Solubility comparison of hexan-1-ol and cyclohexanol in water

Is hexan-1-ol or cyclohexanol more soluble in water? The answer is supposedly cyclohexanol. However, I found a statement in a book:



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> Isomeric alcohols have solubilities in the order: primary alcohol > secondary alcohol > tertiary alcohol, *due to decrease in polar character*.
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By this statement, I guessed the answer to be 1-hexanol, as it is a primary alcohol whereas cyclohexanol is a secondary alcohol. What mistake have I done?




In general, is very difficult to predict solvation some experimental models are used such as [Abraham solvation equation](http://onlinelibrary.wiley.com/doi/10.1002/jps.21922/full).
I suppose that the mistake you have done is linked to the fact that you haven't considered one of the most important factors: the molecule structure.


In this case, you are comparing a cyclic compound with a linear one, if you could have a look at the potential surface of the two molecules you will see something like this:[![enter image description here](https://i.stack.imgur.com/JAQmP.png)](https://i.stack.imgur.com/JAQmP.png)


What makes the alcohol soluble is the hydrophilic hydroxyl group while the lipophilic chain is not. The ratio between hydrophilic and lipophilic areas is greater in cyclohexanol compared to 1-hexanol.


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First I'd have to agree with Karl's comment that 1-hexanol ($\ce{C6H13(OH)}$) and cyclohexanol ($\ce{C6H11(OH)}$) are not isomeric. So when you're compare apples to oranges you can come to unfounded conclusions. 


The gist in this case is however dependent on the "polar surface area" to the total surface area of the molecule. The molecules will have about the same polar surface area, but the cyclohexanol will have less total surface area. Thus cyclohexanol dissolves better in water. 


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