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I realized a reaction of mixed aldolisation between acetone and 2-nitrobenzaldehyde, using a solution of 10% NaOH in ethanol.

I obtained a mixture of products at the end of the reaction. The problem is when I added the ethyl acetate I saw a blue liquid mixed with brown solid. After a night of this reaction I found a lot of white crystals mixed with the blue liquid.

Can someone explain the presence of this blue liquid with the brown solid, knowing that the crystals are soluble in ethanol?

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Well, actually you've made a reaction called the "Baeyer–Drewson indigo synthesis" (original article here), which was route developed in the nineteen century to obtain the indigo dye, nowadays replaced by other methods. I'm not writing the mechanism for this reaction because it is quite long and well-documented elsewhere.

Your blue compound is therefore the indigo dye represented here. As for the brown compound, it's difficult to say without more data (NMR spectra for instance), but the synthesis can give many colored intermediates or by-products such as isatin (bright red).

indigo

I hope this helps.

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In the aldol reaction, we would expect to come up with the following product:

enter image description here

However, this reaction is bound to happen poorly. This is due to the presence of the nitro group, which has so much electron-withdrawing power that you can consider it to be twice as powerful as a carbonyl group (although the nitro group is not as powerful here due to its attachment to the benzene ring). Hence the nitro group then subsequently attacks the ketone in many intramolecular reactions, an example is shown here:

enter image description here

Since there are so many aromatic compounds that are fully conjugated forming, these should all contribute to the blue colour that you are getting.

I believe the brown solid is the result of some polymerization of the ketone with either the aldol or the nitro group.

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  • $\begingroup$ I also have to add that the product I drew happens to be an activated form of pyridine, which is susceptible to electrophilic substitution as well. Hence, electrophiles such as another nitro group can easily attack the pyridine to form polymers. $\endgroup$ – Linus Choy Feb 1 '17 at 8:17

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