15597ch06_099-11210/22/0520:19Page99 EQA 6 Properties and Reactions of Haloalkanes: Bimolecular Nucleophilic Substitution halogen(o caus her serve ent pres onsof the discussed in detail here are fully applicable later on.even though later presentations may not be as compre 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. 64MathansncftephieshesiaKnerc tto cover in detail the mechanism of a specific polar reaction.Alsoa very 6-5 Stereochemistry ying together t reaction with material in Chapter5. 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 99
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
1559T_ch06_099-11210/22/0520:19Pa9e100 EQA 100.Chapter 6 PROPERTES AND REACTIONS OF HALOALKANES:BIMOLECULAR NUCLEOPHILIC SUBSTITUTON Keys to the Chapter 6-1.Physical Properties bond is the ocal point of this chapr spolarihe The carbor physical and che mical behavior of these molecules.Difference appreciate the differences that are presented later. 6-2.Nuclec philic Substitution stage we begin a d ed discussion of our first major class of polar reactions.Note the following 1.All titution 一+m Whenever a new reaction class is presented,analyze the various examples presented from the point of w of the should do this now for all the 2. up.Start geting used tothe variety of species that belong to each sut ons are elec In e ry ca atom in the substrate.Opposite charges attract! r a new reaction class is presented.analyze the example given on the hasis of 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. hde netail when and ow ondin e useful feature of such information is its predicrive value:Mechanistic understanding can tell us whether an unknown new ex- ol。e 6-4,6-5,and 6-6.Mechanism of Nucleophilic Substitution
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
15597ch06099-11210/22/0520:19Pag0101 EQA Keysto the Chapler·1ol istic picture twentieth century.The key features.second order kinetics and inversion of conuration at the substrate carbon,are common to all the reactions in this chapter.The mechanism allows predictions to be made con bles in the of th ns of th in thes swere prdicted ahead of time on thebasis of the loica implications of th nism.Part of you ir job as a student in organic chem ns ass ouns the molecules contain. One fina comment on mee chanisms in general:There exists a common sor of shorthand way of writing indicated by arrws that the movement of pairs of lectrons.For the s process,we have: HO:+CHCCI 一Ho-CH1+C (ofp or leoo8e巴ogas一 nd (2)movement of a pair of from the Mecha 假loge狼 ⊕ using"ctron pushing"in mechanisms as frequently as you can so that you can get used to the technique action of an acid and a base.is shown below. HG:+Hd一Ho-HH,0)+C n the S2 reaction rate:leav of the role each of the thr sion will be:How docs changing this variable affect the activation energ the energy of the transition state relative to that of the starting mate ials?We may not say it in so many words every ti me,but that's what w mean.Evl still b any qu ave,and no actu the S2 tran the en The othe lea work with and is fairly 6-4 Bad"leaving play pivotal roles in organic chemistry
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
