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Chemistry LibreTexts

1.6: Ch. 4 Answers

  • Page ID
    218640
  • 1) \(Cl-Cl + photon (h \nu) \rightarrow 2 Cl \cdot\) (a photon is a unit of light energy)

    2) radicals or free radicals

    3) \(CH_{3}Cl + Cl \cdot \rightarrow \cdot CH_{2}Cl + HCl \)

    \(\cdot CH_{2}Cl + Cl-Cl \rightarrow CH_{2}Cl_{2} + Cl \cdot\)

    4) A (termination steps involve combination of free radicals without generation of new ones)

    5) Use a large excess of methane.

    6)

    1) Initiation

    2) Propagation Ph = Phenyl group (benzene ring attached to a carbon chain)

    7) Amount of energy required to break a covalent bond homolytically.

    8) A

    9) D

    10) A

    11) C

    12) -16 kcal/mol. Hint: remember to look up the correct BDEs for the type of bond, i.e., primary, secondary, tertiary, etc

    13) D

    14) the activation energy or Ea

    15) -4 kcal/mol

     

    16)

    17)

    18) E (1-Bromo-1-ethylcyclohexane)

    19) Three:

    20) The 2o hydrogens are 20 times more reactive than the 1o ones (Refer to p. 156 of your textbook for a similar example).

    21) (CH3)3CCHBrCH3 (2-Bromo-3,3-dimethylbutane). The secondary hydrogens are the most reactive in the reactant. There are no tertiary hydrogens.

    22) 1-Bromo-1-phenylpropane. The benzylic hydrogens are the most reactive in the reactant because the free radical that leads to their formation is resonance stabilized (can you draw it and provide all the resonance structures?) Refer to section 17-4B of your textbook for a very similar example, complete with resonance structures. You should be familiar with these concepts for the test. Don’t let the fact that this example is in ch.17 scare you. You know or have the tools to understand most of the concepts discussed in this section ;-)

    23) (CH3)4C (neopentane or 2,2-dimethylpropane).

    24) E

    25) The first propagation step in free radical bromination is endothermic while the analogous step in free radical chlorination is exothermic. From the Hammond Postulate, this means that the transition state for the bromination is product-like (i.e., radical-like) while the transition state for the chlorination is reactant-like. The product-like transition state for bromination has the C-H bond nearly broken and a great deal of radical character on the carbon atom. The energy of this transition state reflects most of the energy difference of the radical products. This is not true in the chlorination case where the transition state possesses little radical character.

    26) \(CH_{3} \cdot < CH_{3}CH_{2} \cdot < (CH_{3})_{2}CH \cdot < (CH_{3})_{3}C \cdot < CH_{2}=CHCH2 \cdot\) (resonance-stabilized)

    27)

    28) D (refer to section 4-16A, p. 163 for an explanation of concepts such as hyperconjugation and inductive effect).

    29) Cationic center is sp2 hybridized, therefore the CCC bond angle is 120o.

    30) In the bromination of pentane, the lowest energy reaction pathways go through secondary free radical intermediates to produce secondary alkyl bromides (2-bromopentane and 3- bromopentane). Thus 1-bromopentane is a very minor product.

    In cyclopentane, on the other hand, all of the hydrogen atom abstractions lead to the same secondary radical which eventually leads to bromocyclopentane.

    31) 3-bromo-3-methylpentane

    32) 1-bromo-1-methylcyclohexane

    33) \(CH_{3}CH_{3} + Cl \cdot \rightarrow CH_{3}CH_{2} \cdot  + H-Cl \)

    \(CH_{3}CH_{2} \cdot + Cl-Cl \rightarrow CH_{3}CH_{2}Cl + Cl \cdot \)

    34) In an exothermic reaction, weaker bonds are broken and stronger bonds are formed.

    35) E

    36) B

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