1,2-dihydroxybenzene (catechol) shows intramolecular H-bonding. So due to strong bonding, it should be difficult to break the bonds in 1,2-dihydroxybenzene and therefore, 1,2-dihydroxybenzene should have a higher boiling point than 1,3-dihydroxybenzene. Why is it not so?
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asked Apr 4, 2016 at 10:08
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$\begingroup$ Related: Which among phenol and 1,2-dihydroxybenzene has the higher boiling point? $\endgroup$– user7951Commented Apr 4, 2016 at 10:14
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1$\begingroup$ The answer is in your question itself.Intramolecular hydrogen bonding will only reduce the boiling point as intramoleclar H-bond will not allow one molecule to link with another.Intramolecular H-bond two molecules are linked.For boiling up,the molecule-molecule interaction must be broken and not the bonds within a molecule. $\endgroup$– Abhishek P GCommented Apr 4, 2016 at 15:44
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$\begingroup$ @AbhishekPallipparagopakumar Then what factor decides the difference in boiling point among 1,2dihydroxybenzene and 1,3 dihydroxybenzene $\endgroup$– TheNewGenGamerCommented Apr 4, 2016 at 15:46
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1$\begingroup$ Of course as i mentioned above catechol has intramolecular H-bond whereas resourcinol has ** inter**molecular H-bond.Thus there is no interaction between two molecules in catechol.Whereas there is interaction between two molecules of resourcinol. $\endgroup$– Abhishek P GCommented Apr 4, 2016 at 15:52
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1 Answer
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Being more volatile implies that the molecules should be set free from mutual interactions. As 1,2-dihydroxybenzene has intramolecular H bonding, it is easy for it to exist independently, but for 1,3-dihydroxybenzene there might be intermolecular H bonding (interaction between two molecules); hence, it is a bit more difficult to break it apart, thus having a higher boiling point than its counterpart.
