Monday, May 22, 2006

Lecture 038: Exam Review 2

Lecture 038: Exam Review 2

View Lecture

Jean-Claude Bradley: OK, so for the review session, I didn't receive any emails so I'm just going to do the problems we're getting in class here today. The other thing I want to point out is that for chapter nine, the exam will cover chapter one through eight. I was looking at the material that we covered, and the quiz that I gave you, and I didn't quite cover enough ground to be able to do that. But if you're taking 242, anything you did learn in chapter nine, that's where you are going to start, so one to eight for the exam.

[Student1 suggests a problem.]

Jean-Claude Bradley: This one right here? Ok, the SO. All right, so I'm starting off with HN3.

What is the charge of the central atom in HN3?

So available electrons? We're going to have nitrogen to contribute five. Five times three plus one. So, the hydrogen will have 16 and we're going to need eight for each nitrogen plus two for the hydrogen. So that's 26. We're going to share 26 minus 16, that's 10. So we have five bonds.

We have three nitrogen's and one hydrogen's position. Again we have different choices for what we can do. We can consider making triangles or other structures that are cyclic, but we want to avoid that. So if I have three nitrogens and I can't make a triangle out of them, we're going to make a straight line. That's the first thing we're deciding to do.

Now, we have another three bonds to put in a hydrogen so if I notice where else I can put the next two bonds, I pretty much have to do that. Therefore the only place I could logically put the hydrogen would be on the ends, because I already have four bonds on the central nitrogen. So that's the only place that the H could go.

Now I put my five bonds. I have to complete the octets. I'm missing four electrons on the left nitrogen. I'm missing no electrons on the central and one lone pair on the last.

Finally, the charges. So the nitrogen on the left has one, two, three, four, five, six electrons for charges. Nitrogen only has five so I have an excess, minus 1. The nitrogen in the middle: one, two, three, four electrons. Nitrogen has five so I'm going to have a positive charge; it's missing an electron. And the one on the right, one, two, three, four, five is neutral, so the question is what is the charge on the central atom? The central atom is nitrogen with the plus, so the answer here would be plus one. Does that make sense? In fact, this problem is extremely similar to azide, N3-. This is hydrazoic acid. It is just azide with a proton on one end, basically.

I have another question here on charges.

The Lewis Structure of SO, the sulfur has how many lone pairs? So for the available electrons we'll have six plus six equals 12. We'll need 16. It tells us that we have two bonds. With only two atoms, there's not much choice. We have to put the two bonds in the middle. Then we complete the octets. If we look for charges, six electrons with charges on each. Both sulfur and oxygen have six so there's no charges. So the question is, how many lone pairs? It would be two. So the answer you drew on the paper is two. Did you get that wrong?

Student1: Yeah.

Jean-Claude Bradley: Yeah, that's definitely the answer; two lone pairs.

And the Lewis Structure of SO4 two minus, the sulfur atom has what charge?

Ok, so available it would have six for the sulfur, six for each oxygen, so that would be 30. But we also have to add the two electrons from the two minus. We're going to need eight times five. So again, like all sulfur-oxygen compounds that we looked at, we're going to put the sulfur in the middle, surrounded by four oxygen. That uses up our four bonds. I can see here that's where you made your mistake. Is it because you didn't count the two negative?

Student1: Yeah.

Jean-Claude Bradley: So, basically if you count the two negative, you'd end up with only four bonds and then complete the octets. Those will all be negative one. The central sulfur then would be plus two.

Student1 comments on the procedure outlined.

Jean-Claude Bradley: Yes, if you have a plus, you remove that, because you are counting electrons. So the more negative, the higher the number. All right, so this is the charge on the central sulfur so the answer was plus 2.

SO2 is polar, has an average charge of negative one on each oxygen, has a plus two charge on the sulfur, has Van Der Waals as a dominant molecular force or none of the above?

So let's work out the Lewis Structure for SO2. So available we're going to have six times three equals 18. We're going to need 24. Share six. We've got three bonds. It's almost tempting to make a triangle; we know not to do that! We're going to put the sulfur in the middle with the oxygens around it. That uses up only two bonds. We have to use up a third one so we have to create a double bond on one of those oxygen. Complete the octets.

We look at charges. We have a negative on the right oxygen, a plus on the sulfur. Now, that's not the whole picture because that's only one of the resonance forms, so we have to draw the other resonance hybrid.

Ok, now is this molecule polar? SO2? It comes down to the hybridization of the sulfur. The hybridization of the sulfur is sp2. It's got three groups of electrons around it so that means I have 120 degree angles between the two oxygens. So it's bent like this. When we work out the dipoles, remember when you have charges, the charges will be more important than anything else, and you'll go from plus to minus. So I'll have one dipole moment in that direction for that hybrid. I'll have one dipole moment in that direction. Those two arrows can only cancel each other out if they are at 180 degree to each other. They are not, they're at 120, so the dipole moments don't cancel out and therefore, yes, the molecule is polar.

The average charge on each oxygen would be what? How do you figure that out? Negative one half, right. You look at all the resonance hybrids and you average them out. The charge on this oxygen is zero on the left, minus one on the right. Minus one plus zero dided by two is negative one-half, so central sulfur is plus one on the left, plus one on the right. It's average charge will be plus one.

We already know 'A' is the answer, but let's work these out to shed some light here. Is Van Der Waals the dominant intramolecular force? Well, no. It has a dipole moment so therefore will be dipole-dipole so 'A' will be the correct answer. The other questions are on chapter nine, and like I said, chapter nine will not be on the exam. So anything else in chapters one through eight? You guys exhausted everything on Wednesday?

Student2: The last question?

Jean-Claude Bradley: The last question? Yeah, I'm going to ask you to draw out how far you got on it and then come and see me after. OK, anything else, any general questions? Are we all good? So, for some reason I set the exam to start at two o'clock this afternoon, but there's no reason for that. It will start at 10 o'clock if you wanted to take it early. All right. Well, good luck and hopefully I'll see a few of you next term.

