This fact was given as an explanation as to why branched isomers of a compound have lower boiling points.
How are branched isomers less polarisable and how does this relate to London dispersion forces?
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$\begingroup$ Branched isomers have less surface area exposed, thus have weaker Van der Walls forces of attraction. $\endgroup$– ZenixMar 8, 2020 at 20:23
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1$\begingroup$ I was itching to mark this as "homework", but the answer is not so perfectly clear imo. First it should be defined what "polarisable" means exactly. $\endgroup$– KarlMar 8, 2020 at 20:49
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2$\begingroup$ Does this answer your question? Why do branched chain compounds have lower boiling points than the corresponding straight chain isomers? $\endgroup$– MithoronMar 8, 2020 at 22:13
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$\begingroup$ @Karl This is what I don't understand- the definition of 'polarisable'. I was thinking that 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? $\endgroup$– XXb8Mar 9, 2020 at 12:39
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1$\begingroup$ The point is that they might be more or less polarisable but that is not the reason, at least not so easy to be immediately seen. Also in this context as Karl said it is not that clear what is meant... The point is even not the WdW forces to be weaker per se, but their cumulative effect being weaker. Indeed see the answer linked by Mithoron. $\endgroup$– AlchimistaMar 10, 2020 at 8:10
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