Colorant Synthesis

Hello there once again! I have recently started a new chapter in my preparation for the Chemistry Olympiad, the azo compounds, very important for the production of azo colorants, mainly used as food colorants but also as dyes and other kinds of pigments. I was solving some synthesis problems until I got stuck at the synthesis of Naphtol Blue Black (structure shown below) from inorganic and organic substances with no more than two carbon atoms.

From the very beginning it was clear to me that I first had to make the naphthalene ring. Therefore I started from acetylene and made the necessary trimerisation in order to obtain benzene. Next, I used a Friedel- Crafts alkylation and obtained o-diethylbenzene, which through severe dehydrogenation at 400-600 ^oC yields naphthalene.

It is important now to arrive at the H-Acid structure, also called 1-Amino-8- Naphthol-3,6-Disulfonic Acid. In its synthesis it is vital to leave the formation of the amino group at the end so as not to be involved in any unwanted reaction made possible by the increased sensitivity of amines.

In its formation, firstly the naphthalene is nitrated and then trisulphonated. The third acid sulphite group in the α position is used for creating the -OH group through a process known as alkaline melting. The rest of the groups do not undergo this process because they are in less reactive β positions. Nevertheless, the other -SO3H groups suffer neutralization with NaOH and form ionic bonds between oxygen and sodium, thus replacing the hydrogen atom.

Finally the -NO2 group is reduced on iron and hydrochloric acid using 6 [H], produced by the following reactions:

Fe = Fe ³⁺ + 3 e^-

HCl = H⁺ + Cl^-
H⁺ + e^- = [H]

The H-acid is ready. Now comes the tricky part where I got stuck. You now have to do the coupling, but there is the problem of the lack of symmetry within the compound.

The preparing method for the diazonium salts to be used in the coupling reaction. Firstly a nitration followed by reduction on Fe/HCl and then diazotation with NaNO2 and HCl after the reaction:

Ar-NH2 + NaNO2 + 2HCl = Ar-N⁺≡N]Cl^- + NaCl + 2 H2O

If you couple in normal conditions a diazonium salt to our naphthalenic ring there will be two problems. First, it won't couple into the β position, α being definitely preferred by the reaction itself. Second, even if you succeed in coupling a diazonium salt into the β position, it will immediately occupy both β carbons.

But there is one thing that makes the H-Acid so suitable for this multiple coupling. On one of the rings in the naphtalenic cycle, we have phenolic -OH, while on the other we have an amino group -NH2. We can clearly see the difference, one is moderately alkaline while the other is acid.

This does matter because in the coupling we have to choose between acidic or alkaline environment. Therefore, the coupling near the alkaline -NH2 group will take place in acidic conditions whereas the coupling next to the phenolic -OH in an alkaline environment.

The first coupling in acidic conditions

The second coupling in NaOH environment and the final product

A sample of the colorant. Image from http://www.biomed.cas.cz/

Finally, the synthesis is done and we have our Naphtol Blue Black colorant. I considered this useful enough to be posted due to this special property of the H-Acid dictated by the presence of the two different groups that influence drastically the course of the reaction and because it is quite complex enough to bring in front an element of challenge. I'll also try to make it in the lab although I'm not sure I will succeed, being a quite difficult synthesis. See you next time!