1559T_ch07113-13110/22/0520:20Pa9e113 EQA 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 Pay electron pair to m vetoward the ophilic carbon,forming a new bond.As we have emphasized before. CO ing is an impo step to comprehension of the ma imilar e u der several different types of reactions.o of poth the textand 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 s.You are asked to apply your knowledge logically and to pay attention to detail-to reasor Outline of the Chapter 7-1 Solvolysis of Tertiary and Secondary Haloalkanes A surprising reaction of compounds that do not undergo the Sy2 process 7-2,7-3,and 7-4 Mechanism of Solvolysis:Unimolecular Nucleophilic Substitution The explanation.with details and conscquences. 7-5 Substrat displacement reactions. Also,a summary of nucleophilic 7-6 Unimolecular Elimination:E1 7-7 Bimolecular Elimination:E2 Two mechanisms for a new reaction of haloalkanes that allou Keys to the Chapter tho7.Solvolysis:Unimolecular Nucleophilic Substitution anes.although they 113
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
15597ch07113-13110/22/0520:20Pag0114 EQA 114.chapter 7 FURTHER REACTONS OF HALOALKANES:UNIMOLECULAR SUBSTTTUTION AND PATHWAYS OF ELIMINATION completely new mechanism for displacement for which tertiary halides are the best suited substratemoe and st chemical details of experiments that led to the formulation of this mechanism.Note that this is typ ically a two-or three-step reaction,in contrast to the single-step S2 process.It requires a rate-determining aving group) ng to a new reacuve species ca species.Another name in use. efully how the various fea eaving group)ar ely and logically 7-5.Substrate Structure:The Stability of Carbocations The chief requirement for anS mechanism is ease of ionization of the bond between the carbon atom and gical isis formation spondine carbocation.Carbon is not a very elect positive atom.and a carbon with a full positive charge is alkyl groups (as 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 deficiency of the cationic carbon.Therefore,the reactivity of tertiary sub n for methyl and p ary halide (S2 only)vs.tertiary halides (S only).For secondary substrates the behavior is more complicated.De- pen may occu lication of the logical c sequences of the two competing mechanisms.Read the last portion of Section7-5 particularly carefully,be- s on pos reaction pathways 7-6 and 7-7.Elimination Reactions The arbon atom in a carb e-deficient ()cabon nated by a need to n an elect on pair f om any aval source.Th a new bond to carbon However the electron deficiency of cationic carbon is so great that even under typica SyI solvolysis condi of the cation wait te Th e they wil from the cationic center (at the so-called B carbon): it is removed by any Lewis base available in the reaction system,such as solvent or other nucleophile molecules. 2c一*+-c
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 n hond (the involves the same rate-determin its kineti .the nucleophile attach sto the cationi or the prac 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 r occurs(E2 reac with the loss of the leaving group.Examine Figure 7-8 closely.Note the electron motion toward the chlorine bearing carb actually quite simil
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.
15597ch07113-13110/22/0520:20Pag0116 116 chapter 7 FURTHER REACTIONS OF HALOALKANES:UNIMOLECULAR SUBSTTTUTION AND PATHWAYS OF ELIMINATION Substitution or elimination?Analysis: Factor Favors SUBSTRATE is primary:unhindered Substitution NUCLEOPHILE isNH2:strongly basic NUCLEOPHILE is-NHa:sterically unhindered Substitution Result:Substitution,to form CHaCHaCH2CHaNH2 gai Type of nuckeophile Weakly basic Stro Strongly basic ere c Iik CHO like (CHCO No reaction Sl and EI Solutions to Problems OCH2CF3 22.(a)(CHa)sCOCH:CH3 (b)(CHa)2CCH-CHa CHCH,OCH -OCH (e)(CHa)COD mac-0〉 23.(a)For the answer to this part we show each step on a separate line CH3 CH CH3-CCBr CHs CHs
