北京化工大学：《有机化学》课程教学资源（双语习题与答案）Chapter 06 Properties and Reactions of Haloalkanes（Bimolecular Nucleophilic Substitution）.pdf_大学文库

6 Properties and Reactions of Haloalkanes: Bimolecular Nucleophilic Substitution Referring to Chapter 1, recall that polarized covalent bonds are at the heart of most of organic reaction chemistry. Here, for the first time, the properties and chemical behavior of molecules containing a polarized covalent bond are presented in full detail. They are the haloalkanes, containing a polarized carbon()– halogen() bond. Because the reactions here serve as models for many subsequent presentations of the chemistry of organic functional groups, this chapter should be examined particularly closely. Many concepts discussed in detail here are fully applicable later on, even though later presentations may not be as comprehensive. With this and the next chapter we really start covering typical organic chemistry. Now you get your first opportunity to see the “big picture” with respect to one portion of this subject. In addition, much of what you’ve seen up to now will be used to help explain the behavior of haloalkanes. Comprehending this material in a general overall sense is in the long run more useful than remembering every last detail. Outline of the Chapter 6-1 Physical Properties of Haloalkanes 6-2 Nucleophilic Substitution The nature of this functional group, and an introduction to its most characteristic reaction. 6-3 Reaction Mechanisms Describing how reactions occur. A very important section. 6-4 Mechanism of Nucleophilic Substitution: Kinetics The first section in the text to cover in detail the mechanism of a specific polar reaction. Also a very important section. 6-5 Stereochemistry Tying together this reaction with material in Chapter 5. 6-6 Consequences of Inversion in SN2 Reactions Further implications of material in Chapters 4 and 5. 6-7 The Leaving Group 6-8 The Nucleophile 6-9 The Structure of the Substrate The roles of the major variables on the favorability of the single reaction type under discussion. 99 1559T_ch06_099-112 10/22/05 20:19 Page 99

100 • Chapter 6 PROPERTIES AND REACTIONS OF HALOALKANES: BIMOLECULAR NUCLEOPHILIC SUBSTITUTION Keys to the Chapter 6-1. Physical Properties The carbon–halogen bond is the focal point of this chapter. Its polarization is the major feature governing the physical and chemical behavior of these molecules. Differences among the four halogens will affect the degree to which a haloalkane exhibits any given physical or chemical characteristic. This section reveals many qualitative similarities among the haloalkanes. Understand these first; you will then be in a better position to appreciate the differences that are presented later. 6-2. Nucleophilic Substitution At this stage we begin a detailed discussion of our first major class of polar reactions. Note the following features: 1. All nucleophilic substitutions have a similar general appearance, with a similar cast of characters, so to speak. Using the first example in the text table, we have HO CH3Cl 88n CH3OH Cl Nucleophile Substrate Product Leaving group Whenever a new reaction class is presented, analyze the various examples presented from the point of view of the common roles played by the chemical species involved. You should do this now for all the examples in the table in this text section. That is, in each case identify the nucleophile, substrate, product, and leaving group. Start getting used to the variety of species that belong to each category. 2. All nucleophilic substitutions are electrostatically sensible. In every case an electron-rich atom in the nucleophile ultimately becomes attached via a new bond to an electrophilic (positively polarized) carbon atom in the substrate. Opposite charges attract! Again, whenever a new reaction class is presented, analyze the examples given on the basis of electrostatics. Focus on the logical consequences of oppositely charged or polarized atoms attracting each other. Do this now for the examples in the reaction table in the text section. When you analyze the components involved in a new reaction class and understand in a fundamental way why the reaction is reasonable, you have taken the first step toward learning organic chemistry by understanding instead of by memorizing. 6-3. Reaction Mechanisms The mechanism of a reaction describes in detail when and how bonding changes occur. One useful feature of such information is its predictive value: Mechanistic understanding can tell us whether an unknown new example of a reaction is likely to work or not. This knowledge is important in synthesis: the preparation of new molecules from old ones for, for example, medicinal purposes or theoretical study. 6-4, 6-5, and 6-6. Mechanism of Nucleophilic Substitution Section 6-4 describes one of the more common methods of deriving mechanistic information: kinetic experiments. Section 6-5 describes another common method: experiments involving the observation of stereochemO Cl H H H H C Cl HO CH3 Electron-rich atom Electrophilic atom New bond 1559T_ch06_099-112 10/22/05 20:19 Page 100

