In many organic and inorganic molecules which contain a nitro group, the bonding takes place through the nitrogen, which gives the nitrogen "four" bonds, but the $\ce{N-O}$ bonds are delocalized across both oxygens which is obviously quite stabilizing.

That being said, I can't help but notice that if the nitro group bonded through the oxygen like $\ce{R-O-N=O}$, then there is not the same situation where nitrogen has over three bonds and the oxygens only have about 1.5 bonds. Additionally, the $\ce{C-O}$ bond is about $360\ \mathrm{\frac{kJ}{mol}}$ compared to about $310\ \mathrm{\frac{kJ}{mol}}$ for the $\ce{C-N}$ bond.

That means that whatever the explanation is, it needs to make up around $50\ \mathrm{\frac{kJ}{mol}}$ in bond energy. I believe the other bonds present should be pretty similar as far as energy goes.

Finally, when the nitro group bonds through the oxygen, presumably there will be less steric hindrance because the group would be roughly linear. I could imagine that assumption being incorrect though.

I imagine that the answer (when bonded to an aromatic molecule) will have something to do with conjugation of the nitrogen lone pair into the pi system. I would think, however, that the oxygen would have a similar effect.

Looking forward to answers.

  • 3
    $\begingroup$ See, it's not like the nitro group just hops around, thinking "which way should I bond today". The molecules are the way they are, and can't easily change into another form. (In cases when they can, they do.) There are nitro compounds $\ce{R-NO2}$, and there are organic nitrites $\ce{R-O-N=O}$; these are isomers, each with its own unique properties, both relatively stable, synthesized in different ways. $\endgroup$ – Ivan Neretin Apr 4 '16 at 9:30

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