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Chapter 21 Organic Chemistry 483 (b) This is a dehydration reaction, CH₂ CH₂ (c) This is an addition reaction that follows Markovnikov's rule, Br HBr CH₃ CH₃ 21.107 (a) Because there are six primary hydrogen atoms and two secondary hydrogen atoms, if they are equally reactive, we expect a ratio of 6:2 or 3:1. (b) Assume that we generate 100 product molecules. The yield = (# hydrogen atoms) (reactivity). So 1° = 45 = (3) (reactivity 1°) and 2° = 55 = (1) (reactivity 2°). Taking the ratio, 2°:1° = 55:(45/3) = 55:15 = 11:3. The 2° hydrogens are much more reactive. 21.109 The chiral products (that have four different groups on a carbon) are (second carbon), (third carbon), and (second and third carbons). 21.111 The first propagation step for F is very rapid and exothermic because of the strength of the H-F bond that forms. For I, the first propagation step is endothermic and slow because the H-I bond that forms is relatively weak. In addition, the F bond is much stronger than C-I (477 kJ versus 241 kJ). H H H Cl 21.113 The three structures that have no dipole moment C C and , H H H H Cl H CI The structures with dipole moments are C C , and C C H H H H H Conceptual Problems 21.115 For the structure to have only one product after a single bromination, all of the hydrogens must be equivalent. CH₃ CH₃ The structure is To name it, find the longest carbon chain: CH₃CH₃ CH₃CH₃ Because there are four carbons in this chain and it is an alkane, the base name is butane. CH₃CH₃ CH₃ CH₃ There are four methyl substituent groups. It does not matter how the chain is numbered CH₃ CH₃ because it is symmetrically substituted. The methyl groups are assigned the numbers 2, 2, 3, and 3. The name is 2,2,3,3-tetramethylbutane. Copyright © 2017 Pearson Education, Inc.