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Solutions to Problems 51 24. Initiation: Br₂ 2 Br. = +46 kcal mol⁻¹ Propagation: (1) Br. + C₆H₆ HBr + C₆H₅· = +25 kcal (2) C₆H₅ + Br₂ C₆H₅Br + Br. = -35 kcal mol⁻¹ Overall = 10 kcal mol⁻¹ The overall is not very different from those of typical alkane C-H bonds: methane, = -6 kcal mol⁻¹; 1° C-H, = 10; 2° C-H, 13.5; 3° C-H, = 15.5. However, the rate- determining first propagation step in the reaction of benzene is much more endothermic than any of the alkane reactions, due to the exceptional strength of the C-H bonds in benzene. The result is that bromination of benzene by this mechanism is exceedingly difficult (very slow) and does not compete kinetically with bromination reactions of typical alkanes. 25. Qualitatively the same exercise as in Problem 23, but with different bond strength values. First propagation step Second propagation step + Br₂ + HBr: E E = = 35 kcal m +25 kcal + Br. Reaction coordinate Reaction coordinate 26. The diagram on the left shows that the first propagation step has a late transition state, close to the products along the reaction coordinate. In contrast, the second propagation step (right) has an early transition state, close to the starting materials. 27. Unless otherwise assume that no more than one halogen atom attaches to each alkane molecule. (a) No reaction. Iodination of alkanes is endothermic. (b) + CH₃CH₂CH₂F F₂ is not very selective. (c) CH₃ Br Bromination selects the tertiary position whenever possible. See note in answer to Problem 15(d). CH₃ CH₃ CH₃ CH₃ / (d) CH₃ CH₂ CH₃ + CH₂ CH₂ CH₃ Cl CH₃ CH₃ CH₃ CH₃ CH₃ CH₃ + CH₃ CH CH₃ + CH₃ CH₂ CH₂ CH₃ CH₃