Methyl > primary > secondary >> tertiary

Transcription

1 Lecture 1 Notes September 6, 2012 Welcome to CHM 427. For some of you, this will be the last organic chemistry class that you will ever take. So I see it as my personal mission to make sure that you know certain key points of organic chemistry that you may not have focused on before now. Let s go over the syllabus in some detail so that we are all on the same page. There will be two in-class tests, two literature briefs, and periodic problem sets that will not be graded. The problem sets and their answers keys will be posted online. It is in your interest to take the problem sets seriously and actually do the problems. These will be your best preparation for the test. There will also be a final, which may or may not be cumulative. 15% of your grade is for participation and attendance. The procedure for how this is assigned is lifted directly from Dr. Euler s methods. He s been teaching a lot longer than I have. Every class you get a grade of up to 10 for participation. You get a 7 for coming on time and being awake. There are 22 classes that are not tests or Jewish holidays. The total participation grade will be out of 180, which means you have 4 classes that are freebies. Do not cheat or plagiarize. There are not so many of you in the class. Assume I will catch you and it will not end well for you. There is no textbook for this class. I will post my lecture notes on the web. If you are interested in a book, you can get Organic Chemistry, An Intermediate Text by Robert V. Hoffman. I have the second edition. I will generally take a fair number of practice problems from this book. Office hours are officially from 11 am to noon on Tuesdays and Thursdays (right after this class), but you can send me an or drop by my office (251 Pastore Hall, right down this hall) at any time. DEMONSTRATION In the first couple of classes, I am going to go over some material that I hope will be review for most of you. Today we will review nucleophilic substitution reactions. Within this topic, we are going to cover leaving groups, mechanism, and nucleophiles. Here are some examples of substitution reactions: X in these molecules refers to some generic leaving group. You see that you can generate a whole variety of interesting products using these reactions.. Halides are a prime example of a leaving group. The order of reactivity of these halides (i.e. how good are they as leaving groups?) is: I > Br > Cl >> F

2 What is the reason for this order? It has to do with the strength of the carbon-halide bond. X=I, 57.6 kcal/mol X=Br, 72.1 kcal/mol X=Cl, 83.7 kcal/mol X=F, 115 kcal/mol The stronger the C-X bond is, the less likely it is to want to break and be displaced by another atom. The next question (and a deeper level of understanding) is: Why do the C-X bond strengths follow this trend? Think about this and we will discuss it a little next time. In general, a leaving group should be able to stabilize a negative charge. Chemists have developed a variety of halide equivalents i.e. things that act like halides in terms of their ability to act as a leaving group. Think about the following three structures, all of which are leaving groups. Which do you think would be the best leaving group? Triflate is the best leaving group, because the three electronegative fluorine atoms stabilize the negative charge. Among the other two, tosylate can stabilize a negative charge through resonance, whereas the mesylate cannot, so tosylate is the better leaving group. Substitution reactions in general proceed via an SN1 or an SN2 mechanism. SN2 stands for: substitution nucleophilic bimolecular, which means that there are two molecules that are involved in the transition state. You form a new bond at the same time that you are breaking the old bond. In the transition state, the main carbon atom is penta-coordinate i.e. bound to five atoms (in some fashion) simultaneously. The two bonds to the incoming and outgoing substituents are longer and weaker than the other bonds. The rate of the reaction is equal to some constant, multiplied by the two species that are involved in the rate determining step the carbon-halide molecule, and the incoming nucleophile. Rate = k[nu][r-x] There are two options for how the stereochemistry of the reaction works: it either proceeds via retention of configuration or inversion of configuration. Retention: Inversion:

3 The easiest way to figure out the stereochemistry is to look at the configuration of the product compared to the starting material. An example is shown below: This is a pretty obvious example. We are going to see a little later that real-world reactions rarely have such neat and beautiful stereochemistry. I am assuming that people know what R and S mean, and that you know how to determine the chirality of a molecule. If you cannot do that, you should review it ASAP. The rate of an SN2 reaction depends on the steric hindrance around the leaving group. So if you look at the series of alkyl halides shown below, the methyl bromide proceeds the fastest. Methyl > primary > secondary >> tertiary In general, methyl and primary halides always undergo SN2 substitution. Tertiary halides always undergo SN1 substitution. SECONDARY HALIDES occupy some ambiguous middle ground. Real-world scenario: Consider neopentyl bromide (structure shown below). Will this molecule undergo SN2 substitution? Reasons why it would: It is a primary halide, and we just finished saying that primary halides undergo SN2 reactions. Reasons why it would not: It looks pretty bulky. In actuality, it does not undergo SN2 substitution. If you look at the stereochemical models of this compound, you will see that the backside is basically completely blocked by methyl groups.

4 SN1 reactions: SN1 stands for substitution nucleophilic unimolecular it means that there is only one species involved in the transition state. The reaction proceeds via a carbocation intermediate. The overall reaction is shown below: And it occurs via the following mechanism: The rate determining step here is the first one the slow step which is the dissociation of the bromide to form a carbocation intermediate. Again, the rate of the reaction is equal to a constant multiplied by the species involved in the slow step: rate = k[r-x] In general, the rate of a reaction is always determined by its slowest step. It s like going up a series of hills. You can t get to the end of the finish line any faster than it takes you to get up the slowest hill. The reactivity of halides to SN1 substitution depends on how stable the carbocation is that you are leaving behind, which will determine how facile the breaking of the C-X bond is. (Of course it is also going to depend on the strength of the C-X bond and how easily that bond can be broken). In general: 3o carbocations > 2o carbocations > 1o carbocations >> methyl carbocations This trend makes it clear that alkyl groups stabilize carbocations. Why is that? They stabilize carbocations through resonance, which basically means that they have a way to delocalize the positive charge over multiple atoms. The stereochemistry of SN1 reactions is determined by the carbocation this is a planar intermediate. When the nucleophile comes and encounters the planar carbocation, it has an equal probability of attacking from the top face and the bottom face. Therefore, SN1 reactions should lead to 50% inversion of configuration and 50% retention of configuration.

5 Let s consider a real-world example: This reaction proceeds mostly with SN2? If it were SN1, should be 50% retention/ 50% inversion. If it were SN2, should be 100% inversion. It turns out that this is SN1, but the reason it looks like this is b/c once the carbocation forms, the leaving group remains to solvate the carbocation. This makes sense once you think about it the leaving group is negatively charged, and the carbocation is positively charged, so they are likely to have some sort of interaction. This interaction is what prevents us from seeing a complete 50/50 mixture. In general, if you see 100% inversion, that means SN2 mechanism. If you see anything less than 100% inversion, it is likely SN1 even if it is not a 50/50 mixture. Summary from today s lecture: Leaving groups and what makes a good leaving group SN2 mechanism and its affect on stereochemistry SN1 mechanism and how to determine carbocation stability inversion of configuration and a little bit with retention. Is it SN1 or