# Why can't NAD+ undergo single-electron reduction?

$$\ce{FAD}$$ can undergo single-electron reduction to form a stable radical, which can then be reduced again to $$\ce{FADH2}$$. This is supposedly possible due to resonance stability, where the unpaired electron is delocalised amongst the molecule. In $$\ce{NAD+}$$ however, the free radical formed is very unstable, so it can only undergo hydride-mediated (or two-electron) reduction.

This sounds simple by itself, but the diagram I found (shown below), in my opinion, contradicts this claim.

It seems to me that $$\ce{NAD+}$$ is just as capable in forming resonance structures as $$\ce{FAD}$$; the double bonds are adjacent to the unpaired electron and the lone pair on the carbon (in $$\ce{NAD+}$$) should be in a p-orbital, which also allows it to be delocalised. It is true that $$\ce{FAD}$$ is larger and has more aromatic rings, which makes it a better candidate to undergo single-electron reduction; however, $$\ce{NAD+}$$ should ALSO be able to form radicals, albeit not as often. Contrary to that, it is said that the $$\ce{NAD}$$ radical is very rare or even non-existent.

The source I mainly used (due to lack of others) is this Chemistry LibreTexts site. If anyone can point out the reasoning behind this oddity, I would appreciate it very much.

• The difference between “not as often” and “very rare” is semantic. There is also the issue of evolved function. NAD is not utilized by enzymes that catalyze one electron transfers – Andrew Apr 6 at 23:23