Ϡ asked in Science & MathematicsChemistry · 1 decade ago

HELP!!! URGENT ORGANIC CHEMISTRY HELP NEEDED! Will you please help me?

I am stuck on a problem that deals with (E)-going to (Z)-3-hexene. There is a 80:20 ratio of E to Z isomer. But since Z is unfavorable (due to a positive delta G), how come 20% of it is still produced? Thank you very much. I really appreciate your help.

Update:

Sorry about that. There is an equilibrium mixture of E and Z isomers. Thank you so much!!!!

2 Answers

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  • Anonymous
    1 decade ago
    Favorite Answer

    Answering this question would be reasier with more reaction details.

    If the reaction is controlled kinetically, then the free energy is irrelevant because there is no equilibirum between products.

    If the reaction proceeds with thermodynamic control, then the two products form an EQUILIBRIUM MIXTURE as determined by the value of the free energy difference between the products. A product mixture is anticipated.

    Just because one product has a lower absolute free energy does not mean that it is the exclusive product.

    If you give more details, then maybe a more specific answer is possible.

    Source(s): my Ph.D. in organic chemistry
  • 1 decade ago

    delta g is the free energy of the system, or gibbs free energy specifically. a positive delta G denotes an unfavorable reaction. a negative delta G on the other hand is favorable.

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