# Leaving Group Tendencies in Amide Hydrolysis

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 $$\ce{H2N-}$$.

• chem.libretexts.org/Bookshelves/Organic_Chemistry/…. Does this help? – Safdar Faisal Jul 17 '20 at 11:52
• 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). – Waylander Jul 17 '20 at 13:02
• 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. – Zhe Jul 17 '20 at 13:04
• The reversible steps are also driven to completion by the deprotonation of the carboxylic acid that is formed after the loss of -NH2. – user55119 Jul 17 '20 at 15:50

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.