1559T_ch06_099-11210/24/0518:49Pa9e102 ⊕ EQA 102.Chapter 6 PROPERTES AND REACTIONS OF HALOALKANES:BIMOLECULAR NUCLEOPHILIC SUBSTITUTON Consideration of the ncleophile follows The section discusses two nucleophilicity: .Ge into the habit of don this henever you re ons and their kinctic favorability.A similar discussion of solvent Solutions to Problems 27.(a)Chlorocthane (b)1.2-Dibromoethane (c)3-(Fluoromethyl)pentane (d)1-lodo-2.2-dimethylpropane (e)(Trichloromethyl)cyclohexane (f)Tribromomethane 28.(a)CH;CHICHCH2CH3 (b)CHCLCH-CHBICH ⊕ H -CH2Br (d) △ H (e)CH2CICCH2CI 29.Answers for Problem 31 are also given.Stereocenters are marked with an asterisk and numbers of stereoisomers are given in parentheses .CH. opane(2) H:CBrCICH3 ne (2) 2-Bro 2-chloropropan
Consideration of the nucleophile follows. The section discusses two characteristics that can give rise to high nucleophilicity: high basicity, as in hydroxide ion, and large size of the nucleophilic atom, as in iodide ion. The reasons behind this are explored in detail and are relatively straightforward. In attempting to understand these, consider, again, how the characteristics in question influence the relative stability of the transition state of the reaction. Get into the habit of doing this whenever you are faced with totally new concepts relating to reactions and their kinetic favorability. A similar discussion of solvent effects is also presented. Consideration of substrate structure is much simpler, as the major consideration for an SN2 reaction is, simply, How crowded is the “back side” of the atom being attacked? The less crowded, then the less steric hindrance to approach of the nucleophile on route to the transition state, so the better will be the situation for an SN2 reaction. The extreme importance of steric hindrance is emphasized by the fact that tert-butyl and neopentyl halides are virtually incapable of reacting by the SN2 mechanism. Solutions to Problems 27. (a) Chloroethane (b) 1,2-Dibromoethane (c) 3-(Fluoromethyl)pentane (d) 1-Iodo-2,2-dimethylpropane (e) (Trichloromethyl)cyclohexane (f ) Tribromomethane 28. (a) (b) CHCl2CH2CHBrCH3 (c) (d) (e) 29. Answers for Problem 31 are also given. Stereocenters are marked with an asterisk and numbers of stereoisomers are given in parentheses. BrClCHCH2CH3 1-Bromo-1-chloropropane (2) * BrCH2CHClCH3 1-Bromo-2-chloropropane (2) * ClCH2CHBrCH3 2-Bromo-1-chloropropane (2) * CH3CBrClCH3 2-Bromo-2-chloropropane BrCH2CH2CH2Cl 1-Bromo-3-chloropropane Cl CH3 CH2ClCCH2Cl CH CCl3 2Br H H CH2CH2Cl CH2CH3 CH3CHICHCH2CH3 102 • Chapter 6 PROPERTIES AND REACTIONS OF HALOALKANES: BIMOLECULAR NUCLEOPHILIC SUBSTITUTION 1559T_ch06_099-112 10/24/05 18:49 Page 102
15597ch06099-11210/22/0520:19Pag0103 EQA Solutions o Problems103 30.Stereocenters are marked with an asterisk and numbers of stereoisomers are given in parentheses. BrCH2CH-CH-CH2CH3 1-Bromopentane CH, CH3 CH; BrCH2CCH3 CH;CH:CHCHaBr 1-Bromo-2-methylbutane (2) mpopane 31.See 29 and 30. hhe nucleophilie atom in the nucleophile and the cletophilie tom in the Reaction Nucleophile Substrate Leaving group 1. HO:- CH.CI CHO: CHCHaI CH-CHBrCH-CH Br- :N=C: (CH.)CHCH,l CHS:- HB. r :P(CH) CH,Br Br 3.ab2c:一-C i tio4 (b)The N may act atom in cyanide (CN-).The reaction would then proceed as follows cH,dS+i=c:一r+ CH3 [H H,dcH,-2c:一 H2-N=C CH. An organic"isonitrile
