# What is the product of the chemical reaction between phenol and ferric chloride?

The chemical reaction between phenol and ferric chloride is a test for the presence of phenol. They react with each other to produce a violet complex. However, the reaction is given differently in different links:

1. $$\ce{3ArOH + FeCl3 → Fe(OAr)3 + 3HCl}$$

2. $$\ce{6 ArOH + FeCl3 -> [Fe(OAr)6]^3- + 3H+ + 3HCl}$$ source

I guess the 2nd reaction is the appropriate one because it shows the formation of a complex, $\ce{[Fe(OAr)6]^3-}$. How could $\ce{Fe(OAr)3}$ be considered a complex? It should act like a ionic salt like $\ce{Na^+C6H5O^-}$ or $\ce{NaOAc}$.

So, what is the correct product of the reaction; $\ce{[Fe(OAr)6]^3-}$ or $\ce{Fe(OAr)3}$?

• Well you'd need complexation and deprotonation equilibria to describe it properly... – Mithoron Jan 3 '16 at 17:05

After doing some looking around, and going back to check what I thought I knew about coordination complexes, I think I understand why you're seeing two equations. I first recommend reading the pages on Chemguide that talk about coordination chemistry, particularly the first half of this page on acidity, and the pages on reactions with hydroxide and ammonia. Both were helpful to me in understanding an investigation I'm doing on copper (II) complexes, but the same ideas apply to 3+ compounds.

The important concepts to keep in mind are that coordination compounds (in situations like this, at least) will either undergo acid-base reactions or ligand-exchange reactions. Acid-base reactions result in neutralization and a precipitate, ligand-exchanges cause a color change. Phenol is a relatively strong acid (stronger than alcohol at least), and so is ferric chloride. I had to look around but this page says at the very bottom that all oxygen-containing compounds act as bases in the presence of Lewis acids (like ferric chloride).

$\ce{3ArOH + FeCl3 -> Fe(OAr)3 + 3HCl}$ is the acid-base reaction, and you get a ferric phenolate salt as a precipitate.

$\ce{6ArOH + FeCl3 -> [Fe(OAr)6]^3- + 3H+ + 3HCl}$ is the ligand-exchange. It's not immediately obvious, but you can rewrite the two reactions a different way and it becomes clear.

$$\ce{3ArOH + [Fe(H2O)6]^3+ -> Fe(H2O)3(OAr)3 + 3HCl}$$

$$\ce{6ArOH + [Fe(H2O)6]^3+ -> [Fe(OAr)6]^3- + 6H2O + 3H+ + 3HCl}$$

Those three $\ce{H+}$ ions are from the excess phenol. The ligand-exchange occurs only when the phenol is in excess and more water molecules can be replaced, making the iron compound ionic and soluble again.

So depending on the context, you'll see the one with three or the one with six. Both are correct, but $\ce{[Fe(OC6H5)6]^3-}$ is the new complex and $\ce{Fe(OAr)3}$ is an intermediate product.

Hope that helps.