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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.

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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.

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  • $\begingroup$ Related: Why does NADH have 2 peaks in its absorption spectrum but NAD+ has only one? $\endgroup$
    – Mithoron
    Feb 21, 2017 at 18:23
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    $\begingroup$ 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. $\endgroup$
    – MaxW
    Feb 21, 2017 at 20:18

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Essentially NADH has an additional pair of electrons. These additional electrons which are oftein drawn as a lone pair of electrons alters the molecule in a similar way to the way in which many pH indicators change upon protonation.

Here are diagrams of the reduced and oxidized forms of NADH / NAD+

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A typical indicator when protonated appears to have a colour, however upon deprotonation the indicator loses a hydrogen and gains a lone pair of electrons now appearing to have a different colour. The addition of the lone pair of electrons actually lowers the transition energy between the pi bonding and the pi anti-bonding states.

For example for methyl red it can be predicted that the unprotonated form has a HOMO/LUMO gap of 2.175 eV (according to extended Huckel calculations). For the protonated form of methyl red has a HOMO/LUMO gap of 3.275 eV.

This lowered energy correlates with a decreased absorbance energy, ultimately reflecting a higher energy wave of light (Frequency of Blue > Red). NADH's absorbance wavelength his greater than NAD+'s absorbance wavelength, because NADH is absorbing lower energy light. This, like pH indicators, is due to the lone pair lowering pi and pi anti-bonding energy state differences.

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