# Why does magnesium bromide transfers from a carbon atom to nitrogen in piperidine?

In this reaction, why does $$\ce{MgBr}$$ goes to the $$\ce{>NH}$$ group and form a bond with nitrogen by removing $$\ce{-H}$$.

Also where does the negative charge go?

• To substantiate @Safdar's comment further, in the Evans pKa tables (e.g., www2.chem.wisc.edu/areas/reich/pkatable/index.htm), cyclopentane is estimated (section «-Benzyl») by a pKa(DMSO) of ~59, and pyrrolidine (section «-Amines») by a pKa(DMSO) of 44. Ammonia is listed by a pKa(DMSO) of 41. Thus, the differences are considerable. Aug 17 '20 at 6:34
• Acid base reactions are more favoured here. N has an acidic hydrogen hence attack happens. So R-MgBr converts into R-H. Aug 17 '20 at 14:00
• A better title would be, "Why does a hydrogen transfer from nitrogen to carbon in this reaction?" Aug 17 '20 at 19:54

Note that approximate $$\mathrm{p}K_\mathrm{a}$$ values of ring $$\ce{C-H}$$ and $$\ce{N-H}$$ are 50 and 40, respectively. Thus, as soon as Grignard reagent is formed, it exchanges relatively acidic protons of $$\ce{N-H}$$ fast (Remember, Grignard is also a strong base). To visualize the exchange process, I included schematic representation at the bottom of the image.