Which one do you think is a better leaving group?Ĭomparing the p Ka values of alcohols (16) and amines (38), we know that alkoxy groups are much weaker bases and therefore better leaving groups than a conjugate base of an amine. In order to restore the C=O double, either the alkoxy (RO –) or the (RNH –) must be expelled. In the first, we have a nucleophilic addition of the amine to the carbonyl and with a plenty of amine in the reaction mixture, the nitrogen is quickly deprotonated forming a negatively charged tetrahedral intermediate: Going back to the reaction between esters and amines let’s understand how it happens and why this conversion is possible anyway. Remember the limitations of Fischer esterification such as the need for high temperatures and nucleophilicity of the alcohol. And in general, acyl chlorides require milder conditions for nucleophilic acyl substitution than the alternative approaches do. The reaction goes by a nucleophilic addition-elimination mechanism and alkoxy groups (RO –), being poor leaving groups, make this method not as practical as, for example, the reaction of acyl chlorides with amines. Esters can be converted into primary, secondary and tertiary amides by an aminolysis reaction with ammonia, primary amine and a secondary amine respectively:
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