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Long chain of c=c carbon double for room temp superconductivity

  1. Sep 8, 2006 #1
    could you have a long chain of c=c carbon double bonded to carbon

    each carbon would be sp hybridzed, with two pi orbals bonded to adjacent carbons


    going on for millions or even billions of repeating elements

    which would be in resonance

    c-c ---c-c---c
    where --- represents a triple bond

    with the ends capped by hydrogen or maybe oxygen or nitrogen or to a buckminster fullerine.

    with 2 delocalized pi ponds, i would predict room-temperature superconductivity

    i want to thank bananan for his generous donation
  2. jcsd
  3. Sep 8, 2006 #2
    Huh? What donation? And what makes you think this would be superconducting?
  4. Sep 8, 2006 #3
    think more about the resonance... yeah it doesn't break any rules by resonating in the middle, but how would the end groups interact... how does the molecule lower its energy by resonating?
  5. Sep 8, 2006 #4
    (What's this? Another polymer thread? :tongue2:)

    in addition to representing the lowest energy bond arrangements for a given structure, major resonant forms of molecules are (obviously enough) usually equivalent (or, 'close to') in energy. The form, ...C≡C-C≡C-C... is (relatively) much higher in energy than ...C=C=C=C..., and would essentially be a (very) minor resonance form of ...C=C=C=C... (simply put, you will find too little resonance within your polymer to exhibit any significant conductivity).

    On the other hand, a good example of a conductive polymer is polyacetylene (the synthesis and further research of which earned Alan J. Heeger, Alan G MacDiarmid, and Hideki Shirakawa the 2000 Nobel Prize in Chemistry). Not only does polyacetylene (...-CH=CH-CH=CH-...) have a conjugated system (implying stability as well as resonance), its hydrogens can also be substituted with various functional groups. For example, doping polyacetylene with iodine leads to a 108 increase in conductivity (according to Wikipedia).
    Last edited: Sep 8, 2006
  6. Sep 8, 2006 #5


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  7. Sep 8, 2006 #6
    delocalized pi orbitals
  8. Sep 8, 2006 #7
    thank you how the heck did u draw in the triple bond?

    the c---c-c would be in resonance with c-c---c
  9. Sep 8, 2006 #8
    However, the resonance will be too infrequent for your polymer to exhibit any appeciable conductivity (given the great energy difference between the major cumulene and minor polyyne forms). A conductive material must possess delocalized electrons; in your polymer, however, the electrons will remain relatively localized in the cumulene form (the only major resonance form of your polymer).

    Polyacetylene, a similar polymer that yours resembles, does have sufficient resonance (and thus electron delocalization) within its conjugated system to possess significant conductivity.
    Last edited: Sep 8, 2006
  10. Sep 8, 2006 #9
    it is delocalized 2 pi orbitals
  11. Sep 9, 2006 #10
    even with the cumulene structure, you have carbon-carbon that is sp bonded with two pi bonds, you could have resonance as to how those two pi bonds are connected as they are at right angles to one another.

    so, as in benzene and graphite, those pi bonds adjacent to one another would harbor delocalized electrons.

    while i can't draw it out here, imagine c-c-c-c-c-c-c-c-c-c-c

    between each c-c is a sp bond

    at right angles to each c-c are two pi orbitals perpendicular to one another. they would delocalize in the same way pi orbitals do in graphite or benzene, but unlike graphite or benzene, you have 2 sets of pi orbitals, not just one.
  12. Sep 9, 2006 #11
    My previous post is not in this thread?? (Anyhow, thanks for quoting it :smile:)//in post #10
    No, as there are significant differences between graphite/benzene and your polymer.

    ~In benzene and graphite, pi-electrons are transferred from (1σ,1π) to (1σ) bonds. Furthermore, the (only) two (available major) resonant forms are equivalent in energy.

    ~In your polymer, you have two resonant forms quite apart in energy (hence one is major, and one is minor). For resonance, π-electrons must be transferred from (1σ,1π) into (1σ,1π). Why would a double bond transfer electrons into...a neighboring double bond?

    Furthermore, it takes less energy to transfer pi-electrons from (1σ,1π) bond to a neighboring (1σ) bond (as in graphite & benzene), than from a (1σ,1π) bond into another (1σ,1π) bond (as in your polymer). A double bond has greater electron density than a single bond, and offers quite more resistance to formation of an additional π-bond than would a single bond (which is only a lone σ-bond). In addition, you would create a (previously inexistent) imbalance of electron densities around each carbon.

    As previously explained,
    1) More energy is required to transfer π-electrons into double bond than into a single bond. (Double bonds offer more resistance to the formation of an additional π-bond than would single bonds)

    2) Benzene & graphite transfer π-transfer electrons from (1σ,1π) bonds to neighboring (1σ) bonds. There is no energy barrier/difference at all between the resonance forms.
    Your polymer, however, (in order to resonate) must transfer π-electrons from (1σ,1π) bonds to neighboring (1σ,1π) bonds. Not only is more energy required, but you also create a previously inexistent imbalance of electron densities around each carbon (unlike the resonance in benzene and graphite). As you see, the resonant forms (cumulene, with no such imbalance, and polyyne) are quite energetically different.

    3) Given the energy difference between the cumulene and polyyne forms, whatever resonance exists between the forms would not occur frequently enough (your electrons are not sufficiently delocalized) for your polymer to exhibit any significant conductivity.
    Last edited: Sep 9, 2006
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