{"id":10015,"date":"2013-03-29T07:26:47","date_gmt":"2013-03-29T07:26:47","guid":{"rendered":"http:\/\/www.ch.imperial.ac.uk\/rzepa\/blog\/?p=10015"},"modified":"2013-03-29T09:21:09","modified_gmt":"2013-03-29T09:21:09","slug":"a-sideways-look-at-the-mechanism-of-ester-hydrolysis","status":"publish","type":"post","link":"https:\/\/www.ch.ic.ac.uk\/rzepa\/blog\/?p=10015","title":{"rendered":"A sideways look at the mechanism of ester hydrolysis."},"content":{"rendered":"
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The mechanism of ester hydrolysis is a staple of examination questions in organic chemistry. To get a good grade, one might have to reproduce something like the below. Here, I subject that answer to a reality check.<\/p>\n

\"actyl\"<\/p>\n

In this scheme, HA is a general acid, R=Me, and the net result is to break what is called the acyl-oxygen bond (red). The mechanism is actually incomplete, since the label PT designates a proton-transfer (the mechanism for which is left somewhat undefined). Additionally, a lot of charges come and go and five steps or so are involved. So a student might be tempted to “fast-track” the whole process. Below I show two such fast-tracks (I prefer to say simplifications):<\/p>\n

\"acetyl-ester1\"<\/p>\n

In the blue mechanism<\/strong><\/span>, the role of HA is actually played by one water molecule, and a second water is assisting the PT step (a far more thorough analysis of the mechanism can be found in this reference[1]<\/a><\/span>). The reaction is bimolecular in ester and the HA (=water in this case). The third water would make it a termolecular reaction overall, but if the reaction takes place in water itself than [H2<\/sub>O] would be constant. It would correspond to what the text books call AAC<\/sub>2 since we consider one molecule as an acid HA. But, one could look at it differently and consider the second water as a nucleophile generated by concurrent deprotonation (by the first water). This would make it a BAC<\/sub>2 type. It turns out that if one makes the mechanism cyclic, the AAC<\/sub>2 and BAC<\/sub>2 annihilate each other in effect to create a single (peri)cyclic mechanism (which has no well known name, but might be referred to as the co-operative pathway). Such a mechanism can be extended using a third water molecule (magenta diagram<\/span>); I will come to the reason for including that presently.<\/p>\n

Why would one want to even consider such mechanisms? Because, if you look carefully, you will see no charges! Charge separation (= large dipole moment) takes energy. It is normally thought that this energy is more than compensated for by additional solvation (a process which is implicit rather than explicitly shown in text-book diagrams). But if you do not generate charge separation, you might not need that solvation energy. I will turn to quantum mechanics to try to decide what might be viable (I hesitate to use the term “going on”).\u00a0<\/p>\n

A \u03c9B97XD\/6-311G(d,p)\/SCRF=water model (in which solvation is approximately included as a continuum model) calculation yields the following for the blue mechanism.<\/p>\n\n\n\n
\"acyl-ester\"[2]<\/a><\/span><\/td>\n\"acyl-ester\"<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n
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  1. Points to note are that it is concerted, in other words the quantum mechanics tells us that all the bonds CAN make and break in a single concerted process within a single kinetic step.<\/li>\n
  2. The mechanism has an uncanny resemblance to the nucleophilic aromatic substitution<\/a> I reported a couple of posts ago! It resembles an Sn2 displacement at an sp2<\/sup> centre. Such juxtaposition of these two mechanisms is also not found in text-books. Recollect that with such aromatic substitution, it was possible to get both cncerted and stepwise mechanisms, depending on the substituents. Perhaps the same might be possible here?<\/li>\n
  3. However, the energy barrier for the process with the substituents shown above (~45 kcal\/mol) is rather too high (the experimental value is estimated as >22 kcal\/mol[1]<\/a><\/span>). There may be at least three reasons for this;\n