有機化學(xué)結(jié)構(gòu)與功能5th答案sg_ch07

1、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

2、 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 mor

3、e practical importance and a focal point of the chapter is the fact that similar compounds can un- dergo 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 mechan

4、istic understanding of the reactions. You are asked to apply your knowledge logically and to pay attention to detailto 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

5、, and 7-4Mechanism 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 Unimol

6、ecular 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 haloal

7、kanes, 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 te

8、rtiary halides are the best suited substrate mole- cules: 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 typ- ically a two- or three-step

9、 reaction, in contrast to the single-step SN2 process. It requires a rate-determining ionization of the carbonhalogen (or carbonleaving 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 na

10、me for these species. Another name currently in use is carbenium ion. Note carefully how the various fea- tures 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

11、-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 specie

12、s (carbocation), it is logical that the ease of the SN1 mechanism will reflect the ease of generation of the corre- sponding 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 exp

13、lained 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 pres- ent to help alleviate the electron de

14、ficiency of the cationic carbon. Therefore, the reactivity of tertiary sub- strates in SN1 reactions boils down to the ease of their ionization to form the relatively stable tertiary carbo- cation intermediate. Table 7-2 summarizes the distinct modes of substitution for methyl and primary halides (S

15、N2 only) vs. tertiary halides (SN1 only). For secondary substrates the behavior is more complicated. De- pending 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 con- se

16、quences of the two competing mechanisms. Read the last portion of Section 7-5 particularly carefully, be- cause 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 Reacti

17、ons 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

18、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 wont wait to combine with external electron-pair sources. Instead

19、, they will seek available electron pairs within their own molecular structures. The most available of these are electrons in carbonhydrogen 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 t

20、wo products: an alkene and a proton, the result of E1 elimination. As mentioned in the text, the proton doesnt 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?CC? H CC? ? 114 Chapter 7 FURTHER REACTI

21、ONS OF HALOALKANES: UNIMOLECULAR SUBSTITUTION AND PATHWAYS OF ELIMINATION 1559T_ch07_113-131 10/22/05 20:20 Page 114 Note that only the ? carbonhydrogen 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 w

22、ere 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 attach

23、es to and removes a proton. For the prac- tical 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 secon

24、d elimination mechanism with second-order kinetic behavior occurs (E2 reac- tion). 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 t

25、he loss of the leaving group. Examine Figure 7-8 closely. Note the electron motion toward the chlorine- bearing carbon: It is actually quite similar to the electron motion of an SN2 process! The tertiary halide can- not undergo SN2 displacement, but the E2 mechanism is a way for it to move electrons

26、 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

27、 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 pref- erence in most cases can be det

28、ermined 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

29、 favored. Otherwise sub- stitution 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) CC H C C H?C?C? CH bond: H A carbene (Very unstable) Keys to the Cha

30、pter 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 115 116 Chapter 7 FURTHER REACTIONS OF HALOALKANES: UNIMOLECULAR SUBSTITUTION AND PATHWAYS OF ELIMINATION Substitution or

31、 elimination? Analysis: FactorFavors SUBSTRATE is primary: unhinderedSubstitution NUCLEOPHILE is ?NH2: strongly basicElimination NUCLEOPHILE is ?NH2: sterically unhinderedSubstitution Result: Substitution, to form CH3CH2CH2CH2NH2. The text summarizes the preferences for E1, E2, SN1, and SN2 reaction

32、s for 1?, 2?, and 3? haloalkanes, as a function of reaction conditions, in quite a bit of detail. The chart that follows repeats the same material, again somewhat oversimplified for clarity (solvent effects are not included, for instance). SUMMARY CHART Major reactions of haloalkanes with nucleophil

33、es Type of nucleophile Weakly basicStrongly basicStrongly basic Poorgoodunhinderedhindered Type of nucleophilenucleophilenucleophilenucleophile halidelike H2Olike I?like CH3O?like (CH3)3CO? MethylNo reactionSN2SN2SN2 1?No reactionSN2SN2E2 2?Slow SN1SN2E2E2 3?SN1 and E1SN1 and E1E2E2 Solutions to Pro

34、blems OCH2CF3 A 22. (a) (CH3)3COCH2CH3(b) (CH3)2CCH2CH3 (c)(d) (e) (CH3)3COD(f) 23. (a) For the answer to this part we show each step on a separate line. Br?CBr CH3 CH3 CH3 C? CH3 CH3 CH3 (CH3)3C H O C CH3 CH3 OCH O OCH3CH3CH2 1559T_ch07_113-131 10/22/05 20:20 Page 116 In the last step of problems l

