HELP FAST - question on reductive amination

Question

What molecule results from treating 4-oxopentanal with NH4Cl and NaCNBH3? Why? Is this a good, high yield reaction?

Answer 1

The product formed from this reaction is 1-methyl pyrrolidine. This is a multi-step reaction as follows:

1. Ammonium chloride is a source of NH3 in solution, therefore the ammonia would attack the aldehyde of 4-oxopentanal forming the imine intermediate and releasing water as the by product in the process.
2. Next, the imine would undergo reductive amination by reacting with sodium cyano borohydride (NaCNBH3), thus forming a primary amine.
3. The primary amine would attack the ketone carbonyl via an intramolecular cyclization mechanism, and again forming a five membered ring imine intermediate, and releasing another molecule of water.
4. Finally, the imine would undergo reductive amination similar to explanation 2 above.

This is a very well known reaction and yields for this particular method can range from 80-99%.