Redox reaction

NAD+ is getting reduced. It is recieving electrons. Think of NAD+ as a metal that has its charge, or oxidation number, being reduced from 1 to 0. And since NAD+ is being reduced, it must be oxidizing something else. Therefore, NAD+ is the oxidizer.

Where does the electron come from? The reducing agent (which gets oxidized (think -1 to 0)). In this case it is the hydride ion, H-. Which is a bit counterintuitive, I know, at physiological pH. Nevertheless, NADH and NADPH are good hydride donors, often just called electron donors by biochemists who don't want to have to think about hydrides.

 

Ahh, then I'm sure you know the roles of NADH and NADPH in reductive biosynthesis. Which in a sense the opposite of oxidative decarboxylation, or other catabolic energy producing pathways. Short hand rule is that catabolism requires oxidation and biosynthesis needs reduction. And hydride ions are great reducing agents, even us clumsy organic chemists use them. You may recall from your O. Chem. class the use of LiAlH4, lithium aluminum hydride, or NaBH4, sodium borohydride to reduce carbonyls to alcohols. I haven't my biochem textbook in front of me, but I believe the pentose phospate pathway is a good example. All those pyruvates and dihydroxy acetones and what have you have to be anabolized and their carbonyls reduced to alcohols to give you those sedheptuloses and other sugars. And as ATP is the energy currency for many pathways, in the PPP NADPH (either one or the other, I think it's NADPH) is the hydride, or reductive currency. And the other is used in all sort of places where reduction is needed, peroxide metabolism for example.

Now in organic chemist, which is what biochemistry is, to reduce something is to give it electrons. And the hydride ion is H-, basically just a proton with two extra charges. So when it bonds to a compound, the compound gets an electron (reduced) and a hydrogen atom, which is basically just a filler for organic compounds, sawdust in your hamburger, if you will. So hydride donors are often refered to as electron donors, because nobody cares about the proton. You may see why using H- in an aqueous environment has advantage over harsh reducing agents, like heavy metal anions.

If you look at your citric acid cycle, not the main structures but the little arrows coming and going, you'll probably see lot's of H+'s and e-'s and NAD+ and NADH. So you may ask how if NAD+ picks up a proton and goes from being +1 to 0, which is unfortunately how it is often shown, it's because of those electrons flying haphazardly around. It's not two electrons and a proton, it's the marvelous hydride ion.

Does that make it clearer or worse?

 

Oh, that Hydride Ion stuff makes things MUCH better!!!! I always wondered about that... NAD+ AND Hydrogen goes to NADH....But isn't Hydrogen a positive thing? Should a positive ion be reacting with a negative ion?

Now it all makes sense...this Hydrogen has 2 electrons!!!! Of course!!!

Just to clarify though: When we see NADH+ and H- reacting, in reality, is it actually reacting with a single hydronium ion, or is it reacting with a molecule that donates the hydronium? (which is just conveniently left out of the picture?)

 

A bit on terminology: cations (positive charge) usually end with the suffix 'ium' while anions end with 'ide.'

H+ is hydronium while H- is hydride. In water, H+ usually takes the form of H3O+, which can also be called hydronium, for clarity's sake.

The hydrogen atom can exist as a cation, anion, or the neutral radical. Now hydride ions are quite reactive. They're very basic, and will react with water immediately to form hydrogen and hydroxide (-OH). So it won't be floating around by it's self in a physiological environment, but will come from a donor. What ever reduces NAD+ to NADH.

So, in glycolysis it could be glyceraldehyde-3-phosphate (or g-3-p dehydrogenase, I don't know the mechanism) that donates the hydride. You'll note that it is a reducing agent here, and is getting oxidized in the process.

 

There is no reason why it has to have a double bond. Remember, each bond has two electrons, so if an hydrogen atom breaks off of a molecule, it could take both electrons with it. The above example of G-3-P oxidizing to 1,3-BPG works, essentially the negative hydride comes off of the carbonyl and is replaced by the negative oxygen belonging to inorganic phosphate. Of course, it's not that simple mechanistically, but it's still the same end result.