Solutions to Problems • 103 30. Stereocenters are marked with an asterisk and numbers of stereoisomers are given in parentheses. 31. See 29 and 30. 32. In the answers below, the nucleophilic atom in the nucleophile and the electrophilic atom in the substrate are both underlined. Reaction Nucleophile Substrate Leaving group 1. CH3Cl Cl 2. CH3CH2I I 3. CH3CHBrCH2CH3 Br 4. (CH3)2CHCH2I I 5. Br 6. CH3CH2I I 7. CH3Br Br 33. (a) in reaction 4. (b) The N may act as a nucleophilic atom in cyanide (CN). The reaction would then proceed as follows: CH3CCH2 N C H I I CH3 CH3CCH2 N C H CH3 CH3CCH2 H CH3 An organic “isonitrile” N C N C N C P(CH3)3 NH3 CHBr CH3S N C I CH3O HO BrCH2CH2CH2CH2CH3 1-Bromopentane CH2CHBrCH2CH2CH3 2-Bromopentane (2) CH3CH2CHBrCH2CH3 3-Bromopentane * BrCH2CH2CHCH3 1-Bromo-3-methylbutane * CH3 CH3CHBrCHCH3 2-Bromo-3-methylbutane (2) CH3 CH3CH2CBrCH3 2-Bromo-2-methylbutane CH3 * CH3CH2CHCH2Br 1-Bromo-2-methylbutane (2) CH3 BrCH2CCH3 1-Bromo-2,2-dimethylpropane CH3 CH3 1559T_ch06_099-112 10/22/05 20:19 Page 103
1559T_ch06_099-11210/24/0504:49 PM Page104 ⊕ EQA 104.Chapter 6 PROPERTES AND REACTIONS OF HALOALKANES:BIMOLECULAR NUCLEOPHILC SUBSTITUTON 34.Answers are presented in the same manner as for Problem 32.Arrows mark electrophilic atoms. Reaction Nucleophile Substrate Leaving group Product B (c) CH.C 入人TscN 35.Bimolecular displacement is first order in each component. (a)Rate=k[CHCl][KSCN]:2 x 10-*mol L-'s-1 =k(0.1 M)(0.1 M).so k=2x 10-L mol-'s- (b)The three rates are (i)4 (ii)1.2x10-;and (iii)3.2x10-7 mol L-'s,respectively. 36.(a)CH;CH2CH2I (b)(CHa)2CHCH2CN (e)CHOCH(CH3)2 (d)CH;CH2SCH2CH, e○一CHSdCH,c*ar (f)(CH3)2CHN(CH3)*-OSO2CH 37.(a)Starting material is R. H Br- -CHa Product is 5. (b)Starting material is(25.3)-2-bromo-3-chlorobutane. CHA CH. Product is(2R.3R)-2.3-diiodobutane H (c)Starting material is(15.3R)-3-chlorocyclohexanol (the position of the OH group is understood to beCI)
34. Answers are presented in the same manner as for Problem 32. Arrows mark electrophilic atoms. Reaction Nucleophile Substrate Leaving group Product (a) NH2 CH3I I CH3NH2 (b) HS Br (c) I CF3SO3 (d) N3 Cl (e) CH3Cl Cl (f) SeCN I 35. Bimolecular displacement is first order in each component. (a) Rate k[CH3Cl][KSCN]: 2 108 mol L1 s 1 k(0.1 M)(0.1 M), so k 2 106 L mol1 s 1 . (b) The three rates are (i) 4 108 ; (ii) 1.2 107 ; and (iii) 3.2 107 mol L1 s 1 , respectively. 36. (a) CH3CH2CH2I (b) (CH3)2CHCH2CN (c) CH3OCH(CH3)2 (d) CH3CH2SCH2CH3 (e) (f ) (CH3)2CHN(CH3)3 OSO2CH3 37. (a) Starting material is R. (b) Starting material is (2S,3S)-2-bromo-3-chlorobutane. (c) Starting material is (1S,3R)-3-chlorocyclohexanol (the position of the OH group is understood to be C1). CH3 CH3 Product is (2R,3R)-2,3-diiodobutane. H I I H CH3 Product is S. CH2CH3 Br H CH2Se(CH2CH3)2 Cl SeCN I H3C CH3 N CH3 N H Cl N3 H I S O O O CF3 Br SH 104 • Chapter 6 PROPERTIES AND REACTIONS OF HALOALKANES: BIMOLECULAR NUCLEOPHILIC SUBSTITUTION 1559T_ch06_099-112 10/24/05 04:49 PM Page 104
1559r.eh06.099-11210/22/0520:19Page105 EQA Solutions o Problems105 Product is(15.35)-3-hydroxycyclohexylacetate (d)Starting material is (15.3S)-3-chlorocyclohexanol. Product is(1R.35)-3-hydroxycyclohexyl acetate. 0 Notice that in the oduct for (c).e han 38.(36a)::-CH,-CH2-CH2Br (36b):N=C (CH3)CH-CH2C (36c)(CHa)CHO:CH (36d)CH:CH-S:-CH-CH2-Br eacH,ag○a6 (360)(CHN:(CH.CH COSO.CH, H (37a):Br:CH, CH2CH3 CI H Bw年c不c