35、ike this students frequently make the mistake of removing an alkyl group from the oxygen, and giving an alcohol as the final reaction product. That process does not happen in solvolyses in alcohol solvents: Loss of a proton is far more favorable than loss of an unstable alkyl cation (which would be

36、a primary ethyl cation in the example above). (b) For the remaining parts, we still show each step separately, but we connect the steps in a continuous sequence. The full structure of each intermediate is still shown. (c) (d) OK, youre asking, why did that oxygen attach to the carbocation instead of

37、 the other one? Makes it look a lot more complicated, too. The reason is that the species you get is resonance stabilized ?H? ? C C CH3 CH3 O H H O C C CH3 CH3 O H O ?Br? ? C CH3 CH3C Br CH3 CH3 HC HO O ?H? ? CCH3 CH3 OCH2CF3 CH2CH3CCH3 CH3 CH2CH3 CH2CF3H O ?Br? ? CCH3 CH3 CH2CH3C Br CH3 CH3 CH2CH3

38、CF3CH2HO C CH3 CH3 CH3 CH2CH3 H H? ? OCO CH3 CH3 CH3 CH2CH3 C CH3 CH3 CH3 CH2CH3 H C? CH3 CH3CH3CH2OH CH3 ? O Solutions to Problems 117 ?Cl?H? ? ? CH2CH3 CH2CH3 Cl CH3HO CH2CH3O CH3 H CH2CH3O CH3 1559T_ch07_113-131 10/22/05 20:20 Page 117 (see below). When we cover carboxylic acids a dozen chapters

39、from now youll see more of this. Sorryreality is reality. (e) (f) 24. (a) Two steps: CH3 CH3 Br Br? H CH3 CH3 H CH3 CH3 HOCH3 H CH3 CH3 CH3OH H ? ? ? CH3OH ? CH3OH or H? Product with cis CH3s H? Product with trans CH3s CH3 CH3 OCH3 H and CH3 OCH3 CH3 H CCl CH3 CH3 CH3 C? CH3 CH3 CH3 ?Cl?D? DDO C CH3

40、 CH3 CH3 D D ? OC CH3 CH3 CH3 OD ? C C CH3 CH3 O H H O ? ? C C CH3 CH3 O H HH O C C CH3 CH3 O H O 118 Chapter 7 FURTHER REACTIONS OF HALOALKANES: UNIMOLECULAR SUBSTITUTION AND PATHWAYS OF ELIMINATION CCl CH3 CH3 CH3 C? CH3 CH3 CH3 ?Cl? ?D? D H O C CH3 CH3 CH3 D ? OC CH3 CH3 CH3 O H H 1559T_ch07_113-

41、131 11/2/05 16:36 Page 118 The nucleophile can attach to either face of the planar cationic carbon, reactions yielding the two products shown. (b)By reattachment of Br?to the opposite side of cationic carbon. (Reversal of the dissociation step.) 25. Two products: From attachment of nucleophile to ei

42、ther face of the planar cation. 26. (a) H2O will speed up all of the reactions except for 22(d), because it is more polar than any of the other solvolysis solvents. It will also compete for the carbocations, forming alcohols as products. (b) Ionic salts strongly increase polarity and accelerate SN1

43、reactions (see Problem 45, however). The main products will be iodoalkanes. (c) Same as (b); azide ion is a strong nucleophile and products will be azidoalkanes (alkyl azides, RON3). (d) This solvent should reduce polarity and slow down all the solvolyses. 27. 28. (a) (CH3)2CClCH2CH3? (CH3)2CHCHClCH

44、3? (CH3)2CHCH2CH2Cl (3? ? 2? ? 1?) (b)(Order of leaving group ability) (c) 29. (a) A secondary system with an excellent leaving group and a poor nucleophile F SN1 reaction. ? ?OSO2CF3 ? (CH3)2CHOCH2CH3 CH3CH2 (CH3)2CH (CH3)2CHOCH2CH3 OSO2CF3(CH3)2CH OH H H? ? CH3 Cl ?Br Cl RCl ? ROCCH3 ? ROH O CH2H

