Sciencemadness Discussion Board

H2 and negatively charged molecules

Electra - 10-1-2014 at 13:49

Excuse the basicness of this question. A concept so simple I don't hear it discussed much.

If there is a molecule that gets reduced in a solution, and it is to be hydrogenated by only one hydrogen molecule, if it comes into contact with an H2 molecule in the solution, will one of the Hydrogen leave to hydrogenate the negatively charged molecule? I've read the H2 bond is very weak, but is it weak enough to be snatched away by a just reduced molecule, leaving the other hydrogen ion floating in the solution?

I would assume so but I wanted to get a second opinion. This does seem to be the basis behind all reduction/hydrogenation reactions. Many reduction reactions release more than one hydrogen at a time, and it is inevitable that some of those hydrogen will bond together, but that bond quickly breaks in the presence of an negatively charged molecule, correct?


[Edited on 10-1-2014 by Electra]

Zyklon-A - 10-1-2014 at 20:53

H2 is a molecule, H is just an atom.


Electra - 11-1-2014 at 10:36

I realize that... What I was asking is whether or not an H atom(proton) can be easily liberated from an H2's covalent bond in the presence of a molecule that has just been reduced/gained an electron. Or is it necessary for the H atom/proton to already be separated in the H+ form before it can be accepted. I would think the attraction forces on the reduced molecule would need to be stronger than the H2 covalent bond, though, how likely is this to happen in most cases?

leu - 11-1-2014 at 12:06

You can use:

http://www.sciencemadness.org/talk/search.php?fid=12

to find answers to such questions on this web site ;) You could also read articles on the internet encyclopedia:

http://en.wikipedia.org/wiki/Organic_reduction

http://en.wikipedia.org/wiki/Hydrogenation

and

http://en.wikipedia.org/wiki/Transfer_hydrogenation

so you might gain better understanding of what you're inquiring about :P The end results from the effort applied :cool: