Question #73391

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Question #73391

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Answer 1 · 2016-02-17T01:01:25

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

Heat the amide with $NaOH$ $"NaOH"$ , and ${OH}^{-}$ $"OH"^(-)$ can attack the carbonyl carbon, even though it is less electropositive than that of a ketone. The heat helps.
A proton transfer must be made to the ${NH}_{2}$ $"NH"_2$ because ${NH}_{2}^{-}$ $"NH"_2^(-)$ is a terrible leaving group; its pKa is way higher than $36$ $36$ . On the other hand, the pKa of ${NH}_{3}$ $"NH"_3$ is naturally lower than that of $O^{2 -}$ $"O"^(2-)$ , so ${NH}_{3}$ $"NH"_3$ can favorably leave in this scenario of high heat.
Finish up the proton transfer.
Tetrahedral collapse.

And we do indeed form a sodium alkanoate and release ${NH}_{3} (g)$ $"NH"_3(g)$ .

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

Two ways it might go, if at all (ignoring ${Na}^{+}$ $"Na"^(+)$ )...

The amine could grab the proton off of ${OH}^{-}$ $"OH"^(-)$ .

${RNH}_{2} + {OH}^{-} ⇌ {RNH}_{3}^{+} + O^{2 -}$ $"RNH"_2 + "OH"^(-) rightleftharpoons "RNH"_3^(+) + "O"^(2-)$
pKa: $~ 36$ $~36$ $color(white)()$ pKa: $a a a$ $color(white)(aaa)$ pKa: $~ 11$ $~11$
$a a a a a a$ $color(white)(aaaaaa)$ $Large$ $"Large"$

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

${RNH}_{2} + {OH}^{-} ⇌ {RNH}^{-} + H_{2} O$ $"RNH"_2 + "OH"^(-) rightleftharpoons "RNH"^(-) + "H"_2"O"$
pKa: $~ 36$ $~36$ $a a a a a a a a a a a a a a a a$ $color(white)(aaaaaaaaaaaaaaaa)$ pKa: $~ 15.7$ $~15.7$

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 $N - H$ $"N"-"H"$ 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 ${OH}^{-}$ $"OH"^(-)$ AND stay protonated.

On the second one, the amine is the weaker acid in comparison to water, so the $N - H$ $"N"-"H"$ bond is stronger in comparison to the $O - H$ $"O"-"H"$ bond in water. Hence, it is difficult to break the $N - H$ $"N"-"H"$ bond using ${OH}^{-}$ $"OH"^(-)$ .

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

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Question #73391

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