# What makes diazo compounds so unstable and explosive?

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.

• Diazo compounds can easily liberate Nitrogen gas, a pretty good leaving group. – Pritt Balagopal May 11 '17 at 4:08

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):

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.

• @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... – user95 May 5 '12 at 15:57
• @Fx Since you only mentioned it, I explained a bit more why $\ce{N2}$ is more stable at STP. – CHM May 5 '12 at 17:59
• @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. – CHM May 5 '12 at 18:02

$\Delta G = \Delta H - T\Delta S$
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.
• 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. – F'x May 5 '12 at 18:36