# Determination of brominated aromatics by 13C and 1H NMR spectroscopy

How can I identify 4-bromoazobenzene, 4-bromoazoxybenzene, and 4,4’-dibromo-$\ce{N,N}$’-diphenylhydrazine products using $^1\ce{H}$ and $^{13}\ce{C}$ NMR?

I’m really confused by the general shape of NMR peaks of aromatic protons. I want to distinguish each product from the other products, knowing that the three products are synthesized starting by the reaction of 4-bromonitrobenzene with sodium borohydride.

• Do you want to distinguish between those substances in a single mixture or one by one? – pH13 - Yet another Philipp Aug 14 '15 at 0:20
• I doubt the peak shapes are going to help you, since these are all para-substituted aromatics. You could probably distinguish by chemical shift. Try to find authentic spectra of each. They should be available. – jerepierre Aug 14 '15 at 0:21
• one by one in itself – Nado Aug 14 '15 at 19:07

## 1 Answer

I don't find any experimental spectra online, but I have been suggesting NMRDB.org to students and colleagues. I don't know how accurate the predictions are, but the site is easy to use and for the compounds you're suggesting, rule-based methods should be fairly reliable.

In the 4,4’-dibromo-N,N’-diphenylhydrazine, you should have two sets of "double doublets" (i.e., four peaks in each set of equal height).

In the 4-bromoazobenzene, you break symmetry, so there should be four signals which integrate to two protons, and one additional signal integrating to one proton (i.e., all ~7.3-7.8 ppm, so a big mess).

In the $\ce{^1H}$ spectra of 4-bromoazoxybenzene, the protons closest to the azoxy -N+-O- should be deshielded a bit (maybe ~7.8-7.9 ppm?) compared to the aza analogue. Certainly it's also going to be a fairly congested spectrum.

I'll leave the $\ce{^13C}$ NMR as an exercise (or to others to answer).