Monday, May 22, 2006

Lecture 023: Alkyl Halides 3

Lecture 023: Alkyl Halides 3

View Lecture

Jean-Claude Bradley: We have one part from chapter six left to do, which involves some of the rules with respect to minor and major products of elimination reactions so let's do one compound which encapsulates a lot of the concepts we've been working with.

Jean-Claude Bradley: We did this reaction between this halide last class and hydroxide or chloride and we did SN1 and SN2, now what we're going to do is E1 and E2 and again make sure we understand how to tell the major products from the minor. If I tell you to do this via an e2 reaction, then that means its second order, its an elimination, so you are going to get an alkene and its going to happen in one step just like the SN2, the E2 happens in one step. Now what you have to eliminate is HBr and to know which hydrogen is actually eligible to be removed, a good way to do that is to count to three from the bromine. So if I start on the Br I can count 1, 2, three and find the hydrogen atom, so i can do an E2 reaction and eliminate that hydrogen.

Jean-Claude Bradley: Is there another hydrogen in that molecule that we can do? The one over here? Yes, one of these other three hydrogen also can be eliminated. Is that it? Are there just two possibilities?

Jean-Claude Bradley: The first step in doing these problems is to determine how many different ways that we can lose HBr so, count to three from the halide and identify these 2.

Jean-Claude Bradley: I'm going to draw two mechanisms for each of these eliminations. In the E2 the base comes in, grabs the proton, we lose the C-H bond and form a carbon-carbon double bond where that arrow is going into and at the same time we lose a Br-. So E2 is always going to have those three arrows in that exact order. Now when you draw the products from that E2 or any alkene a good suggestion is always start to draw the carbon-carbon double bond and then draw the others around it, because the carbon-carbon double bond is the only one that had a rigid geometry and its a lot easier to start with the SP2 hybridized template than it is to try to fit it in when you draw the molecule. So if we're forming the carbon-carbon double bond then the molecule's going to look different, so that is the other reason why you have to be careful writing the product out. You have to determine what are the two carbons where the double bond is forming. I'm going to draw a yellow rectangle around them. That's where the double bond is going.

Jean-Claude Bradley: What you're asking is what's left on the left and what's left on the right. I remove the hydrogen and on the left two methyl groups are left. On the right I remove the bromine so what's left is a methyl group and a hydrogen.

Jean-Claude Bradley: Now one of the things we have to use is when we did exercises to figure out if alkenes have cis/trans isomers this is where you apply it. Basically if you have cis/trans isomers you have to write them out. This compound does not have that, it has two methyl groups on one side so there is only one product for this pathway. If you had cis and trans you need to put both in. This is one way to do the elimination. The other way, let's look at the methyl group and see what happens.

Jean-Claude Bradley: Same arrows, same pattern, three bonds from the bromine. Again, draw the carbon-carbon double bond template and look at what's connected to it. If I draw a yellow box around the carbon-carbon double bond, one side has two hydrogens that are left and on the other we have a hydrogen and an isopropyl group. We check for cis and trans and we see that again this compound does not show cis/trans isomerism, so we only have one compound. We're almost at the end of this. We identified all possible HBr patterns that can be removed by E2, we worked out these two solutions, we worked out all the products. Only thing left is to determine the major product and minor product. What is going to determine that is degree of branching of the alkenes? An alkene is said to be branched if it has anything other than a hydrogen is on it. So the top alkene has three methyl groups and one hydrogen, so it has a branching of 3. Does not matter how big the group is it counts as one branch. I would write here branching is equal to three.

Jean-Claude Bradley: What would be the branching of the lower one based on that definition? Just one, one alcohol group, all other hydrogens, has a branching of one. What we're going to use is Saytzeff's Rule?? the most highly branched. I said I would give you very unambiguous definitions for major and minor product in each reaction. In this reaction the products that have the highest branching are going to be major. If we compare we have a branching of three and a branching of one. The one that has a branching of three would be the major product in this case and anything else would be a minor product.

