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15.5.10: Chapter 10

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    482464
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    Problem 10-1
    (a) 1-Iodobutane
    (b) 1-Chloro-3-methylbutane
    (c) 1,5-Dibromo-2,2-dimethylpentane
    (d) 1,3-Dichloro-3-methylbutane
    (e) 1-Chloro-3-ethyl-4-iodopentane
    (f) 2-Bromo-5-chlorohexane
    Problem 10-2
    (a) A six-carbon chain with a chlorine on C 2 and two methyl groups on C 3.
    (b) A six-carbon chain with a methyl group on C 2 and two chlorines on C 3.
    (c) A five-carbon chain with an ethyl group and a bromine on C 3.
    (d) The structure of 1,1-dibromo-4-isopropylcyclohexane.
    (e) A nine-carbon chain with a chlorine on C 2 and a four-carbon chain, connected at the second carbon, on C 4.
    (f) The structure of 1,1-dibromo-4-tert-butylcyclohexane.
    Problem 10-3 Chiral: 1-chloro-2-methylpentane, 3-chloro-2-methylpentane, 2-chloro-4-methylpentane Achiral: 2-chloro-2-methylpentane, 1-chloro-4-methylpentane
    A condensed formula for 1-chloro-2-methylpentane.
    A condensed formula for 2-chloro-2-methylpentane.
    A condensed formula for 3-chloro-2-methylpentane.
    A condensed formula for 2-chloro-4-methylpentane.
    A condensed formula for 1-chloro-4-methylpentane.
    Problem 10-4 1-Chloro-2-methylbutane (29%), 1-chloro-3-methylbutane (14%), 2-chloro-2-methylbutane (24%), 2-chloro-3-methylbutane (33%)
    Problem 10-6 The intermediate allylic radical reacts at the more accessible site and gives the more highly substituted double bond.
    Problem 10-7
    (a) 3-Bromo-5-methylcycloheptene and 3-bromo-6-methylcycloheptene
    (b) Four products
    Problem 10-8
    (a) 2-Methyl-2-propanol + HCl
    (b) 4-Methyl-2-pentanol + PBr3
    (c) 5-Methyl-1-pentanol + PBr3
    (d) 3,3-Dimethyl-cyclopentanol + HF, pyridine
    Problem 10-9 Both reactions occur.
    Problem 10-10 React Grignard reagent with D2O.
    Problem 10-11
    (a) 1. NBS; 2. (CH3)2CuLi
    (b) 1. Li; 2. CuI; 3. CH3CH2CH2CH2Br
    (c) 1. BH3; 2. H2O2, NaOH; 3. PBr3; 4. Li, then CuI; 5. CH3(CH2)4Br
    Problem 10-12
    (a) Four compounds arranged in order of increasing oxidation level: cyclohexane is lowest, cyclohexanone and 1-chloro-1-cyclohexene are equal and intermediate, and benzene is highest.
    (b) Three compounds arranged in order of increasing oxidation level: ethylamine is lowest, ethylenediamine intermediate, and acetonitrile is highest.
    Problem 10-13
    (a) Reduction
    (b) Neither

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