Transcription by CastingWords

Lecture 037: exam review 1

Lecture 037: exam review 1

View Lecture

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.

Transcription by CastingWords

Lecture 035: alkynes

Lecture 035: alkynes

View Lecture

Jean-Claude Bradley: Let's get started with chapter nine, which are alkynes, carbon-carbon triple bonds. So, this chapter is going to be pretty similar to the alkene chapter. Alkynes and alkenes are pretty similar, but there are some notable differences, but you'll notice that a lot of the reactions are very, very similar. You're likely going to redo chapter nine if you take 242, so this will be kind of an introduction. We're not going to go in as much depth as you will when you redo this chapter later.

Let's start off with the basics, nomenclature. So the simplest alkyne would be just two carbons and hydrogens. So, this molecule is acetylene, but you can also name it by UPAC standards, where you would take the 'eth', for two carbons, and put a 'yne' at the end to show that it's an alkyne, so I can call that ethyne. It also has a common name ethylene, no actually that's fine ethyne or acetylene.

Okay, so you don't have a numbering system here because there's no other place a double bond could be, same thing with three carbons. There's only one place that I can have a triple bond, so there I would have propyne. I wouldn't have to use a numbering system.

Now, when I have four carbons, just like with alkenes, I have to specify where the triple bond is. You do it the same way as you do for alkenes. You start at the position where the triple bond starts. So, this would be position two, so this one would be 2-butyne. It's also simpler than alkenes because there's no scissor trans. We have two sp hybridizeds centers, so the acetylenes are always going to be straight. The nomenclature is simpler for that reason as well.

We spent some time in the first week looking at the orbitals that are involved here, so you have your sp hybridized carbon and you have two leftover p orbitals on each carbon. You have one sigma bond and two pi bonds that constitute that.

It turns out that having that triple bond there, having a high percentage s character on the carbon, actually can stabilize the negative charge on the carbon. It turns out if I treat a terminal acetylene - let me get this nomenclature out of the way here - this will be a terminal alkyne, meaning that it has at least one end with a hydrogen. If I had two alkyl groups, I would have an internal alkyne. That's our definitions for internal and terminal alkyne.

Now, back to the acidity, internal alkynes don't have hydrogen, so they're not going to be acidic, but if I have a terminal alkyne, then it is possible to remove that proton and put a negative charge. When I say that it's acidic, that's relative to an alkene or relative to an alkane, it's certainly not acidic in the sense of water or any carboxylic acid even. These are way beyond all these, so we need to use an extremely strong base. The traditional base that we use here is called sodamide, NaNH2. That's basically ammonia that acted as an acid. We removed the proton from ammonia, so you end up with NH2-. That's extremely basic and it is basic enough to be able to abstract the proton on terminal alkynes. The way this works is like this, you just abstract that.

This is advantageous because by putting a negative charge on the acetylide, we can make it nucleophyllic. Now, this can act just like any other nucleophile, like Cl- or OH-, yet we can do SN2 reactions in the same way. An example of that would be, let's say I react this with methyl bromide. That would give you a classic SN2 reaction. So that's a good way to synthesize alkynes. That would fall under the first preparation of alkynes.

Another way we can make alkynes, if we don't have one to start with, is to dehydrohalogenate, much in the same way that we made alkenes. But, instead of losing one HBr molecule, we're going to have to lose two HBr molecules to make the alkyne. Losing one HBr molecule isn't very difficult to create an alkene, however if we want to put that second double bond in, it turns out to be a little bit more difficult. We need to use a strong base, and in this case, we will use sodamide and furthermore, we're going to do it at high temperature.

So we're going to do it at 150 degrees. Previously, when we used sodamide, we were generally doing that in liquid ammonia, so about -33, this is different, this is much more intense. What that will do is remove two HBr molecules and we can create a triple bond.

There's a lot more to this reaction, and like I said, if you follow along to 242, you're going to start with this chapter. You'll go into a lot more detail about how this whole process happens.

Let's look at a few reactions of alkynes. The first is hydrogenation. If we use the same catalytic system as we did for alkenes, like palladium and hydrogen. Here 'xs' means literally that you have an excess of hydrogen, so you use up as much hydrogen as you need in the reaction. What will happen is that you will use up two equivalents of hydrogen to reduce the triple bond to a single bond, so you'll form a alkane if you use a standard catalyst like palladium.

If you want to do this reaction and you want to stop at the double bond and you don't want to go further, one tactic that you could use, which you'll see used again and again in chemistry, is to take a really good catalytic system and to make it less efficient. What happens is that it becomes less and less able to do all the competing reactions, and if you're lucky, you'll get the energies just right so it will do the reaction that you want, but not the ones you don't want.

Luckily, it's easier to reduce the triple bond to a double bond, than it is to reduce the double bond to a single bond. That means if we poison the catalyst, add a poison like quinoline, you can actually make the catalyst unable to reduce a double bond, but still reduce be able to reduce a triple bond.

Let's take a look at that. Same alkyne, palladium on barium sulfate and quinoline. This mixture is also called Lindlar's catalyst. What will happen is it will reduce to a double bond and it will stay there. That will be an example of partial hydrogenation.

Another thing we can do is add halogens across triple bonds. Adding an excess of bromine, we'll add the first bromine molecule across the triple bond to the double bond, and then all the way to having four bromine atoms added across. The mechanism is a little bit different than the addition onto an alkene. If you remember, when you added on an alkene, you would get trans addition, here you end up getting mixtures of cis and trans initially, but it doesn't matter because if you have excess bromine, you're only going to get one product.

Let me show you what I mean by that. If we get an addition like this - I'm going to put these in square brackets to indicate that those are not isolated; they're just transiently produced in the reaction mixture - both of these alkenes will act like normal alkenes and we will add bromine across the double bond and they will give the same product.

