November 25, 2024, 02:27:29 PM
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Topic: Why do london dispersion forces increase with carbon chain length?  (Read 2930 times)

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Offline Adityax26

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Hello,

I was wondering why london dispersion forces increase with carbon chain length. For instance, decanol has more L.D. forces with other decanol molecules than ethanol has with other ethanol molecules. Therefore, the intermolecular force for decanol is larger than that of ethanol.

However, why is this? I researched this and found that as carbon chain length increases (keeping everything else like position/number of OH groups constant), the molecular mass increases, which means that there will be more 'instantaneous dipoles' present due to some 'electron cloud'. I didn't quite get what all this really meant, and would love if someone could elaborate on it..

Offline mana

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Re: Why do london dispersion forces increase with carbon chain length?
« Reply #1 on: June 10, 2018, 08:25:08 AM »
there is a direct relationship btw the surface of the molecules and L.D

Offline Corribus

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Re: Why do london dispersion forces increase with carbon chain length?
« Reply #2 on: June 10, 2018, 09:28:09 AM »
Although nonpolar molecules like alkanes are on average charge neutral, sporadic, random fluctuations in electron density create temporary charge imbalances (electric dipoles). When a molecule with a short lived electric dipole is near another molecule, it creates an "induced" dipole in the nearby molecule, and the two molecules stick together a little bit. These forces are variously called London forces, or van der Waals forces. The bigger the molecules are, the more electrons they have, the more sticky they get get. Therefore there is a correlation between molecular size (related in turn to molecular weight) and the amount of intermolecular forces there are. Which is why heavier molecules - all things equal - tend to have higher boiling points.
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

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