[mir]

Attached is a MS spectra of a biproduct from a reduction of 2-nitrobiphenyl to 2-aminobiphenyl in THF.
The biproduct and amine was formed in 10% yield. A large amount of the nitrocompund wasnt converted.

I did a library search, but the result didnt match at all (the library just cover 2400 spectra).

I have come to the following conclusions:
- The molecular formula is: C15H12N2 or C16H12O
- The biproduct has logically a some sort of a biphenyl structure
- The biproduct also is more polar than the amine and the nitro compund.

Could someone give me a hint? :-)

[Yggdrasil]

If the M peak is 161, wouldn't the nitrogen rule tell you that there is an odd number of nitrogens in the compound?

[mir]

If the M peak is 161, wouldn't the nitrogen rule tell you that there is an odd number of nitrogens in the compound?

You have to scroll the image. The Molecular Ion peak is 220. There is only large oddnumbered OE-peaks in the  spectra, wich rule out two nitrogen.

[Dude]

I think you should describe your conditions in more detail if this is really of interest. or if not, try another process to go from the niitro group to the amino group;.

1.  Was it a simple hydrogenation or did you use a metal (ie tin chloride) and what metal? 
2.  Was the experiment run under nitrogen and at what temperature (THF reflux?)?
3.  What are the properties of the byproduct?  Is it very dark or white crystals?  Can you run an NMR to establish if all of the carbons or hydrogens are aromatic?  Can you measure N content, mp?


The easiest explanation (although likely incorrect if no other analysis is done) would be to assume scenario 2 (C16H12O) and that the THF somehow coupled to the biphenyl along with loss of N and two H's.  This would most consistent if a metal reductant was used.  There would probably need to be an alcohol functional group (ie THF ring opening) to explain a higher polarity (presumably that is defined by TLC migration rates).