0
$\begingroup$

enter image description here

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?

$\endgroup$
3
  • 2
    $\begingroup$ 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. $\endgroup$
    – Buttonwood
    Aug 17 '20 at 6:34
  • $\begingroup$ Acid base reactions are more favoured here. N has an acidic hydrogen hence attack happens. So R-MgBr converts into R-H. $\endgroup$ Aug 17 '20 at 14:00
  • $\begingroup$ A better title would be, "Why does a hydrogen transfer from nitrogen to carbon in this reaction?" $\endgroup$
    – user55119
    Aug 17 '20 at 19:54
5
$\begingroup$

There is no step 1 and step 2 in this reaction. Both are consecutive and second one is fast because it is an acid-base reaction:

Acid-base exchange in Grignard reagent

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.

$\endgroup$
6
  • 2
    $\begingroup$ If the Grignard's reaction with the halogen followed by the exchange with the nitrogen-H, would it be reasonable to call the product of the first reaction the kinetic, the later one the thermodynamic product? $\endgroup$
    – Buttonwood
    Aug 17 '20 at 18:23
  • 1
    $\begingroup$ Buttonwood : I think the reaction rate depends on preparation of Grignard reagent in either way because it is the RDS. The proton exchange is much faster to my knowledge. Do you have any other thoughts? $\endgroup$ Aug 17 '20 at 19:07
  • 2
    $\begingroup$ Mathew is correct. There is no step 1 and step 2. There is just an acid/base reaction as the "Grignard reagent" is formed. To say there is a second step is to say, in the reaction of NaOH and HCl, the first step is the formation of HO-H. True. And the second step is the formation of NaCl. I think not. $\endgroup$
    – user55119
    Aug 17 '20 at 19:50
  • 2
    $\begingroup$ @MathewMahindaratne I recognize now that my argumentation was (too much) influenced by a recent re-read about kinetic / thermodynamic deprotonations and metallations on halogenated pyridines. So it isn't sensible to this line of thought on saturated heterocyclic substrates as this one here. $\endgroup$
    – Buttonwood
    Aug 17 '20 at 20:50
  • 1
    $\begingroup$ @Soham Chatterjee: I used ChemDraw. $\endgroup$ Feb 17 at 0:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.