Question

Question #73391

Answer 1

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)`.

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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"`.