116 • Chapter 7 FURTHER REACTIONS OF HALOALKANES: UNIMOLECULAR SUBSTITUTION AND PATHWAYS OF ELIMINATION Substitution or elimination? Analysis: Factor Favors SUBSTRATE is primary: unhindered Substitution NUCLEOPHILE is NH2: strongly basic Elimination NUCLEOPHILE is NH2: sterically unhindered Substitution Result: Substitution, to form CH3CH2CH2CH2NH2. The text summarizes the preferences for E1, E2, SN1, and SN2 reactions 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 nucleophiles Type of nucleophile Weakly basic Strongly basic Strongly basic Poor good unhindered hindered Type of nucleophile nucleophile nucleophile nucleophile halide like H2O like I like CH3O like (CH3)3CO Methyl No reaction SN2 SN2 SN2 1 No reaction SN2 SN2 E2 2 Slow SN1 SN2 E2 E2 3 SN1 and E1 SN1 and E1 E2 E2 Solutions to Problems 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 C Br CH3 CH3 CH3 C CH3 CH3 CH3 (CH3)3C H O C CH3 CH3 OCH O CH3CH2 OCH3 1559T_ch07_113-131 10/22/05 20:20 Page 116
1559T_ch07-113-13110/22/0520:20Pa9e117 EQA Solutions to Problems.117 CH3-c+ CH.CH.OH CH,CH-CH →CH,-C--CHCH+H CHs CH.CHs CHa CBr CHC-CHCH,CH-CH,CH CH ⊕ OCH2CF3 CH-C-CH;CH,CH-C-CHCH, CH.CH CH3-:CH.CHs HO CH -CH
In the last step of problems like 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 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, you’re asking, why did that oxygen attach to the carbocation instead of 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 C CH3 Br CH3 CH3 C H HO O H CH3 C CH3 OCH2CF3 CH3 C CH2CH3 CH3 CH2CH3 H CH2CF3 O Br CH3 C CH3 C CH2CH3 Br CH3 CH3 CH2CH3 CF3CH2 O H C CH3 CH3 CH3 CH2CH3 H H O C O CH3 CH3 CH3 CH2CH3 C CH3 CH3 CH3 CH2CH3 H C CH3 CH3 CH3CH2OH CH3 O Solutions to Problems • 117 Cl H Cl CH2CH3 CH2CH3 CH3 O H O CH2CH3 CH3 H O CH2CH3 CH3 1559T_ch07_113-131 10/22/05 20:20 Page 117
15597.eh07.113-13111/2/0516:36Page118 EQA 118.chapter 7 FURTHER REACTONS OF HALOALKANES:UNIMOLECULAR SUBSTTTUTION AND PATHWAYS OF ELIMINATION below).When we cover carboxyne csyou see more of this. 2--0 .0 CH. D (e)CH3 G-CH H CH CHs CH,- 56H D:CH3- CH (a)Two steps CH.OHCH CH H
(see below). When we cover carboxylic acids a dozen chapters from now you’ll see more of this. Sorry—reality 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 CH3’s –H Product with trans CH3’s CH3 CH3 OCH3 H and CH3 OCH3 CH3 H C Cl CH3 CH3 CH3 C CH3 CH3 CH3 Cl D D D O C CH3 CH3 CH3 D D O C CH3 CH3 CH3 O D C C CH3 CH3 O H H O C C CH3 CH3 O H H H O C C CH3 CH3 O H O 118 • Chapter 7 FURTHER REACTIONS OF HALOALKANES: UNIMOLECULAR SUBSTITUTION AND PATHWAYS OF ELIMINATION C Cl CH3 CH3 CH3 C CH3 CH3 CH3 Cl D D H O C CH3 CH3 CH3 D O C CH3 CH3 CH3 O H H 1559T_ch07_113-131 11/2/05 16:36 Page 118
1559T_ch07_113-13111/2/0516:06Page119 ⊕ EQA Solutions to Problems.119 ecan attach to either face of the planar cationic carbon,reactions yielding the two Br CHa By reattachment of Brto the opposite side of cationic carbon.(Reversal of the dissociation step.) 25.Two products: and From attachment of nucleophile to either face of the planar cation olarity and accelerate S reactions (see Problem 45.however).The ()(b)e ooadprodtstee (allye ⊕ (d)This solvent should reduce polarity and slow down all the solvolyses. Tertiary） Secondary Primary 28.(a)(CH3)CCICH-CH3>(CH3)2CHCHCICH>(CH)2CHCHCH2CI (3>2>1) 0 (b)RCI>ROCCHs>ROH (Order of leaving group ability) 29.(a)A secondary system withan excellent leaving group and a poor nucleophile reaction (CHCHCose0,cE,一os0,C乐，+(cH,dqHa H (CH).CHOCH.CH,-H*+CHa)CHOCH-CH
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 either 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 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)2CHCHClCH3 (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 (CH3)2CH OSO2CF3 OH H H CH3 Cl Br Cl RCl ROCCH3 ROH O H CH2 (Primary) H CH3 (Secondary) CH3 (Tertiary) C6H5 CH3 C6H5 H CH3 CH3CH2O C6H5 C6H5 CH3 H CH3 OCH2CH3 and CH3 CH3 Br H Solutions to Problems • 119 1559T_ch07_113-131 11/2/05 16:06 Page 119
