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 "NH"_2 because "NH"_2^(-) is a terrible leaving group; its pKa is way higher than 36. On the other hand, the pKa of "NH"_3 is naturally lower than that of "O"^(2-), so "NH"_3 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 "NH"_3(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"^(-).

"RNH"_2 + "OH"^(-) rightleftharpoons "RNH"_3^(+) + "O"^(2-)
pKa: ~36 color(white)() pKa: color(white)(aaa) pKa: ~11
color(white)(aaaaaa) "Large"

Or, it could donate a proton to "OH"^(-).

"RNH"_2 + "OH"^(-) rightleftharpoons "RNH"^(-) + "H"_2"O"
pKa: ~36 color(white)(aaaaaaaaaaaaaaaa)pKa: ~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 "N"-"H" 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 "N"-"H" bond is stronger in comparison to the "O"-"H" bond in water. Hence, it is difficult to break the "N"-"H" bond using "OH"^(-).

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