Chapter 9 - Instructor Supplemental Solutions to Problems...

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Instructor Supplemental Solutions to Problems © 2010 Roberts and Company Publishers Chapter 9 The Chemistry of Alkyl Halides Solutions to In-Text Problems 9.1 (b) The product is ethylammonium iodide. 9.2 (b) As in part (a), because there are two types of ± - hydrogens, two alkenes can be formed: 9.3 (b) Methyl iodide, H 3 C—I, can form only a substitution product, dimethyl ether, H 3 C—O—CH 3 . (d) (Bromomethyl)cyclopentane can form one substitution product and one elimination product. 9.4 (b) Because iodide ion is a weaker base than chloride ion, the equilibrium lies to the left. (d) Because methoxide ion is a much stronger base than chloride ion, the equilibrium lies to the right. 9.5 (b) The reaction is first order overall, and first order in alkyl halide. The rate constant has the dimensions of sec 1 . 9.6 (b) We transform Eq. 9.22c to get the difference between the standard free energies of activation. ± G ° A ± G ° B = 2.30 RT log k B k A ² ³ ´ ´ µ · · = (5.71)log(450) = 15.2 kJ mol –1 Therefore, reaction A has the higher ± by 15.2 kJ mol –1 . 9.9 Because the S N 2 mechanism involves a molecule of alkyl halide and a molecule of nucleophile in a bimolecular reaction, the expected rate law is second order, first order in alkyl halide and first order in cyanide: rate = k [C 2 H 5 Br][ CN] 9.11 There is more than enough sodium cyanide to react with both the acid HBr and the alkyl halide. Therefore, the products are sodium bromide (NaBr, 0.2 M, half from the reaction with HBr and half from the reaction with ethyl bromide), “ethyl cyanide” (propionitrile, CH 3 CH 2 CN, 0.1 M ), and unreacted sodium cyanide (0.8 M ). However, 0.1
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INSTRUCTOR SUPPLEMENTAL SOLUTIONS TO PROBLEMS • CHAPTER 9 2 M NaBr is formed instantaneously, and the rest of the NaBr as well as the nitrile are formed much more slowly, because Brønsted acids reacts much more rapidly with bases than alkyl halides. 9.12 The reaction is an S N 2 reaction with inversion of configuration. Because the relative priorities of the groups attached to the asymmetric carbon are not changed, the product has the S configuration. 9.14 (a) The products of the S N 2 reaction between potassium acetate and ethyl iodide: (b) Potassium acetate is a better nucleophile in acetone because ethanol is a protic solvent and reduces the nucleophilicity of potassium acetate by hydrogen-bond donation. Consequently, potassium acetate in acetone reacts more rapidly with ethyl iodide than a solution of the same nucleophile in ethanol. 9.17 (a) The stepwise process involves formation of a methyl cation, which is very unstable. The instability of this cation, by Hammond’s postulate, raises the energy of the transition state and retards the reaction. The concerted mechanism avoids formation of this high-energy intermediate. (b)
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Chapter 9 - Instructor Supplemental Solutions to Problems...

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