# Why does the reduction of NAD+ to NADH change the absorbance so much, and cause fluorescence?

This is an updated question:

I don't have a very strong background in biophysics, but I want to understand the theory behind $$\ce{NAD+}$$ and $$\ce{NADH}$$ absorbance and fluorescence.

Background: I understand that the additional hydrogen that $$\ce{NAD+}$$ gains to form the reduced $$\ce{NADH}$$ results in the molecule being able to absorb light at $$\pu{340nm}$$. And that only the reduced $$\ce{NADH}$$ is able to produce fluorescence. This is also the property which is explored when studying, e.g., enzyme kinetics - if the enzyme binds $$\ce{NADH}$$.

Question: The oxidized form, $$\ce{NAD+}$$, still has an aromatic ring - why is it that it is not able to fluoresce? Why, in a nutshell, is the transfer of the hydrogens causing fluorescent in $$\ce{NADH}$$ and such a big change in absorbance spectrum?

Also, perhaps a bit naive question, but why is it fluorescence better to measure than absorbance?

I have tried to look for articles (the old and original) that explores these properties, but I can't seem to find any that purely speak about the spectroscopic properties.

• Feb 21, 2017 at 18:23
• In a nutshell absorption is caused by photon interacting with an electron in an occupied orbital and exciting the molecule by "transitioning" the electron to an unoccupied molecular orbital. This happens over a wide range of photon energies. When NAD is protonated the energies and bonding simultaneously change so different transitions occur. The different transitions may be different molecular orbitals entirely (new types of bonds), shifts in energies of the same molecular orbitals (same bonds), or changes in intensity of the transitions.
– MaxW
Feb 21, 2017 at 20:18