Another reaction we will revisit is Markovnikov addition of HBr. It will also work with alkynes. Let's say that we do Markovnikov addition of HBr, we just add HBr to this. If we have an excess, because it is hard to stop it after the first stage, so if you get just one product, we'll have an excess. We will first give it a Markovnikov addition, which means that the H will go where there are more hydrogens. In this case there is one hydrogen on the left, zero hydrogens on the right, so the H will go on the left and the Br will go in the middle. That is a transient product. That will also undergo an Markovnikov addition. The H, again, will go where there are more hydrogens on the left, so we will get 2Br bromopropane from that.

If we do Markovnikov addition of water, it's a little different. We're going to use an acidic aqueous solution again, the traditional mixture used is mercuric sulfate, HgSO4, and sulfuric acid, H2SO4, with water. Basically, you need to translate this into a Markovnikov addition with water. It's not exactly the same as the previous case because something interesting happens. The first step will be exactly as you would expect, Markovnikov addition of water, where we get H on the left and the OH in the middle. The thing is, is that this is not a normal alcohol; it is an alcohol that is on an alkene. That's a special kind of relationship called an enol. The enol doesn't survive very long, it basically will tautomerize, which is a word that refers to rearrange, in a sense of going from an enol to a ketone. In the mixture, what will happen is the hydrogen that is on the oxygen will leave from there and go on the double bond, on the carbons on the double bond. That's basically the movement of a proton.

All we did here is we moved the double bond to the oxygen and moved the proton from the OH onto the carbon, down here. The end result of the hydration of an alkyne is you end up with a ketone. If you are looking for the final product, what you need to remember is that you are going to get a methyl ketone in this case. If you have a terminal alkyne, you always get a methyl ketone like this.

If you want to do anti Markovnikov addition of water, you do something similar as what we did for alkenes. For alkenes, we use borane, BH3, and then we use peroxide to oxidize away the boron. Here we are going to do something similar, except that if we were to use borane, there would be a chance that we would add twice across the triple bond, because borane is a pretty small molecule, so I would end up with a alkene and then I would end up with two borons on that structure. To avoid that, what we use is a huge boron reagents that has two pentol groups on it, and the specific pentol groups are called siamyl. The traditional reagent will be this, sia 2BH, siamylborane. Again, the siamyl groups are just pentol groups; they are just really, really big, so because of this spheric hindrance, I can only put one.

If you remember, after you add the boron, you got oxidized, so we would do exactly the same thing in this case, H2O2 basic. Anti Markovnikov addition of water, the OH will go on the left, but again, we have an enol, so that is not the product we isolate. We end up with an aldehyde in this case. You can see that the reactions are very analogous, but there are slight twists to each and every one of these reactions. Any questions on this addition of water?

Next we have the permanganate oxidations. Cold end dilute will be similar to alkenes in that there is not any breakage of the carbon-carbon bond, and it is also similar in that we deliver oxygens, but the difference is instead of getting alcohols, we get ketones, or, I should say, carbon enol. We just replace the triple bond with two carbon enol groups. That's called end dilute. Warm and concentrated would be very similar to alkenes, in that we break the carbon-carbon triple bond and then oxidize each carbon as far as it can go. If we break this triple bond, then on the left we have one carbon. Whenever you have one carbon with concentrated permanganate, it oxidizes all the way to CO2. On the right, that carbon will be oxidized to carboxylic acid. That's very similar to the alkenes.

That's basically the theory that I want to cover for chapter nine. Did anyone bring their books?

Transcription by CastingWords

Lecture 034: review for test2 make-up

Lecture 034: review for test2 make-up

View Lecture

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

Lecture 033: test 2 review

Lecture 033: test 2 review

View Lecture

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...

Transcription by CastingWords

Lecture 030: Alkenes 3

Lecture 030: Alkenes 3

View Lecture

Jean-Claude Bradley: So we left off with the Anti-Markovnikov addition of HBr to an alkene last time, so this time we'll talk about the mechanisms, so let's do that.

The way that works is if we've got an alkene that's unsymmetrical, in this sense, where we've got a methyl group on one side, we have two hydrogens on the other" there are a different number of hydrogens on either side. So anti-Markovnikov addition, the hydrogen would go on the side there are fewer hydrogens; it's the opposite of the Markovnikov rule. And to do that, we mention that we need peroxide" any peroxide will do.

"R" could be an alkyl group, "R" could be a hydrogen, it's all going to work.

If the H goes on the side were there are fewer hydrogens, the H will go in the middle. And the Br will go at the end.

The mechanism for this, as I said, is similar to one we've seen before, which involves free radicals. So let's take a look at that.

As in all free radical mechanisms, we're going to have initiation step, propagation, and termination. This time, there's more than one initiation step. There's no light in this reaction. So it's not like bromine separates into two bromine radicals. So what's happening in this case is that the peroxide is first going to break into radicals, and that doesn't need light. That just happens with heat. If you take a solution of hydrogen peroxide and heat it, you'll see oxygen coming out of it very quickly. So that reaction is pretty easy. And I'm going to use hydrogen peroxide as an example, but it would be the same with dimethyl peroxide or any other peroxide.

This is just a homolytic cleavage of the oxygen-oxygen single bond, this is actually really weak. Okay, so that's the first step, it's going to generate the OH radicals. Now, this OH radical is pretty reactive in itself, and it's going to react with the HBr.

When you're breaking a bond with free radicals, you're always going to have the same three arrow pattern, that we've seen before. So a good way to remember this is that you're making something really stable, in the second step, you're making water. The only way to make water from the reagents is by reacting the peroxy radical with the HBr. So it's the Br radical that is actually going to add onto the alkene. So those two steps are what we consider to be the initiation.

Next, the bromine radical is going to add onto the alkene. Now, that the alkene is unsymmetrical, we're going to end up with a radical, or two different possibilities for a radical. One of these radicals is going to be more stable than the other, and that's what going to decide the product. So this step right here.