(d) Starting material is (1S,3S)-3-chlorocyclohexanol. Notice that in the product for (c), exchanging the OH and groups does not change the molecule. However, in the product for (d), such an exchange turns the molecule into its enantiomer. 38. (36a) (36b) (36c) (36d) (36e) (36f) (37a) (37b) H3C CH3 H Br Cl H I I “ “ Cl CH2CH3 Br CH3 H (CH3)3N (CH3)2CH OSO2CH3 (CH3CH2)2Se CH2 Cl CH3CH2S CH3 CH2 Br (CH3)2CHO CH3 I N C (CH3)2CH CH2 I I CH3 CH2 CH2 Br OCCH3 O OCCH3 O 1 2 3 HO Product is (1R,3S)-3-hydroxycyclohexyl acetate. OCCH3 O HO Product is (1S,3S)-3-hydroxycyclohexyl acetate. Solutions to Problems • 105 1559T_ch06_099-112 10/22/05 20:19 Page 105
1559T_ch06_099-11211/2/0516:05Pa9e106 ⊕ EQA 106.Chapter 6 PROPERTES AND REACTIONS OF HALOALKANES:BIMOLECULAR NUCLEOPHILIC SUBSTITUTION (37e)H:C-C 9 39.(a)No reaction,although CH3CH2CH2OH would eventually form(after a few centuries).H2O is a very poor nucleophile. (b)No reaction.HSO.is a st acid.a ver weak base,and a very,very poor nucleophile!Its conjugate base.HSOis a very weak nucleophile as well. (e)CHCH2CH2OH (d)CH;CH2CH2I (e)CH.CH.CH.CN use for (g)CH:CH,CH2S(CH)Br (h)CH,CH2CH2NH,*Br- No reaction..Ho (j)No (or y 40.(a)CH:CH2CH2CH2OH (b)CH:CH2CI -CH2OCH2CH, (d)(CHCHCHI (rather slowly) (e)Chch-CH-SCN (f)No reaction.F-isa very poor leaving group. (g)No reaction.OH is an even worse leaving group (h)CHSCH (i)No reaction.OCHCH,is not a reasonable leaving group. (CH.CH.OCCH OSO2CH3 N3 41.(a)(R-CHCHCH.CH+Na'N (5)-CH CHCH.CH
(37c) (37d) 39. (a) No reaction, although CH3CH2CH2OH would eventually form (after a few centuries). H2O is a very poor nucleophile. (b) No reaction. H2SO4 is a strong acid, a very weak base, and a very, very poor nucleophile! Its conjugate base, HSO4 , is a very weak nucleophile as well. (c) CH3CH2CH2OH (d) CH3CH2CH2I (e) CH3CH2CH2CN (f ) No reaction. As in (b), HCl is also a very weak base and therefore a dreadful excuse for a nucleophile. Being a strong acid, however, HCl can dissociate in appropriate solvents to release Cl ions, which are moderately nucleophilic, so it would not be wrong to answer, “Slow formation of CH3CH2CH2Cl.” (g) CH3CH2CH2S(CH3)2 Br (h) CH3CH2CH2NH3 Br (i) No reaction. However, in the presence of heat or light, radical chlorination will occur, giving a mixture of products. (j) No (or very slow) reaction. F is a rather poor nucleophile. (Note: F is a much better nucleophile in aprotic solvents.) 40. (a) CH3CH2CH2CH2OH (b) CH3CH2Cl (c) (d) (CH3)2CHCH2I (rather slowly) (e) CH3CH2CH2SCN (f ) No reaction. F is a very poor leaving group. (g) No reaction. OH is an even worse leaving group. (h) CH3SCH3 (i) No reaction. OCH2CH3 is not a reasonable leaving group. (j) The halide ions that are released in reactions (a) through (e), (h), and (j) (Cl, Br, and I) are all good leaving groups. 41. (a) (R)-CH3CHCH2CH3 (S)-CH3CHCH2CH3 OSO2CH3 Na N3 N3 CH3OH CH3CH2OCCH3 O CH2OCH2CH3 H3C C O O Cl HO H3C C O O Cl HO 106 • Chapter 6 PROPERTIES AND REACTIONS OF HALOALKANES: BIMOLECULAR NUCLEOPHILIC SUBSTITUTION 1559T_ch06_099-112 11/2/05 16:05 Page 106
