O-Chem

Discussion in 'MCAT Study Question Q&A' started by hansen44, May 6, 2008.

  1. hansen44

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    I am having a problem finding information on the differences between kinetic stability and thermodynamic stability. I know its crucial to O chem and probably gen chem. Can someone please elaborate what this is exactly, I cant remember it from O chem. You don't have to write an essay but just the main ideas would be most appreciated. thanks
     
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  3. TheGreatHunt

    TheGreatHunt High Performance

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    Thermodynamic stability is the most energetically favorable, where you are not losing any excess energy to needless repulsions that cause strain on the molecule. This is not always the case though, especially when you initially have steric hindrance on a molecule you are trying to react. Let's say you have a couple tert-butyl groups hanging around a more active site(normally), even if it is the more stable reaction after the reaction is complete, the amount of energy it would take to overcome the steric hindrance initially is going to be more than reacting on an alternate part of the reactant, and so you might get the more kinetically favourable reaction instead of the thermodynamically favourable one.
     
  4. ssnickerer

    ssnickerer A Thinker

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    I think I agree with what TheGreatHunt was saying (though it took me a while to wrap my head around his interesting wording :)).

    I like to think of it in terms of the reaction coordinate diagram: something is kinetically stable if the energy barrier (Ea) is waaay up high. On the other hand, something is thermodynamically stable if one product is much, much, much lower in energy than another. Related but separate.
     
  5. osumc2014

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    To elaborate upon the last 2 posters:

    Thermodynamic products are more stable at the end, usually forming the more substituted product, and formed slowly, this is usually done at high temps because it takes more energy to get over the high Ea and thus it is harder to be reversed thus more stable.

    Kinetic products on the other hand, have less stable final products, but have more stable intermediates, such as more stable carbocations. These products are formed faster, usually due to availability to react and because of the low Ea required. These products are thus more easily reversible and convert to thermodynamic products at higher temperatures (Ea overcome).

    Hope that was clear :)
     
  6. Zerconia2921

    Zerconia2921 Bring your A-game!

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    If I remember right its from Gen chem.

    Thermodynamic stability depends on delta G delta H and delta S ie. free energy, enthalpy, and entropy. for a reaction to occur thermodynamically speaking a - delta G occurs. If delta G is zero we are at equilb. Also if delta H is neg and delta S is positive from the formula delta G = delta H-TdetaS the reaction will be spont at all temperatures.

    Kinetic stability is different from thermo becaus you are talking about enzymes which increase the rate of a reaction. They do not relate to thermodynamics.
     
  7. TheGreatHunt

    TheGreatHunt High Performance

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    Sorry, upon further examination, I realize that mine needed a picture, so here it is:
    [​IMG]

    As you can see, from the above, at 80 degrees C 3-Bromo 1-Butene is created faster, but at 40 degrees C, 1-Bromo-2Butene is created. When there is sufficient energy in the system, the system will go to create the fastest product, not the most thermodynamically stable product(It is better to have the double bond within the molecule than on a primary carbon.
     
  8. Foghorn

    Foghorn Guest

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    This is wrong on at least two levels. First, the numbering system on the allylic cation ions are incorrectly numbered. Given above is an example of a 1,2 & 1,4 addition reaction to a conjugated diene, resulting in allylic cation intermediates, at high and low temperature reaction conditions. By standard convention numbering for 1,2/1,4 addition reaction to a conjugated diene should be 4, 3, 2, 1 on the allylic cation intermediates as shown by resonance structures, from left to right.

    Second, at the higher temperature 80C, (1, 4 addition) 1-bromo-2-butene will be the thermodynamically favored, more stable major product. At lower temperature 40C (1,2 addition) 3-bromo-1-butene is the kinetically favored major product.
     
  9. BerkReviewTeach

    BerkReviewTeach Company Rep & Bad Singer
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    Foghorn is 101.738% correct on this one. The author of your example screwed up their numbering of the intermediate and have the explanation completely backwards. I'd like to elaborate a bit.

    A good way think about kinetic control versus thermodynamic control is to think of it in terms of economics. You have money to invest (similar to thermal energy floating around a system that can be used to overcome the activation barrier) and a desired return on your investment (energy released following reaction.)

    If you have a low temperature, then there is very little energy to invest. You are forced to choose the smallest investment, which is the one with the lowest activation energy. Because activation energy dictates the rate of a reaction, this pathway is refered to as the kinetic pathway. Hence, as Foghorn has pointed out, the lower temperature leads to the kinetic product.

    If you have a high temperature, then there is a large amount of energy to invest. You can choose either investment, so it is most probable that you will choose the one with the greatest return (most exergonic). Because heat/free energy dictates the favorablility of a reaction, this pathway is refered to as the thermodynamic pathway. Hence, as Foghorn has pointed out, the higher temperature leads to the thermodynamic product.

    The conjuagated diene and the addition to the alpha-carbon of a carbonyl are classic examples of this.

    {{{{{Opinionated sidenote}}}}} Whatever source you are using, you should be very careful, because this is a fundamental error that anyone on the caliber of material author should not make or at the very least catch upon a cursory editting. There are materials out there, some of them quite popular, that I find myself correcting many of their errors for during my open office hours. {{{{{end of Opinionated sidenote}}}}}
     
  10. hansen44

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    ok thanks for all your guys help this really helped.
     
  11. BlackSails

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    While we are on the topic - Is the Hammond postulate on the MCAT?
     
  12. BerkReviewTeach

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    From what I recall, in the distant past, they had a passage on activation and deactivation that included a Hammond plot in the passage. So many of their passages have text, equations, and a graph that appear a bit daunting compared to the what the level of the test is supposed to be, but in the end it turn out not to be a difficult passage afterall. So, it'll probably be a case where the questions are distilled down to basic resonance types of questions. And given that they don't test benzene reactions any longer, it seems like if the subject showed up, it would focus on mostly electronic properties.
     
  13. TheGreatHunt

    TheGreatHunt High Performance

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    Yeah I grabbed it verbatim off the page... Thank you Google.
     
  14. TJames

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    Kinetic stability = The product of a reaction that can produce 2 or more products and is non reversible. It is the product that forms fastest and is not necessarily more thermodynamically stable.
     
  15. BlackSails

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    Kinetic products occur in both reversible and non reversible reactions.
     
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  17. renordw

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    In the simplest possible terms:

    The kinetic product has the lowest activation energy. It tends to be more reversible.
    The thermodynamic product is the lowest energy product. It tends to be more stable.
     
  18. aldol16

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    These people are all likely in med school and couldn't care less which one is which. Also, the thermo product not only "tends" to be more stable, but it is, by definition, the most stable.
     
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