Hi,
iwhat do I get if reduce two aromatic nitro-groups under strong HCl (6 M) / SnCl2 condition to get an aromatic amine if I have a Boc proteced secondary amine in the molecule as well. I expect the Boc-protective group to be cleaved.
Let's say, I isolate the product under slightly basic condition (--> neutral molecule). If I reprotonate it (for example excess 2M HCL) do I get a 3+ ion (2x aromatic amine + secondary amine) or is just the former boc-amine group protonated?
Can anybody give me a hint? Thank you very much for answers!