北京化工大学：《有机化学》课程教学资源（习题与答案）Chapter 07 Further Reactions of Haloalkanes：Unimolecular Substitution and Pathways of Elimination.pdf_大学文库

7 Further Reactions of Haloalkanes: Unimolecular Substitution and Pathways of Elimination In Chapter 7 we continue and complete a discussion of major reaction types and mechanisms for haloalkanes. Three new processes are discussed. Pay close attention to how the mechanisms of each make electrostatic sense: Just like the SN2 mechanism in the last chapter, each of these new processes provides a means for an electron pair to move toward the electrophilic carbon, forming a new bond. As we have emphasized before, conceptual understanding is an important step to comprehension of the material. Of more practical importance and a focal point of the chapter is the fact that similar compounds can undergo several different types of reactions. A goal of both the text and this guide is to show you how to choose which process is most likely to occur under any given set of conditions based on a mechanistic understanding of the reactions. You are asked to apply your knowledge logically and to pay attention to detail—to reason like a scientist. Outline of the Chapter 7-1 Solvolysis of Tertiary and Secondary Haloalkanes A surprising reaction of compounds that do not undergo the SN2 process. 7-2, 7-3, and 7-4 Mechanism of Solvolysis: Unimolecular Nucleophilic Substitution The explanation, with details and consequences. 7-5 Substrate Structure: The Stability of Carbocations A new kind of reactive intermediate in organic chemistry. Also, a summary of nucleophilic displacement reactions. 7-6 Unimolecular Elimination: E1 7-7 Bimolecular Elimination: E2 Two mechanisms for a new reaction of haloalkanes. 7-8 Substitution or Elimination? Factors that allow prediction of favored reaction pathways. Keys to the Chapter 7-1 through 7-4. Solvolysis: Unimolecular Nucleophilic Substitution Tertiary haloalkanes, although they do not react via the SN2 mechanism (Chapter 6), still undergo very rapid nucleophilic displacement under certain reaction conditions. This reaction is due to the appearance of a 113 1559T_ch07_113-131 10/22/05 20:20 Page 113

completely new mechanism for displacement for which tertiary halides are the best suited substrate molecules: unimolecular nucleophilic substitution, or the SN1 mechanism. Sections 7-2 and 7-3 present the kinetic and stereochemical details of experiments that led to the formulation of this mechanism. Note that this is typically a two- or three-step reaction, in contrast to the single-step SN2 process. It requires a rate-determining ionization of the carbon–halogen (or carbon–leaving group) bond, leading to a new reactive species called a carbocation (pronounced car-bo-cat-ion). You may occasionally encounter the term carbonium ion, an older name for these species. Another name currently in use is carbenium ion. Note carefully how the various features of this reaction as described in this section (e.g., rate, stereochemistry, sensitivity to solvent polarity and leaving group) are closely and logically derived from the nature of the rate-determining first step. 7-5. Substrate Structure: The Stability of Carbocations The chief requirement for an SN1 mechanism is ease of ionization of the bond between the carbon atom and the leaving group. Because the initial result of this is formation of a positively charged (cationic) carbon species (carbocation), it is logical that the ease of the SN1 mechanism will reflect the ease of generation of the corresponding carbocation. Carbon is not a very electropositive atom, and a carbon with a full positive charge is, in general, not very stable, and, therefore, difficult to generate. As explained here, however, alkyl groups (as opposed to hydrogen atoms) stabilize cationic carbon centers. Thus, cations in which the positive charge is on a tertiary carbon atom will be the most stable and the easiest to generate, because three alkyl groups are present to help alleviate the electron deficiency of the cationic carbon. Therefore, the reactivity of tertiary substrates in SN1 reactions boils down to the ease of their ionization to form the relatively stable tertiary carbocation intermediate. Table 7-2 summarizes the distinct modes of substitution for methyl and primary halides (SN2 only) vs. tertiary halides (SN1 only). For secondary substrates the behavior is more complicated. Depending on the specific situation, either SN1 or SN2 reaction, or both, may occur. Prediction of the pathway a secondary substrate will follow is another direct application of the logical consequences of the two competing mechanisms. Read the last portion of Section 7-5 particularly carefully, because it presents a classic example of the use of mechanistic information to explain (and predict) the effects of reaction variables on possible reaction pathways. 7-6 and 7-7. Elimination Reactions The positively charged carbon atom in a carbocation is an extremely electron-deficient (electrophilic) carbon. As such, its behavior is dominated by a need to obtain an electron pair from any available source. The SN1 reaction illustrates the most obvious fate of a carbocation: combination with an external Lewis base, forming a new bond to carbon. However, the electron deficiency of cationic carbon is so great that even under typical SN1 solvolysis conditions, surrounded by nucleophilic solvent molecules, some of the cations won’t wait to combine with external electron-pair sources. Instead, they will seek available electron pairs within their own molecular structures. The most available of these are electrons in carbon–hydrogen bonds one carbon removed from the cationic center (at the so-called carbon): Attraction of this electron pair toward the positively charged carbon leads to two products: an alkene and a proton, the result of E1 elimination. As mentioned in the text, the proton doesn’t just “fall off.” Actually, it is removed by any Lewis base available in the reaction system, such as solvent or other nucleophile molecules. H C H C C C H C C 114 • Chapter 7 FURTHER REACTIONS OF HALOALKANES: UNIMOLECULAR SUBSTITUTION AND PATHWAYS OF ELIMINATION 1559T_ch07_113-131 10/22/05 20:20 Page 114

