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At my lab we perform a basic extraction of seized MDA with $\ce{NaOH}$ $\pu{2M}$ prior to GC/MS analysis. $\ce{MDA + NaOH -> MDA- + Na+ + H2O}$. This reaction should favor the products since $\mathrm{p}K_\mathrm{a}$ of MDA is about 10 and $\mathrm{p}K_\mathrm{a}$ of $\ce{H2O}$ is about 16. All of the MDA should be converted to its ionic form which is soluble in the aqueous layer. Why is there still enough of MDA in the organic layer ($\ce{CHCl3}$) to be detected by GC/MS?

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MDA is an alkaloid, i.e., basic. When you talk about its $\mathrm{p}K_\mathrm{a}$ value, you're really talking about the $\mathrm{p}K_\mathrm{a}$ of the protonated ammonium ion.

The ammonium ion is water soluble, allowing us to potentially separate it from the other organic components of the reaction. After making the aqueous solution basic, the ammonium ion is deprotonated to form the free amine, which is much more soluble in the organic phase than the aqueous phase. Then, it is back extracted into clean organic solvent. This isolates (hopefully) just the free amine in the organic layer during the extraction.

So in answer to your question, what you have in the organic layer is the free amine form of MDA, not the ion.

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