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Solutions to Problems 113 33. Answers for Problem 35 are also given. Stereocenters are marked with an asterisk and numbers of stereoisomers are given in parentheses. * * BrCH₂CHCICH₃ 1-Bromo-1-chloropropane (2) 1-Bromo-2-chloropropane (2) * CICH₂CHBrCH₃ 2-Bromo-1-chloropropane (2) 2-Bromo-2-chloropropane BrCH₂CH₂CH₂Cl 1-Bromo-3-chloropropane 34. Stereocenters are marked with an asterisk and numbers of stereoisomers are given in parentheses. * BrCH₂CH₂CH₂CH₂CH₃ CH₃CH₂CHBrCH₂CH₃ 1-Bromopentane 2-Bromopentane (2) 3-Bromopentane CH₃ CH₃ CH₃ * BrCH₂CH₂CHCH₃ CH₃CH₂CBrCH₃ 1-Bromo-3-methylbutane 2-Bromo-3-methylbutane (2) 2-Bromo-2-methylbutane CH₃ CH₃ BrCH₂CCH₃ CH₃CH₂CHCH₂Br * CH₃ 1-Bromo-2-methylbutane (2) 1-Bromo-2,2-dimethylpropane 35. See 33 and 34. 36. In the answers below, the nucleophilic atom in the nucleophile and the electrophilic atom in the substrate are both underlined. Reaction Nucleophile Substrate Leaving group 1. HO:- CH₃Cl 2. CH₃CH₂I :i:- 3. Br⁻ 4. CHBr 5. 6. :NH₃ 7. :P(CH₃)₃ Br⁻