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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:

the intermediate & what my teacher and books say

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 $\ce{H2N-}$.

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  • $\begingroup$ chem.libretexts.org/Bookshelves/Organic_Chemistry/…. Does this help? $\endgroup$
    – Safdar Faisal
    Jul 17, 2020 at 11:52
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    $\begingroup$ You are generally correct that NH2- is a worse leaving group the HO-. It is, however, not so bad that it never happens, which is why basic hydrolysis of amides is not a great reaction requiring forcing conditions and long reaction times. What makes it viable is that H2N- immediately grabs a proton from solvent and the NH3 boils out of the reaction mixture (see LeChateliers Principle). $\endgroup$
    – Waylander
    Jul 17, 2020 at 13:02
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    $\begingroup$ The hydroxide leaves all the time in a non-productive reaction. But eventually, you'll get the amide to leave. This is why basic hydrolysis of amides frequently requires heating to speed up the reaction. $\endgroup$
    – Zhe
    Jul 17, 2020 at 13:04
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    $\begingroup$ The reversible steps are also driven to completion by the deprotonation of the carboxylic acid that is formed after the loss of -NH2. $\endgroup$
    – user55119
    Jul 17, 2020 at 15:50

1 Answer 1

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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.

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