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Chapter 17 Solutions

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    1115
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    In-chapter exercises

    E17.1:

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    E17.2:

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    E17.3:

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    End-of-chapter problems

     

    P17.1:

    To simplify matters, we'll use 'R' to abbreviate the methyl ester group:

    image1036.png

    The first two propagation steps, forming a acrylamide dimer, are shown below:

    image1038.png

    The polymer is represented by:

    image047b.png

    P17.2:

    a)


    image049b.png

    image051b.png

    image053b.png

    b) The bis-acylamide molecule has two alkene groups, at either end, that can participate in radical chain elongation reactions.  This allows it to tie two growing polyacrylamide strands together (ie. to form a cross-link):

    image055b.png

    Here is a more detailed mechanism.  We start with a growing polyacrylamide strand (strand 1) reacting in a chain propagation reaction with a bis-acrylamide molecule.  In the next, step, the remaining alkene group on bis-acrylamide reacts adds to a second growing polyacrylamide strand (strand 2).

    image1050.png

    image1052.png

    image1054.png

    image1056.png

    P17.3:

    image1058.png

    Notice that the radical form of resveratrol shown above is more stable, compared to a radical species in which the unpaired electron is located on one of the 'lower' phenoxy groups.  The extra stability is due to resonance - the unpaired electron can be delocalized over both of the aromatic rings.  (Consider the two other  alternative radical species below - in these cases, the unpaired electron cannot be delocalized over both rings!  It all comes back to para vs. meta positioning on the aromatic ring.)

    image067b.png

    P17.4:

    image069b.png

    P17.5:

    image071b.png

    The driving force for homolytic cleavage is the formation of nitrogen gas, which is very entropically favorable.

    P17.6:

    a)

    image073b.png

    b) Butane could form from an alternate chain termination step:

    image075b.png

    c)

    image1070.png

    d,e) Radical intermediate B is the most stable (it is tertiary), but the resulting alkyl chloride is not necessarily the most abundant product. Even though radical A is less stable than radical B, there are six possible pathways for its formation (the chlorine radical could abstract six different hydrogens to form radical A), as opposed to only one pathway for the formation of B.  There are two pathways to radical C, and three to D.

    P17.7:

    a) The peroxide contaminant acts as a radical chain initiator:

    image079b.png

    The regioselectivity arises from the higher stability of the secondary alkyl radical intermediate (compared to the alternate possibility, which would be a primary radical).

    image1074.png

    b)

    image1076.png

    P17.8:

    image085b.png

     

    P17.9:

    a)

    image087b.png

    b)

    image089b.png

    P17.10:

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    P17.11:

    image093b.png

     

     

    Contributors

    • Organic Chemistry With a Biological Emphasis, Tim Soderberg (University of Minnesota, Morris)