Imine/Amine with NaBH4/H3O+ question w/ pics

[rfig08d]

Please note: this is NOT a homework question. I am reviewing my notes for next week's quiz and I need some clarification.

Hi everyone! I'm a little bit stumped on this as I cannot see the reaction!

Does anyone know a useful link for imine to amine reduction mechanism using sodium borohydride and hydronium ( 1) NaBH4 2) H3O+ ) as reagents? Specifically this mini-synthesis (in screenshot). I know how NaBH4 works with most compounds, but for some reason I cannot see how the nitrogen obtained another hydrogen.

[PLAIN]http://img829.imageshack.us/img829/720/imine.jpg
(Ignore the mechanism, look at the reactions below it)
How does the final product obtain the hydrogen attached to the nitrogen? What's the 1st step without hydronium acting on it?
[PLAIN]http://img832.imageshack.us/img832/5369/imine2.jpg

I hate to ask, but a picture would be 100% more helpful than words right now. If I can see it once, then I have it set... so if anyone has a link or can draw a quick mechanism (doesn't have to be detailed, I just want to see how the hydrogen attaching on the nitrogen gets there), I would REALLY appreciate it. I only need 1 reaction explained, I simply provided the other ones as extra.

Thank you!

 

[DDTea]

Well I can't draw a picture right now, so I hope you can settle with words. Draw a picture to follow along, you'll learn it better if you're comfortable drawing it. In water, Sodium Borohydride (NaBH₄) dissociates according to the following reaction:

NaBH₄ ---> NaBH₃⁺ + H^-

So now you've got your hydride acting as a nucleophile. It's the same picture as all the other nucleophilic attacks on carbonyl groups: H- bonds to the carbonyl group's carbon, shifting electron density up to the nitrogen atom. There is now a lone pair on the nitrogen atom, giving it a formal negative charge. But remember, you've got NaBH₃⁺ in solution still; as such, it complexes with the formally negative nitrogen atom.

The first step of the reaction is now done, but the product is not the amine that you are after! To liberate the amine, you have to hydrolyze the complex under acidic conditions--that's why you use H₃O⁺. THIS is where that extra hydrogen comes from. Remember, what's the fastest reaction in organic chemistry? Proton transfer. So when you've got that nitrogen with that extra lone pair on it under acidic conditions, it's going to grab the proton off the H₃O⁺.

By the way, that third picture reminds me of my final in my second semester of O. Chem. We were asked to write a synthesis of that substance (whose discussion is not permitted on that forum) from that very precursor (the ketone).

 

[rfig08d]

Remember, what's the fastest reaction in organic chemistry? Proton transfer. So when you've got that nitrogen with that extra lone pair on it under acidic conditions, it's going to grab the proton off the H3O+.

Ahh thank you.

This is what I've been looking for!

I drew out the picture based on your words:

[PLAIN]http://img9.imageshack.us/img9/8369/imines3.jpg
(The scratched out lines in part 2 is a single bond... accidentally put a double bond)

Thank you very much, I really appreciate your help! :-)

 

[DDTea]

I want to add a bit more of an explanation here and also a suggestion for if you're asked to draw this mechanism on your quiz. It's likely that a mixture of ionic complexes form after the nucleophilic addition of the hydride to the carbonyl group. It may simply be better to write "M⁺" instead of NaBH₃⁺. The point is that in order to remove that cation, you must perform acid hydrolysis (which is where you extra proton comes from, again).

Kind of a minor detail, but I thought it was worth mentioning.