Quantum Chemistry 11.7 - Pi Resonance

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  • Опубликовано: 9 ноя 2024

Комментарии • 16

  • @someguy1576
    @someguy1576 5 лет назад +2

    Why this video has a dislike is beyond me.

  • @matteorinaldi4613
    @matteorinaldi4613 10 месяцев назад

    @TMPChem the fact that we have interaction between the p_z orbitals 2 and 3 means that the molecule will have a multireference character ?

  • @albericvigne871
    @albericvigne871 10 месяцев назад

    Hello, is there a reason you put on your butadiene diagram an alpha on each carbon and a beta between them ? does it means something?

  • @VyvienneEaux
    @VyvienneEaux 5 лет назад

    Aww, I was hoping you would solve the secular equations to find the orbital coefficients! I have no idea how to do that or if indeed that is how you're supposed to go about it.

  • @josephsifakis5274
    @josephsifakis5274 8 лет назад

    Great work with all these videos! Just curious how you factored out beta to get beta^4

    • @TMPChem
      @TMPChem  8 лет назад +3

      Great question. One way the determinant can be defined is through all possible permutations of the columns of a matrix. In every permutation, there is a unique set of diagonal elements. Each term in a determinant expansion is the product of all the elements in one of these diagonals. The sign of the term is determined by whether it took an even or odd number of exchanges to get there. So in every term for a 4x4 determinant, we have a product of 4 elements. If we want to factor out a beta from each of them, that requires beta^4, because the terms are being multiplied. If this was 2x2 it would be beta squared, 3x3 beta cubed, et cetera. A general NxN determinant requires beta to the N.

    • @josephsifakis5274
      @josephsifakis5274 8 лет назад

      Wow. Fast reply love it! Thanks for that, although one last question, if it were an N*N matrix where N is an odd number would it be (-)beta^N

    • @TMPChem
      @TMPChem  8 лет назад

      It would still be (+)beta^N. The positive or negative sign stays with each individual term, just as it would if you were factoring a polynomial with terms of varying signs.

    • @josephsifakis5274
      @josephsifakis5274 8 лет назад +1

      Ok perfect. I reread your original reply and had misunderstood. All clear now! Thanks again!

  • @johngalt517
    @johngalt517 7 лет назад

    So can it be said that to find the resonance energy of any conjugated Pi bonding system one must take the difference of the system in question and the C2H4 system? So if one wanted to find this value for cyclohexene one would multiply the C2H4 system by 3? Does the ring change this formula?

    • @johngalt517
      @johngalt517 7 лет назад

      Ha I prematurely asked this question. I see that your next video (Aromaticity) has the answer to my "cyclo" question.

    • @TMPChem
      @TMPChem  7 лет назад

      Indeed, the resonance energy here is merely defined as the energy stabilization of CC double bonds due to conjugation from adjacent CC double bonds.

  • @jonathanyara1064
    @jonathanyara1064 6 лет назад

    why is there only 4 electrons in the MO diagram? also shouldnt there be 4 electrrons on the degenerate left side?

    • @TMPChem
      @TMPChem  6 лет назад +1

      There are only 4 electrons in the MO diagram because there are only 2 pi bonds, and thus 4 pi electrons in the molecule. This diagram only applies to pi orbitals.

    • @jonathanyara1064
      @jonathanyara1064 6 лет назад

      doesnt a covalent bond have a pair of electrons and a double bond 2 pairs so 4 electrons. So two pi bonds should be 8 electrons?

    • @TMPChem
      @TMPChem  6 лет назад +2

      A double bond typically consists of one sigma bond and one pi bond, each with 2 electrons. Thus there are 2 pi electrons from each double bond.