Here's my alkene. The bromine could add onto the left or the right. So let's add it to the left, and see what happens.

Again, I'm using the same three arrow pattern, but now I'm actually breaking a double bond. So I haven't completely cleaved the carbon-carbon bond. I still have a single bond left there. But I have created a carbon-bromine bond on the left. So that generates that radical.

Now, let's consider the alternative, which I'll call three-prime (3'). The alternative is that the bromine would add on the right side. Okay, so there are our two alternatives. So you can see from this, the first reaction, we generate a secondary radical. And in the 3" we generate a primary radical.

Going back to the stability order, from the previous chapter, the secondary radical is always going to be more stable than the primary.

Because of this, step 3' simply doesn't happen. So it turns out you're going to get anti-Markovnikov products whenever you're faced with the difference in different kinds of radicals. So when you do anti-Markovnikov addition, you count the hydrogens on each side of the double bond, but this is ultimately the reason why it happens.

Basically, the product is decided. We just have to finish it. So in step four I'm going to take the radical that I generated in step 3, and now we're going to react it with HBr.

Actually we've generated the anti-Markovnikov product in this step. And we've also generated a bromine radical. So steps three and four then are the propagation steps, because the radical generated in step four will then be used in another step 3. So if you didn't run out of material or you didn't have any termination steps, then you wouldn't need any more initiation once you had that stuff gone through, right? That's how propagation works.

Of course, you do have termination steps, and that's going to work the same way we did free radicals. Each free radical coming together would be a termination step. So what do we have available here?

We have a bromine radical, and we have a carbon radical that has a bromine on it. That's pretty much all we have to work with in terms of terminations. We have two possibilities, two possible radicals I should say. The easiest one is the two bromines coming together.

Another example would be the bromine-carbon radical combining with a bromine radical. And finally we can have two carbon radicals coming together. Yes?

Student 1: [inaudible question]

Jean-Claude Bradley: Yeah, basically the termination involves terminating something involved in the propagations steps. So as you get it started, if you capture a hydroxyl radical, yeah if a hydroxyl radical reacted with a bromine, yes, that would be a termination. But that would create something unstable, so. Just focus on the one's that we did, that are analogous. Yes, any two radicals involved in the propagation.

Again with the termination, you get extremely small amounts of these compounds, you will get just trace amounts because so many cycles of propagation relative to termination, but that's what they would be.

Any questions on anti-Markovnikov addition of HBr?

Now, though you can add HCl Markovnikov, you can't add HCl anti-Markovnikov just by dumping in some peroxide. So this particular reaction is for HBr. And the reason for that has to do with step number 2. It's that the energies are right for the hydroxyl radical to make a bromine radical. It won't be right for HCl or HF.

Remember that when you're doing problems. If you add peroxide, you're not going to do much. Just Markovnikov addition for HCl. And we're going to do some of that in the problem set.

Let's talk about Markovnikov addition of water: Instead of adding like HBr, we want to add an H and an OH, we can still do that on an alkene. For adding water, all I have to do is add a reasonably strong acid.

If I have water, and I'll just put H+ here, but that can be sulfuric acid, that can be any strong acid you add to this, you just need to catalyze this process. So acid and water will give you Markovnikov addition of H2O. This means that the H goes where there are already more hydrogens, In this molecule on the left, it means the H would go in the middle.

That's all we mean by Markovnikov.

The mechanism for this is going to be really similar for what we did for HBr: We have the proton that can go either on the left or right, we already did that, so we're just going to show the one that happens. The proton is going to go on the carbon here to generate the secondary carbon cation. You form the secondary carbon cation, you look for the one that shifts, we don't have any in this example, and then we add the water to this.

You can't add OH- at this point because you're under acidic conditions. So, there will be very very little OH- floating around. Okay, so the nucleophile in this case will be water. Generally water is not that great of a nucleophile, but a carbyl cation is very tantalizing for a poor nucleophile like water, so that's what will happen.

So that means we need an additional step to complete the reaction, we now have to lose the proton that's on the alcohol.

Now, we have the neutral alcohol. So that's how Markovnikov addition with water works, similar to what we had with HBr.

One of the problems with this Markovnikov addition, using acid and water, is that if you do have the possibility of a 1, two shift, then you've got a problem, because you're not going to get the addition that you want. So it turns out there's a way around that, that you will get Markovnikov addition no matter, even if there's a possibility of a 1, two shift or not.

And that's using an oxymercuration approach, so let's do that one next.

We're not going to look at the full mechanism of this, we're just going to look at the reagents.

So the first step will be mercuric acetates and water, and the second step will be sodium borohydrate. So the OAC here stands for acetate, which is just that. But that's how we're typically going to abbreviate that when we do reactions. So those two steps, again when you have one and two on the same arrow, that means you have to do this sequentially, those two steps will always give you Markovnikov addition of water, no matter what. So that means I will end up with OH in the middle and not have to worry about 1, two shifts.

Any questions on that?

Student 2: [inaudible question]

Jean-Claude Bradley: Yes, the mechanism's in the chapter, that's not one I'm going to expect you to know. But if you look, the mercury adds, there's not a possibility, you never get a full carbyl cation. Some, but not enough to promote a 1, two shift.

We're going to look at reactions where we go through every step of the mechanism, like the one we just did this morning, and then you're increasingly going to see reactions where we don't look at every step of the mechanism, but you are expected to know what it does. I think a lot of these, if you're really curious, will be in your book. If you want to get down and see what it's like.

The only thing we haven't covered is anti-Markonikov addition of water.

It would be tempting to just think we could add peroxide to the acidic water, but that doesn't work -- again -- because the energies don't make any sense. The peroxide trick will work only for HBr.

So, in this case, we're going to be using again two sets of reactions that we're not going to look at the mechanism of, but that will always give you anti-Markovnikov addition of water.