1559r.ah07.113-13110/22/0520:20Page120 120 chapter 7 FURTHER REACTONS OF HALOALKANES:UNIMOLECULAR SUBSTTTUTION AND PATHWAYS OF ELIMINATION (b)A tertiary halide in a polar solventS reaction r○X (c)A primary halide with a good nucleophile in an aprotic solvent2 reaction CH;CH.CH2CH2TBr+(C.Hs)p-CH.CHCH.CHaP(CoHa)3Br (d)Similar to (e).except a secondary halidestill an S2 reaction. CH,CH,CHCH,CH3+厂→CH,CH,CHICH,CH solvents because of the greater stabilization of the transition state for cation-anion dissociation. (a)Primary substrateSx2 to give,CN:best in aprotic solvent. 82e血ot (e)Tertiary substrateS substitution to form(CH)CSCH,CH:best in protic solvent. 31.Two successive S2 inversion steps are necessary to give the desired net result of stereochemical retention: (R)-2-chlorobutane (5--bromobutane-2-azdobutane 0 32.(1)Racemie CH,CH2CH(OCH)CH3 will be formed via an SI process (a solvolysis).The solvent (a carboxylic acid)is very polar and protic.but a weak nucleophile. (2)(R)-CH,CH,CH(OCH)CH,will be formed via an S.2 process (good nucleophile.aprotic solvent) Note the very different conditions
(b) A tertiary halide in a polar solvent F SN1 reaction (c) A primary halide with a good nucleophile in an aprotic 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 in polar protic solvents because of the greater stabilization of the transition state for cation–anion 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; 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 protic 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)CH3 will 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) (R)-CH3CH2CH(OCH)CH3 will 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 CH3CH2CHCH2CH3 CH3CH2CHICH2CH3 Cl I CH3CH2CH2CH2P(C6H5)3 Br CH3CH2CH2CH2 Br (C6H5)3P S Br CH3 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
1559T_ch07-113-13110/22/0520:20Pa9e121 EQA Solutions to Problems.121 happens in the second case but not in. aps we can see why the sulfur.Since butyl is primary.we are limited to the S2 mechanism.The simplest way to get there is to use the product molecule itself as a nucleophile Ch:Ch-CH-CH CH.CH.CH.CH.+CH.CH.CH-CH Brr CH.CHCH.CH. Then loss of Hfrom sulfur completes the sequence.Why doesn't the same thing happen in the first as nucleophiles to carry out 2 displacements.whil thiols can achieve this ormation readily CHCH,CH,CH SH CH;:-+H* followed by CH.CH.CH,CH.+CH.CH-CH CHBr CH;CH2CH2CH2 CH;CH2CH2CH2 plausible but because the p&.of the thiol sHt competitive:The concentration of the conjugate base is too low. 34.(la)(CHg)zC-CH (1b)CH2=C(CH3)CH2CH;CH;CH=C(CH3)2 te)CH.CH-CH.CH (ld)(CH3)C= CH2=C(CH3)- (le),(If)Same as (la) CH,CH2、 CH
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
1559r.ah07.113-13110/22/0520:20Page122 122 chapter 7 FURTHER REACTONS OF HALOALKANES:UNIMOLECULAR SUBSTTTUTION AND PATHWAYS OF ELIMINATION This product forms no matter which of the six possible protons next to the reactive carbon is lost CH. CH:CH2 一NH+ +Br CH,CHCH.CH CH:CH2 CH (b)E2 CH,CH、 H -(CH3)C-OH CH;CH2CH=CH2 Cl (CH3)C-6: (c)E2 ⊕ (d)Either EI or E2 can occur.and two products can form from either ○atw○at Example of El mechanism: Example of E2 mechanism a+a 一aot+cr+○a
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 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 CH3 CH3OH Cl Cl O CH2 CH3 CH3 H CH3 Cl Cl H CH3 and CH2. H2O H Br OH H C C H Br OH CH3CH2CH H CH3CH2 (CH3)3C CH2 O Cl (CH3)3C C C Cl H H H C CH3CH2 CH3CH2 CH2CH3 CH3CH2 CH3 H CH3 C Br C NH3 C Br H H H2N 122 • Chapter 7 FURTHER REACTIONS OF HALOALKANES: UNIMOLECULAR SUBSTITUTION AND PATHWAYS OF ELIMINATION 1559T_ch07_113-131 10/22/05 20:20 Page 122