Chemical Forums
Chemistry Forums for Students => Organic Chemistry Forum => Topic started by: XXb8 on March 11, 2020, 01:33:20 PM
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I thought it was because even if there was an induced dipole in a branched isomer, all the surrounding branches would experience an oppositely induced dipole, so cancelling out the overall induced polarity of the isomer. This would not happen in a straight-chain isomer, because there are no branches to provide an opposite dipole. Is this wrong?
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You should be more specific concerning isomers - eg. both n-octane and isooctane (2,2,4-trimethylpentane) show dipole moment equal to 0.
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Why do you think branched isomers are less polarizable?
This paper says otherwise:
https://www.brown.edu/research/labs/mittleman/sites/brown.edu.research.labs.mittleman/files/uploads/NickelCPL.pdf
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A molecule is polar when possessing dipole moment.
But the magnitude of the molecular dipole moment depends on the distance between the positive and the negative charge, which is shorter in branched molecules.
Besides, dipole moment is a vector unity and strongly depends on the axis angle between charges.
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OP is talking about polarisability, not polarity.
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Thank you for the clarification.
Anyway, polarisability is the ability of temporal polarity that is induced by an external electric field. As a consequence, polarisability depends on the same (or at least, similar) factors as polarity, as long as the electrostatic induction lasts.