Leaving Group Tendencies in Amide Hydrolysis

Question

When we do hydrolysis of amides in basic conditions, we come across a mechanism in which the $ce{OH-}$ ion attacks the carbonyl forming an intermediate as given in the photo:

When the carbonyl gets formed again it ejects off $ce{H2N-}$-that’s what my teacher says. But I have a problem that why this happens, why not the $ce{OH-}$ group leaves off when the carbonyl gets formed again? After all, $ce{OH-}$ is a better leaving group than $ce{H2N-}$.

Mathew Mahindaratne · Accepted Answer

You can imagine following situation in this hydrolysis:
$$ce{R-C(=O)NH2 + OH- <=>[$k_1$][$k_2$] R-C(NH2)(OH)O- \->[$k_3$] R-C(=O)OH + H2N- ->[$k_4$] R-C(=O)O- + NH3}$$
Keep in mind that:
$$ce{R-C(=O)NH2 + OH- ->[$k_1$] R-C(NH2)(OH)O- }$$
$$ce{R-C(NH2)(OH)O-  ->[$k_2$] R-C(=O)NH2 + OH-}$$

In this reaction, $k_1 gt k_2$ because of LeChateliers principle since $[ce{OH-}]$ is in excess thus forward reaction is in favor.
Also, $k_2 gt k_3$ because of the reason given by OP ($ce{OH-}$ is better leaving group than $ce{H2N-}$).
Since acid-base reaction is much faster than any other reaction, we can conclude that $k_4 gt gt k_3$.

Therefore, as soon as $ce{R-C(=O)OH + H2N- }$ formed they would convert to $ce{R-C(=O)O- + NH3}$ quickly so that $ce{R-C(=O)OH + H2N- }$ can not able to go back to $ce{ R-C(NH2)(OH)O-}$ intermediate. That means, when boiling with extra basic solutions, amide will be hydrolyze slowly to its corresponding carboxylate.

Leaving Group Tendencies in Amide Hydrolysis

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