45、? (Primary) CH3H (Secondary) ? CH3 (Tertiary) ? ? C6H5 C6H5CH3 H CH3 CH3CH2O C6H5 C6H5CH3 H CH3 OCH2CH3 and CH3 CH3 Br H Solutions to Problems 119 1559T_ch07_113-131 11/2/05 16:06 Page 119 (b) A tertiary halide in a polar solvent F SN1 reaction (c) A primary halide with a good nucleophile in an apro

46、tic solvent F SN2 reaction. (d) Similar to (c), except a secondary halide F still an SN2 reaction. 30. First decide what the most likely mechanism is for each reaction. Then write the product. Finally, recall that SN2 reactions are faster in polar aprotic solvents, whereas SN1 reactions are faster i

47、n polar protic solvents because of the greater stabilization of the transition state for cationanion dissociation. (a) Primary substrate F SN2 to give CH3CH2CH2CH2CN; best in aprotic solvent. (b) Branched, but still primary, and nucleophile is not a strong base F SN2 again. Product is (CH3)2CHCH2N3;

48、 aprotic solvent is again best. (c) Tertiary substrate F SN1 substitution to form (CH3)3CSCH2CH3; best in protic solvent. (d) Secondary substrate with an excellent leaving group and a weak nucleophile F SN1 is the most likely mechanism from this combination, forming (CH3)2CHOCH(CH3)2; fastest in pro

49、tic solvent. 31. Two successive SN2 inversion steps are necessary to give the desired net result of stereochemical retention: O B 32. (1) Racemic CH3CH2CH(OCH)CH3will be formed via an SN1 process (a solvolysis). The solvent (a carboxylic acid) is very polar and protic, but a weak nucleophile. O B (2

50、) (R)-CH3CH2CH(OCH)CH3will be formed via an SN2 process (good nucleophile, aprotic solvent). Note the very different conditions. (R)-2-chlorobutane (R)-2-azidobutane KBr, DMSO (S)-2-bromobutane NaN3, DMSO CH3CH2CHICH2CH3CH3CH2CHCH2CH3 Cl I? ? CH3CH2CH2CH2P(C6H5)3 Br?CH3CH2CH2CH2Br ? (C6H5)3P S? Br C

51、H3 CH3 SCH3 CH3 CH3 CH3 H ? Br? H? HSCH3 120 Chapter 7 FURTHER REACTIONS OF HALOALKANES: UNIMOLECULAR SUBSTITUTION AND PATHWAYS OF ELIMINATION 1559T_ch07_113-131 10/22/05 20:20 Page 120 33. The first reaction is an uncomplicated SN2 displacement. The second is SN2 as well, but there is a complicatio

52、n. Lets 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 t

53、he 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 comple

54、tes the sequence. Why doesnt 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

55、 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 pKaof the thiol SH

56、 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)CH2CH3CH3CHPC(CH3)2 (1c) (1d) (1e), (1f) Same as (1a) 35. (a) The ba

57、se is a very strong one (the pKaof NH3is 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)2CCH2C(CH3) CH3CHCH3CH2 CH3CH2CH2 CH3CH2CH2CH2 CH3CH2CH2CH2 CH2Br? ?Br? SCH3CH2CH2CH2S ? H?CH3CH2CH2CH2S ? CH3CH2CH2CH2S

58、H? CH3CH2CH2CH2SHCH3CH2CH2 CH3CH2CH2CH2 CH3CH2CH2CH2 CH2Br? ? ?Br? SH Solutions to Problems 121 1559T_ch07_113-131 10/22/05 20:20 Page 121 This product forms no matter which of the six possible protons next to the reactive carbon is lost: (b) E2 (c) E2 (d) Either E1 or E2 can occur, and two products

59、 can form from either: Example of E1 mechanism: Example of E2 mechanism: 36. As in Problems 29 and 30, first things first: Categorize the substrate (primary, secondary, etc.) in order to identify the mechanistic pathways available to it. 1-Bromobutane is an unbranched primary haloalkane; CH2 H ? CH3CH3OH ? Cl? Cl CH2O ? CH3CH3CH3?H? ? Cl? Cl H CH3CH2.and H2O H Br? ? OH ?CCH H Br CH3CH2CHOH H CH3CH2 CH2(CH3)3C O Cl? (CH3)3C ?C CCl H H H ? C CH3CH2 CH3CH2CH3CH2CH2CH3 CH3 H CH3 C? Br?NH3C C Br H H H2N ? 122 Chapter 7 FURTHER R