1559r_eh06.099-11210/22/0520:19Page107 Solutions o Problems107 (b)()where the ou are asked it is necessary to devise two-in up of wo S2 reactions to get the prope stereochemic +Na+I-Acetone -H -H CH:O- H Na+-CN DMSO H -CN 一H (d)Does this ooThis showto which u are asked toh roup an alkylation reaction.This is just the same as an S2 reaction.but now appropriate haloalkane substrate to react with the nucleophile,instead of the other way around N CH,CH CH 42.(a))H0->CH,C02>H20 (2H0>CH,C02>H,0 (3)H0>CH:C02->H0 (b)(l)F->CI->Br->I- Larger size stabilizes negative charge,making base weake (2)1>Br->C>F galop6a (3)厂>Br>C>F Reverse order of (1)
Solutions to Problems • 107 (b) In contrast to (a), where inversion of configuration at the stereocenter was required, you are asked here to substitute Br with CN with retention. Because each SN2 reaction proceeds with inversion, it is necessary to devise a two-inversion scheme made up of two SN2 reactions to get the proper stereochemical result. The first SN2 reaction must be done with a nucleophile that is also a good leaving group (I fits this description): (c) Notice that an inversion is required: In the substrate the Br is trans to the ring fusion hydrogens, whereas in the product the SCH3 group is cis to them. Use one SN2 reaction: (d) Does this look strange? This shows a nucleophile, to which you are asked to attach an alkyl group: an alkylation reaction. This is just the same as an SN2 reaction, but now you need to supply an appropriate haloalkane substrate to react with the nucleophile, instead of the other way around: 42. (a) (1) HO CH3CO2 H2O (2) HO CH3CO2 H2O (3) H2O CH3CO2 HO (b) (1) F Cl Br I (2) I Br Cl F (3) I Br Cl F Reverse order of (1). Larger size decreases solvation and increases polarizability, increasing nucleophilicity. Larger size stabilizes negative charge, making base weaker. Leaving-group ability is inversely related to basicity. For a single atom, nucleophilicity parallels basicity. Basicity increases with charge and decreases with charge stabilization. N CH3 CH3 CH3 N CH3I CH3CH2OH I Na Br SCH3 H H CH3OH SCH3 H H Second SN2 inversion Na CN CH3 CH3 H CN CH3O H DMSO CH3 CH3 First SN2 inversion Na I H Br H CH3O CH3 CH3 I H CH3O H Acetone 1559T_ch06_099-112 10/22/05 20:19 Page 107
1559T_ch06_099-11210/22/0520:19Pa9e108 ⊕ EQA 108.Chapter 6 PROPERTES AND REACTIONS OF HALOALKANES:BIMOLECULAR NUCLEOPHILIC SUBSTITUTON (c)(1)-NH2 >-PH2 NH3 r base than "NH:lack of (2)PH,>NH,>NH Size puts "PH2 first:Lack of charge puts NH,last. (3)NH3 >-PH2 >-NH2 Reverse of (1) (d)(1)-OCN >-SCN Size (smaller atom is more basic) (2)-SCN >-OCN Size (langer atom is more nucleophilic) (3)-SCN >-OCN Reverse of (1). (e)(1)HO->CH3S->F llsie vrswhereas (②)CHS>H0>F Large size of CH,S takes precedence for nucleophilicity. (3)F->CHS->HO They'reall bad.Order s revere of(). )()NH3>H,0>H2S Electronegativity,then size. (2)HS>NH>H,0 Sie,thenedlectronegativi (3)H2S>H20>NH Reverse of (1). 43.(a)No reaction.Starting material is an alkane.Alkanes don't react with nucleophiles (b)CH,CH2OCH H (O)CH Q CH (d)CH.CH. CH (e)No reaction.The leaving group would beOH,a strong base.Strong bases are very poor leaving groups. (f)No (or extremely slow)reaction.Now the leaving group is good (CH SO).but HCN is a weak acid and therefore a very poor source of CN-nucleophiles. (g)Finally. up (CHSO3-)and a (h)(CH,),CHCH,CH,SCN.Leaving group is -0 (i)No reaction.NH-is a bad leaving group. (j)CH;NH2
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
