Question #73391

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"#.