Question

The mechanism for the Fischer esterification is shown on page 762 of your textbook. The acid catalyst is specifically used in twosteps – protonation of 10 to make 11 and in the formation of 13 (it is unlikely that 12 directly goes to 13 – there is probably a deprotonation of 12 then a reprotonation to make 13).Explain mechanistically what the purpose of these two protonations are.

Answer 1

Mechanistic aspects of Fischer esterification:

1. In acidic medium carboxylic acid get protonate at carbonyl O (C=O). [Not at O of OH of carboxyl group because electron lone pair on this O is in conjugation with C=O and hence there is less probability of finding this lone pair to be protonated. Whereas lone pair on carbonyl O is not involved in conjugation rather it is easily available for donation as this O in resonance get the extra electron density}
2. C=O group is already a polar group ( where C has partial +? and O get partial –? charge) but such protonation makes C=O bond more polar and this enhance the electrophilicity of carbonyl C. Displacement of ? electron density between C=O bond suffice a potential carbocation)
3. This carbocation being highly electrophilic get attacked by lone pair on O of an alcohol. Alcohol a neutral nucleophile and hence weak actually(even in acidic condition). And its weakness is enhanced by acidic condition but strong electrophilicity of carbocation generated more than compensate and successful nucleophilic attack of alcohol takes place.
4. Nucleophilicity of alcohol can’t be increased beyond certain limit and hence the only way to speed up the reaction is to make carbonyl C more electrophilic and hence the importance of this first protonation step.
5. Now in 5^th step structure E there is protonation of -OH group forming –OH₂⁺ grouping. We can see in 5^th step we H2O molecule a stable , neutral species as leaving group. If this OH would not have protonated they we would have OH^- (hydroxyl group) – charged species as leaving group. We know that –vely charged species ( a conjugate base) is more nucleophilic than neutral (or corresponding acid) species or we can say less good leaving group than neutral species. i.e. H2O is good leaving and less nucleophilic group where OH^- is bad leaving group and good nucleophile. Hence being having protonated OH (i.e. –OH2⁺) as leaving group forward reaction goes well and this increase reaction rate. Hence the importance of this protonation step.

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