The first of these is to add borane, BH3, and then we'll add a peroxide. The peroxide here is acting in a completely different way from the peroxide we had with HBr. Actually here it's a base, so, typically, that's written H2O2, and hydroxide, you have to be in a basic environment.

Those two steps will give you anti-Markovnikov addition of water, so that means the H will go where there are fewer Hs. So the H will go in the middle and the OH on the end.

Again, with the anti-Markovnikov additions, there is no need to worry about 1, two shifts. Because you know that radicals don't shift. Even though a secondary radical may be more stable than a primary radical, it's just not possible to do 1, two shifts as with carbyl cations. So again we don't have to worry about any kind of rearrangement.

That covers the reactions with alkenes, but there are additional reactions we are going to look at. The next one is catalytic hydrogenation.

Basically involves adding hydrogens across an alkene. For that you actually use molecular hydrogen, H2, but you need a catalyst. Here, we'll think of a heavy metal like platinum or palladium or iridium or something like that. That's pretty simple to do: It will happen for any normal, unconjugated alkene.

If I just have a carbon-carbon double bond, that will go fine, but if I have a hyper-stabilized double bond, like in an aromatic, that won't be enough. You'll have to use partial reduction.

For example, if I had styrene, and I used a normal catalyst, it's not going to do this, not the catalyst we're going to look at. But if we just took hydrogen and palladium, you would selectively reduce the carbon-carbon double bond that's outside the ring.

I'm writing one atmosphere here to show normal conditions, we're not trying to force it, then you get selectivity.

The Simmons-Smith reaction is the way of making a cyclopropane. So you would use methylene iodide, CH2I2, in the presence of zinc. And that CH2 unit would then become attached to the double bond, and make a three-member ring. So that's one way to make a cyclo propane.

Next is halogenation. Yes?

Student 3: [inaudible question]

Jean-Claude Bradley: You mean in the product? Let's make it clear by putting in the hydrogens. So you've got a double bond and the CH2 getting stuck on the top or bottom of the ring for that matter.

So, halogenation: Hydro-halogenation means to add Hx, halogenation just means to add x2. So, for example, adding Br2. I can add bromine across a double bond, just simply by adding it. You have to be careful with these reactions because this is in the dark: We know that bromine can react with basically alkyl groups, if you have light, so you have a complication there.

If you have a double bond, and you have other positions that would create stable radicals, then if you really want to do this, you have to do this in the dark. Without any light.

In this case, it wouldn't make any difference, because there are no stable radicals. If it just says Br2, and it doesn't say if it's light or dark, you can assume that it's in the dark. When we want to add light, we'll add it on purpose.

We are going to look at the mechanism of this reaction. This is kind of interesting: What happens here instead of opening the double bond, to the left or to the right, the way that we did a proton, we're actually going to go directly from the middle. We'll do something like this.

What happens here, we form this, and end up with bromine in a three-membered ring, with two other carbons: That's what's known as a bromonium ion. This is reactive as an electrophile, and what's going to attack it is going to be the Br- I generated at this step. So I kick out the Br- and the Br- comes back and reacts with the bromonium ion.

What is interesting about this mechanism is that, because it involves an Sn2 reaction, Sn2 reactions are always a backside attack.

That means that the two Brs are always going to end up on opposite sides, they'll always end up trans to each other, if there's a cis and trans possibility.

In ethylene, of course, there is no cis and trans in the final product, so regardless of what the mechanism would be I would end up with 1, two dibromyl ethane. So let's look at the possibility where you do have cis and trans, like in a ring.

Reacting cyclohexene to bromine in the dark, I would end up with two bromines on the opposite sides. So trans addition because of the mechanism.

With this mechanism, also, there are some competitive reactions that will happen. If I mix bromine with another nucleophile, like let's say water, the first step will happen, I'll still get the bromonium ion, but then I'll have a competition between the Br- and the water. So if we do the same reaction in the presence of water, we will end up with competitive reactions at the same time.

Let's take a look at that: We have considerably more water than bromine in this example, because we want competition to be one-sided in favor of water.

Let's take a look at the mechanism we would end up with: First, formation of the bromonium ion. Then, the water would come in and attack the electrophile from the opposite side of the ring. That's the final result: You get Br and OH added. Again, trans.

Lots of other things we could do with alkenes. We can also put an oxygen on it, the same way we made a bromonium ion.

In order to do that, we need what is called a peracid, it has CO3H. It's kind of a hybrid between a carboxylic acid and a peroxide. So whenever you see an oxygen-oxygen single bond, you know that's weak, so it's going to be easy to break that bond. And what the peroxy acid is going to do, it's going to deliver one of the oxygens on top of the carbon-carbon double bond.

That's called an epoxide. Epoxide is just a three-membered ring.

The thing about epoxides is that why don't we just call them ethers? Call a three-membered ring an ether? Because the nomenclature in organic chemistry is based upon behavior.

That means it has a different name because it doesn't act like a normal ether.

The fact that I have a three-membered ring means that it's strange. I have three SP3-hybridized atoms there, the angle wants to be 109.5 degrees, but it's forced into a triangle. So it's forced into being 60 degrees. For that reason, it's unstable. So epoxides can open up with a variety of nucleophiles. I think in the problem set we do a couple examples of that.

The "R" group here can really be anything, so as long as you have the CO3H, you can expect you're going to get epoxidation. In practice, typically you would use a compound called McPBA, which is metachloroperoxybenzoic acid. That would basically be an example of a peracid. It's common enough I expect you to recognize it, McPBA, in a problem. For epoxidation.

The next thing we'll look at a reaction with permanganate. There are two conditions we'll look at. The first one is cold and dilute. So that means you have potassium permanganate, but the conditions are very mild. And that will do something very different from when you heat it up. So in the cold, you're simply going to add OH groups on the double bond.

