Lecture 033: test 2 review

Lecture 033: test 2 review

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Jean-Claude Bradley: I'll start off with some questions that were requested for the test. I think these are all quiz questions.


2s-3s and 2-3 dibromopentane and 2s-3r. The question has to do with the relationship. It's either enantiomers, diastereomers, geometric isomers, the same molecule or none of the above. Pretty classic based on some of the problems we did from that chapter.


The first thing we do is we will write out the molecule an official projection. We have three clockwise. Lowest priority group is in the front, so the top one here is s. Going clockwise, clockwise is normally r, but the lowest priority group is in the front so that's s. Going counter clockwise, counterclockwise is normally s, but the lowest priority group is in the front so that's r. What I drew in this isomer is the 2s3r, basically, this one here. To draw the other one, both of the positions are inverted so I just need to switch the two bromine positions. This one would be r3s. Because each s becomes an r and each r becomes an s, we know that basically it's either going to be the same molecule or they will be enantiomers, so which is it in this case?

Student: [unintelligible]

Jean-Claude Bradley: Same molecule?

Student: [unintelligible]

Jean-Claude Bradley: How can we tell? If it's the same molecule I have to be able to rotate 180 degrees and can I do that and superimpose these two? I can't because I have an ethyl group on the bottom and an methyl group on top. If I had methyl groups on top and bottom then yes. I could rotate and superimpose it, so that's not the case. The answer in this case is that they are enantiomers. They are mirror images.


Another question has to do with the number of molecules that can be drawn for 2-3 dibromobutane. The first thing here is we need to determine is we've got two chiral centers in this molecule. If I have two chiral centers, that means if I do all the possible combinations there will be four structures that I can draw. A good way to draw them is as mirror images, it will make it easy to do the final count that the end. These two are mirror images, and then I can have these two possibilities. How many molecules can be drawn for 2-3 dibromobutane?

Student: Three.

Jean-Claude Bradley: Three, right. Part of the process you have to go through is after you draw everything, you need to determine if any two of these are identical. You'll notice the first two are the same molecule, because I can rotate 180 degrees and superimpose it, so the answer is three molecules.

Student: [unintelligible]

Jean-Claude Bradley: The first one, you mean this? Actually we can do it more easily in the second problem. A diastereomer would be any two molecules that would have the same framework. Like there would be two three dibromobutane but would not be an enantiomer. Any pair, for example, if I have A B C, A and C would be diastereomers. Now we have an sn2 reaction question. One bromo two 2-dimethylbutane reacts with chloride by an sn2 reaction.


We're going to draw out one bromo two 2-dimethylbutane. Sn2 mechanism is a short one, it's only one step. One of the lone pairs from the nucleophile will attack the electrolytic carbon connected to the bromine and simultaniously kick out the O-. When you do an sn2 reaction, you need to be wary about the Walden inversion. If I had a chiral center there, I would get an inversion, but I don't, so it's just going to replace the Br with a Cl.


The various answers we have are one chloro two 2-dimethylbutane, one chloro two 2-methylpentane, two chloro two methylpentane, three chloro three methylpentane. This is obviously one chloro two 2-dimethylbutane, so it would be A in this case. Be wary when you do these problems, there are many problems that look very similar, but they're not, so make sure that you carefully look at each answer and make sure it's really the molecule you've drawn.


We have an e1 example. Reacts with hydroxide via e1. If we're doing e1 there's going to be multiple steps. First step is always the generation of the carbocation. The second step is once you make the carbocation, you need to investigate whether or not you can get a 1-2 shift. Can we in this case?

Student: [unintelligible]

Jean-Claude Bradley: Right, so we had a secondary carbocation, there's only one shift that we can do that will give us a tertiary, and that is the one that we'll do. The 1-2 shift, I would have to move this hydrogen over. If I moved the methyl or the ethyl group, I would still have a secondary carbo cadine which you don't do, you always go more stable in these problems. We start drawing the arrow in the middle of the CH bond and end it on the carbon. I'm drawing out the carbon hydrogen bonds that I need to worry about at this point because we're doing an e1. At this point, the only thing left to do after the shift is to lose a proton and to generate all the various alkenes that are possible. This particular question asks you for the major product, but you could just as easily get the minor project, so let's to do the whole problem and see what we end up with.


In order to figure out which hydrogen we can lose we're going to count to two starting from the carbocation. When the H is on the CH2 on either the left ethyl group or the right ethyl group will give you the same product. The two ethyl groups are identical. The other possibility is one of the hydrogens from the methyl group. We have two sets of alkenes that we're going to make for that. I'll use different colors to show the two different mechanisms. In red, we would lose that proton. Actually, the mechanish states hydroxide, but of course, any base would do here.


We are going to make an alkene. The easiest way to do that is to first draw the two sps hybridized centers, then put the groups that are left. I'm losing this hydrogen, so what's left on the left hand side would be a hydrogen and a methyl group. On the right hand side I have an methyl group and an ethyl group.


Now, whenever you make an alkene you need to worry about cis-trans isomers. In this case we actually do have two products we're going to get from this. You need to remember that, especially if you're counting products.

Student: [unintelligible] one of the groups are cis and trans, you would consider it two products?

Jean-Claude Bradley: If you can have cis-trans isomers, you're going to get cis and trans. Or I should say e and z. It's not specifically cis and trans, it's just if you have geometric isomers.


Let's pick a different color to do the other side, blue. We want to grab the methyl group. Going the blue route, we would have another alkene. Lets say that we put the left side of the of the alkene from the bottom, what we have left are two hydrogens on one side. On the other side we have an ethyl group and an ethyl group. No cis-trans isomers for this product, so that means in total we would have three products from the e1 reaction. This specific question asks for the major products, so we need to look at the branching. Branching on the left structures is one, two, three. On the right structure the branching is only two. What we have are two major products and one minor product.


When you see this kind of question it will typically say which one of the following is a major product, so there may be five different answers to that question. It's really important you step through the whole problem to make sure you don't miss any. Any other questions on that? E1 is the longest, probably the most difficult problems. If you can do that you should be fine with all these problems.


Last one here, 1-butene reacts with mercury. In this problem we notice that we have a mercury reagent. Do we know what this product is going to be? When do we normally use the mercury reagent? I think we only used it one time in this class. Remember?

Student: The guaranteeing of the Markovnikov addition of water?

Jean-Claude Bradley: Of water, right. To guarantee the Markovnikov addition of water we use mercury. Now the only problem with this particular question is that the reagents are wrong in the second step. First, we use mercuric acetate, but after that we need to use sodium borohydride. If this question was asked in this way, it would be none of the above because we did look at the mechanism. This second one should be NaBH, the Markovnikov addition of water. The time we use the peroxide and sodium hydroxide is when we do anti Markovnikov addition of water, where we first we use borane and then we use peroxide.


That is what I have from the review problems, any other questions from chapters one through eight? Let me wrap up the ones that we...

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