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Relative stability of ortho and para isomers

  1. Jan 15, 2015 #1
    Generally, in organic reactions, the para isomer is found to be more stable (as it is symmetrical) because of which it is produced in greater amount than the ortho isomer. However, there are some exceptions also, for example, in some reactions the less sterically hindered ortho isomer is produced in more amount than the corresponding para isomer.

    If all this is true, then,

    1) In Kolbe's reaction (where phenol is first treated with NaOH and then with CO2), why is ortho isomer of Hydroxybenzoic acid (salicylic acid) - which is 2-Hydroxybenzoic acid produced in greater amount than 4-Hydroxybenzoic acid?

    There's absolutely no steric hindrance present at the para position!

    2)Also, when phenol is treated with dilute HNO3 at 293K, para isomer (p-Nitrophenol) is produced in greater amounts than o-Nitrophenol. Why?

    PS - I'm a high school student, so please answer accordingly.

    Thanks in advance!
  2. jcsd
  3. Jan 16, 2015 #2


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    Has there been any discussion in class or textbook regarding "electron withdrawing groups" or "electronegativity" or "resonance structure/stabilization?"
  4. Jan 20, 2015 #3

    Quantum Defect

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    You might try Googling about kinetic and thermodynamic control of reactivity. "Stability" usually refers to thermodynamics. You can produce less thermodynamically favorable products if a reaction is "kinetically" controlled -- generally means the energy landscape is such that you cannot get to the lowest energy product.

    For example, suppose you know that para amino benzoic acid is more stable than ortho amino benzoic acid. (I don't know if this is the case, but suppose it is). Now suppose that you start with an ortho ester of amino benozic acid. If you hydrolyze the ester, you will form the ortho product. Why don't you form the more stable para amino benzoic acid?
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