[hy07jr]

Hi!

Ive attempted to write an equation for this reaction and have got this far:

1.   Ag+ (aq) + H- (aq) → AgH (s)         (Reduction)
2.   (CH3)3CNH2.BH3 (aq) + ? → (CH3)3CNH2.BH2+ (aq) + H- (aq) (Oxidation)

I'm confused as to whether the BH3 loses a hydride anion, or just an electron - as im pretty sure it is silver and not silver hydride that precipitates.

Also, is there another substance (ie water or NO3-) that is involved in the reaction?

Some help would be much appreciated! Cheers.

[cth]

I would think (CH₃)₃CNH₂.BH₃ releases H₂ when reacting. I have never used t-butylamine borane, but I know that sodium borohydride releases hydrogen when in water:
NaBH₄ + 4H₂O  NaB(OH)₄ + 4H₂
I don't know what are your reaction conditions. Is there a solvent? Do you do the reaction in the solid state? Anyway, I think it is similar in your case.

Then, H₂ is a reducing agent. So, I would expect with silver:
2Ag⁺ + H₂  Ag_(s) + 2H⁺
So, I am not surprised you get silver metal at the end.

[hy07jr]

Cheers for that!

The t-butylamine borane was dissolved in sodium nitrate solution and warmed gently. Taking the solvent to be water, then, I have come up with this:

(CH3)3CNH2.BH3 + 3H2O   (CH3)3CNH2 + B(OH)3 + 3H2

6Ag+ + 3H2   6Ag + 6H+

I'm basically assuming that the adduct betwen borane and the amine becomes broken, leaving BH3 to become oxidised by water.

[cth]

Yeah, it sounds fine. Well done

The only thing is that I would keep (CH₃)₃CNH₂ and B(OH)₃ linked together. The bond between N and B is covalent and strong. It is named a dative bond http://www.chemguide.co.uk/atoms/bonding/dative.html. I would expect it to remain intact.