This question already has an answer here:
Looking at examples of these two molecular interactions makes me wonder what the differences or similarities are exactly.
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marked as duplicate by bon, M.A.R., Klaus-Dieter Warzecha, ron, Ben Norris Jul 8 '15 at 19:10
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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$\begingroup$ Welcome to chemistry.SE! If you had any questions about the policies of our community, please visit the help center. $\endgroup$ – M.A.R. Jul 7 '15 at 19:17
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$\begingroup$ Consider looking at this link chemistry.stackexchange.com/questions/10210/… . I think it may have what you are looking for. $\endgroup$ – Andy Jul 7 '15 at 19:20
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To summarize the excellent but rather lengthy and detailed Stack Exchange answer referred to by Andy above, hydrophobic interactions do not require London forces between the hydrocarbon chains. They are a result of the very strong attraction of water for itself.
I would further quote the following comment to that entry (by https://chemistry.stackexchange.com/users/5633/thomij), which I think sums it up nicely:
"This is a good answer, but the last part about van-der waals forces being required to pull the hydrophobic solvent together is incorrect. If you had particles with no attractive forces between them in a solvent that has attractive forces, as long as the enthalpy drop for phase separation is larger than the entropic penalty, they will separate. Van der waals forces help to make the enthalpy drop larger, but they aren't necessary. – thomij Jun 23 '14 at 19:02"
