Respostas a Problemas de Química

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Solutions to Problems 41 51. It is usually a good idea to make use of the given information-in fact, write it all out so that it's sitting there in front of you-before trying to answer the question. (a) Br + I⁻ I + Br⁻ Br + I + Br⁻ This reaction is 10,000 times slower than the one above. (b) The reaction sites are indicated by dots in the structures above. They are both primary carbon atoms because they are each attached directly to exactly one other carbon atom. (c) Electrostatics suggests that the negative iodide ion will be attracted to the positively polarized carbon atom in the C-Br bond. However, because this carbon atom already has a closed shell, the iodide cannot actually bring in an electron pair for bonding unless some other atom, such as the bromine, leaves and takes an electron pair away. The second-order kinetics does not support a sequence in which the bromide ion leaves before the iodide ion comes in. So most likely both events occur together: I C Br The big rate reduction in the second example above, relative to the first, suggests that the increased size of the alkyl group has gotten in the way of the iodide ion in its attempt to bond to the carbon atom (an example of steric hindrance; see text Section 2-8, page 84). This makes the most sense if for some reason the iodide ion needs to pass close to this alkyl group to form a bond, perhaps in a trajectory that looks like the following sketch: H (d) H C Br CH₃ Physical bulk of the alkyl CH₃ group, as indicated by this arc, interferes with the approach of the iodide ion from this side. CH₃ 52. (d) Two secondary hydrogens (on the CH₂ group) and one tertiary (the CH) in 53. (b) Products are lower in energy than starting materials, so energy came out. 54. (b) It's an alkane. All the angles are 109.5°. 55. (c) See Figure 2-12 on text page 84. 56. (e) The C=0 group is part of an ester.