Vollhardt Capítulo 7 (Haloalcanos 2)
Disciplina:Química Orgânica II963 materiais • 31.941 seguidores
is a good solvent for SN1 reactions (Section 7-4). 2. We learned in Section 7-4 that different nucleophilic power has no effect on the rates of unimo- lecular processes. The rates should be (and, experimentally, are) identical. 3. This part requires a bit more thought. According to Table 7-4 and Section 7-8, elimination by the E1 pathway always accompanies SN1 displacement. However, increasing the base strength of the nucleophile may “turn on” the E2 mechanism, increasing the proportion of elimination product. Referring to Tables 6-4 and 6-7, we see that chloride is more basic (and less nucleo- philic) than iodide. More elimination is indeed observed with chloride than with iodide. The mechanism is as shown in Figure 7-7, with chloride acting as the base to remove a proton from the carbocation. b. The table in the margin presents data for the reactions that take place when the chloro compound shown here is dissolved in acetone containing varying quantities of water and sodium azide, NaN3: A H3C CH Cl A OH A N3 CH3 CH3 H3C� H2O, NaN3, acetone CH CH3H3C CH In the table, H2O% is the percentage of water by volume in the solvent, [N3 2] is the initial concentra- tion of sodium azide, RN3% is the percentage of organic azide in the product mixture (the remainder is the alcohol), and krel is the relative rate constant for the reaction, derived from the rate at which the starting material is consumed. The initial concentration of substrate is 0.04 M in all experiments. Answer the following questions. 1. Describe and explain the effects of changing the percentage of H2O on the rate of the reaction and on the product distribution. 2. Do the same for the effects of changing [N3 2]. Additional information: The reaction rates shown are the same when other ions — for example, Br2 or I2 — are used instead of azide. H2O [N3 2] RN3 krel % % 10 0 M 0 1 10 0.05 M 60 1.5 15 0.05 M 60 7 20 0.05 M 60 22 50 0.05 M 60 * 50 0.10 M 75 * 50 0.20 M 85 * 50 0.50 M 95 * *Too fast to measure. C h a p t e r 7 277 SOLUTION 1. We begin by examining the data in the table, specifi cally rows 2 – 5, which compare reactions in the presence of different amounts of water at constant azide concentration. The rate of substitution increases rapidly as the proportion of water goes up, but the ratio of the two products stays the same: 60% azide and 40% alcohol. These two results suggest that the only effect of increasing the amount of water is to make the solvent environment more polar, thereby speeding up the initial ionization of the substrate. Even when the proportion of water is only 10%, it is present in great excess and is trapping carbocations as fast as it can, relative to the rate at which azide ion reacts with the same intermediates (Section 7-2). 2. We note from rows 1 and 2 in the table that the reaction rate rises by about 50% when NaN3 is added. Without further information, we might assume that this effect is a consequence of the occurrence of the SN2 mechanism. Were that to be the case, however, other anions should affect the rate differently. However, we were told that bromide and iodide, far more powerful nucleo- philes, affect the measured rate in exactly the same way as does azide. We can explain this obser- vation only by assuming that displacement is entirely by the SN1 mechanism, and added ions affect the rate only by increasing the polarity of the solution and speeding up ionization (Section 7-4). In rows 5 – 8 of the table, we note that increasing the amount of azide ion increases the amount of azide-containing product that is formed. At the higher concentrations, azide, a better nucleo- phile than water, is better able to compete for reaction with the carbocation intermediate. New Reactions 1. Bimolecular Substitution — SN2 (Sections 6-2 through 6-9, 7-5) Primary and secondary substrates only }0 GC