Now if you do the same thing under hot conditions, you will actually break the carbon-carbon double bond. "Conc" stands for concentrated, so if you have hot, concentrated potassium permanganate, you will break the carbon-carbon double bond, and furthermore you will oxidize either end as far as it can be oxidized.

In this example, when we break the double bond, first we replace C-double bond-C [C=C] with C-double bond-O [C=O]. So the initial products we get would be this. I'm going to put these in square brackets, this means that they're going to be formed transiently. In the reaction, you never actually isolate them, because soon as they're formed, they get oxidized. Aldehydes are easily oxidized through carboxylic acids.

So if you form an aldehyde at this step, you need to carry it through all the way to the carboxylic acid.

Permanganate is a pretty strong oxidizer, and it's going to do that. Okay, there are various permutations here, if you produce ketones, ketones will not be oxidized by permanganate. We'll do some of that when we do the problems.

While we're on this topic here, this top reaction, this hydroxylation reaction, can either be done with potassium permanganate under cold conditions, or it can be done with osmium salt and hydrogen peroxide.

Finally, if we treat the alkene with first ozone O3, and then dimethyl sulfide, you'll do pretty much what you did with the hot potassium permanganate, but you will not oxidize the carboxylic acids. So that means any double bond, you will break it here, and put oxygens on either side.

That's basically it for chapter 8, so we're at the problems which we'll do the next time we have class.

Transcription by CastingWords

Lecture 029: Alkenes 2

Lecture 029: Alkenes 2

View Lecture

Jean-Claude Bradley: We looked at SN1, SN2 reactions. Those were called nucleophilic substitution reactions, because the reagents had a pair of electrons that was attacking the compound. In our case it was the alkyl halide. Those were called nucleophilic reactions. In the case of an alkene, it is the opposite. The double bond is actually electron rich. So instead of having nucleophilic reactions, we'll have electrophilic reactions. [writing]

Jean-Claude Bradley: And one of those would be electrophilic addition. So if I want to add HBr to an alkene, the way that happens is by first taking the electron out of the double bond and putting a proton on there. Let's take a look at an example of that. [drawing] If I react ethylene with HBr, let's think of the HBr as being dissociated, so the first thing, just like the proton goes on an oxygen in an alcohol, the proton will go on the alkene like this. [drawing]

Jean-Claude Bradley: I end up with a carbonation, and, again, if you make a carbonation, you've got to worry about 1, two shifts, so the molecule I chose here won't have that problem. The next step is going to be the bromine coming in and finishing the job. [drawing] So in two steps we have the electrophilic addition of HBr to an alkene.

Jean-Claude Bradley: Now by using ethylene, there was only one product that I could get from that, because it didn't really matter which way the HBr added. Let's take a look at an example where there is a difference. There are two different ways you can add it. If I have something unsymmetrical, like propane, then I can think that the Br could either go on the end of the molecule, or it could go in the middle. Let's look at both scenarios.

Jean-Claude Bradley: Now when I'm drawing the proton coming in on the alkene, you want to think of it almost like opening a door. The door will open either to the left or it will open to the right. If I put the proton on the left-hand side to show that the carbonation is ending up on the right, I'm going to be opening it up like this. [drawing] And I've got my carbonation in the middle. And then as a last step, we add the bromine, so then we get 2-bromo-propane. Ok the alternative is to do it the other way. [drawing] So now the carbonation would go on the end. [drawing]

Jean-Claude Bradley: Okay, so let's stop at this point right here. Which one of these two is more likely, do you think? [inaudible response] The one on the left? [inaudible response] Yes, it just comes down to everything we've seen before: primary, secondary, tertiary. So that's secondary, this is primary. So what ends up happening is that given that choice, you really don't get any of the right-hand side. All of the product will come from that secondary carbonation intermediate. [drawing]

Jean-Claude Bradley: So, you know, we can go through the mechanism every time and show, you know, I get secondary vs. Primary or tertiary vs. Secondary, whatever you want, but the pattern that actually comes out of that, and that has to do with where the hydrogen ends up going. If you notice in this case, as I'm adding HBr, the H goes where there are already more hydrogens. In other words, in this double bond on the left-hand side I have two hydrogens; on the right-hand side I have one hydrogen. So the rule is that in electrophilic addition, the H will go where there are already more H's. That's an empirical rule and that's called Markovnikov's rule. [writing]

Jean-Claude Bradley: So that's a way of explaining things without having to invoke the mechanism. I'll give you another example here. We want to apply Markovnikov's rule. Ok so if I have this compound, and I react it with HBr. So the first thing you need to recognize is that that is an electrophilic addition, that's the first step. Then you determine, how is that HBr going to add? You can go through the mechanism, but if you don't want to go through all that, you simply look at how many hydrogens are on each side of the double bond. On the top portion of the double bond, I have zero hydrogens. On the bottom portion, I have one hydrogen. So to apply Markovnikov's rule, we simply add the hydrogen where there are more hydrogens on the bottom. So I can draw the product then [drawing] by doing that. Okay, that's following Markovnikov's rule. Everybody clear on that? You're going to have a bunch of examples to do anyway, in the problem set for that. So Markovnikov's rule is pretty useful but there is one caveat, and that is if you have 1, two shifts, you're going to have a problem with applying Markovnikov's rule, so you need to be a little bit careful about that. But as long as you don't have any 1, two shifts, the rule will apply exactly like this.

Jean-Claude Bradley: So as I said?? [question from student, inaudible] Right, that's why I didn't finish that one, because I didn't want to get into that whole debate about the shift. I mean, basically what we're deciding at this point is just, where is the proton going to go? Is it going to go on the left or the right? So you're right, after we put it on there, other things can happen. Normally, though, you're not likely to get a shift a lot of the time, simply because you already have a selection of a more stable carbonation. But it is possible to design a problem where you will have going from secondary to tertiary afterwards. But typically because you have that initial choice, you're going to get probably the most stable carbonation, which is why Markovnikov's rule will apply almost all of the time. But that is a good question, and that is why I stopped at that point.

