July 08, 2020, 07:34:21 AM
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Topic: Why placing in ice bath decrease compound's affinity to aqueous layer?  (Read 1226 times)

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Offline K.z_z

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It was emphasized to do extraction of diisopropylamine with NaOH in an ice bath because lower temperature decrease the compound's affinity to the aqueous layer. Can someone explain the relationship between this? Also there was gas formation when adding in NaOH, is that hydrogen? Thank you.

Offline Esrevinu

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Re: Why placing in ice bath decrease compound's affinity to aqueous layer?
« Reply #1 on: February 15, 2015, 02:09:25 PM »
What do you know about the relationship between solubility and temperature. Is it a linear relationship? Does it depend on the solute and solvent?

At what stage did you add the NaOH and what were you adding it too? What are all of the components? Obviously NaOH is a base. What is dipa?

Offline K.z_z

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Re: Why placing in ice bath decrease compound's affinity to aqueous layer?
« Reply #2 on: February 15, 2015, 11:47:51 PM »
I added NaOH to an aqueous solution that contains protonated diisopropylamine ion. So there was an acid base reaction happened that cause diisopropylamine (DIPA) to return to its basic form and then form its own organic layer. There was heat and gas released. The aqueous layer was cloudy which means there was still DIPA dissolved in it and then then turned clear. Now I am thinking it's just the concept of lowering temperature in general decrease solubility, so when cooled on ice solubility of DIPA in the aqueous solvents decreases. Am I right?

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