[Krkann]

I'm trying to modify a published derivatization procedure by using a different derivatization reagent. LogP of the derivative in the publication is approx 4.07. LogP of my derivative is much lower, around 2.05. In the published procedure, a mixture of pentane-ether (1:3) was used to extract the compound from aqueous phase (both original and my derivatization is done in water).

Which extraction solvent or a mixture might be a good substitute for pentane-ether, given that my product has lower logP? The initial reaction mixture is always aqueous. Many thanks!

[pgk]

Roughly, logP = 4 means that per 10,000 molecules in the organic solvent, 1 molecule is in an equal volume of water. In contrast, logP = 2 means that per 100 molecules in the organic solvent, 1 molecule is in an equal volume of water.
In other words, if you extract with the same solvent system in equal volumes, you lose 1% of your product after one exctraction; if you extract with the same solvent system at 2/1 v/v-o/w ratio, you lose 0.5%; if you extract with the same solvent system at 4/1 v/v-o/w ratio, you lose 0.25% and so on…, which cannot be considered as high losses, especially after a couple of repeated extractions that is the common practice.
But the selection of the extraction solvent, also depends on the compatibility with the structure of the new derivatization agent, e.g. if containing a longer alkyl chain, hexane/ ether is preferred; if containing aromatic rings, extractions with toluene are preferred; if containing hydroxyl, methoxy, halide, etc. substituents, dichloromethane might be preferred, and so on...

[Krkann]

v/v-o/w ratio

Can you explain the abbreviations?

Thanks for the previous answer.

[Babcock_Hall]

What about doing a second extraction with the same solvent?  You could keep the two organic layers separate, and quantify how much material is in the second layer.

[pgk]

I apologize for the incomprehensive abbreviations.
v/v-o/w means: volume per volume of organic solvent/water layers.

[pgk]

Besides and apart the partition coefficient theory, starting materials, side products and other impurities are distributed in both phases during extractive purification, which change the solubility parameters of the solvent. But the most of water soluble impurities will be removed by the first extraction, which increases the solubility capacity of the solvent and leads to a higher overall extraction yield.

[rolnor]

Pure ether could be a start, then EtOAc if you need more polar. DCM is also very efficient in many cases.