Keys to the Chapter • 101 ical changes. By combining information from these and other types of experiments, the mechanistic picture of the bimolecular nucleophilic displacement, or SN2 reaction, was developed over the first 30-odd years of the twentieth century. The key features, second-order kinetics and inversion of configuration at the substrate carbon, are common to all the reactions in this chapter. The mechanism allows predictions to be made concerning the effects of changing any of a number of variables in the SN2 reaction. The last four sections of the chapter explore the most important of these. It should be pointed out that many of the observations described in these sections were predicted ahead of time on the basis of the logical implications of the SN2 mechanism. Part of your job as a student in organic chemistry will be to develop the ability to predict the result of a reaction of molecules you may never have seen before on the basis of your knowledge of reaction mechanisms associated with the functional groups the molecules contain. One final comment on mechanisms in general: There exists a common sort of shorthand way of writing organic reaction mechanisms. First, each step is written separately. Second, bonding changes in each step are indicated by arrows that represent the movement of pairs of electrons. For the SN2 mechanism, a one-step process, we have: The two arrows show (1) movement of a pair of electrons from oxygen to carbon to form the new C— O bond and (2) movement of a pair of electrons from the C—Cl bond to chlorine to form the chloride ion. Mechanism arrows depict the movement of pairs of electrons, not the movement of atoms. This is a logical result of the fact that chemical reactions are due to changes in bonds, and bonds are made out of electrons. Practice using “electron pushing” in mechanisms as frequently as you can so that you can get used to the technique. Note that proper use of electron-pair arrows automatically results in the correct Lewis structures for the products of a reaction, including charges, if any. An example of its application to an even simpler reaction, the reaction of an acid and a base, is shown below. 6-7, 6-8, and 6-9. The SN2 Reaction in Depth These sections explore the effects of changing three variables on the SN2 reaction rate: leaving group, nucleophile, and substrate structure. All the material in these sections derives logically from a knowledge of the mechanism and an awareness of the role each of the three variables can play. In each of these sections the effect of the variable on the rate of reaction is considered. This should tell you that the point of each discussion will be: How does changing this variable affect the activation energy—the energy of the transition state relative to that of the starting materials? We may not say it in so many words every time, but that’s what we mean. Even if the discussion is totally qualitative, and no actual rate data are given, the focus of such a discussion will still be the effect of the variable in question on relative transition state energy. Consideration of the leaving group is fairly simple. Because the leaving group is “beginning to leave” in the SN2 transition state, the energy of the transition state will reflect to some extent the stability of the leaving group. Stable leaving groups are therefore better (interpret: “faster”) leaving groups. The parallel between leaving-group ability and nonbasic character drawn in the chapter is the easiest one to work with and is fairly reliable. “Good” leaving groups are usually the conjugate bases of strong acids (Table 6-4). “Bad” leaving groups are strong bases, i.e., they are the conjugate bases of weak acids. Later you will learn how to manufacture good leaving groups out of bad ones. Notice how such fundamental concepts as acid/base strength can play pivotal roles in organic chemistry. Cl HO H Cl HO H (H2O) Cl HO CH3 Cl HO CH3 1559T_ch06_099-112 10/22/05 20:19 Page 101

