How to prepare pentabromobenzene from aniline?
1 Answer
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
- 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. - 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. - 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. - 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. - 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" . - 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.