I have to obtain 1-nitro-2,4,5-triaminobenzene from 1,3-dichlorobenzene. I have thought of:
- Reaction with $\ce{Mg}$
- Rc with $\ce{CO2}$, then $\ce{H2O}$. By this point I have izophthalic acid.
- $\ce{PCl5}$, then $\ce{NH3}$. I have formed the diamide.
- Reaction with $\ce{NaOBr}$, which turns the $\ce{-CONH2}$ groups into $\ce{-NH2}$. (Hoffmann degradation) now I have 1,3-phenilendiamine.
- Rc with $\ce{Ac2O}$, then nitration, reduction with $\ce{Fe + HCl}$, and again, $\ce{Ac2O}$.
- Nitration and then, getting rid of $\ce{AcO-}$ groups with $\ce{H2O}$. Now I should have reached the target molecule.
I am dissatisfied with this synthesis, though. It is a bit too long and I feel like I miss something. I have also thought of nitration or aromatic nucleophilic substitutions with $\ce{NaNH2}$ as the first step of another possible synthesis, but I can't reach the end.
Is the synthesis I have thought of viable? Should I try sth different?
Thank you in advance.
LE: my bad, the target molecule is 1-nitro 2,4,5-triaminobenzene. Thank you orthocresol for pointing out a few things.
LLE: Thanks for your answer, Klaus, good work. You just gave me another idea for a synthesis! Even though there is something I don't like about it.
-double nitration, then rc with 2 moles of $\ce{NaNH2}$. I should have 1,3-dinitro-4,6-diaminobenzene by doing so.
-Zinin reduction, which will transform one $\ce{-NO2}$ group into an $\ce{-NH2}$ one (it does not matter which).
I wonder if 1,3-chlorobenzene after the rc with the first mole of $\ce{HNO3}$ will react with another mole of $\ce{HNO3}$, since it will be very deactivated. It looks too simple to be good.