Jean-Claude Bradley: Okay, so we can call these products Markovnikov products, and that's what that means. [writing] If we can have Markovnikov products, then if you want to do the opposite, if I wanted to actually put the bromine on the bottom of the ring and put the H next to the methyl, that would be called an anti-Markovnikov addition. And you just simply can't do it with this particular approach, there's no way that just by adding HBr you can do it. But there is a trick, and that is if you add peroxide, you can actually have anti-Markovnikov addition. So let's take a look at that. [drawing]

Jean-Claude Bradley: Key things about this: the peroxide that I add does not have to be equimolar. I can have catalytic amounts of peroxide. I can have a trace. I think often in the problems, they write "trace" or something like that. It won't hurt it if you have more peroxide, but you don't need a full equivalent. The other thing about the peroxide is that it doesn't have to be hydrogen peroxide, it can be any R-O-O-R, where R can be any alkyl group. So sometimes you'll have dimethyl peroxide, or whatever. It doesn't matter, as long as you have a peroxide. What will happen is you will get the anti-Markovnikov addition. [drawing]

Jean-Claude Bradley: So anti-Markovnikov addition means the H will go where there are fewer hydrogens. There are fewer hydrogens on the top, so that's where the H will go, and the Br will go on the bottom. [drawing] Also this reaction will really only work with HBr, so you can't do it with HCl. If you wanted to put a chlorine in there instead, then you'd have to use some of the other knowledge you have, like you could do an SN2 reaction, on that alkyl halide, and put a chlorine, but you do need to remember that this only applies to HBr. This mechanism is actually pretty extensive, and it's going to be similar to another one we did previously. And I think that is where I will start the next class, at the mechanism for this anti-Markovnikov. That's it.

Transcription by CastingWords

Lecture 027: Alkenes 1

Lecture 027: Alkenes 1

View Lecture

Jean-Claude Bradley: Just before we look at different ways of making alkenes, let's look at a bit of nomenclature. An alkene will be any compound that has a "Carbon=Carbon" double bond, as we've seen many times. And the way we name these is, if we just have a simple alkene, we start with the alkane. The corresponding alkane to this would be ethane, so I change the A to an E and I have ethene. So you can always do that, but a lot of the smaller alkenes have more common names. So this one would be the same as ethylene. But the proper IUPAC name would be ethene for this compound. Same thing for this, IUPAC name would be propene, but it could also be called propylene. When we have more carbons, then we have the possibility of the double bond in different locations. There's no reason to number the ethylene and propene, because there's only one place the double bond can be. Now we have four carbons - four carbons, so it would be butene - so I have to specify where the double bond is. I'm going to start counting from the side that gives me the lower number. Now so far the compounds I've drawn don't have cis/trans isomers, which we spent some time evaluating, so there's no other descriptor we need to add to the name.

Now when we get to compounds which do have cis/trans isomers, then we absolutely have to specify whether the methyl groups are on the same side or opposite sides. The only way we have to do that right now is using cis/trans, so for the top compound, we're looking for trans. How do we know it's trans? We have two groups that are the same on opposite sides. Another way of saying that is I can say that the methyl groups are trans and the hydrogens are trans, and that's unambiguous so we can say it's trans. And specifically it's trans-2-butene. Same thing down here, the two H are on the same side, or the two methyl groups are the same side, so this would be cis-2-butene. We also came across some examples where you did have cis/trans isomers and we couldn't name them. The best we could do was to say this group was cis, but that's not really a way of naming a compound if you're looking for it. So let's see what happens in those kinds of situations.

All right, so if I have a compound like this, it has four different groups on it - I, Cl, F, Br - this is ethene, it's name is actually bromo-1, chloro-2, fluoro-2, iodoethene. So that compound would be two different possibilities, because we do have cis/trans isomers. So we do something similar to what we did to R/S. We divide the alkene into two parts, and we determine the relative priority of each group. So on the left we have I and F, so the I has higher priority than the F. We use the same rules as for R and S. So that would be one and then 2. So now we're done with the left side, we forget about it and go to the right side. Between the Br and a Cl, the Br will be higher. So you're always going to end up with two possibilities, either the two high priority groups are on the same side or different sides. We can't use cis and trans for this because they have the condition that two same groups are on same or opposite sides. We don't have the same groups here. So instead we use an E and Z nomenclature. So where the two highest priority groups are on the same side, that would be Z. If they're on the opposite side, that would be E. So now you can name any alkene unambiguously. Now the E and the Z actually derive from German words so it's hard to remember how to remember. So one of the tricks we can use is if you look at the E, if you picture the top part of the E flipping back, now you can see they have to be on opposite sides. So that's kind of an unusual thing to remember, so you're likely to remember. It would be nice if the letters look the same, but it doesn't. It's the same with Z, it's got two groups pointing in opposite directions, but it's really on the same side. So that's basically all about how you name alkenes.

The other issue about alkenes that you need to worry about that we did discuss - the eight carbon rule, that is- in a ring, we can't have a trans double bond unless we have at least eight carbons. So this would be the smallest cycloalkene that you could make that would be trans, otherwise it's not possible. Okay, so this would be, if you wanted to name this guy, it would be trans-cyclooctene. And questions about that?

So let's actually get into the preparation. There are five methods we have to prepare alkenes. Some of these will be review, but we'll put them in one place: dehydrohalogenation. Let's break down the word. So halogene would be X (F, I, Br, Cl), hydro means H or water, in this case H, and de- means to remove. So dehydrohalogenation is an easy elimination reaction, so E1 or E2. We spent a lot of time on this, so let me just put one example. You need a base, typically to do E1 or E2. And you would get your alkene out. And there's the HBr that you lose. So we spent a lot of time on that, enough to say that you have to keep it in mind, because one of the things you're responsible for here - we're not going to look at the mechanisms of all of these, but you have five ways of making alkenes, so you could have a questions where I draw an alkene and that could be made from a bunch of different things. So even if you didn't see the mechanism, you should know five ways of making alkenes. So I could make an alkene from ethyl bromide.

