Hieronim Piotr Janecki TU Radom ......Kaffee auf meinem Schreibtisch
CH3COOH + H2NC2H5 a CH3COHNC2H5 + H2O
sollte richtig sein, wo -OH Gruppe mit-NHR Gruppe substituiert ist und gleichzeitig H2O entsteht
man sagt aber, das Acylierung von Aminen leichter mit Hilfe anderer Anfangsstoffe vorlautf:
This is a somewhat simplifed mechanism.
The formation of an ester, for instance the acetylation of an alcohol, also follows this basic mechanism.
Other conditions required (solvent, bases, acids, protecting groups...etc.) are reaction dependent.
"Have you heard about the agnostic dyslexic insomniac who lies awake in bed all night wondering if there really is a Dog?"
Typically, in order for one an amine to add to a carboxylic acid, the incoming amine (the nucleophile), must be counter-balanced by another group that is leaving. In acetic acid (CH3-C(=O)-OH), if the electrons of nucleophile attack the electropositive carbonyl group (C=O), a tetrahedral intermediate is formed as one of the two bonds between the carbon and the oxygen is replaced by the newly formed bond to the nitrogen, resulting in a formal negative charge on the oxygen. This tetrahedral intermediate is very unstable though, and must rearrange itself. The electrons that migrated to the oxygen now collapses back to the carbon they were originally bound to, and now a different pair of electrons must leave that carbon, otherwise it would share more than four bonds, and something that carbon can only very, very rarely do. The only possibilities for a leaving group then are the methyl group (CH3-), which would never happen; the nitrogen, which would lead back to your starting compounds; or the hydroxyl group, (-OH), which can be a leaving group. However, if you measure the stability in solution of a hydroxide ion versus a free amine, the free amine is far more stabile, meaning it is a better leaving group, meaning that this reaction will not proceed as desired.