Esterification is a condensation reaction that combines alcohol and acid to produce an ester. Sulfuric acid acts as a proton donor, increasing the reaction rate and preventing “desterification.” It is not used for tertiary alcohols, which undergo dehydration without ester formation. Lactones can also be produced through esterification within a single molecule.
Esterification, or the combination of an alcohol with an acid to produce an ester, is a form of a condensation reaction, as water is removed in the process. The reverse reaction can also occur: the ester can recombine with water to produce the alcohol and acid. In some cases this “desterification” can be prevented by introducing a small amount of sulfuric acid into the reaction vessel. It helps by combining with the product water and effectively binding it. Primarily, the advantage of sulfuric acid in esterification is that it acts as a proton donor, increasing the reaction rate between the acid and the alcohol; when the acid used is a carboxylic acid, the reaction is sometimes called Fischer-Speier esterification.
Carboxylic acids (R-COOH, where R is an organic attachment) may be too weak to be used alone for an esterification reaction. A strong proton donor is needed for the carboxylic acid to act as if it were, in itself, a good source of protons. The sulfuric acid in esterification performs the task by injecting a proton into the carboxylic acid structure through the reaction H2SO4+R-COOH→HSO4-+R-C+(OH)2. The alcohol molecule, R′-OH, with its electron-rich oxygen atom, is attracted to this protonated carboxyl structure and forms a complex conglomerate, R-C+(OH)OR′+HSO4-→RC(O)- R′ .
This arrangement of atoms and charge is not very stable, so it undergoes a proton shift (H+), namely RC(OH)(O(H2)+)-OR′. In this state, it is easy for the clearly identifiable water molecule to move away, giving greater stabilization and leaving behind the energetically more favorable species, R-C+(OH)OR′. Finally, the regeneration of the sulfuric acid completes the process: R-C+(OH)OR′+HSO4-→RC(O)-R′. Because sulfuric acid is regenerated but not consumed by the reaction, it is considered a catalyst, not a reactant.
Interestingly, esterification does not require separate alcohol and acid molecules, but in some cases a reaction can occur within a single molecule containing both functional molecular parts or groups. A few conditions must be met: Both hydroxyl and carboxyl groups must be spatially free and able to go through each step of the process unaffected. An example of a molecule that can undergo this type of esterification is 5-hydroxypentanoic acid, HO-CH2CH2CH2CH2COOH. The ester produced by this form of esterification, which results in ring closure, is called a lactone, in this case δ-valerolactone. The positioning of the oxygen ring (-COC-) with respect to the carbonyl group (C=O) is that indicated by the Greek letter, delta.
Sulfuric acid in esterification is generally not used in connection with tertiary alcohols, those which have their hydroxyl-containing carbon atom attached to three other carbon atoms. Dehydration without ester formation occurs in tertiary alcohols in the presence of sulfuric acid. For example, tertiary butyl alcohol, (CH3)3C-OH, when combined with sulfuric acid, produces isobutylene, (CH3)2=CH2+H2O. In this case, the alcohol is what gets protonated, followed by the departure of a water molecule. The use of sulfuric acid in esterification is not a viable methodology for preparing tertiary esters.
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