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The $\ce{NAD^+}$ molecule has an electron-poor carbon susceptible to a nucleophilic attack by a hydride ion: It's the C4 of the pyridine ring, who loses its aromaticity in this way. Usually, it is written that $\text{NAD}$ + is reduced with the acquisition of two hydrogen atoms, so the reduced form of $\text{NAD}^+$ would be $\text{NADH} + \ce{H}^+ $. But the $\ce{H}^+$ ion, where does it come from?

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  • $\begingroup$ It comes from one of the two hydrogen atoms that you just mentioned. $\endgroup$ Oct 19, 2022 at 20:25
  • $\begingroup$ One of them is the hydride ion who binds to C4 in the pyridine ring. And the other? $\endgroup$ Oct 19, 2022 at 20:36
  • $\begingroup$ You just said it all yourself. We are given two (2) neutral hydrogen atoms. One becomes hydride ion and goes to NADH. Another becomes H+. $\endgroup$ Oct 19, 2022 at 20:45

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Here is the oxidation half reaction:

$$\ce{NADH -> NAD+ + H+ + 2e- }$$

When you balance the half reaction, you put a hydrogen ion on the product side, and this would go into aqueous solution.

And here is the full redox reaction with formaldehyde reacting to methanol (small molecules for simplicity; you could also use pyruvate and lactate):

$$\ce{NADH + HCHO + H+ -> NAD+ + CH3OH}$$

Notice that the $\ce{H+}$ appears on the opposite side. When NADH reduces a carbonyl, it provides the hydride anion, and the hydrogen ion (i.e. cation) comes from the solution.

The reaction with molecule hydrogen would be written like this (but I am not sure if it is observed in nature - it is certainly not part of the primary animal or plant metabolism):

$$\ce{NADH + H+ -> NAD+ + H2}$$

Again, NADH provides the hydride, and the aqueous solution provides the hydrogen ion.

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