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If NADPH returns to being NADP+, then that means one proton and two electrons have been incorporated into 1,3-BPG. If only one proton and one electron were attached to the carbon in 1,3-BPG (removing phosphate), I’d understand, since there’s a single covalent bond between carbon and hydrogen which establishes stable octets. But there’s another electron hanging around. What happens to this one? Or am I totally mistaken about the whole process?

This is the chemical formula for G3P:

enter image description here

The Lewis diagram below shows the top part:

enter image description here

As stated, NADPH gave off 2 electrons, yet only one is needed to bond with carbon (orange). Where did the other electron disappear to?

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  • $\begingroup$ NADPH is always accompanied by H+, and the electrons reduce as a pair. Otherwise, there will be no balance of charges. $\endgroup$
    – March Ho
    Apr 25, 2015 at 9:06
  • $\begingroup$ And after NADPH reduces 1,3-BPG, it goes back to being NADP+. It loses two electrons. In 1,3-BPG, the hydrogen bonds with carbon, contributing one electron to the single covalent bond. My question is where the another electron has gone, if it’s not in 1,3-BPG and not in NADP+. $\endgroup$ Apr 25, 2015 at 9:26
  • $\begingroup$ Sometimes the electrons react w/ oxygen to form superoxide, or react with surrounding hydrogen to form water $\endgroup$
    – CKM
    Apr 25, 2015 at 14:38
  • $\begingroup$ @Kendall So the electron does go elsewhere, and is not incorporated in G3P, correct? $\endgroup$ Apr 26, 2015 at 2:01

1 Answer 1

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Single covalent bond is a shared pair of electrons. For stable bond there should be 2 atoms with unfilled higher orbitals and at least a pair of electrons that is "distributed" among those slots.

UC Davis have hilarious page on that.

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    $\begingroup$ I have edited my question to show what I’m talking about better. A proton and one electron may be equivalent to a hydrogen atom, which can bond with a carbon atom. But there’s an extra electron that NADPH gives. What happens to it? $\endgroup$ Apr 25, 2015 at 3:34

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