Jean-Claude Bradley: The branching would be any groups on the alkene that are not hydrogen, would count as one branch. These are probably the longest problems you'll have to do in this course, you need to take your time and there are several questions on them. I ask you to go through and draw out the problems and then I ask you questions about it. Some of the questions might be like how many total products? We have two products total here. Sometimes you'll have more than one major product. If you have two products with the same branching-I had another here with a branching of 3, then you'd have two major products. Basically no shortcuts you really do have to work out all of these products in their entirety to answer questions about them.

Jean-Claude Bradley: There is one more we haven't done yet, the nastiest one of all, the E1. Let's take look at this example and see what happens. The E1 and SN1 are similar; they start out the exactly same its just how they end up is different. The 1st step is always what? E1 and SN1. That's the one where we have carbocations. In order to have carbocations we have to first loose the halide. Always do the same in these cases. We are going to Br- Br- first.

Jean-Claude Bradley: We just made a carbocation so before we do anything else we need to worry about rearrangements. Is there a 1-2 shift possible in this case? This is a secondary carbocation, so the only shift we need to worry about are the ones that would create a tertiary carbocation. To find out what groups could potentially do the 1-2 shift We start at the carbocation and count to two; so I can go one, two and think about moving the methyl group over. If I move that methyl group I would have generated a secondary carbocation so it wouldn't work. I could move the hydrogen and it would make the same situation even worse in fact, I would go from secondary to primary carbocation. Only one left is 1-2 moving this hydrogen over, that would generate a tertiary carbocation. So that is the one we must do before we finish the problem. The 1-2 shifts you start in middle of the carbon-hydrogen bond and end up on the carbocation. Move the hydrogen and its two electrons over. So now we have the tertiary carbocation. Now we are finished doing the 1-2 shift; this is the same for SN1 and E1 so far. Now is where the difference comes in. If I had an SN1 the nucleophile would attack C+ and it would be done, in this case I'd have an alcohol because I'm using hydroxide. But we're doing an E1 so that means we're going to lose a proton instead and generate an alkene. Because there's going to be more than one way to lose proton from this I can get a lot more products. If you want to use a counting strategy to figure out what you could lose, start on the C+ count to two to see what hydrogen is left. Let's say in yellow I will circle the candidates here?? 1,2. I could lose that proton to generate an alkene. Recognize that this methyl group is the same as this methyl group. They're on the same carbon. It wouldn't matter which of these six hydrogens I lost, I would still end up at the same alkenes. Otherwise I can go to the right and count 1,2, I could lose a proton and make a different alkene going to this side. That's it, so there are two pathways we have to work out.

Jean-Claude Bradley: So I could lose a proton on the left methyl group, and if I do that the only thing I left to do to make the double bond is to move the electrons between the CH to between the CC. So again the E2 is really, you can think of it - the E1 is like an E2 done in two steps, sort of, except that you potentially having a rearrangement in between. But we're doing the same thing?? we're losing a bromine and losing a hydrogen. So to draw this product. First we draw our SP2 template. What do we have left? On the left we have two hydrogen left and on the right we have a methyl and an ethyl group. Now I do the same thing with the one on the right. On the left I have two methyl groups left and on the right hand side I have a methyl group and a hydrogen left.

Jean-Claude Bradley: We're careful to check for cis/trans isomer. Again, these two compounds don't have any cis/trans isomers. Total we would have two products from this E1. We want to determine major and minor, that is done in exactly the same way. I look at the carbon-carbon double bond, the number of alkyl groups on the top one is two so the branching is two. On the bottom we have three alkyl groups and one hydrogen, so the branching is three. Using Saytzeff's rule the alkene with the highest branching is the major product; that would be the bottom one in this case.

Jean-Claude Bradley: So that is basically it. This example is good because it has everything really - SN1, SN2, E1, E2, has rearrangement and major and minor products. That is it for chapter 6, any questions?

Transcription by CastingWords


Post a Comment

<< Home