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Objectively, what makes the solubility of $\ce{C5H11OH}$ greater than the $\ce{C5H11Cl}$ in water?

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I believe the $\ce{C5H11OH}$ is more soluble because it has dipole dipole, H-bonding and London Dispersion. The $\ce{C5H11Cl}$ only has dipole dipole and London Dispersion. The H-bonding is highly soluble in water

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  • $\begingroup$ Yes you perfectly right, $\ce{Cl}$ never does H-bonding ;) $\endgroup$ – ParaH2 May 7 '15 at 22:20
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In the case of the water solubility what you want to know is how powerful the secondary interaction of the compound that you are going to dissolve with water?If the compound that you wish to dissolve have hydrogen bonding then it's definitely dissolve in water .It does not mean others are insoluble in water.

But the compounds do not have hydrogen bonding are less soluble in water when compared to the ones which have hydrogen bonding.Because ones which have hydrogen bonding interact with water molecules easily than others which do not have hydrogen bonding

Now take a look at the image below,

enter image description here

It shows two water molecules which posses the hydrogen bonding interaction.

Now give the eye on compounds that you have got,

The compounds that you going to dissolve may form interaction force with water as above image. If that interaction force is hydrogen bonding ($\ce{C5H11OH}$) then it's solubility is greater than ones which do not have hydrogen bonding($\ce{C5H11Cl}$) (dipole dipole interactions)

So because of that solubility of $\ce{C5H11Cl}$ is less than the solubility of ($\ce{C5H11OH}$) with water.

Then the bottom line is ,the solubility of a compound is vary according the secondary interaction between the solute and solvent.In the case of water acting as the solvent solubility is highest in compounds which having H bonding , than which haaving dipole-dipole forces and weak dispersion forces.

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