H2 elimination and addition in L-lactate dehydrogenase?
Greetings, unfortunately I'm stuck with the following question:
The following excerpt from Wikipedia is given:
The reaction catalyzed by LDH (L-lactate dehydrogenase): Pyruvate (left) is reduced to lactate (right) with the cofactor NADH. The reaction is reversible.
Which of the terms below is a correct name for the reaction taking place (forward or reverse reaction)
- H 2 elimination or H 2 addition
- Substitution or elimination
- H 2 O elimination or H 2 O addition
- Oxidation or reduction
The terms highlighted in bold are correct (H2 elimination or H2 addition, and oxidation or reduction). However, I don't understand where exactly H2 elimination or H2 addition takes place. If it were a 2H+ elimination or 2H+ addition, I could at least partially understand it, but like this…?
I was wondering whether "substitution or elimination" might be a correct answer, since H is eliminated. But it isn't replaced by another atom. So that would also be wrong.
So, I thought only the last answer was correct. Is that correct? If not, why is this an H2 elimination or addition?
Jensek81's greetings from Jensek
(3 votes)
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Count the electrons before and after. So from O=C and H-O-C-H. Once you have 4 electrons in a double bond and once 6 electrons in 3 single bonds. So you lose 2 electrons and 2 H+ particles formally from H-O-C-H to O=C. Makes formal an H2 elimination or Addition in the other direction.
Maybe the NAD / NADH has confused you too. This is the following:
NAD+ + H+ + 2e -> NADH
However, since you still have 1 H+ left, the NADH H+ reaction equation is written because you have an NADH and an additional H+.
Thank you.
Have also seen Google that there is no such thing as “2 H+ elimination” (at least I didn’t find anything about it). Then the correct term seems to have been H2 elimination. Makes sense, as you explained. Merci beaucoup!