It could form the enamine, but in the presence of acid, there will be equilibrium between the enamine and the iminium. So long as the iminium is formed reduction can occur.

Thank you. I”m making a note to research this step and correct.

Glad it was useful to you Josh, cheers!

The electrophile is the carbon, correct? And the more partial positive charge there is on that carbon, the better an electrophile it is, correct? Now draw the resonance form for the imine where you move the pair of electrons in the C-N pi bond to the nitrogen. Next, draw the conjugate acid (iminium) and do the same. Which resonance form do you think will be more stable between the two? Therein lies your answer.
Bottom line, protonation of the nitrogen results in a greater contribution from the resonance form where there is a positive charge on the carbon, and it’s more electrophilic.

I would look it up in the Encyclopedia of Reagents for Organic Synthesis. The original reference I see is Borch. R, F. et. al. JACS, 1971, 93, 2897. https://pubs.acs.org/doi/10.1021/ja00741a013

Hey Britney – you want to use excess to make sure that the amine has a chance to react with all of the aldehyde/ketone first, before adding your hydride source. Once the hydride source is added any aldehyde/ketone that remains will be reduced to an alcohol rendering it useless for the process.

Thanks for this, I’m adding it to my correction queue! Although I may be mistaken, I see pKa’s of protonated carbonyls at about -2 or so. Still, that’s about 15 orders of magnitude to traverse, so you’re correct on this point.