108 • Chapter 6 PROPERTIES AND REACTIONS OF HALOALKANES: BIMOLECULAR NUCLEOPHILIC SUBSTITUTION (c) (1) NH2 PH2 NH3 (2) PH2 NH2 NH3 (3) NH3 PH2 NH2 (d) (1) OCN SCN (2) SCN OCN (3) SCN OCN (e) (1) HO CH3S F (2) CH3S HO F (3) F CH3S HO (f ) (1) NH3 H2O H2S (2) H2S NH3 H2O (3) H2S H2O NH3 43. (a) No reaction. Starting material is an alkane. Alkanes don’t react with nucleophiles. (b) CH3CH2OCH3 (c) (d) (e) No reaction. The leaving group would be OH, a strong base. Strong bases are very poor leaving groups. (f ) No (or extremely slow) reaction. Now the leaving group is good (CH3SO3 ), but HCN is a weak acid and therefore a very poor source of CN nucleophiles. (g) Finally, everything is right for a reaction to occur. Both a good leaving group (CH3SO3 ) and a good nucleophile (CN) are present, so the product (CH3)2CHCN forms readily. (h) (CH3)2CHCH2CH2SCN. Leaving group is (i) No reaction. NH2 is a bad leaving group. (j) CH3NH2 44. (a) BMIM is polar (very! It is an ionic salt!) and aprotic (it lacks hydrogens attached to electronegative atoms), and, therefore, cannot serve as a source of hydrogens for hydrogen bonding. CH3 S O O O CH3CH2 CH3S CH3 C H CH3 CH3 I A good conformation of the product after inversion of the reacting carbon. Immediately following SN2 displacement, the molecule possesses an eclipsed shape. Draw it! Reverse of (1). Size, then electronegativity. Electronegativity, then size. They’re all bad. Order is reverse of (1). Large size of CH3S takes precedence for nucleophilicity. HO stronger than CH3S because of size, and stronger than F because of electronegativity difference. Comparison between CH3S and F hard to make as small size favors F, whereas lower electronegativity favors CH3S. Reverse of (1). Size (larger atom is more nucleophilic). Size (smaller atom is more basic). Reverse of (1). Size puts PH2 first; lack of charge puts NH3 last. Larger size makes PH2 a weaker base than NH2; lack of charge makes NH3 the weakest. 1559T_ch06_099-112 10/22/05 20:19 Page 108

(b) Being a polar, aprotic solvent, BMIM should enhance the rate of any nucleophilic substitution reaction that proceeds via the SN2 mechanism. 45. The four diastereomers of 2-bromo-3-hydroxy-4-methylpentanoic acid are shown below, with the configuration of each stereocenter identified. The target molecule is the 2S,3S isomer of 3-hydroxyleucine. To prepare this particular diastereomer from one of the bromo compounds above we need to consider both the reaction that we need to use and its stereochemical consequences. Replacement of bromine by nitrogen is needed; we have been introduced to ammonia, NH3, as a suitable nitrogen-containing nucleophile. So the necessary reaction is identified. But which stereoisomeric starting material is the best? The mechanism gives us the answer: SN2, which proceeds with stereochemical inversion. This insight tells us that we must begin with the stereoisomer of the starting compound that will give us the product we need after stereochemical inversion at the site of reaction. Therefore, we should choose the 2R,3S isomer (at the lower right from the group of four, above). It already contains the correct configuration at the hydroxy carbon (C3); that stereocenter is not the site of reaction so it remains unchanged. Meanwhile, C2 will undergo inversion from R to S as the substitution takes place, giving the required outcome: The lesson to take from a problem such as this is one that recurs time and time again: Always look to the mechanism for the insight to solve problems in full detail. 46. Iodide ion is both a good nucleophile and a good leaving group. Given the presence of excess iodide ion and lots of time, the initial (desired) product, (R)-2-iodoheptane, can react with iodide via backside displacement again. Iodide displaces iodide! The product is the enantiomer (S)-2-iodoheptane. This process can occur as long as the reaction mixture, with all its ingredients, is left intact. By the time the reaction is finally stopped by separation of the organic products from iodide ion, all the iodoalkanes have experienced many such displacements. Statistically, about half undergo an even number of iodide substitutions to finish with the R stereochemistry (same as they started with), and the other half react an odd number of times, ending up S. So a racemic mixture results. 47. (a) This is the best way you know, even though a mixture is formed. (b) Making use of the selectivity of bromination for secondary COH bonds leads to the best route: CH3CH2CH2 CH3CHBrCH3 CH3CHClCH3 Br2, hv KCl, DMSO CH3CH2CH3 CH3CH2CH2Cl CH 3CHClCH3 Cl2, 25C, h 43% 57% OH S S R S NH2 OH Br NH3 OH – H+ O OH O OH Br OH OH S S S R R S R R Br OH O Br OH O OH O OH O Br OH Solutions to Problems • 109 1559T_ch06_099-112 10/24/05 19:19 Page 109