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. $\endgroup$– Ivan NeretinOct 19, 2022 at 20:25
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$\begingroup$ One of them is the hydride ion who binds to C4 in the pyridine ring. And the other? $\endgroup$– user3713179Oct 19, 2022 at 20:36
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$\begingroup$ 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+. $\endgroup$– Ivan NeretinOct 19, 2022 at 20:45
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1 Answer
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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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1$\begingroup$ See also: web.mit.edu/7.01x/7.014/documents/Redox.pdf $\endgroup$– Karsten ♦Oct 19, 2022 at 22:27