My notes say that hydrogen bonding occurs between a hydrogen atom covalently bonded to a highly electronegative atom ($\ce{N, O, or F}$) and the lone pair of a highly electronegative atom of another molecule. I am wondering if hydrogen bond would form between a molecule with hydrogen atom covalently bonded to $\ce{N/O/F}$ and a molecule with $\ce{N/O/F}$ that has lone pair but is not bonded to hydrogen. For example, it seems like oxygen, ozone, carbon dioxide are soluble in water so they can form hydrogen bond with water though the $\ce{O}$ atom is not bonded to $\ce{H}$. How about $\ce{CH3CH2COCH2CH3}$, the $\ce{O}$ atom has a lone pair, but it seems like it is not soluble in water and cannot form hydrogen bonding with water.

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    $\begingroup$ Why, it can. It's just that these hydrogen bonds are not enough to overcome the influence of the hydrophobic part. $\endgroup$ Dec 30, 2015 at 8:25

1 Answer 1


The molecule your example, 3-pentanone, actually has a water solubility of $\pu{50 g/L}$ at $\pu{20^oC}$. The water solubility of it's pure hydrocarbon analog, n-pentane, is $\pu{40 mg/L}$ at $\pu{20^oC}$. A simplistic interpretation of this comparison is that the addition of a single hydrogen-bond acceptor functionality to n-pentane increases it's water solubility by greater than three orders of magnitude! So the answer to your question is pretty clearly that yes, hydrogen bonding is happening in this scenario.


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