superimpose them, you have to be able to rotate them. When you have a ring, typically the kind of rotation you'll do is a flip. If you can flip the ring over according to an axis and generate the other structure, then that's what will happen. If you look at the top two structures, if I put an axis right here, and if I flip it 180 degrees, two bromines that are in the front go in the back. The first top two structures is really the same molecule. If I try to the same thing on the bottom molecule, when I flip them I end up with the same thing. I can't interconvert; the two bottom structures are different. This is just another example of having a meso compound, and then a set of enantiomers; so you would have three different molecules possible.


R-2-bromobutane reacts with hydroxide. Let's see, that's three, so that's R; and it reacts by an sn1 reaction with hydroxide. The first thing is we will lose the Br-; generate a secondary carbocation. If you check around there's no 1-2 shifts possible, so that will stay like that, but when we generated the carbocation, the carbocation is flat, it's not a chiral structure, we've lost the chiral information in that first step. Now this cation will react by hydroxide, hydroxide will come either from the top or the bottom; there's no reason to expect one to be more likely than the other, so we'll get a 50-50 mixture of R and S.


2-methyl-propane reacts with HBr. How does 2-methyl-propane react with HBr? It'll react nicely with an alkene, but there's no reason to expect a reaction between an alkane and HBr, so there's no reaction.


Reaction of bromine with cyclohexane in the presence of light? How many alkyl halide products are expected, not counting enantiomers and diastereomers? We need to determine how many different hydrogens we have. In cyclohexane how many different hydrogens? How many different ways can I put a single bromine? There's only one way. There's only one bromocyclohexane. There're no chiral centers in this molecule; well, we're not even counting chiral centers, but even if we were, there's just one way to put it, so that would be one product.


I'm seeing some repetition here, E-2-chloro-2-butene has the following two groups; which of the following two groups are trans? If I have E-2-chloro-2-butene, I compare chlorine versus carbon, chlorine has higher priority than carbon. On the right hand side, carbon is higher priority than hydrogen. My two highest priority groups are in opposite directions, so this is definitely the E isomer. Once you've drawn it correctly, let's take a look at our options. Which of the following two groups are trans? Methyl and ethyl; well, we don't even have an ethyl group here, so that's not it. Chlorine and butyl; we don't have a butyl group. H and Cl; the H and the Cl are cis. H and methyl; I see the H and the methyl group are trans, so that would be A.


(2R, 3R) 2, 3-dibromopentane, we actually could have done this problem really without drawing them, but let's take a look at them anyway. If you take a mirror image of the (2R, 3R) 2, 3-dibromopentane, you will definitely get (2S, 3S). So they're going to be either enantiomers or the same molecule. They clearly can't be the same molecule; the only time that kind of thing can happen is if you have meso compounds, in which case I would have to have a symmetrical molecule, which I don't. I drew them down here, you can see that there's no way I can rotate one into the other. These are definitely going to be enantiomers. 2, 3-dibromopentane; how many molecules? We're going to determine we have two chiral centers. Any of these the same molecule? That's what you have to do after you draw them. First of all, it's asymmetrical, so any of them that I attempt to rotate, the methyl group will end up on the bottom and the ethyl group would end up on top. The answer here is that we have four different molecules.


1-butene reacts with palladium and hydrogen. Pd H2, those are standard hydrogenation conditions; you will add hydrogen across the double bonds in a molecule. We would end up with butane, which is D in this particular example.


How many different molecules can be drawn for 1, 2 dibromoethane? First we determine the number of chiral centers, which is what for this one? No chiral centers, so there's only one way you can draw it. One molecule.


How many molecules can be drawn for 1, two dibromobutane? So that would be the same as the dibromopropane. Go back to that example.


1-bromo-2-methylbutane reacts by chloride by an SN1 reaction. Let's draw out the connection here first. 1-bromo-2-methyl. The R will refer to the chiral center that is the second carbon. I put CH2Br. That's the R isomer; so we do an SN1 reaction with hydroxide. We generate a carbocation that's next to a chiral center, we look for 1-2 shift; is there a 1-2 shift possible in this case? There is only one way that I can make a tertiary carbocation out of this, so I will. Now that I've made a carbocation on my carbon, that carbon is no longer chiral, so I can't use a Fischer projection to display it anymore. So, what do I have? I have two methyl groups and an ethyl group and I'm going to attack it with hydroxide at the end. There is an equal probability of attacking it from the top or the bottom, however even that doesn't matter in this case because I don't end up with a chiral product. I have two methyl groups that are the same. I would have CH3, COH, another methyl; so the answer in this case would be 2-chloro-2-methyl-butane, D, just replace the OH with a CO.


SN2 reaction, hydroxide. Again, you want to be happy that you have the two mechanism; only one step. In this particular example I did not give the chirality, so you don't have to worry about that. Let's see what we have; CH3, CO, H, H, C, so it looks like the answer is A, in this example.


1-butene can best be made from dehydrohalogenation of which alkyl halide? What we want to do in these kinds of problems is do an E-2 reaction, and see which one of the reactions will only give us one product. There are two ways I can start; I can start with the bromine on the first position or on the second position. If I have 1-bromobutane, that would certainly give me the product, and there would be no other side products from that. I would just get the alkene that I want, so that would be a good solution. Let's see if the other one is just as good. Unfortunately, the other possible starting material, 2-bromobutane, would generate the product that we want by losing HBr from the left, but we would also get some contamination with this elimination product. I get the top and the bottom, and because I didn't generate exclusively the product that I wanted, the right one would not be suitable. The only suitable one would be this, which is answer A.


Which one of the following compounds shows cis-trans isomerism? A methane, no. 1, 2-dibromocyclobutane? With 1, 2-dibromocyclobutane, I can either have the bromines on the same side or on the opposite. We're not looking at enantiomers on this particular question; this is from a previous chapter. We only want to look at cis-trans isomerism, so B would be the answer.


1-butene reacts with mercuric acetate; water followed by sodium borohydride. Those are standard conditions for Markovnikov addition of water without the possibility of rearrangement. That's all you do; you do Markovnikov addition of water. 2-butenol would be the product, which is B, in this question.


I think I covered all of the questions; is there anything else? That's it? Good luck on the test; we'll see you next week.

Transcription by CastingWords