Lecture 037: exam review 1

Lecture 037: exam review 1

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Jean-Claude Bradley: Let's do this review session for the test. The average was 79 percent, so if you want to situate yourself relative to that. I received a number of questions both online and in class, the thing is you also have to be proactive in looking at the problems we did in class, because many of the problems are already done. So, if we see that we have done an identical problem on Monday, I am going to skip those. In the future I am going to require you, if you want me to go over a question, to show me the work you have done on that question. So save your papers after you come out of the test and bring it to me, because it really does not do anybody any good to repeat what has already been recorded. However if I can help you work out where you were blocking, I would certainly be happy to do that.

Jean-Claude Bradley: Let me do what I can here, with the problems we have. How many different molecules for 1, 3-dibromocyclohexane? This one looks awfully familiar, I think we've already done it, but let me go through this. So, 1, 3-dibromocyclohexane, worked out we got two chiral centers, we draw out all the molecules that might be possible, and then we need to determine if any two are the same. So, if I have an axis of rotation, I can rotate 180 degrees about that axis, I see that the first two structures are in fact the same. They can be superimposed. So, total, we would have 1, 2, 3, so we got three molecules for that one.

Jean-Claude Bradley: Reaction of two butene with water in the presence of acids, so this says it's cis or trans, it does really matter in this case. So, here we're going to do what, Markovnikov addition of water. Now, one of the carbons has one hydrogen, the other side has one hydrogen, so what happens in that case? You can't follow the Markovnikov rule, so what actually happens? I can think about the mechanism as the wide Markovnikov rules exists, because you generate a more stable carbocation, if you're going to generate a secondary carbocation, secondary in the right, you're just going to get random. So you apply the Markovnikov rule when you know you have a difference between the left and the right, but if you don't have a difference in the hydrocarbons between left and right, then you would get random distribution. Now in this particular case, the left and the right are the same, so it's not going to make any difference; but if you come across a problem where you have different groups and they're both going to generate secondary carbocations, then you would end up with both. So in this case we would just put the OH on the two positions.

Jean-Claude Bradley: Two butene, or propene, react with MCPBA. MCPBA, meta-Chloroperbenzoic acid, so that is an epoxidizing agent, it will put on epoxide. Reaction of two butene with ozone, then dimethylsulfide reacts, results in what? Again, it doesn't matter if it is cis or trans. First ozone then methylsulfide is standard conditions for ozonalysis, and we don't get further oxidation with ozone, so we just end up with ethaldehyde. Okay, one butene reacts with methyliodide and zinc, so that's the Simmons-Smith reaction, we make a cyclopropane from that reaction. We'd have ethyl-cyclopropane in this case. We are doing SN1 with hydroxide. So first step with SN1 is always the same, generate a secondary carbocation, and then we see we are going to do a 1-2 shift, and we generate a tertiary carbocation. The final step is the attack of the nucleophile.


Student: Does it matter if the molecules are chiral? From the perspective of the carbocation?

Jean-Claude Bradley: Is it chiral?


Student: It looks the same on the left and the right, does it matter?

Jean-Claude Bradley: Where do you have four different groups?


Student: Does it matter that the SN1 has to be chiral?

Jean-Claude Bradley: Well, SN1, if the chiral center is on the carbon that has a bromine, then when you make the carbocation you lose the chiral information, because you go from four to three groups; if the chiral center was somewhere else, then you retain it. It really depends where the chirality is. And I think we did a couple of problems with a few chiral molecules.

Jean-Claude Bradley: Again we have an SN1 reaction. I have secondary carbocation, I see there's one 1-2 shift that I can do to generate a tertiary carbocation, so I'm going to do that one. That's the only product of that. CH2 CH2 CHBr. Hydroxide E2. We look for the various possibilities, 1, 2, three anywhere I try to count three on the right, I can't find a proton to remove, so I have to go on the left. 1, 2, 3, so there's only one proton in this case that is suitable for an E2. We draw this one that we see that they are cis/trans isomers, so we don't need to worry about that. Because both of these have the same branching, we have a branching with two, a branching with two, they would be the major product, it would be the only product, there would be no minor products in this case.

Jean-Claude Bradley: With the synthesis of 1-butene, I could put the Widding up on the left, and it'll end up like this. Or I can put the Widding on the right. So far as the possible starting materials you could have bromomethene, propanal, 1-bromopropane, formaldehyde.

Jean-Claude Bradley: E2-butene, the following two groups are trans. So here's E2-butene, and we see that the methyl and the methyl group are trans; also the hydrogen are trans. 2-butene reacts with warm, concentrated permanganate. So we create the carbon-carbon double bond and then oxidize each carbon, as far as it will go, which is a carboxylic acid, so acetic acid will be the product. Are there any other questions that were not brought to me on paper or email?

Jean-Claude Bradley: That one we did on Monday, come see me after class. Anything else? Okay, so what I'd like you to do on Friday if you have any questions for me, definitely work out the problems as best as you can, and then it will be a lot better for you to understand how to solve these things. All right, I will hang around until the end of the class. If any of you want to take the make up or have extra questions I will be happy to answer them.

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