Let's see how else we could make an alkene: dehalogenation. That's one we haven't looked at yet. So we break down the word: halogene could be X (F, Cl, Br, I), and de- means remove, so here we're removing only a halogen. The way we do that is by actually having two halogens, and there are two reagents we're going to look at, one of them is zinc, or you could use I-, and that's specific to I-, you can't substitute Br- or something, it's something special about the I itself. So with those reagents, you lose the two Br's and form an alkene. So it's not any dihalide that will do this, it's only dihalides that are vicinal. Vicinal means a 1, two relationship, so these two Br's are vicinal because they have a 1, two relationship. If they're on the same carbon, it's not vicinal, if they're another carbon away, it's not vicinal - they have to be next to each other. That's the only condition for this kind of dehalogenation reaction. Next we have the dehydration of alcohols. So dehydration means lose of water. Okay, we have just covered this today at the beginning of class. So if we want to lose an OH, we can't do that directly, we have to protonate the OH, then we can do E1 or E2. Typically you use a strong acid, usually sulphuric acid. You can use other acids, but this one have the advantage of not being nucleophilic very much. So if you're were to use something like HBr, you have the complication of the Br doing SN1 or SN2 on your compound. So if you don't want the complication of substitution reactions, you use something like sulphuric acid, which is an excellent acid but a poor nucleophile. So we actually looked that the mechanism of this earlier. Bottom lines if you end up with your alkene. You do have to be wary though, because you could have 1, two shifts. Because if you do an elimination by E1, you have to make sure there's not any carbocation that can rearrange. In this example there isn't any, but that is one small thing you need to worry about. And we revisit something we saw first week, when we look at reactions of alkanes, when we could break alkanes into smaller pieces, and some of those pieces were alkenes. So we're going to add this as a way of making alkenes. Use the same example as I used earlier. So propane in some catalyst, palladium, platinum, whatever, some heavy metal catalyst, and high temperature and usually high pressure, will actually break this molecule into smaller pieces. One of the pieces would be methane, and the other piece will be ethylene. The reason I get an alkene in this case if because if we break a C-C bond, we can't put a H there because we don't have enough H's, so you notice that breaks in such a way that we have the same number of H's on both sides. Industrially this is a pretty good way to make small alkenes like ethylene, but in the lab it's impractical. For one thing, the conditions are very harsh, so if you have anything else on this molecule on this molecule besides the alkane, this would destroy it. And it's not very specific, so if I have eight carbons, it would be random where it would break, so that's not good if you're making a specific alkene. But industrially it's very good because these are cheap processes, so everything has its place, and this is not going to have a place in the lab where you're going to be working. On the other side of that, we have a reaction that's much more expense, but is very specific, so if a chemist wants to make an alkene, this is one of the best ways to do it. This is called the Wittig synthesis. And what you need is an alkyl halide and a ketone or aldehyde. So let's start with methyl bromide. I'll take you through the whole mechanism; in the future I'll summarise it, but in the beginning let's go through it carefully. This is actually going to be an SN2 reaction, we have triphenylphosphene and methyl bromide. P is right under N, so it has a lone pair, still reasonable nucleophilic and is going to attack an alkyl halide the same way you would use and amine to attack it. So we end up with - these phi symbols are benzene rings, same as Ph or drawing out the whole thing. The reason we write triphenylphosphene like that is because it's really a pain to have to draw three benzene rings, so typically it's abbreviated like this. So just like the reaction of an amine with an alkyl halide, I have a positive charge on the P, and I've also lost Br- here. What we want to do is basically make the carbon that's attached to the P nucleophilic, so we're going to remove a proton from it. If I remove a proton from this, I put a negative charge on a C. Normally it would be difficult to put a negative charge on C, but here we have positive P here that would stabilise it a lot. You still can't use a mild base, but you can use a base that's available, and that base is typically N-butyl lithium. Now the way you want to think about N-butyl lithium is like this, you have a CH2- and a Li+. Now there is a covalent bond between the C and the Li, but it acts as if we have a negative charge on the C. So we draw the arrow, the CH2 is extremely basic, so it's going to extract a proton from the C next to the P. So now I have positive charge on P, negative on the C, and when you have that kind of situation, where there is a negative charge next to a positive charge, that is called an ylide. This is your Wittig reaction, we spent time preparing it. When we talk about the Wittig reaction, it's going to be the phosphorous ylide. So one way to draw the ylide is like this, with the + on the positive and the - on the carbon, and you'll notice that this follows the octet rule nicely. But I'm also going to draw it in a way where you might see it in problems, where the octet rule is not followed. So you can also draw it with five bonds on the P, and you don't have to draw a charge, but it's understood that if you're going draw a correct Lewis structure for this, you're going to draw this octet, but this is more convenient because you don't have to draw the charges. So I've created the Wittig reagent. The Wittig reagent is actually half of your alkene. The other half is going to come from either an aldehyde or a ketone. So let's say we just use an aldehyde as an example. The Wittig reaction will react with the aldehyde in such a way that the C=O will be removed, and exchanged for C=C. And a side product from the Wittig reaction proper will be triphenylphosphene oxide. What makes this a particularly useful synthesis is that it's extremely general to pretty much any alkene you can draw. So you'll have a problem where you have an alkene and try to figure out what original alkyl halide you could use, what original aldehyde or ketone you can use to make the alkene. And when you work backwards with the Wittig, is you just need to remember you stitched the double bond in the end. So you break it off, and you see what groups you need in the alkyl halide and what groups you need in the ketone or aldehyde. So that's something you're going to need to know fairly where; there are a couple of quiz questions on that alone. So that's pretty much it for the theory part of chapter 7, and we're going to practise this next time doing problems.

Transcription by CastingWords