Question

How to prepare pentabromobenzene from aniline?

Answer 1

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 `"FeBr"_3` catalyst is required, and the three `"Br"` atoms add easily.
2. The next `"Br"` needs to go on meta to the `"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 `"Br"` to get on.
3. We attempt to get a `"Br"` onto carbon-3 or 5 whichever wants to occur. Either one could occur. This time we do need an `"FeBr"_3` catalyst.
4. This is the first step to removing the `"NH"_2`. We wanted to try limiting steric clutter to maximize the chances of a fifth `"Br"` atom making it on. So, we add some acid and hydrolyze the amide bond back into `"NH"_2` and a carboxylic acid side product.
5. The first step here is the conversion of `-stackrel(..)("N")"H"_2` into `-stackrel(+)"N"-="N":`, a diazonium cation. The second step removes it completely, using hypophosphorous acid to replace it with an `"H"`.
6. The last step would be a hopeful `"Br"` EAS onto one of the remaining carbons. It doesn't matter which one since it's a symmetric molecule.