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

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Main Question or Discussion Point

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

c=c=c=c=c=c=c=c=c=c=c=c=c=c=c=c=c=c=c=c

going on for millions or even billions of repeating elements

which would be in resonance

c-c ---c-c---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
 

Answers and Replies

  • #2
Huh? What donation? And what makes you think this would be superconducting?
 
  • #3
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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?
 
  • #4
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(What's this? Another polymer thread? :tongue2:)

Bananan,
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).
 
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  • #6
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Einstein Mcfly said:
Huh? What donation? And what makes you think this would be superconducting?
delocalized pi orbitals
 
  • #7
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bomba923 said:
(What's this? Another polymer thread? :tongue2:)

Bananan,
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).
thank you how the heck did u draw in the triple bond?

the c---c-c would be in resonance with c-c---c
 
  • #8
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bananan said:
the c---c-c would be in resonance with c-c---c
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.
 
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  • #9
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bomba923 said:
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.
it is delocalized 2 pi orbitals
 
  • #10
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bomba923 said:
I don't think you understand the problem:
There is a sufficiently large energy barrier between the cumulene and polyyne resonance forms to suppress the resonance frequency below what is required for (likely) any significant conductivity.
Your electrons will remain relatively localized in cumulene (the major resonance form).

-The cumulene resonance form: ...-C-(1σ,1π)-C-(1σ,1π)-C-(1σ,1π)-C-(1σ,1π)-...
-The polyyne resonance form: ...-C-(1σ)-C-(1σ,2π)-C-(1σ)-C-(1σ,2π)-...

Can you explain why (you believe) your polymer's pi electrons should suddenly dislocate themselves from the cumulene's continuous (1σ,1π) system--with (essentially-->) a perfectly even electron density all throughout--into a neighboring (1σ,1π) bond (so as to obtain polyyne's (1σ,2π) bonds), and (relatively speaking-->) greatly imbalance the electron densities between consecutive carbons (which, unlike in the polyyne, are equivalent in the cumulene)?

Or, at least, with enough frequency for your polymer to exhibit any significant conductivity?
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.
 
  • #11
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My previous post is not in this thread?? (Anyhow, thanks for quoting it :smile:)//in post #10
bananan said:
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 π-bonds adjacent to one another would harbor delocalized electrons.
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.

bananan said:
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.
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.
 
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