[sharbeldam]

So i was thinking about this and i need a confirmation about my thoughts.

So basically all the reactants are reacting with a secondary amine and NaBH3CN just like the first reaction.

*in the first reaction, when a secondary amine reacts with a carbonyl we get enamine  (the first reaction) and then the double bond would be reduced by NaBH3CN, this is the normal reaction.

but in the second and third one we cant get enamine obviously because we'd get two double bonds aside, so i guess it would add to the beta carbon in this case? but the NaBH3CN would not reduce anything basically because it cant reduce the carbonyl and we have no double bonds.

in the fourth one we should not get a reaction at all.

am i right? thank you!

[Meter]

In order for an enamine to form, there must a "vacant" alpha carbon, which can participate in double bonding. To be honest, I haven't seen NaBH₃CN before now, and the only method for reductive amination I was introduced to was enamine formation followed by catalytic hydrogenation.

However, I can only find sources that utilize NaBH₃CN for reduction of imines to amines?