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Solubility comparison of hexan-1-ol and cyclohexanol in water
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Lonnie Dinsmore
Solubility comparison of hexan-1-ol and cyclohexanol in water
In general is very difficult to predict solvatation some experimental models are used such as Abraham solvation equation.
I suppose that the mistake you have done is the fact that you haven't consider one of the most important factor: the molecule structure.
In this case you are comparing a cyclic compound with a linear one, if you could have a look to the potential surface of the two molecules you will see something like this:
What makes the alchol soluble is the hydrophilic hydroxyl group while the lipophilic chain is not. The ratio between idrophilic and lipophilic area is greater in the cyclohexanol compared to 1-hexanol.
In general is very difficult to predict solvatation some experimental models are used such as Abraham solvation equation.I suppose that the mistake you have done is the fact that you haven't consider one of the most important factor: the molecule structure.
In this case you are comparing a cyclic compound with a linear one, if you could have a look to the potential surface of the two molecules you will see something like this:
What makes the alchol soluble is the hydrophilic hydroxyl group while the lipophilic chain is not. The ratio between idrophilic and lipophilic area is greater in the cyclohexanol compared to 1-hexanol.
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.
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.
In general is very difficult to predict solvatation some experimental models are used such as Abraham solvation equation. I suppose that the mistake you have done is the fact that you haven't consider one of the most important factor: the molecule structure.
In this case you are comparing a cyclic compound with a linear one, if you could have a look to the potential surface of the two molecules you will see something like this:
What makes the alchol soluble is the hydrophilic hydroxyl group while the lipophilic chain is not. The ratio between idrophilic and lipophilic area is greater in the cyclohexanol compared to 1-hexanol.
In general is very difficult to predict solvatation some experimental models are used such as Abraham solvation equation.I suppose that the mistake you have done is the fact that you haven't consider one of the most important factor: the molecule structure.
In this case you are comparing a cyclic compound with a linear one, if you could have a look to the potential surface of the two molecules you will see something like this:
What makes the alchol soluble is the hydrophilic hydroxyl group while the lipophilic chain is not. The ratio between idrophilic and lipophilic area is greater in the 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.
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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