I once had an Orgo TA refer to a diazo compound as "diazo-boom-boom" (the technical term). I have always been curious as to the reason behind the instability and reactivity.

According to Wikipedia

Some of the most stable diazo compounds are α-diazoketones and α-diazoesters since the negative charge is delocalized into the carbonyl. In contrast, most alkyldiazo compounds are explosive

What is it about the alkyldiazos that makes them so much more unstable? There doesn't appear to be any bond strain or other factors. Naively, I can say that having a resonance structure should make it marginally more stable.

enter image description here

  • 2
    $\begingroup$ Diazo compounds can easily liberate Nitrogen gas, a pretty good leaving group. $\endgroup$ May 11, 2017 at 4:08

2 Answers 2


Well, your question is equivalent to “what is it about α-diazoketones that makes them so much more stable?”, which is easier to see. Compared to an alkyldiazo, the α-diazoketone has a resonance structure in which the negative charge goes to the ketone’s oxygen (and far away from the positively-charged nitrogen atom):

resonance structures

Because the oxygen is a quite electronegative element, the resonance form is quite stable and explains the extra stability of α-diazoketones. It is for the same reason that the protons in position α to the ketones are always more acidic than alky chain protons.

Coming back to alkyldiazo compounds, you have to realize that merely being able to write a resonance structure does not intrinsically imply stabilization: the resonance structure has to have some intrinsic stability factor. In the alkyldiazo, the resonance form you wrote is a carbanion, which is considered quite unfavourable unless it has a further stabilizing factor. Moreover, the most common reactions gives N2, which is a very stable compound… the reaction is thermodynamically very favourable.

  • 1
    $\begingroup$ @jonsca: from an entropy viewpoint, the production of a gaseous product like nitrogen is very favorable indeed (what could be more "chaotic" than a gas?). That's why your diazo group does quite well as a leaving group... $\endgroup$
    – user95
    May 5, 2012 at 15:57
  • $\begingroup$ @Fx Since you only mentioned it, I explained a bit more why $\ce{N2}$ is more stable at STP. $\endgroup$
    – CHM
    May 5, 2012 at 17:59
  • $\begingroup$ @JM Chaotic isn't the right word. 1 mole of $\ce{N2}$ in the gas phase in a 1L volume simply has more populated energy levels than the same amount of $\ce{N2}$, but frozen. In other words, there are more microstates associated with its macrostate, than when frozen, say. $\endgroup$
    – CHM
    May 5, 2012 at 18:02

Very short answer:

$\Delta G = \Delta H - T\Delta S$

As this is an org-chem question, no need to solve the equation numerically. As you might know, $\ce{N2_{(g)}}$'s standard formation enthalpy ($\Delta H_\text{f}^\circ$) is 0 – at standard $T$ and $p$, $\ce{N2_{(g)}}$ is gaseous. Therefore, its enthalpic contribution is 0. Its $S^\circ$ is $191.32\ \mathrm{\frac{J}{mol\ K}}$, which indicates that, at 273.15 K, $\Delta G^\circ_\text{f}$ is not only negative, but also large.

Thermodynamically, $\ce{N2_{(g)}}$ is favored. Take a look at nitrogen's phase diagram, generated by Wolfram Alpha.


I've just noticed you're considering diazonium ions, not azo compounds. For the reasons F'x explained, the ions are more prompt to explosion than are the azo compounds. I think azo compounds are used in the fabrication of CDs, or so my teacher told us... They're supposedly light sensitive.

  • $\begingroup$ Well, I wouldn't say anything about the sign of the reaction from a single formation Gibbs free energy… you don't need to show that N2's $\Delta_f G$ is negative, but write a reaction free energy, which is much harder. $\endgroup$
    – F'x
    May 5, 2012 at 18:36
  • $\begingroup$ well, I think as currently written it only generates confusion… also, the basis for your calculation that N2 is stable is that N2 is the reference state of nitrogen, which it is because it is more stable… Well, I've given my opinion, I'll leave it at that. $\endgroup$
    – F'x
    May 5, 2012 at 18:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.