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#1 2010-10-05 13:01:33


Describe the electron transfer steps in forming sodamide

How did we obtain negative a hydrogen in the following:

       2Na + 2NH3 -> 2NaNH2 + H2(g)

How can sodium pull H2 from NH3? In general, H(-) is difficult to get in organic chemistry. I think Na added an electron pair to NH3, the molecule was unstable, it ejected H(-). It appears that in the same way that NH4(+) can give NH3 + H(+), NH3(-) can eject H(-). CORRECT ME IF I AM WRONG> adebayo

This is a very good question. I think there are two possible answers. Let us start with a single electron transfer reaction.

In the bromination of isobutane, the longest bond (which are the most available electrons) accepts an electron. This must give an intermediate with three electrons which decomposes to HBr and (CH3)3C radical. I cannot exclude a direct attack on a proton by an electron, but two things suggest that is not the case. One, if an electron were to attack a proton, then the proton most easily attacked should be the more acidic hydrogen (the longest proton-electron pair distance). That would correspond with the shortest bond. A bond becomes shorter as the electrons are pulled away from carbon. Acetylene has short C-H bonds and are more acidic. Secondly, I read somewhere that the trajectory of attack on radical reactions with C-H bonds is to the electrons. That makes sense so the most available electrons should be attacked most rapidly. That is, if carbon is an electron donor, then it should release electrons and the electrons attached to the tertiary carbon are the most easily attacked.

If sodium donates single electrons, then a single electron could be added to the non-bonded electrons or one of the N-H bonds. If it were to the N-H bond, this could decompose to either NH2 + H(-) or NH2(-) + H. In the single electron transfer reaction, the latter is probably the more stable. An H could either accept an electron from sodium to form Na(+) and H(-) or it could react with NH3 to give NH2 + H2.

An alternate mechanism could be a two electron transfer process. If 2Na donated 2e(-), then a direct reaction of two electrons would occur on the hydrogen of an N-H bond. Remember, the charge of a hydrogen is unchanged. It is always positive. This reaction could be equivalent to a proton transfer between HCl and NaOH. The electrons of HO(-) are donated to the hydrogen of HCl to give H2O and Cl(-). If that same reaction occurred with NH3, then NH2(-) and H(-) would be produced. The H(-) would react further with another NH3 to give a second NH2(-) and H2.

I cannot say that there is overwhelming evidence or arguments that I can think of that can eliminate one or the other. That is, there certainly are single electron transfer reactions as the bromination reaction shows. I have searched for evidence of radical products in the Birch reduction (Li/NH3) of cumene. The cumyl radical is notably stabilized and thus one might anticipate some products occurring from it if a single electron transfer mechanism were in operation. (I would not be able to distinguish between two rapid single electron transfer reactions that lead directly to a dianion and direct addition of two electrons leading to the same intermediate.) I thought if a single electron transfer radical anion formed and it gained a proton to form a radical, then I ought to find other products if that radical can couple or be involved in other radical reactions. From literature reports, it does not appear that the Birch reduction of cumene gives any reportable amount of radical products.

The driving force for single electron transfer reactions appears to form stable electron pairs. If that is the driving force, it is difficult to eliminate electron pair reactions that follow a path that simply leads to the weakest base as the product.


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