$\ce{FAD}$ 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.

fad nad

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.

  • $\begingroup$ 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 $\endgroup$ – Andrew Apr 6 at 23:23

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