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The boiling point of esters increases with molecular weight (and therefore, number of carbon atoms) because of more dispersion forces. But if you had two different esters with the same number of carbon atoms but different structures (e.g. methyl-ethanoate and ethyl-methanoate), what determines which one will have a higher boiling point? Does it have something to do with the number of carbon atoms in the alkyl group in comparison to the acyl group?

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  • $\begingroup$ Well, of course, as that is the only difference allowed in your question! :-/ It will depend on which one is more polar, as usual with polar organic compounds. I'll guess that a shorter alkyl chain on the carbonly side will give the higher bp. Not bets. $\endgroup$ – Karl Sep 4 '16 at 13:01
  • $\begingroup$ Honestly: This is a homework question, right? And you're too lazy to figure it out yourself? $\endgroup$ – Karl Sep 4 '16 at 13:10
  • $\begingroup$ Not at all. I tabulated the different boiling points for groups with the same number of carbons and found that, for some, the longer alkyl groups had higher boiling points and vice versa for other compounds. That's what got me thinking... @Karl $\endgroup$ – E.BELL Sep 4 '16 at 22:34
  • $\begingroup$ Why didn't you say so? Please put that table into your question! $\endgroup$ – Karl Sep 5 '16 at 8:44
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In this case you have consider the degree of branching of the isomer, this influences the strength of intermolecular forces. Think of how tree logs with lots of branches sticking out cannot stack together with each other as easily as a pile of logs with no branches. Branched chain isomers have less contact and therefore weaker intermolecular forces and lower boiling point and straight chained isomers have better contact and thus stronger molecular forces and higher boiling point.

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  • $\begingroup$ "In this case .." ?!? This is a very general comment, which is true, applicable to the problem, but does not address the original question. $\endgroup$ – Karl Sep 5 '16 at 8:46

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