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Why is cis isomers polar but not trans isomers? watch

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    Like for example, a cis-dichloroethene and a trans-dichloroethene.
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    (Original post by CookieHero)
    Like for example, a cis-dichloroethene and a trans-dichloroethene.
    Are you sure that's right? I mean, I don't know, I wasn't explicitly taught this but I don't see why E-dichloroethene wouldn't be polar? It's just that there would be more than one pole?
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    Polarity is due to a net dipole in the molecule.

    Consider Cis-1,2-dichloroethene.
    Both Cl atoms are locked on the same side of the molecule. As you know from electronegativity, both of these bonds will be polar. When you look at the polarity, you will see that the vectors reinforce one another perpendicular to the double bond. Hence, there is an overall dipole.

    In the trans version, you see the polarity vectors cancelling one another out, hence there is no overall dipole.
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    (Original post by Plagioclase)
    Are you sure that's right? I mean, I don't know, I wasn't explicitly taught this but I don't see why E-dichloroethene wouldn't be polar? It's just that there would be more than one pole?
    What is it that you think a dipole is?
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    (Original post by Infraspecies)
    What is it that you think a dipole is?
    Wouldn't there be a quadrupole in the trans isomer?
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    (Original post by Plagioclase)
    Wouldn't there be a quadrupole in the trans isomer?
    No.

    The electrical dipole moment is calculated as the sum of all the vector products of the partial charges with the displacements of all nuclei in the molecule.
    If you will:

    iSumN (qi x ri)

    This yields a vector answer, as is wanted; the electric dipole moment is a measure of net charge distribution.


    Electronic quadrupoles are not what you think they are.
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    (Original post by Infraspecies)
    No.

    The electrical dipole moment is calculated as the sum of all the vector products of the partial charges with the displacements of all nuclei in the molecule.
    If you will:

    iSumN (qi x ri)

    This yields a vector answer, as is wanted; the electric dipole moment is a measure of net charge distribution.


    Electronic quadrupoles are not what you think they are.
    That diagram you linked earlier in the thread does seem to make sense. So there would be no local effects at all resulting from the electronegativity of the chlorine?
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    (Original post by Plagioclase)
    That diagram you linked earlier in the thread does seem to make sense. So there would be no local effects at all resulting from the electronegativity of the chlorine?
    The C-Cl bond still is polar, yes. It serves to make the allyl carbons even more susceptible to nucleophiles.
    However, there is no net molecular dipole.
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    (Original post by Infraspecies)
    The C-Cl bond still is polar, yes. It serves to make the allyl carbons even more susceptible to nucleophiles.
    However, there is no net molecular dipole.
    Would the δ+ end of a water molecule still be attracted to the δ- chlorine (and vice versa for the carbon)?
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    (Original post by Plagioclase)
    Would the δ+ end of a water molecule still be attracted to the δ- chlorine (and vice versa for the carbon)?
    If you're asking whether there is a charge density difference the Cl-Cl diagonal and the H-H diagonal, then yes, there is. And this is a quadrupole, yes (I had taken you to mean, by "quadrupole" the existence of two relevant opposing dipoles having an effect). However, the contextually relevant part of this is that it does not generate a molecular solvation effect based on the polar C-Cl bonds. Especially not as in the cis isomer.
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    (Original post by Infraspecies)
    If you're asking whether there is a charge density difference the Cl-Cl diagonal and the H-H diagonal, then yes, there is. And this is a quadrupole, yes (I had taken you to mean, by "quadrupole" the existence of two relevant opposing dipoles having an effect). However, the contextually relevant part of this is that it does not generate a molecular solvation effect based on the polar C-Cl bonds. Especially not as in the cis isomer.
    Okay, thank you very much!
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    (Original post by Plagioclase)
    Okay, thank you very much!
    To labour the point, as I so love to do, the boiling point of trans-1,2-dichloroethene is some 30 degrees lower than that of the alternate geometric isomer.
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    Because, Caitlyn Jenner.
    Spoiler:
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    sorry. the voice in my head made me do it.
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    Molecules with a centre of inversion cannot posses a permanent dipole
 
 
 
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