Note that only the carbon–hydrogen bond (the one next to the positive carbon) is susceptible to cleavage in this manner. Other COH bonds would not give such stable products if they were to be broken: Because the E1 process involves the same rate-determining step as the SN1 reaction, its kinetics are the same: first order. E1 elimination almost always accompanies SN1 substitution. The difference is simple: In SN1, the nucleophile attaches to the cationic carbon; in E1, it attaches to and removes a proton. For the practical purposes of synthesis, the presence of the E1 “side reaction” can limit the usefulness of SN1 substitution. If, however, elimination is desired, it may be achieved by addition of strong base to tertiary halides. At high concentrations, in fact, a second elimination mechanism with second-order kinetic behavior occurs (E2 reaction). Section 7-7 describes its details pictorially. As in the E1 process, the electrons in a COH bond move toward the electrophilic carbon; in the E2 reaction, however, this electron movement occurs simultaneously with the loss of the leaving group. Examine Figure 7-8 closely. Note the electron motion toward the chlorinebearing carbon: It is actually quite similar to the electron motion of an SN2 process! The tertiary halide cannot undergo SN2 displacement, but the E2 mechanism is a way for it to move electrons in a similar manner, getting them from a COH bond instead of from an external nucleophile. As the rest of this section shows, any haloalkane with a COH bond can undergo E2 elimination. In the case of 1 and 2 halides, E2 and SN2 reactions compete. However, as the following sections in the text and below will show, it is very easy to predict the favored products in these cases. 7-8. Substitution vs. Elimination: General Guidelines for Prediction The key consideration for synthetic purposes is elimination versus substitution, and, as the text shows, the preference in most cases can be determined by answering the following three questions: 1. Is the nucleophile a strong base? 2. Is the nucleophile sterically very bulky? 3. Is the substrate sterically hindered (i.e., 3, 2, or 1 with branching)? If the answer to at least two out of these three questions is “yes,” elimination will be favored. Otherwise substitution will predominate. Check the reactions in the chapter to see how these guidelines work. Here is one more for practice. Na NH2 ? Liquid NH3 CH3CH2CH2CH2I H C C H bond: C A cyclopropane (Strained) C C H C C H C C H bond: H C A carbene (Very unstable) Keys to the Chapter • 115 Both of these are very uncommon processes. Their occurrence is limited to situations where normal elimination cannot take place. 1559T_ch07_113-131 10/22/05 20:20 Page 115

33. The first reaction is an uncomplicated SN2 displacement. The second is SN2 as well, but there is a complication. Let’s take a look at how its product can react further, and then perhaps we can see why it happens in the second case but not in the first. The pathway to the byproduct, (CH3CH2CH2CH2)2S, must involve reaction between the product of the first displacement, CH3CH2CH2CH2SH, and another molecule of the initial starting material CH3CH2CH2CH2Br: This is the only reasonable way to get a second butyl group on the sulfur. Since butyl is primary, we are limited to the SN2 mechanism. The simplest way to get there is to use the product molecule itself as a nucleophile: Then loss of H from sulfur completes the sequence. Why doesn’t the same thing happen in the first reaction, the formation of the alcohol? What do you know about the differences in nucleophile strength between S and O? Sulfur is far better, especially in protic solvent (Section 6-8). Alcohols are too weak as nucleophiles to carry out SN2 displacements, while thiols can achieve this transformation readily. Did you consider an alternative mechanism, one in which the SH group of the thiol is deprotonated before nucleophilic attack? That is, , followed by This mechanism is qualitatively plausible, but because the pKa of the thiol SH bond is around 10, in the absence of base the equilibrium of the initial deprotonation is too unfavorable for this sequence to be competitive: The concentration of the conjugate base is too low. 34. (1a) (CH3)2CPCH2 (1b) CH2PC(CH3)CH2CH3 CH3CHPC(CH3)2 (1c) (1d) (1e), (1f ) Same as (1a) 35. (a) The base is a very strong one (the pKa of NH3 is way up there, around 35; it is a very weak acid), and that favors E2 rather than E1. The exclusive product is C CH3CH2 CH3CH2 H CH3 C (CH3)2C CH2 C(CH3) CH3CH CH3CH2 CH3CH2CH2 CH3CH2CH2CH2 CH3CH2CH2CH2 CH2 Br Br CH3CH2CH2CH2S S H CH3CH2CH2CH2S CH3CH2CH2CH2SH CH3CH2CH2CH2SH CH3CH2CH2 CH3CH2CH2CH2 CH3CH2CH2CH2 CH2 Br Br S H Solutions to Problems • 121 1559T_ch07_113-131 10/22/05 20:20 Page 121