Question #73391

1 Answer
Feb 17, 2016

First, let's make sure that we get the amide mechanism.

  1. Heat the amide with NaOH, and OH can attack the carbonyl carbon, even though it is less electropositive than that of a ketone. The heat helps.
  2. A proton transfer must be made to the NH2 because NH2 is a terrible leaving group; its pKa is way higher than 36. On the other hand, the pKa of NH3 is naturally lower than that of O2, so NH3 can favorably leave in this scenario of high heat.
  3. Finish up the proton transfer.
  4. Tetrahedral collapse.

And we do indeed form a sodium alkanoate and release NH3(g).


An amine, however, has no electrophilic carbonyl carbon. Instead, the only way it could interact with NaOH is through an acid/base reaction.

Two ways it might go, if at all (ignoring Na+)...

The amine could grab the proton off of OH.

RNH2+OHRNH+3+O2
pKa: ~36 pKa: aaa pKa: ~11
aaaaaa Large

Or, it could donate a proton to OH.

RNH2+OHRNH+H2O
pKa: ~36 aaaaaaaaaaaaaaaapKa: ~15.7

The equilibrium lies on the side of the weaker acid.

On the first one, the protonated amine is a stronger acid than the original amine (lower pKa), so that NH bond that just formed is weaker in comparison to that of the protonated amine. Hence it is difficult to make the protonated amine grab a proton from OH AND stay protonated.

On the second one, the amine is the weaker acid in comparison to water, so the NH bond is stronger in comparison to the OH bond in water. Hence, it is difficult to break the NH bond using OH.

Neither of these is expected to occur. I would not expect a reaction with an amine and NaOH.