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Free Radical Stability

  1. Jan 29, 2016 #1
    WP_20160129_12_53_22_Pro.jpg 1. The problem statement, all variables and given/known data


    2. Relevant equations


    3. The attempt at a solution
    option 1: 1 degree C
    option 2: 1 degree C
    option 3: 3 degree C
    option 4: 1 degree C

    SO the answer should be option 3?
     
  2. jcsd
  3. Jan 29, 2016 #2

    Suraj M

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    Is that the only stabilizing factor involved here?
    Any polar effect involved?
    PS- always thought saying tertiary or secondary carbon was a more universal naming system
     
  4. Jan 29, 2016 #3
    Why would there be a polar effect? Carbon and hydrogen have very similar electronegativities.
    Yeah tertiary and primary are the universal terms, somehow this book doesn't refer to them as tertiary and primary carbons but as 1 degree, 2 degree and 3 degree carbons.
     
  5. Jan 29, 2016 #4

    Suraj M

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    By polar effect, I am referring to mesmeric effect in this case
     
  6. Jan 29, 2016 #5
    If the mesomeric effect is considered, then option 3 and 4 are ruled out. But I don't understand why I can't just see which carbon is tertiary and say that's the most stable.
     
  7. Jan 29, 2016 #6

    Suraj M

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    Mesmeric effect ALWAYS dominates here
     
  8. Jan 29, 2016 #7

    Suraj M

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    Decide between 1 and 2 then
     
  9. Jan 31, 2016 #8
    I suggest thinking about what types of resonance give the greatest stability. Also consider drawing out resonance structures. Radical stability trends closely resemble carbocation trends (tertiary > secondary > primary). Think about where aromaticity fits into this trend.
     
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