CI Direct backside displacement with 100% inversion of configuration H3C CH3 CH2CH3 CH2CH3 I�� H H O NuO }0 GNu�ð 2. Unimolecular Substitution — SN1 (Sections 7-1 through 7-5) Secondary and tertiary substrates only A OCH3 CH3 A CH3 Through carbocation: Chiral systems are racemized CBr A OCH3 CH3 A CH3 CNuOCH3 �Br� Nu�ð CH3 CH3 C� G D 3. Unimolecular Elimination — E1 (Section 7-6) Secondary and tertiary substrates only A OCH3 CH3 A CH3 Through carbocation CCl OCH3 �Cl� B�ð CH2 CH3H3C C B GD BH� CH3 CH3 C� G D 4. Bimolecular Elimination — E2 (Section 7-7) � �BH I�PCH3CH2CH2I CH3CH CH2 Simultaneous elimination of leaving group and neighboring proton B�ð Important Concepts 1. Secondary haloalkanes undergo slow and tertiary haloalkanes undergo fast unimolecular substitution in polar media. When the solvent serves as the nucleophile, the process is called solvolysis. 2. The slowest, or rate-determining, step in unimolecular substitution is dissociation of the C – X bond to form a carbocation intermediate. Added strong nucleophile changes the product but not the reaction rate. I m p o r t a n t C o n c e p t s 278 C h a p t e r 7 F u r t h e r R e a c t i o n s o f H a l o a l k a n e s 3. Carbocations are stabilized by hyperconjugation: Tertiary are the most stable, followed by sec- ondary. Primary and methyl cations are too unstable to form in solution. 4. Racemization often results when unimolecular substitution takes place at a chiral carbon. 5. Unimolecular elimination to form an alkene accompanies substitution in secondary and tertiary systems. 6. High concentrations of strong base may bring about bimolecular elimination. Expulsion of the leaving group accompanies removal of a hydrogen from the neighboring carbon by the base. The stereo chemistry indicates an anti conformational arrangement of the hydrogen and the leaving group. 7. Substitution is favored by unhindered substrates and small, less basic nucleophiles. 8. Elimination is favored by hindered substrates and bulky, more basic nucleophiles. Problems 25. What is the major substitution product of each of the following solvolysis reactions? H3C H3C Br H CH3OH /∑ /∑ (a) A A CH3 CH3 CH3CBr CH3CH2OH (b) A Br (CH3)2CCH2CH3 CF3CH2OH (c) CH2CH3 CH3OH Cl (d) A A O B C Br CH3 CH3 HCOH O (e) A A CH3 CH3CCl CH3 D2O (f) A A CH3 CH3CCl CH3 H OD 26. For each reaction presented in Problem 25, write out the complete, step-by-step mechanism using curved-arrow notation. Be sure to show each individual step of each mechanism separately, and show the complete structures of the products of that step before going on to the next. 27. Write the two major substitution products of the reaction shown in the margin. (a) Write a mechanism to explain the formation of each of them. (b) Monitoring the reaction mixture reveals that an isomer of the starting material is generated as an intermediate. Draw its structure and explain how it is formed. 28. Give the two major substitution products of the following reaction. CH3CH2OH H3C OSO2CH3 C6H5 C6H5H3C H 29. How would each reaction in Problem 25 be affected by the addition of each of the following substances to the solvolysis mixture? (a) H2O (b) KI (c) NaN3 (d) CH3CH2OCH2CH3 (Hint: Low polarity.) 30. Rank the following carbocations in decreasing order of stability. � � CH3CH3H CH2�H C h a p t e r 7 279 31. Rank the compounds in each of the following groups in order of decreasing rate of solvolysis in aqueous acetone. (a) A CH3 CH3CHCH2CH2Cl A A CH3 Cl CH3CHCHCH3 A A CH3 Cl CH3CCH2CH3 (b) B OCCH3 O OHCl (c) H3C ClClBr 32. Give the products of the following substitution reactions. Indicate whether they arise through the SN1 or the SN2 process. Formulate the detailed mechanisms of their generation. CH3CH2OH (a) (CH3)2CHOSO2CF3 uuuy (b) CH3 Excess CH3SH, CH3OH Br (C6H5)3P, DMSO NaI, acetone (c) CH3CH2CH2CH2Br uuuuuy (d) CH3CH2CHClCH2CH3 uuvuy 33. Give the product of each of the following substitution reactions.