# How to prepare pentabromobenzene from aniline?

Oct 23, 2016

I can't think of an easy synthesis for this; Ernest Z. suggested more or less what I would have done, but noted that sterics would play into making the yield quite poor.

EDG = electron-donating group
EWG = electron-withdrawing group
wrt = with respect to
EAS = electrophilic aromatic substitution

1. The amine group is an EDG, so it activates the aromatic ring wrt EAS, allowing the ring to behave as a nucleophile (remember, EAS is where benzene reacts with an electrophile, but isn't an electrophile itself). That means no ${\text{FeBr}}_{3}$ catalyst is required, and the three $\text{Br}$ atoms add easily.
2. The next $\text{Br}$ needs to go on meta to the ${\text{NH}}_{2}$, so we have to transform it into a meta-director. One way to do it is to perform a Friedel-Crafts Acylation to generate a protecting group.
This becomes an EWG, which deactivates the ring wrt EAS. It would be a meta-director so that helps. Despite that, having the four EWGs on the ring makes it difficult for another $\text{Br}$ to get on.
3. We attempt to get a $\text{Br}$ onto carbon-3 or 5 whichever wants to occur. Either one could occur. This time we do need an ${\text{FeBr}}_{3}$ catalyst.
4. This is the first step to removing the ${\text{NH}}_{2}$. We wanted to try limiting steric clutter to maximize the chances of a fifth $\text{Br}$ atom making it on. So, we add some acid and hydrolyze the amide bond back into ${\text{NH}}_{2}$ and a carboxylic acid side product.
5. The first step here is the conversion of -stackrel(..)("N")"H"_2 into $- \stackrel{+}{\text{N"-="N}} :$, a diazonium cation. The second step removes it completely, using hypophosphorous acid to replace it with an $\text{H}$.
6. The last step would be a hopeful $\text{Br}$ EAS onto one of the remaining carbons. It doesn't matter which one since it's a symmetric molecule.